Maths solved problems.pdf

1 120 maths solved Questions 2

2 451 Question and Answer 154

3 Age Calculation 245

4 Area 268

5 Averages 320

6 Bankers Discount 347

7 Boat and Streams 374

8 Calendar 414

9 Chain Rule 448

10 Mixture and Allegations 499

11 Pipes and Cistern 534

12 Time and Distance 562

13 Time and Work 596

14 Time 630

15 Train sum 679

Maths Solved ProblemsComplied by - RangaRakes

Index

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120 Maths solved Questions

• Factorial

Let n be a positive integer. Then n factorial (n!) can be defined as n! = n(n-1)(n-2)...1

Examples

i. 5! = 5 x 4 x 3 x 2 x 1 = 120

ii. 3! = 3 x 2 x 1 = 6

Special Cases

iii. 0! = 1

iv. 1! = 1 • Permutations

Permutations are the different arrangements of a given number of things by taking some or all at a time

Examples


i. All permutations (or arrangements) formed with the letters a, b, c by taking three at a time are (abc, acb, bac, bca, cab, cba)

ii. All permutations (or arrangements) formed with the letters a, b, c by taking two at a time are (ab, ac, ba, bc, ca, cb)

• Combinations

Each of the different groups or selections formed by taking some or all of a number of objects is called a combination

Examples

i. Suppose we want to select two out of three girls P, Q, R. Then, possible combinations are PQ, QR and RP. (Note that PQ and QP represent the same selection)

ii. Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR

• Difference between Permutations and Combinations and How to Address a Problem

Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.


Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.

If order is important, PQ will be different from QP , PR will be different from RP and QR will be different from RQ

If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ

Hence, If the order is important, problem will be related to permutations. If the order is not important, problem will be related to combinations.

For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc

For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc.

Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of


geometrical figures , distribution of items (there are exceptions for this) etc will be related to combinations.

• Repetition

The term repetition is very important in permutations and combinations.

Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R.

If repetition is allowed, the same object can be taken more than once to make a sample.

i.e., if repetition is allowed, PP, QQ, RR can also be considered as possible samples.

If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples

Normally repetition is not allowed unless mentioned specifically.

• pq and qp are two different permutations ,but they represent the same combination.

• Number of permutations of n distinct things taking r at a time

Number of permutations of n distinct things taking r at a time can be given by nPr = n! (n−r)! =n(n−1)(n−2)...(n−r+1)where 0≤r≤n


If r > n, nPr = 0

Special Case: nP0 = 1

nPr is also denoted by P(n,r). nPr has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)r or nr

Examples

i. 8P2 = 8 x 7 = 56

ii. 5P4= 5 x 4 x 3 x 2 = 120 • Number of permutations of n distinct things taking all

at a time

Number of permutations of n distinct things taking them all at a time = nPn = n!

• Number of Combinations of n distinct things taking r at a time

Number of combinations of n distinct things taking r at a time ( nCr) can be given by nCr = n! (r!)(n−r)! =n(n−1)(n−2)⋯(n−r+1) r! where 0≤r≤n

If r > n, nCr = 0


Special Case: nC0 = 1

nCr is also denoted by C(n,r). nCr occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by (n r )

Examples

i. 8C2 = 8×7 2×1 = 28

ii. 5C4= 5×4×3×2 4×3×2×1 = 5

1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? A. 24400 B. 21300 C. 210 D. 25200

Here is the answer and explanation

Answer : Option D

Explanation :

Number of ways of selecting 3 consonants out of 7 = 7C3 Number of ways of selecting 2 vowels out of 4 = 4C2


Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = 7C3 x 4C2

=(7×6×5 3×2×1 )×(4×3 2×1 )=210

It means that we can have 210 groups where each group contains total 5 letters(3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120 Hence, Required number of ways = 210 x 120 = 25200

2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? A. 159 B. 209 C. 201 D. 212



Answer : Option B

Explanation :

In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there. Hence we have 4 choices as given below We can select 4 boys ------(Option 1). Number of ways to this = 6C4 We can select 3 boys and 1 girl ------(Option 2) Number of ways to this = 6C3 x 4C1 We can select 2 boys and 2 girls ------(Option 3) Number of ways to this = 6C2 x 4C2 We can select 1 boy and 3 girls ------(Option 4) Number of ways to this = 6C1 x 4C3 Total number of ways = (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3) = (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) [Applied the formula nCr = nC(n - r) ]

=[6×5 2×1 ]+[(6×5×4 3×2×1 )×4]+[(6×5 2×1 )(4×3 2×1 )]+[6×4]

= 15 + 80 + 90 + 24


= 209

3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? A. 624 B. 702 C. 756 D. 812


Answer : Option C

Explanation :

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men. Hence we have the following 3 choices We can select 5 men ------(Option 1) Number of ways to do this = 7C5 We can select 4 men and 1 woman ------(Option 2) Number of ways to do this = 7C4 x 6C1 We can select 3 men and 2 women ------(Option 3) Number of ways to do this = 7C3 x 6C2 Total number of ways = 7C5 + [7C4 x 6C1] + [7C3 x 6C2]


= 7C2 + [7C3 x 6C1] + [7C3 x 6C2] [Applied the formula nCr = nC(n -

r) ]

=[7×6 2×1 ]+[(7×6×5 3×2×1 )×6]+[(7×6×5 3×2×1 )×(6×5 2×1 )]

= 21 + 210 + 525 = 756

4. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? A. 610 B. 720 C. 825 D. 920


Answer : Option B

Explanation :

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA). Hence we can assume total letters as 5. and all these letters are different. Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120


All The 3 vowels (OIA) are different Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6 Hence, required number of ways = 120 x 6 = 720

5. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? A. 47200 B. 48000 C. 42000 D. 50400


Answer : Option D

Explanation :

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO). Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and rest of the letters are different.

Number of ways to arrange these letters = 7! 2! =7×6×5×4×3×2×1 2×1 = 2520


In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.

Number of ways to arrange these vowels among themselves = 5! 3! =5×4×3×2×1 3×2×1 =20

Hence, required number of ways = 2520 x 20 = 50400

6. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? A. 1 B. 126 C. 63 D. 64


Answer : Option C

Explanation :

We need to select 5 men from 7 men and 2 women from 3 women Number of ways to do this = 7C5 x 3C2 = 7C2 x 3C1 [Applied the formula nCr = nC(n - r) ]

=(7×6 2×1 )×3

= 21 x 3 = 63


7. In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together? A. 9800 B. 100020 C. 120960 D. 140020


Answer : Option C

Explanation :

The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS(AEAI). Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters are different.

Hence,number of ways to arrange these letters = 8! (2!)(2!) =8×7×6×5×4×3×2×1 (2×1)(2×1) =10080

In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.

Number of ways to arrange these vowels among themselves = 4! 2! =4×3×2×1 2×1 =12


Hence, required number of ways = 10080 x 12 = 120960

8. There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed? A. 10420 B. 11 C. 11760 D. None of these


Answer : Option C

Explanation :

We need to select 5 men from 8 men and 6 women from 10 women Number of ways to do this = 8C5 x 10C6 = 8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]

=(8×7×6 3×2×1 )(10×9×8×7 4×3×2×1 )

= 56 x 210 = 11760

9. How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?


A. 720 B. 420 C. None of these D. 5040


Answer : Option A

Explanation :

The word 'LOGARITHMS' has 10 different letters. Hence, the number of 3-letter words(with or without meaning) formed by using these letters = 10P3 = 10 x 9 x 8 = 720

10. In how many different ways can the letters of the word 'LEADING' be arranged such that the vowels should always come together? A. None of these B. 720 C. 420 D. 122


Answer : Option B

Explanation :

The word 'LEADING' has 7 letters. It has the vowels 'E','A','I' in it and


these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. that is, LDNG(EAI). Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters = 5! = 5 x 4 x 3 x 2 x 1 = 120 In the 3 vowels (EAI), all the vowels are different. Number of ways to arrange these vowels among themselves = 3! = 3 x 2 x 1= 6 Hence, required number of ways = 120 x 6= 720

11. A coin is tossed 3 times. Find out the number of possible outcomes. A. None of these B. 8 C. 2 D. 1


Answer : Option B

Explanation :

When a coin is tossed once, there are two possible outcomes - Head(H) and Tale(T) Hence, when a coin is tossed 3 times, the number of possible


outcomes = 2 x 2 x 2 = 8 (The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT )

12. In how many different ways can the letters of the word 'DETAIL' be arranged such that the vowels must occupy only the odd positions? A. None of these B. 64 C. 120 D. 36


Answer : Option D

Explanation :

The word 'DETAIL' has 6 letters which has 3 vowels (EAI) and 3 consonants(DTL) The 3 vowels(EAI) must occupy only the odd positions. Let's mark the positions as (1) (2) (3) (4) (5) (6). Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order. Hence, number of ways to arrange these vowels = 3P3 = 3! = 3 x 2 x 1 = 6 Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order


Hence, number of ways to arrange these consonants = 3P3 = 3! = 3 x 2 x 1 = 6 Total number of ways = number of ways to arrange the vowels x number of ways to arrange the consonants = 6 x 6 = 36

13. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw? A. 64 B. 128 C. 32 D. None of these


Answer : Option A

Explanation :

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there. Hence we have 3 choices as given below We can select 3 black balls --------------------------(Option 1) We can select 2 black balls and 1 non-black ball------(Option 2) We can select 1 black ball and 2 non-black balls------(Option 3)


Number of ways to select 3 black balls = 3C3 Number of ways to select 2 black balls and 1 non-black ball = 3C2 x 6C1 Number of ways to select 1 black ball and 2 non-black balls = 3C1 x 6C2 Total number of ways = 3C3 + (3C2 x 6C1) + (3C1 x 6C2) = 1 + (3C1 x 6C1) + (3C1 x 6C2) [Applied the formula nCr = nC(n - r) ]

=1+[3×6]+[3×(6×5 2×1 )]

= 1 + 18 + 45 = 64

14. In how many different ways can the letters of the word 'JUDGE' be arranged such that the vowels always come together? A. None of these B. 48 C. 32 D. 64


Answer : Option B

Explanation :


The word 'JUDGE' has 5 letters. It has 2 vowels (UE) in it and these 2 vowels should always come together. Hence these 2 vowels can be grouped and considered as a single letter. That is, JDG(UE). Hence we can assume total letters as 4 and all these letters are different. Number of ways to arrange these letters = 4!= 4 x 3 x 2 x 1 = 24 In the 2 vowels (UE), all the vowels are different. Number of ways to arrange these vowels among themselves = 2! = 2 x 1 = 2 Total number of ways = 24 x 2 = 48

15. In how many ways can the letters of the word 'LEADER' be arranged? A. None of these B. 120 C. 360 D. 720


Answer : Option C

Explanation :

The word 'LEADER' has 6 letters.


But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters = 6! 2! =6×5×4×3×2×1 2×1 =360

16. How many words can be formed by using all letters of the word 'BIHAR'? A. 720 B. 24 C. 120 D. 60


Answer : Option C

Explanation :

The word 'BIHAR' has 5 letters and all these 5 letters are different. Total words formed by using all these 5 letters = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120

17. How many arrangements can be made out of the letters of the word 'ENGINEERING' ? A. 924000 B. 277200 C. None of these D. 182000



Answer : Option B

Explanation :

The word 'ENGINEERING' has 11 letters. But in these 11 letters, 'E' occurs 3 times,'N' occurs 3 times, 'G' occurs 2 times, 'I' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters = 11! (3!)(3!)(2!)(2!) =[11×10×9×8×7×6×5×4×3×2×1 (3×2×1)(3×2×1)(2×1)(2×1) ]=277200

18. How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated? A. 20 B. 16 C. 8 D. 24


Answer : Option A

Explanation :

A number is divisible by 5 if the its last digit is a 0 or 5

We need to find out how many 3 digit numbers can be formed from the 6 digits (2,3,5,6,7,9)


which are divisible by 5. Since the 3 digit number should be divisible by 5, we should take the digit 5 from the 6 digits(2,3,5,6,7,9) and fix it at the unit place. There is only 1 way of doing this

1

Since the number 5 is placed at unit place, we have now five digits(2,3,6,7,9) remaining. Any of these 5 digits can be placed at tens place

5 1

Since the digits 5 is placed at unit place and another one digits is placed at tens place, we have now four digits remaining. Any of these 4 digits can be placed at hundreds place.

4 5 1

Required Number of three digit numbers = 4 x 5 x 1 = 20

19. How many words with or without meaning, can be formed


by using all the letters of the word, 'DELHI' using each letter exactly once? A. 720 B. 24 C. None of these D. 120


Answer : Option D

Explanation :

The word 'DELHI' has 5 letters and all these letters are different. Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once = Number of arrangements of 5 letters taken all at a time = 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120

20. What is the value of 100P2 ? A. 9801 B. 12000 C. 5600 D. 9900


Answer : Option D

Explanation :

100P2 = 100 x 99 = 9900


21. In how many different ways can the letters of the word 'RUMOUR' be arranged? A. None of these B. 128 C. 360 D. 180


Answer : Option D

Explanation :

The word 'RUMOUR' has 6 letters. But in these 6 letters, 'R' occurs 2 times, 'U' occurs 2 times and rest of the letters are different.

Hence, number of ways to arrange these letters = 6! (2!)(2!) =6×5×4×3×2×1 (2×1)(2×1) =180

22. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? A. 3200 B. None of these C. 2400 D. 3600


Answer : Option D

Explanation :


We have 6 periods and need to organize 5 subjects such that each subject is allowed at least one period. In 6 periods, 5 can be organized in 6P5 ways. Remaining 1 period can be organized in 5P1 ways. Total number of arrangements = 6P5 x 5P1 = (6 x 5 x 4 x 3 x 2 ) x (5) = 720 x 5 = 3600

23. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once? A. 720 B. 360 C. 1420 D. 1680


Answer : Option D

Explanation :

The first two places can only be filled by 3 and 5 respectively and there is only 1 way of doing this Given that no digit appears more than once. Hence we have 8


digits remaining(0,1,2,4,6,7,8,9) So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways Total number of ways = 8P4 = 8 x 7 x 6 x 5 = 1680

24. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair? A. 100 B. 80 C. 110 D. 64


Answer : Option B

Explanation :

He has has 10 patterns of chairs and 8 patterns of tables Hence, A chair can be arranged in 10 ways and A table can be arranged in 8 ways Hence one chair and one table can be arranged in 10 x 8 ways = 80 ways

25. 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?


A. None of these B. 600 C. 576 D. 625


Answer : Option B

Explanation :

He can go in any bus out of the 25 buses. Hence He can go in 25 ways. Since he can not come back in the same bus that he used for travelling, He can return in 24 ways. Total number of ways = 25 x 24 = 600

26. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours? A. 62 B. 48 C. 12 D. 24


Answer : Option D

Explanation :


1 red ball can be selected in 4C1 ways 1 white ball can be selected in 3C1 ways 1 blue ball can be selected in 2C1 ways Total number of ways = 4C1 x 3C1 x 2C1 =4 x 3 x 2 = 24

27. A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 8 from part P and 4 from part Q, in how many ways can he do that? A. None of these B. 6020 C. 1200 D. 9450


Answer : Option D

Explanation :

Number of ways to choose 8 questions from part P = 10C8 Number of ways to choose 4 questions from part Q = 10C4 Total number of ways = 10C8 x 10C4 = 10C2 x 10C4 [Applied the formula nCr = nC(n - r) ]

=(10×9 2×1 )(10×9×8×7 4×3×2×1 )


=45 x 210 = 9450

28. In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate? A. 2880 B. 1400 C. 1200 D. 3212


Answer : Option A

Explanation :

In a circle, 5 boys can be arranged in 4! ways Given that the boys and the girls alternate. Hence there are 5 places for girls which can be arranged in 5! ways Total number of ways = 4! x 5! = 24 x 120 = 2880

29. Find out the number of ways in which 6 rings of different types can be worn in 3 fingers? A. 120 B. 720 C. 125 D. 729



Answer : Option D

Explanation :

The first ring can be worn in any of the 3 fingers => There are 3 ways of wearing the first ring Similarly each of the remaining 5 rings also can be worn in 3 ways Hence total number of ways

=3×3×3×3×3×6=3 6 =729

30. In how many ways can 5 man draw water from 5 taps if no tap can be used more than once? A. None of these B. 720 C. 60 D. 120


Answer : Option D

Explanation :

1st man can draw water from any of the 5 taps 2nd man can draw water from any of the remaining 4 taps 3rd man can draw water from any of the remaining 3 taps 4th man can draw water from any of the remaining 2 taps 5th man can draw water from remaining 1 tap


5 4 3 2 1

Hence total number of ways = 5 x 4 x 3 x 2 x 1 = 120

31. How many two digit numbers can be generated using the digits 1,2,3,4 without repeating any digit? A. 4 B. 10 C. 12 D. 16


Answer : Option C

Explanation :

We have four digits 1,2,3,4 The first digit can be any digit out of the four given digits

4

Now we have already chosen the first digit. Since we cannot repeat the digits, we are left with 3 digits now. The second digit can be any of these three digits


4 3

Since the first digit can be chosen in 4 ways and second digit can be chosen in 3 ways, both the digits can be chosen in 4 × 3 = 12 ways. [Reference : Multiplication Theorem] i.e., 12 two digit numbers can be formed

32. There are three places P, Q and R such that 3 roads connects P and Q and 4 roads connects Q and R. In how many ways can one travel from P to R? A. 8 B. 10 C. 12 D. 14


Answer : Option C

Explanation :

The number of ways in which one can travel from P to R = 3 × 4 = 12 [Reference : Multiplication Theorem]

33. There are 10 women and 15 men in an office. In how many ways can a person can be selected?


A. None of these B. 50 C. 25 D. 150


Answer : Option C

Explanation :

The number of ways in which a person can be selected = 10 + 15 = 25 [Reference : Addition Theorem]

34. There are 10 women and 15 men in an office. In how many ways a team of a man and a woman can be selected? A. None of these B. 50 C. 25 D. 150


Answer : Option D

Explanation :

Number of ways in which a team of a man and a woman can be selected = 15 × 10 = 150 [Reference : Multiplication Theorem]

35. In how many ways can three boys can be seated on five


chairs? A. 30 B. 80 C. 60 D. 120


Answer : Option C

Explanation :

There are three boys. The first boy can sit in any of the five chairs (5 ways)

5

Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways)

5 4

Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways)

5 4 3


Hence, the total number of ways in which 3 boys can be seated in 5 chairs = 5 × 4 × 3 = 60

36. There are 6 persons in an office. A group consisting of 3 persons has to be formed. In how many ways can the group be formed? A. 30 B. 10 C. 40 D. 20


Answer : Option D

Explanation :

Number of ways in which the group can be formed = 6C3

=6×5×4 3×2×1 =20

37. There are 5 yellow, 4 green and 3 black balls in a bag. All the 12 balls are drawn one by one and arranged in a row. Find out the number of different arrangements possible. A. 25230 B. 23420 C. 21200 D. 27720



Answer : Option D

Explanation :

[Reference : Permutations : Special Case 2 : Permutation of Like Things] The number of different arrangements possible

=12! 5! 4! 3! =12×11×10×9×8×7×6×5×4×3×2×1 (5×4×3×2×1)(4×3×2×1)(3×2×1) =12×11×10×9×8×7×6 (4×3×2×1)(3×2×1) =12×11×10×9×8×7 (4×3×2×1) =11×10×9×8×7 (2) =11×10×9×4×7=252×11×10=27720

38. In how many ways can 7 boys be seated in a circular order? A. 60 B. 120 C. 5040 D. 720


Answer : Option D

Explanation :

[Reference : Circular Permutations: Case 1] Number of arrangements possible = (7-1)! = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

39. In how many ways can 7 beads can be arranged to form a necklace?


A. 720 B. 360 C. 120 D. 60


Answer : Option B

Explanation :

[Reference : Circular Permutations: Case 2 : when clockwise and anticlockwise arrangements are not different] Number of arrangements possible

=1 2 (7−1)!=1 2 ×6!=1 2 ×6×5×4×3×2×1=360

40. In how many ways can a team of 5 persons can be formed out of a total of 10 persons such that two particular persons should be included in each team? A. 56 B. 28 C. 112 D. 120


Answer : Option A

Explanation :

----------------------------------------------------------------------------------------- Solution 1 : Using the Principles ------------------------------------------------------------------------------


----------- Two particular persons should be included in each team i.e., we have to select 5-2 = 3 persons from 10-2 = 8 persons Hence, the required number of ways = 8C3

=8×7×6 3×2×1 =8×7=56 ----------------------------------------------------------------------------------------- Solution 2 : Using the Formula ----------------------------------------------------------------------------------------- [Reference : Case 1: When s particular things are always to be included]

Number of combinations of n different things taking r at a time, when s particular things are always to be included in each selection, is (n-s)C(r-s)

Here n = 10, r = 5, s = 2 Hence, the number of ways = (n-s)C(r-s) = 8C3

=8×7×6 3×2×1 =8×7=56

41. In how many ways can a team of 5 persons can be formed out of a total of 10 persons such that two particular persons should not be included in any team? A. 56 B. 112


C. 28 D. 128


Answer : Option A

Explanation :

----------------------------------------------------------------------------------------- Solution 1 : Using the Principles ----------------------------------------------------------------------------------------- Two particular persons should not be included in each team i.e., we have to select 5 persons from 10-2 = 8 persons Hence, the required number of ways = 8C5 = 8C3[∵ nCr = nC(n - r)]

=8×7×6 3×2×1 =8×7=56 ----------------------------------------------------------------------------------------- Solution 2 : Using the Formula ----------------------------------------------------------------------------------------- [Reference : Case 3: When s particular things are never included]


Number of Combinations of n different things taking r at a time, when s particular things are never included in any selection, is (n-s)Cr

Here n = 10, r = 5, s = 2 Hence, the number of ways = (n-s)Cr = 8C5 = 8C3[∵ nCr = nC(n - r)]

=8×7×6 3×2×1 =8×7=56

42. How many triangles can be formed by joining the vertices of an octagon? A. 56 B. 28 C. 112 D. 120


Answer : Option A

Explanation :

[Reference : Number of triangles formed by joining the angular points of a polygon]

The number of triangles that can be formed by joining the angular points of a polygon of n sides as vertices are n(n−1)(n−2) 6


Here n = 8 Hence, the number of triangles that can be formed by joining the vertices of an octagon

=n(n−1)(n−2) 6 =8(8−1)(8−2) 6 =8.7.6 6 =56

43. If there are 9 horizontal lines and 9 vertical lines in a chess board, how many rectangles can be formed in the chess board? A. 920 B. 1024 C. 64 D. 1296


Answer : Option D

Explanation :

[Reference : Number of rectangles formed by using horizontal lines and vertical lines]

The number of rectangles that can be formed by using m horizontal lines and n vertical lines are mC2 × nC2

Here m = 9, n = 9 Hence, The number of rectangles that can be formed = mC2 × nC2 = 9C2 × 9C2 = (9C2)

2


=(9×8 2×1 ) 2 =36 2 =1296 (To save the time, you don't need to really calculate the actual value of 362. You know that 362 is a number whose last digit is 6. From the given choices, 1296 is only one number which has 6 as its last digit. Hence it is the answer)

44. Find the number of diagonals of a decagon? A. 16 B. 28 C. 35 D. 12


Answer : Option C

Explanation :

[Reference : Number of diagonals formed by joining the vertices of a polygon]

The number of diagonals that can be formed by joining the vertices of a polygon of n sides are n(n−3) 2

Here n = 10 Hence, The number of diagonals


=n(n−3) 2 =10(10−3) 2 =10×7 2 =5×7=35

45. Find the number of triangles that can be formed using 14 points in a plane such that 4 points are collinear? A. 480 B. 360 C. 240 D. 120


Answer : Option B

Explanation :

[Reference : Number of triangles formed by joining n points out of which m points are collinear]

Consider there be n points in a plane out of which m points are collinear. The number of triangles that can be formed by joining these n points as vertices are nC3 -

mC3

Here n = 14, m = 4 Hence, The number of triangles = nC3 -

mC3 = 14C3 - 4C3

= 14C3 -

4C1 [∵ nCr = nC(n - r)]

=14×13×12 3×2×1 −4=(14×13×2)−4=360


46. What is the sum of all 4 digit numbers formed using the digits 2, 3,4 and 5 without repetition? A. 93324 B. 92314 C. 93024 D. 91242


Answer : Option A

Explanation :

[Reference : Sum of all numbers formed from given digits]

If all the possible n digit numbers using the n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! × (Sum of the n digits) × (111 ... n times)

Here n=4. Hence the sum of all 4 digit numbers formed using the digits 2, 3,4 and 5 without repetition = (4-1)! (2 + 3 + 4 + 5)(1111) = 3! × 14 × 1111 = 6 × 14 × 1111 = 93324

47. In a birthday party, every person shakes hand with every other person. If there was a total of 28 handshakes in the party, how many persons were present in the party? A. 9 B. 8


C. 7 D. 6


Answer : Option B

Explanation :

[Reference : Number of persons and handshakes]

Assume that in a party every person shakes hand with every other person. Let n = the total number of persons present in the party h = total number of handshakes Then, h=n(n−1) 2

Here h = 28

h=n(n−1) 2 ⇒28=n(n−1) 2 n(n-1) = 28 × 2 =>n(n-1)= 56 => n = 8 To find out the value of n from the equation n(n-1) = 56, use any of the following methods


Method 1: Trial and error method Just substitute the values given in the choices in the equation to see which value satisfies the equation. n(n-1) = 56 If n = 6, n(n-1) = 6 × 5 ≠ 56 If n = 7, n(n-1) = 7 × 6 ≠ 56 If n = 9, n(n-1) = 9 × 8 ≠ 56 If n = 8, n(n-1) = 8 × 7 = 56 . Hence n= 8 is the answer. Method 2: By Factoring [Reference : Quadratic Equations and How to Solve Quadratic Equations] n(n-1) = 56 n2 - n – 56 = 0 (n-8)(n + 7) = 0 n = 8 or -7 Since n cannot be negative, n = 8 Method 3: By Quadratic Formula [Reference : Quadratic Equations and How to Solve Quadratic Equations]


n(n-1) = 56 n2 - n – 56 = 0

n=−b±b 2 −4ac − − − − − − − √ 2a =1±(−1) 2 –4×1×(−56) − − − − − − − − − − − − − − − − − √ 2×1

=1±1+224 − − − − − − √ 2 =1±225 − − − √ 2 =1±15 2 =16 2 or −14 2 =8 or −7 Since n cannot be negative, n = 8

48. There are 8 points in a plane out of which 3 are collinear. How many straight lines can be formed by joining them? A. 16 B. 26 C. 22 D. 18


Answer : Option B

Explanation :

[Reference : Number of straight lines formed by joining n points out of which m points are collinear]

Consider there be n points in a plane out of which m points are collinear. The number of straight lines that can be formed by joining these n points are nC2 -

mC2 + 1


Here n=8, m=3 The required number of straight lines = nC2 -

mC2 + 1 = 8C2 -

3C2 + 1 = 8C2 -

3C1 + 1 [∵ nCr = nC(n - r)]

=8×7 2×1 −3+1=28−3+1=26

49. How many quadrilaterals can be formed by joining the vertices of an octagon? A. 60 B. 70 C. 65 D. 74

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| Discuss


Answer : Option B

Explanation :

[Reference : Number of quadrilaterals formed by joining the vertices of a polygon]

The number of quadrilaterals that can be formed by joining the vertices of a polygon of n sides are n(n−1)(n−2)(n−3) 24 where n>3


Here n=8 The required number of quadrilaterals

=n(n−1)(n−2)(n−3) 24 =8(8−1)(8−2)(8−3) 24 =8×7×6×5 24

=8×7×5 4 =2×7×5=7×10=70

50. How many straight lines can be formed by joining 12 points on a plane out of which no points are collinear? A. 72 B. 66 C. 58 D. 62


Answer : Option B

Explanation :

[Reference : Number of straight lines formed by joining n points out of which no points are collinear]

Consider there be n points in a plane out of which no points are collinear. The number of straight lines that can be formed by joining these n points are n(n−1) 2

Here n=12 The required number of straight lines


=n(n−1) 2 =12×11 2 =6×11=66

51. If nC8 = nC27 , what is the value of n? A. 35 B. 22 C. 28 D. 41


Answer : Option A

Explanation :

[Reference : More Useful Relations - Combinations]

If nCx = nCy then either x = y or (n-x) = y

nC8 = nC27 => n – 8 = 27 => n =27 + 8 = 35

52. In how many ways can 10 students can be arranged in a row? A. 9! B. 6! C. 8! D. 10!



Answer : Option D

Explanation :

10 students can be arranged in a row in 10P10 = 10! ways

53. Find the number of triangles which can be drawn out of n given points on a circle? A. (n+1)C1 B. nC1 C. (n+1)C3 D. nC3


Answer : Option D

Explanation :

[Reference : Number of triangles formed by joining n points out of which no three points are collinear]

Consider there be n points in a plane out of which no three points are collinear. The number of triangles that can be formed by joining these n points are n(n−1)(n−2) 6

Since all these m points are on a circle, no three points are collinear.

Hence the required number of triangles =n(n−1)(n−2) 6 = nC3


54. In how many ways can 10 books be arranged on a shelf such that a particular pair of books should always be together? A. 9! × 2! B. 9! C. 10! × 2! D. 10!


Answer : Option A

Explanation :

We have a total of 10 books. Given that a particular pair of books should always be together. Hence, just tie these two books together and consider as a single book. Hence we can take total number of books as 9. These 9 books can be arranged in 9P9 = 9! Ways We had tied two books together. These books can be arranged among themselves in 2P2 = 2! Ways Hence, the required number of ways = 9! × 2!

55. In how many ways can 10 books be arranged on a shelf such that a particular pair of books will never be together? A. 9! × 8 B. 9!


C. 9! × 2! D. 10! × 2!


Answer : Option A

Explanation :

Total number of ways in which we can arrange 10 books on a shelf = 10P10 = 10! -------(A) Now we will find out the total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will always be together. We have a total of 10 books. If a particular pair of books need always be together, just tie these two books together and consider as a single book. Hence we can take total number of books as 9. These 9 books can be arranged in 9P9 = 9! Ways We had tied two books together. These books can be arranged among themselves in 2P2 = 2! Ways Hence, total number of ways in which 10 books can be arranged on a shelf such that


a particular pair of books will always be together = 9! × 2! ------(B) From (A) and (B), Total number of ways in which 10 books can be arranged on a shelf such that a particular pair of books will never be together = 10! – (9! × 2!) = 10! – (9! × 2) = (9! × 10) - (9! × 2) = 9!(10-2) = 9! × 8

56. Arun wants to send invitation letter to his 7 friends. In how many ways can he send the invitation letter if he has 4 servants to carry the invitation letters A. 16384 B. 10801 C. 14152 D. 12308


Answer : Option A

Explanation :

The 1st friend can be invited by any of the 4 servants. Similarly each of the remaining 6 friends can be invited by any of the 4 servants


Hence total number of ways = 47 = 16384 (In this question, you do not want to waste time by expanding 47. We know that any power of 4 can only end with 4 or 6 . (Because 4 × 4 = 16, 6×4 = 24, ...) In the given choices, there is only one answer which is ending with 4 which is 16384. Hence, this must be the answer. You can refer the section 'numbers' for more information on this)

57. How many three digit numbers divisible by 5 can be formed using any of the digits from 0 to 9 such that none of the digits can be repeated? A. 108 B. 112 C. 124 D. 136


Answer : Option D

Explanation :

A number is divisible by 5 if the its last digit is a 0 or 5 read more …

We need to find out how many 3 digit numbers divisible by 5 can be formed from the 10 digits (0,1,2,3,4,5,6,7,8,9) without repetition.


Since the 3 digit number should be divisible by 5, we can take the digit 0 or 5 from the 10 digits(0,1,2,3,4,5,6,7,8,9) fix it at the unit place. We will deal take these as two cases Case 1 : Number of three digit numbers using the 10 digits (0,1,2,3,4,5,6,7,8,9) ending with 0 We take the digit 0 and fix it at the unit place. There is only 1 way of doing this

1

Since the number 0 is placed at unit place, we have now 9 digits(1,2,3,4,5,6,7,8,9) remaining. Any of these 9 digits can be placed at tenth place.

9 1

Since the digit 0 is placed at unit place and another one digits is placed at tenth place, we have now 8 digits remaining. Any of these 8 digits can be placed at hundredth place.


8 9 1

Total number of 3 digit numbers using the digits (0,1,2,3,4,5,6,7,8,9)ending with 0 = 8 × 9 × 1 = 72 -----------------(A) Case 2 : Number of three digit numbers using the 10 digits (0,1,2,3,4,5,6,7,8,9) ending with 5 we take the digit 5 and fix it at the unit place. There is only 1 way of doing this.

1

Since the number 5 is placed at unit place, we have now 9 digits(0,1,2,3,4,6,7,8,9) remaining. But, from the remaining digits, 0 cannot be used for hundredth place. Hence any of 8 digits (1,2,3,4,6,7,8,9) can be placed at hundredth place.

8 1

Since the digit 5 is placed at unit place and another one digits is placed at hundredth


place, we have now 8 digits remaining. Any of these 8 digits can be placed at tenth place.

8 8 1

Total number of 3 digit numbers using the digits (0,1,2,3,4,5,6,7,8,9) ending with 5 = 8 × 8 × 1 = 64 -----------------(B) Hence, required number of 3 digit numbers = 72 + 64 = 136 (∵ from A and B)

58. How many numbers, between 100 and 1000, can be formed with the digits 3, 4, 5, 0, 6, 7? (Repetition of digits is not allowed) A. 142 B. 120 C. 100 D. 80


Answer : Option C

Explanation :

Here we can take only 3 digit numbers which will be between 100 and 1000. We have 6 digits (3, 4, 5, 0, 6, 7). But in these 6 digits, 0 cannot


be used at the hundredth place . Hence any of the 5 digits (3, 4, 5, 6, 7) can be placed at hundredth place.

5

Since one digit is placed at hundredth place, we have 5 digits remaining. Any of these 5 digits can be placed at unit place.

5 5

Since one digit is placed hundredth place and another digit is placed at unit place, we have 4 digits remaining. Any of these 4 digits can be placed at tenth place.

5 4 5

Total number of 3 digit numbers using the digits (3, 4, 5, 0, 6, 7) = 5 × 4 × 5 = 100 Hence, required number = 100

59. A telegraph has 10 arms and each arm can take 5 distinct


positions (including position of the rest). How many signals can be made by the telegraph? A. 10P5 B. 510 - 1 C. 510 D. 910P5 - 1


Answer : Option B

Explanation :

The 1st arm can take any of the 5 distinct positions Similarly, each of the remaining 9 arms can take any of the 5 distinct positions Hence total number of signals = 510 But there is one arrangement when all of the arms are in rest. In this case there will not be any signal. Hence required number of signals = 510 - 1

60. There are two books each of 5 volumes and two books each of two volumes. In how many ways can these books be arranged in a shelf so that the volumes of the same book should remain together? A. 4! × 5! × 2! B. 4! ×14! C. 14! D. 4! × 5! × 5! × 2! × 2!



Answer : Option D

Explanation :

1 book : 5 volume 1 book : 5 volume 1 book : 2 volume 1 book : 2 volume Given that volumes of the same book should remain together. Hence, just tie the same volume books together and consider as a single book. Hence we can take total number of books as 4. These 4 books can be arranged in 4P4 = 4! Ways The 5 volumes of the 1st book can be arranged among themselves in 5P5 = 5! Ways The 5 volumes of the 2st book can be arranged among themselves in 5P5 = 5! Ways The 2 volumes of the 3rd book can be arranged among themselves in 2P2 = 2! Ways The 2 volumes of the 4th book can be arranged among themselves


in 2P2 = 2! Ways Hence total number of ways = 4! × 5! × 5! × 2! × 2!

61. In how many ways can 11 persons be arranged in a row such that 3 particular persons should always be together? A. 9! × 3! B. 9! C. 11! D. 11! × 3!


Answer : Option A

Explanation :

Given that three particular persons should always be together. Hence, just group these three persons together and consider as a single person. Hence we can take total number of persons as 9. These 9 persons can be arranged in 9P9 = 9! Ways We had grouped three persons together. These three persons can be arranged among themselves in 3P3 = 3! Ways Hence, the required number of ways = 9! × 3!


62. In how many ways can 9 different colour balls be arranged in a row so that black, white, red and green balls are never together? A. 146200 B. 219600 C. 314562 D. 345600


Answer : Option D

Explanation :

Total number of ways in which 9 different colour balls can be arranged in a row = 9P9 = 9! ------- (A) Now we will find out the total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are always together. We have total 9 balls. Since black, white, red and green balls are always together, group these 4 balls together and consider as a single ball. Hence we can take total number of balls as 6. These 6 balls can be arranged in 6P6 = 6! Ways We had grouped 4 balls together. These 4 balls can be arranged


among themselves in 4P4 = 4! ways Hence, total number of ways in which 9 different colour balls be arranged in a row so that black, white, red and green balls are always together = 6! × 4! ------(B) From (A) and (B), Total number of ways in which 9 different colour balls can be arranged in a row so that black, white, red and green balls are never together = 9! – (6! × 4!) = (6! × 7 × 8 × 9) - (6! × 4!) = 6! (7 × 8 × 9 – 4!) = 6! (504 – 24) = 6! × 480 = 720 × 480 = 345600

63. A company has 11 software engineers and 7 civil engineers. In how many ways can they be seated in a row so that no two of the civil engineers will sit together? A. 12! B. 11!×12! 5! C. 11! D. 12×12! 5!


Answer : Option B

Explanation :

The 11 software engineers can be arranged in 11P11 = 11! Ways ---(A)


Now we need to arrange civil engineers such that no two civil engineers can be seated together. i.e., we can arrange 7 civil engineers in any of the 12 (=11+1) positions marked as * below * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 *10 * 11 * (Where 1, 2… 11 represents software engineers) This can be done in 12P7 ways ---(B) From (A) and (B), the required number of ways = 11! × 12P7

=11!×12! 5!

64. A company has 11 software engineers and 7 civil engineers. In how many ways can they be seated in a row so that all the civil engineers do not sit together? A. 18! – (12! × 7!) B. 18P411 - 2! C. 18P4 × 11 D. 18! – (11! × 7!)


Answer : Option A

Explanation :


Total number of engineers = 11 + 7 = 18 Total number of ways in which the 18 engineers can be arranged in a row = 18P18 = 18! ---(A) Now we will find out the total number of ways in which the 18 engineers can be arranged so that all the 7 civil engineers will always sit together. For this, group all the 7 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 12. (∵ 11 + 1 ) These 12 engineers can be arranged in 12P12 = 12! ways We had grouped 7 civil engineers. But these 7 civil engineers can be arranged among themselves in 7P7 = 7! ways Hence, total number of ways in which the 18 engineers can be arranged so that the 7 civil engineers will always sit together = 12! × 7! ---(B) From (A) and (B), Total number of ways in which 11 software engineers and 7 civil engineers can be seated in a row so that all the civil engineers will not sit together = 18! – (12! × 7!)


65. In how many ways can 11 software engineers and 10 civil engineers be seated in a row so that they are positioned alternatively? A. 7! × 7! B. 6! × 7! C. 10! × 11! D. 11! × 11!


Answer : Option C

Explanation :

The 10 civil engineers can be arranged in a row in 10P10 = 10! Ways ---(A) Now we need to arrange software engineers such that software engineers and civil engineers are seated alternatively. i.e., we can arrange 11 software engineers in the 11 positions marked as * below * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 *10 * (Where 1, 2… 10 represents civil engineers) This can be done in 11P11 = 11! ways ---(B) From (A) and (B), The required number of ways = 10! × 11!


66. In how many ways can 10 software engineers and 10 civil engineers be seated in a row so that they are positioned alternatively? A. 2 × (10!)2 B. 2 × 10! × 11! C. 10! × 11! D. (10!)2


Answer : Option A

Explanation :

The 10 civil engineers can be arranged in a row in 10P10 = 10! Ways ---(A) Now we need to arrange software engineers such that software engineers and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers either in the 10 positions marked as A,B,C,D,E,F,G,H,I,J or in the 10 positions marked as B,C,D,E,F,G,H,I,J,K, as shown below

10 software engineers can be arranged in the 10 positions marked as A,B,C,D,E,F,G,H,I,J in 10P10 = 10! Ways


10 software engineers can be arranged in the 10 positions marked as B,C,D,E,F,G,H,I,J,K in 10P10 = 10! Ways 10 software engineers can be arranged in the 10 positions marked as A,B,C,D,E,F,G,H,I,J or in the 10 positions marked as B,C,D,E,F,G,H,I,J,K in 10! + 10! = 2 × 10! Ways ---(B) From (A) and (B), The required number of ways = 10! × (2 × 10!) = 2 × (10!)2

67. Kiran has 8 black balls and 8 white balls. In how many ways can he arrange these balls in a row so that balls of different colours are alternate? A. 8! × 7! B. 2 × 8! × 7! C. 2 × (8!)2 D. (8!)2


Answer : Option C

Explanation :

The 8 black balls can be arranged in 8P8 = 8! Ways ---(A) Now we need to arrange white balls such that white balls and black balls are positioned alternatively. i.e., we can arrange 8 white balls either in the 8


positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I as shown below

8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H in 8P8 = 8! Ways 8 white balls can be arranged in the 8 positions marked as B,C,D,E,F,G,H,I in 8P8 = 8! Ways 8 white balls can be arranged in the 8 positions marked as A,B,C,D,E,F,G,H or in the 8 positions marked as B,C,D,E,F,G,H,I in 8! + 8! = 2 × 8! Ways ---(B) From (A) and (B), the required number of ways = 8! × 2 × 8! = 2 × (8!)2

68. A company has 11 software engineers and 7 civil engineers. In how many ways can they be seated in a row so that all the civil engineers are always together? A. 18! × 2 B. 12! × 7! C. 11! × 7! D. 18!



Answer : Option B

Explanation :

All the 7 civil engineers are always together. Hence, group all the 7 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 12. (∵ 11 + 1 ) These 12 engineers can be arranged in 12P12 = 12! Ways ---(B) We had grouped 7 civil engineers. These 7 civil engineers can be arranged among themselves in 7P7 = 7! Ways ---(B) From (A) and (B), The required number of ways = 12! × 7!

69. A company has 10 software engineers and 6 civil engineers. In how many ways can they be seated in a round table so that no two of the civil engineers will sit together? A. 15! B. 9!×10! 4! C. 10!×11! 5! D. 16!



Answer : Option B

Explanation :

10 software engineers can be arranged in a round table in (10-1)! = 9! ways ---(A) [Reference : Circular Permutations: Case 1] Now we need to arrange civil engineers such that no two civil engineers can be seated together. i.e., we can arrange 6 civil engineers in any of the 10 positions marked as * below

This can be done in 10P6 ways ---(B) From (A) and (B), The required number of ways = 9! × 10P6

=9!×10! 4!

70. A company has 10 software engineers and 6 civil engineers. In how many ways can they be seated in a round table so that all the civil engineers do not sit together?


A. 16! – (11! × 6!) B. 15! – (10! × 6!) C. 16! D. 15!


Answer : Option B

Explanation :

Total number of engineers = 10 + 6 = 16 Total number of ways in which the 16 engineers can be arranged in a round table = (16 – 1)! = 15!---(A) [Reference : Circular Permutations: Case 1] Now we will find out the total number of ways in which the 16 engineers can be arranged in a round table so that all the 6 civil engineers will always sit together. For this, group all the 6 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 11. (∵ 10 + 1 ) These 11 engineers can be arranged in a round table in (11-1)! = 10! ways We had grouped 6 civil engineers. These 6 civil engineers can be arranged among themselves in 6P6 = 6! ways


Hence, total number of ways in which the 16 engineers can be arranged in a round table so that the 6 civil engineers will always sit together = 10! × 6! ---(B) From (A) and (B), Total number of ways in which 10 software engineers and 6 civil engineers can be seated in a round table so that all the civil engineers do not sit together = 15! – (10! × 6!)

71. In how many ways can 10 software engineers and 10 civil engineers be seated in a round table so that they are positioned alternatively? A. 9! × 10! B. 10! × 10! C. 2 × (10!)2 D. 2 × 9! × 10!


Answer : Option A

Explanation :

The 10 civil engineers can be arranged in a round table in (10-1)! = 9! Ways ---(A) [Reference : Circular Permutations: Case 1] Now we need to arrange software engineers the round table such that software engineers


and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers in the 10 positions marked as * as shown below

This can be done in 10P10 = 10! Ways ---(B) From (A) and (B), The required number of ways = 9! × 10!

72. A company has 10 software engineers and 6 civil engineers. In how many ways can they be seated in a round table so that all the civil engineers are together? A. 10! × 6! B. 11! × 6! C. (10!)2 D. 9! × 6!


Answer : Option A

Explanation :


We need to find out the total number of ways in which 10 software engineers and 6 civil engineers can be arranged in a round table so that all the 6 civil engineers will always sit together. For this, group all the 6 civil engineers and consider as a single civil engineer. Hence, we can take total number of engineers as 11. (∵ 10 + 1 ) These 11 engineers can be arranged in a round table in (11-1)! = 10! ways ---(A) [Reference : Circular Permutations: Case 1] We had grouped 6 civil engineers. These 6 civil engineers can be arranged among themselves in 6P6 = 6! ways ---(B) From (A) and (B), The required number of ways = 10! × 6!

73. How many 8 digit mobile numbers can be formed if any digit can be repeated and 0 can also start the mobile number? A. 10P8 B. 108 C. 10P7 D. 10P6



Answer : Option B

Explanation :

Here the digits(0,1,2,3,4,5,6,7,8,9) can be repeated and 0 can also be used to start the mobile number. Hence, any of the 10 digits can be placed at any place of the 8 digit number

10 10 10 10 10 10 10 10

Hence, the total number of 8 digit mobile numbers that can be formed using all the digits (0,1,2,3,4,5,6,7,8,9) (with repetition of the digits and 0 can also be used to start the number) = 108

74. How many 8 digits mobile numbers can be formed if at least one of their digits is repeated and 0 can also start the mobile number? A. 108 - 10P7 B. 107 C. 108 D. 108 - 10P8


Answer : Option D


Explanation :

Initially we will find out the number of 8 digits mobile numbers that can be formed if any digit can be repeated (with 0 can also start the mobile number) The digits can be repeated and 0 can also be used to start the mobile number. Hence, any of the 10 digits(0,1,2,3,4,5,6,7,8,9) can be placed at any place of the 8 digit number

10 10 10 10 10 10 10 10

Hence, the total number of 8 digit mobile numbers that can be formed using all the digits (0,1,2,3,4,5,6,7,8,9) if any digit can be repeated (with 0 can also start the mobile number) = 108 ---(A) Now we will find out the number of 8 digits mobile numbers that can be formed if no digit can be repeated (with 0 can also start the mobile number) In this case, any of the 10 digits can be placed at the 1st position. Since one digit is placed at the 1st position, any of the remaining


9 digits can be placed at 2nd position. Since 1 digit is placed at the 1st position and another digit is placed at the 2nd position, any of the remaining 8 digits can be placed at the 3rd position. So on …

10 9 8 7 6 5 4 3

i.e., the number of 8 digits mobile numbers that can be formed if no digit can be repeated (with 0 can also start the mobile number) = 10P8 ---(B) (In fact you should directly get (A) and (B) without any calculations from the definition of permutations itself) From(A) and (B), the number of 8 digits mobile numbers that can be formed if at least one of their digits is repeated and 0 can also start the mobile number = 108 - 10P8


75. How many 8 digits mobile numbers can be formed if at least one of their digits is repeated and 0 cannot be used to start the mobile number? A. 108 - 10P7 B. 107 C. 9 × 107 - 9 × 9P7 D. 108 - 10P8


Answer : Option C

Explanation :

Initially we will find out the number of 8 digits mobile numbers that can be formed if any digit can be repeated and 0 cannot be used to start the mobile number The digits can be repeated. 0 cannot be used to start the mobile number. Hence, any of the 9 digits (∵ any digit except 0) can be placed at the 1st position. Then, any of the 10 digits can be placed at any of the the remaining 7 positions of the 8 digit number

9 10 10 10 10 10 10 10


Hence, the total number of 8 digit mobile numbers that can be formed using all the digits (0,1,2,3,4,5,6,7,8,9) if any digit can be repeated and 0 cannot be used to start the mobile number = 9 × 107---(A) Now we will find out the number of 8 digits mobile numbers that can be formed if no digit can be repeated and 0 cannot be used to start the mobile number Here, any of the 9 digits (∵ any digit except 0) can be placed at the 1st position. Since one digit is placed at the 1st position, any of the remaining 9 digits can be placed at 2nd position. Since 1 digit is placed at the 1st position and another digit is placed at the 2nd position, any of the remaining 8 digits can be placed at the 3rd position. So on …

9 9 8 7 6 5 4 3


i.e., the number of 8 digits mobile numbers that can be formed if no digit can be repeated and 0 cannot be used to start the mobile number = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 9 × 9P7 ---(B) From(A) and (B), the number of 8 digits mobile numbers that can be formed if at least one of their digits is repeated and 0 cannot be used to start the mobile number = 9 × 107 - 9 × 9P7

76. How many signals can be made using 6 different coloured flags when any number of them can be hoisted at a time? A. 1956 B. 1720 C. 2020 D. 1822


Answer : Option A

Explanation :

Given that any number of flags can be hoisted at a time. Hence we need to find out the number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.


The number of signals that can be made using 1 flag = 6P1 = 6 The number of signals that can be made using 2 flags = 6P2 = 6 × 5 = 30 The number of signals that can be made using 3 flags = 6P3 = 6 × 5 × 4= 120 The number of signals that can be made using 4 flags = 6P4 = 6 × 5 × 4 × 3 = 360 The number of signals that can be made using 5 flags = 6P5 = 6 × 5 × 4 × 3 × 2 = 720 The number of signals that can be made using 6 flags = 6P6 = 6 × 5 × 4 × 3 × 2 × 1 = 720 Total number of signals = 6 + 30 + 120 + 360 + 720 + 720 = 1956 i.e., the required number of signals = 1956

77. How many possible outcomes are there when five dice are rolled in which at least one dice shows 6? A. 65 - 55 B. 66 - 56 C. 65 D. 56


Answer : Option A


Explanation :

Initially we will find out the total number of possible outcomes when 5 dice are rolled. Outcome of first die can be any number from (1,2,3,4,5,6). i.e, outcome of first die can happen in 6 ways Similarly outcome of each of the other 4 dice can also happen in 6 ways

6 6 6 6 6

Hence, total number of possible outcomes when 5 dice are rolled = 65 ---(A) Now we will find out the total number of possible outcomes when 5 dice are rolled in which 6 does not appear in any dice. In this case, outcome of first die can be any number from (1,2,3,4,5). i.e, outcome of first die can happen in 5 ways. Similarly outcome of each of the other 4 dice can also happen in 5 ways


5 5 5 5 5

Hence, total number of possible outcomes when 5 dice are rolled in which 6 does not appear in any dice = 55 ---(B) From (A) and (B), the total number of possible outcomes when five dice are rolled in which at least one dice shows 6 = 65 - 55

78. A board meeting of a company is organized in a room for 24 persons along the two sides of a table with 12 chairs in each side. 6 persons wants to sit on a particular side and 3 persons wants to sit on the other side. In how many ways can they be seated? A. 12P5 × 12P2 × 14! B. 12P5 × 12P2 × 15! C. 12P6 × 12P3 × 15! D. 12P6 × 12P3 × 14!


Answer : Option C

Explanation :

First, arrange the 6 persons in the 12 chairs on the particular side. The 6 persons can sit in the 12 chairs on the particular side in 12P6 ways. ---(A)


Now, arrange the 3 persons in the 12 chairs on the other side. The 3 persons can sit in the 12 chairs on the other side in 12P3 ways. -----(B) Remaining persons = 24 – 6 – 3 = 15 Remaining chairs = 24 – 6 – 3 = 15 i.e., now we need to arrange the remaining 15 persons in the remaining 15 chairs. This can be done in 15P15 = 15! ways. -----(C) From (A), (B) and (C), Required number of ways = 12P6 × 12P3 × 15!

79. How many numbers not exceeding 10000 can be made using the digits 2,4,5,6,8 if repetition of digits is allowed? A. 9999 B. 820 C. 780 D. 740


Answer : Option C

Explanation :

Given that the numbers should not exceed 10000


Hence numbers can be 1 digit numbers or 2 digit numbers or 3 digit numbers or 4 digit numbers Given that repetition of the digits is allowed. A. Count of 1 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) The unit digit can be filled by any of the 5 digits (2,4,5,6,8)

5

Hence the total count of 1 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) = 5 ---(A) B. Count of 2 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) Since repetition is allowed, any of the 5 digits(2,4,5,6,8) can be placed in unit place and tens place.

5 5

Hence the total count of 2 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) = 52 ---(B)


C. Count of 3 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) Since repetition is allowed, any of the 5 digits (2,4,5,6,8) can be placed in unit place , tens place and hundreds place.

5 5 5

Hence the total count of 3 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) = 53 ---(C) D. Count of 4 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) Since repetition is allowed, any of the 5 digits (2,4,5,6,8) can be placed in unit place, tens place, hundreds place and thousands place

5 5 5 5

Hence the total count of 4 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed) = 54 ---(D) From (A), (B), (C), and (D), total count of numbers not exceeding 10000 that can be made using the digits 2,4,5,6,8 (with repetition of digits)


= 5 + 52 + 53 + 54 =5(5 4 –1) 5−1 [∵ Reference: Sum of first n terms in a geometric progression (G.P.) ]

=5(625–1) 4 =5(624) 4 =5×156=780

80. How many 5 digit numbers can be formed using the digits 1,2,3,4,… 9 such that no two consecutive digits are the same? A. None of these B. 9 × 84 C. 95 D. 85


Answer : Option B

Explanation :

Here, no two consecutive digits can be the same The ten thousands place can be filled by any of the 9 digits (1,2,3,4,… 9)

9

Repletion is allowed here. Only restriction is that no two consecutive digits can be the same. Hence the digit we placed in the ten thousands place cannot be used at


the thousands place. Hence thousands place can be filled by any of the 8 digits.

9 8

Similarly, hundreds place, tens place and unit place can be filled by any of the 8 digits

9 8 8 8 8

Hence, the required count of 5 digit numbers that can be formed using the digits 1,2,3,4,… 9 such that no two consecutive digits are same = 9 × 84

81. In how many ways can 5 blue balls, 4 white balls and the rest 6 of different colour balls be arranged in a row? A. 15! B. 15! 5!×4! C. 15P6 D. 15P7


Answer : Option B

Explanation :


[Reference : Permutations of Objects when All Objects are Not Distinct]

The number of ways in which n things can be arranged taking them all at a time, when p1 of the things are exactly alike of 1st type, p2 of them are exactly alike of a 2nd type, and pr of them are exactly alike of rth type and the rest of all are distinct is

n! p 1 ! p 2 ! ... p r !

Here, all the balls are not different. Total number of balls= 5 + 4 + 6 = 15 Number of blue balls= 5 Number of white balls= 4 Rest 6 balls are of different colours From the above given formula, the required number of arrangements

=15! 5!×4!

82. A company has 10 software engineers and 6 civil engineers. In how many ways can a committee of 4 engineers be formed from them such that the committee must contain exactly 1 civil engineer?


A. 800 B. 720 C. 780 D. 740


Answer : Option B

Explanation :

The committee should have 4 engineers. But the committee must contain exactly 1 civil engineer. Hence, select 3 software engineers from 10 software engineers and select 1 civil engineer from 6 civil engineers Total number of ways this can be done = 10C3 × 6C1

=10×9×8 3×2×1 ×6=10×9×8=720

83. A company has 10 software engineers and 6 civil engineers. In how many ways can a committee of 4 engineers be formed from them such that the committee must contain at least 1 civil engineer? A. 1640 B. 1630 C. 1620 D. 1610


Answer : Option D

Explanation :


The committee should have 4 engineers. But the committee must contain at least 1 civil engineer. Initially we will find out the number of ways in which a committee of 4 engineers canbe formed from 10 software engineers and 6 civil engineers. Total engineers = 10 + 6 = 16 Total engineers in the committee = 4 Hence, the number of ways in which a committee of 4 engineers can be formed from 10 software engineers and 6 civil engineers = 16C4 --------------(A) Now we will find out the number of ways in which a committee of 4 engineers can be formed from 10 software engineers and 6 civil engineers such that the committee must not contain any civil engineer For this, select 4 software engineers from 10 software engineers. Hence the number of ways in which a committee of 4 engineers can be formed from 10 software engineers and 6 civil engineers such that the committee must not contain any civil engineer


= 10C4 --------------(B) From (A) and (B), The number of ways in which a committee of 4 engineers can be formed from 10 software engineers and 6 civil engineers such that the committee must contain at least 1 civil engineer = 16C4 -

10C4

=16×15×14×13 4×3×2×1 −10×9×8×7 4×3×2×1 =4×15×14×13 3×2 −10×9×2×7 3×2

=4×5×14×13 2 −10×3×2×7 2

=2×5×14×13−10×3×7=1820 –210=1610

84. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. In how many ways can the committee be formed if two of the men refuses to serve together? A. 1020 B. 640 C. 712 D. 896


Answer : Option D

Explanation :


Let the men be X and Y who refuses to serve together Let's find out the number of ways in which the committee can be formed by excluding both X and Y We excluded both X and Y. Hence we need to select 3 men from 4 men (=6-2) and 3 women from 8 women. The number of ways in which this can be done = 4C3 × 8C3 ---(A) Now let's find out the number of ways in which the committee can be formed where exactly one man from X and Y will be present. i.e., we need to select one man from two men(X and Y), remaining 2 men from 4 men(=6-2) and 3 women from 8 women. The number of ways in which this can be done = 2C1 × 4C2 × 8C3 ---(B) From (A) and (B), The number of ways in which a committee be formed if two of the men refuses to serve together = 4C3 × 8C3 + 2C1 × 4C2 × 8C3 = 8C3(

4C3 + 2C1 ×4C2)

= 8C3(

4C1 + 2C1 × 4C2) [∵ nCr = nC(n - r)]


=(8×7×6 3×2×1 )[4+2(4×3 2×1 )]

=(8×7)[4+(4×3)]=56[4+12]=56×16=896

85. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. In how many ways can the committee be formed if two of the women refuses to serve together? A. 1020 B. 1000 C. 712 D. 896


Answer : Option B

Explanation :

Let the women be X and Y who refuses to serve together Let's find out the number of ways in which the committee can be formed by excluding both X and Y We excluded both X and Y. Hence we need to select 3 women from 6 women (=8-2) and 3 men from 6 men. The number of ways in which this can be done = 6C3 × 6C3 ---(A) Now let's find out the number of ways in which the committee can be formed where


exactly one woman from X and Y will be present. i.e., we need to select one woman from two women(X and Y), remaining 2 women from 6 women(=8-2) and 3 men from 6 men. The number of ways in which this can be done = 2C1 × 6C2 × 6C3 ---(B) From (A) and (B), The number of ways in which a committee be formed if two of the women refuses to serve together = 6C3 × 6C3 + 2C1 × 6C2 × 6C3 = 6C3(

6C3 + 2C1 × 6C2 )

=(6×5×4 3×2×1 )[(6×5×4 3×2×1 )+2(6×5 2×1 )]

=(8×7)[4+(4×3)]=20[20+30]=20×50=1000

86. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. In how many ways can the committee be formed if one man and one woman refuses to serve together? A. 722 B. 910 C. 612 D. 896


Answer : Option B

Explanation :


Let the man be X and woman be Y who refuses to serve together Let's find out the number of ways in which the committee can be formed by excluding both X and Y We excluded both X and Y. Hence we need to select 3 men from 5 men (=6-1) and 3 women from 7 women(=8-1). The number of ways in which this can be done = 5C3 × 7C3 ---(A) Now let's find out the number of ways in which the committee can be formed where X is present and Y is not present. Since X is present, we need to select 2 more men from 5 men (=6-1). Since Y is not present, we need to select 3 women from 7 women (=8-1). The number of ways in which this can be done = 5C2 × 7C3 ---(B) Now let's find out the number of ways in which the committee can be formed where Y is present and X is not present. Since X is not present, we need to select 3 men from 5 men (=6-1).


Since Y is present, we need to select 2 more women from 7 women (=8-1). The number of ways in which this can be done = 5C3 × 7C2 ---(C) From (A),(B) and (C), The number of ways in which a committee be formed if one man and one woman refuses to serve together =5C3 × 7C3 + 5C2 × 7C3 + 5C3 × 7C2 = 5C2 × 7C3 + 5C2 × 7C3 +

5C2 × 7C2 [∵ nCr = nC(n - r)] = 5C2(

7C3 + 7C3+7C2)

=(5×4 2×1 )[(7×6×5 3×2×1 )+(7×6×5 3×2×1 )+(7×6 2×1 )]

=10[35+35+21]=10×91=910

87. A box contains 20 balls. In how many ways can 8 balls be selected if each ball can be repeated any number of times? A. 20C7 B. None of these C. 20C8 D. 27C8


Answer : Option D

Explanation :


It is a question of combination with repetition [Reference : Combinations with Repetition]

Number of combinations of n distinct things taking r at a time when each thing may be repeated any number of times is (n+r-1)Cr

Here, n=20, r=8 Hence, require number of ways = (n+r-1)Cr = (20+8-1)C8 = 27C8

88. A box contains 12 black balls, 7 red balls and 6 blue balls. In how many ways can one or more balls be selected? A. 696 B. 728 C. 727 D. 896


Answer : Option C

Explanation :

[Reference : Total Number of Combinations : Case 2]

Number of ways of selecting one or more than one objects out of S1 alike objects of one kind, S2 alike objects of the second kind and S3 alike objects of the third kind is

(S1 + 1) (S2 + 1)(S3 + 1) - 1

Hence, require number of ways = (12 + 1)(7 + 1)(6 + 1) – 1 = (13 × 8 × 7) – 1 = 728- 1 = 727


89. A box contains 12 different black balls, 7 different red balls and 6 different blue balls. In how many ways can the balls be selected? A. 728 B. 225 - 1 C. 225 D. 727


Answer : Option B

Explanation :


Total number of combinations is the total number of ways of selecting one or more than one things from n distinct things . i.e., we can select 1 or 2 or 3 or … or n items at a time.

Total number of combinations = nC1 + nC2 + ... + nCn = 2n - 1

It is explicitly stated that 12 black balls are different, 7 red balls are different and 6 blue balls are different. Hence there are 25(=12+ 7+ 6) different balls. We can select one ball from 25 balls, two balls from 25 balls, … 25 balls from 25 balls. Hence, required number of ways = Number of ways in which 1 ball can be selected from 25 distinct balls


+ Number of ways in which 2 balls can be selected from 25 distinct balls + Number of ways in which 3 balls can be selected from 25 distinct balls . . . + Number of ways in which 25 balls can be selected from 25 distinct balls = 25C1 + 25C2 + ... + 25C25 = 225 - 1

90. There are 12 copies of Mathematics, 7 copies of Engineering, 3 different books on Medicine and 2 different books on Economics. Find the number of ways in which one or more than one book can be selected? A. 3421 B. 3111 C. 3327 D. 3201


Answer : Option C

Explanation :

[Reference :Total Number of Combinations : Case 3]

Number of ways of selecting one or more than one objects out of S1 alike objects of one kind, S2 alike objects of the second kind and rest p different objects is

(S1 + 1) (S2 + 1)2p - 1


12 copies of mathematics are there. These 12 copies can be considered as identical. 7 copies of Engineering are there. These 7 copies can be considered as identical. 3 different books on Medicine and 2 different books on Economics are these. i.e., there are 5 (=3+2) different books also Hence, the required number of ways = (12 + 1)(7 + 1)25 - 1 = 13 × 8 × 32 – 1 = 3328 – 1 = 3327

91. A box contains 4 different black balls, 3 different red balls and 5 different blue balls. In how many ways can the balls be selected if every selection must have at least 1 black ball and one red ball? A. (24 - 1)( 23 - 1) (25 - 1) B. (24 - 1)( 23 - 1) 25 C. 212 - 1 D. 212


Answer : Option B

Explanation :



Total number of combinations is the total number of ways of selecting one or more than one things from n distinct things . i.e., we can select 1 or 2 or 3 or … or n items at a time.

Total number of combinations = nC1 + nC2 + ... + nCn = 2n - 1

It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects. Initially let's find out the number of ways in which we can select the blackballs Given that at least 1 black ball must be included in each selection. Hence, we can select 1 black ball from 4 black balls or 2 black balls from 4 black balls or 3 black balls from 4 black balls or 4 black balls from 4 black balls Hence , the number of ways in which we can select the black balls = 4C1 + 4C2 + ... + 4C4 = 24 - 1 ---(A) Now let's find out the number of ways in which we can select the red balls


Given that at least 1 red ball must be included in each selection. Hence, we can select 1 red ball from 3 red balls or 2 red balls from 3 red balls or 3 red balls from 3 red balls Hence , the number of ways in which we can select the red balls = 3C1 + 3C2 + 3C3 = 23 - 1 ---(B) Now let's find out the number of ways in which we can select the blue balls Here, there is no specific condition given. Hence, we can select 1 blue ball from 5 blue balls or 2 blue balls from 5 blue balls or 3 blue balls from 5 blue balls or 4 blue balls from 5 blue balls or 5 blue balls from 5 blue balls. Also, there is one more possibility that we can select 0 blue balls from 5 blue balls (ie,only black and red balls are there) Hence, the number of ways in which we can select the blue balls = (5C1 + 5C2 + … + 5C5) + 5C0 = (25 - 1) + 1 = 25 ---(C) From (A), (B) and (C), the required number of ways = (24 - 1)( 23 - 1) 25


92. There are 10 different books and 20 copies of each book in a library. In how many ways can one or more than one book be selected? A. 2110 - 1 B. 2200 C. 2200 - 1 D. 2110


Answer : Option A

Explanation :


Number of ways of selecting one or more than one objects out of S1 alike objects of one kind, S2 alike objects of the second kind and S3 alike objects of the third kind is

(S1 + 1) (S2 + 1)(S3 + 1) - 1

There are 10 different books. Each book has 20 copies and all the copies of each particular book can be considered as identical. Hence, required number of ways = [(20 + 1)(20 + 1) … 10 times] – 1 = [21 × 21 × … 10 times] – 1 = 2110 - 1


93. In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls? A. 54 - 1 B. 54 C. 45 - 1 D. 45


Answer : Option B

Explanation :

-------------------------------------------------------------------------- Solution 1 : Using Formula -------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 1] Here n = 5, k = 4. Hence, as per the above formula, the required number of ways = nk = 54 -------------------------------------------------------------------------- Solution 2 : Using Concepts -------------------------------------------------------------------------- Here both balls and boxes are different. Hence the order of selection is important. 1st ball can be placed into any of the 5 boxes. 2nd ball can be placed into any of the 5 boxes. 3rd ball can be placed into any of the 5 boxes. 4th ball can be placed into any of the 5 boxes.


Hence, the required number of ways = 5 × 5 … (4 times) = 54 (If you got the answer as 45 , it is wrong. Let's see why 45 is wrong answer. One might have taken the following reasoning to have got the answer as 45 1st box can contain any number of balls from 4 balls 2nd box can contain any number of balls from the 4 balls … 5th box can contain any number of balls from the 4 balls Hence, the required number of ways = 4 × 4 … (5 times) = 45 But this reasoning is wrong. For instance, if the 1st box contains all the 4 balls, other boxes can not contain any balls. Due to such dependencies, we cannot apply the multiplication theorem in this way because multiplication theorem states that If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways Hence 54 is the correct answer, not 45)


94. In how many ways can three different balls be distributed among two different boxes when any box can have any number of balls? A. 23 - 1 B. 23 C. 32 - 1 D. 32


Answer : Option B

Explanation :

-------------------------------------------------------------------------- Solution 1 : Using Formula -------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 1] Here n = 2, k = 3. Hence, as per the above formula, the required number of ways = nk = 23 -------------------------------------------------------------------------- Solution 2 : Using Concepts -------------------------------------------------------------------------- Here both balls and boxes are different. Hence the order of selection is important. 1st ball can be placed into any of the 2 boxes 2nd ball can be placed into any of the 2 boxes 3rd ball can be placed into any of the 2 boxes

Hence, the required number of ways = 2 × 2 × 2 = 23


(If you got the answer as 32, it is wrong. Let's see why 32 is wrong answer. One might have taken the following reasoning to have got the answer as 32 1st box can contain any number of balls from 3 balls 2nd box can contain any number of balls from the 3 balls

Hence, the required number of ways = 3 × 3 = 32 But this reasoning is wrong. For instance, if the 1st box contains all the 3 balls, other boxes can not contain any balls. Due to such dependencies, we cannot apply the multiplication theorem in this way because multiplication theorem states that If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways Hence 23 is the correct answer, not 32)

95. In how many ways can 5 distinguishable balls be put into 8 distinguishable boxes if no box can contain more than one ball?


A. None of these B. 8P5 C. 85 D. 58


Answer : Option B

Explanation :

--------------------------------------------------------------------------------- Solution 1 : Using Formula --------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 2] Here n = 8, k = 5. Hence, as per the above formula, the required number of ways = nPk = 8P5 --------------------------------------------------------------------------------- Solution 2 : Using Concepts --------------------------------------------------------------------------------- 1st ball can be placed into any of the 8 boxes Since no box can contain more than one ball, we have 7 boxes remaining. 2nd ball can be placed into any of these 7 boxes Similarly 3rd ball can be placed into any of the remaining 6 boxes


4th ball can be placed into any of the remaining 5 boxes 5th ball can be placed into any of the remaining 4 boxes Hence, the required number of ways = 8 × 7 × 6 × 5 × 4 = 8P5

96. In how many ways can 8 distinguishable balls be put into 5 distinguishable boxes if no box can contain more than one ball? A. 0 B. 85 C. 8P5 D. 58


Answer : Option A

Explanation :

--------------------------------------------------------------------------------- Solution 1 : Using Formula --------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 2] Here n = 5, k = 8. Hence, as per the above formula, the required number of ways = nPk = 5P8 = 0 ---------------------------------------------------------------------------------


Solution 2 : Using Concepts --------------------------------------------------------------------------------- Clearly the answer is 0. Given that no box can contain more than one ball. Hence, the 1st ball can be put into any of the 5 boxes, 2nd ball can be put into any of the remaining 4 boxes, 3rd ball can be put into any of the remaining 3 boxes, 4th ball can be put into any of the remaining 2 boxes, 5th ball can be put into the remaining 1 box. But where will you put the remaining 3 balls? Clearly no way is there to make such a placement. Hence the answer is 0.

97. In how many ways can 8 distinguishable balls be put in to 5 distinguishable boxes if any box can contain more than one ball? A. 85 B. 8P5 C. None of these D. 58



Answer : Option D

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes : formula 1] Here n = 5, k = 8. Hence, as per the above formula, the required number of ways = nk = 58 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Here both balls and boxes are different. Hence the order of selection is important. Given that any box can contain more than one ball. Hence, 1st ball can be placed into any of the 5 boxes 2nd ball can be placed into any of the 5 boxes 3rd ball can be placed into any of the 5 boxes ... 8th ball can be placed into any of the 5 boxes Hence, the required number of ways = 5 × 5 ... (8 times) = 58


(If you got the answer as 85 , it is wrong. Let's see why 85 is wrong answer. One might have taken the following reasoning to have got the answer as 85 1st box can contain any number of balls from 8 balls 2nd box can contain any number of balls from the 8 balls … 5th box can contain any number of balls from the 8 balls Hence, the required number of ways = 8 × 8 ... (5 times) = 85 But this reasoning is wrong. For instance, if the 1st box contains all the 8 balls, other boxes can not contain any balls. Due to such dependencies, we cannot apply multiplication theorem in this way because multiplication theorem states that If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways Hence 58 is the correct answer, not 85)


98. In how many ways can 7 different balls be distributed in 5 different boxes if any box can contain any number of balls and no box is left empty? A. 16800 B. 12400 C. 22000 D. 19700


Answer : Option A

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes : formula 3] Here n = 5, k = 7. Hence, as per the above formula, the required number of ways = S(k,n) × n!

=∑ i=0 n−1 (−1) i n C i (n−i) k =∑ i=0 4 (−1) i 5 C i (5−i) 7 =5 C 0 (5) 7 −5 C 1 (4) 7 +5 C 2 (3) 7 −5 C 3 (2) 7 +5 C 4 (1) 7 =(5) 7 −5(4) 7 +10(3) 7 −10(2) 7 +5(1) 7 =78125−81920+21870−1280+5=16800


---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 1, 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 4 boxes) A box (in which 3 balls are put) can be selected in 5C1 ways Now, the three balls can be selected in 7C3 ways Remaining 4 balls can be arranged in 4! Ways Hence, total number of ways = 5C1 × 7C3 × 4! ---(A) Case B : 1, 1, 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in each of the remaining 3 boxes) The two boxes (in each of them, two balls are put) can be selected in 5C2 ways. Now, two balls for the first selected box can be selected in 7C2 ways


Two balls for the second selected box can be selected in 5C2 ways Remaining 3 balls can be arranged in 3! Ways Hence, total number of ways = 5C2 × 7C2 × 5C2 × 3! ---(B) From (A) and (B), The required number of ways = (5C1 × 7C3 × 4!) + (5C2 × 7C2 × 5C2 × 3!) = (5 × 35 × 24) + (10 × 21 × 10 × 6) = 4200 + 12600 = 16800

99. In how many ways can 7 identical balls be distributed in 5 different boxes if any box can contain any number of balls? A. 200 B. 330 C. 410 D. 390


Answer : Option B

Explanation :

[Reference : Distribution of k balls into n boxes: formula 5] Here n = 5, k = 7. Hence, as per the above formula, the required number of ways = (n+k-1)Ck = 11C7 = 330


100. In how many ways can 7 different balls be distributed in 5 different boxes if box 3 and box 5 can contain only 1 and 2 balls respectively and rest of the boxes can contain any number of balls? A. 10100 B. 6200 C. 8505 D. 12800


Answer : Option C

Explanation :

One ball for Box 3 can be selected in 7C1 ways Two balls for Box 5 can be selected in 6C2 ways Remaining balls = 4. Remaining boxes = 3 In these 4 balls, 1st ball can be put in any of these 3 boxes Similarly 2nd ball can be put in any of these 3 boxes 3rd ball can be put in any of these 3 boxes 4th ball can be put in any of these 3 boxes i.e., these 4 balls can be arranged in 3 × 3 × 3 × 3 = 34 ways The required number of ways = 7C1 × 6C2 × 34


= 7 × 15 × 81 = 8505

101. In how many ways can 7 different balls be distributed in 5 different boxes if any box can contain any number of balls except that ball 3 can only be put into box 3 or box 4? A. 2 × 56 B. 56 C. 65 D. 2 × 65


Answer : Option A

Explanation :

1st ball can be put in any of these 5 boxes 2nd ball can be put in any of these 5 boxes Given that Ball 3 can only be put into box 3 or box 4. Hence, 3rd ball can be put in any of the 2 boxes 4th ball can be put in any of these 5 boxes 5th ball can be put in any of these 5 boxes 6th ball can be put in any of these 5 boxes 7th ball can be put in any of these 5 boxes


Hence, the required number of ways = 2 × (5 × 5 × ... 6 times) = 2 × 56

102. In how many ways can 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 cannot be put in the same box? A. 16800 B. 15000 C. 17200 D. 16400


Answer : Option B

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- Initially let's find out the total number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls and no box can be empty [Reference : Distribution of k balls into n boxes : formula 3] Here n = 5, k = 7. Hence, as per the above formula,


the total number of ways = S(k,n) × n! =

=∑ i=0 n−1 (−1) i n C i (n−i) k =∑ i=0 4 (−1) i 5 C i (5−i) 7 =5 C 0 (5) 7 −5 C 1 (4) 7 +5 C 2 (3) 7 −5 C 3 (2) 7 +5 C 4 (1) 7 =(5) 7 −5(4) 7 +10(3) 7 −10(2) 7 +5(1) 7 =78125−81920+21870−1280+5=16800

Now let's find out the total number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls , no box can be empty and ball 3 and ball 5 are in the same box For this, tie ball 3 and ball 5 and consider it as a single ball. Hence, we can consider the total number of balls as 6. Using the same formula mentioned above, with n = 5 and k = 6, the total number of ways = S(k,n) × n! =

=∑ i=0 n−1 (−1) i n C i (n−i) k =∑ i=0 4 (−1) i 5 C i (5−i) 6 =5 C 0 (5) 6 −5 C 1 (4) 6 +5 C 2 (3) 6 −5 C 3 (2) 6 +5 C 4 (1) 6 =(5) 6 −5(4) 6 +10(3) 6 −10(2) 6 +5(1) 6 =15625–20480+7290–640+5=1800


Required Number of ways = (Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty) - (Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box) = 16800 – 1800 = 15000 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Initially let's find out the total number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls and no box can be empty Since no box can be left empty, there can be only two cases Case 1 : 1 , 1, 1, 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 4 boxes)


A box (in which 3 balls are put) can be selected in 5C1 ways Now, the three balls can be selected in 7C3 ways Remaining 4 balls can be arranged in 4! Ways Hence, total number of ways = 5C1 × 7C3 × 4! ---(A) Case 2 : 1, 1, 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in each of the remaining 3 boxes) The two boxes (in each of them, two balls are put) can be selected in 5C2 ways. Now, two balls for the first selected box can be selected in 7C2 ways Two balls for the second selected box can be selected in 5C2 ways Remaining 3 balls can be arranged in 3! Ways Hence, total number of ways = 5C2 × 7C2 × 5C2 × 3! ---(B) From (A) and (B), The total number of ways = (5C1 × 7C3 × 4!) + (5C2 × 7C2 × 5C2 × 3!) = (5 × 35 × 24) + (10 × 21 × 10 × 6) = 4200 + 12600 = 16800


Now let's find out the total number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box For this, tie ball 3 and ball 5 and consider it as a single ball. Hence, we can consider the total number of balls as 6. Now, with no box can be left empty, there can be only one case as given below. 1 , 1, 1, 1, 2 (i.e., 2 balls are put in 1 box and 1 ball is put in each of the remaining 4 boxes) A box (in which 2 balls are put) can be selected in 5C1 ways Now, the two balls can be selected in 6C2 ways Remaining 4 balls can be arranged in 4! Ways Hence, total number of ways = 5C1 × 6C2 × 4! = 5 × 15 × 24 = 5 × 15 × 12 × 2 = 10 × 15 × 12 = 1800


Required Number of ways = (Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty) - (Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box) = 16800 – 1800 = 15000

103. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if no box can be empty and all balls and boxes are different? A. 220 B. 150 C. 120 D. 190


Answer : Option B

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ------------------------------------------------------------------------------


---- [Reference: Distribution of k balls into n boxes: formula 3] Here n = 3, k =5. Hence, as per the above formula, the required number of ways = S(k,n) × n!

=∑ i=0 n−1 (−1) i n C i (n−i) k =∑ i=0 2 (−1) i 3 C i (3−i) 5 =3 C 0 (3) 5 −3 C 1 (2) 5 +3 C 2 (1) 5 =(3) 5 −3(2) 5 +3(1) 5 =243−96+3=150

---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 2 boxes) A box (in which 3 balls are put) can be selected in 3C1 ways Now, the three balls can be selected in 5C3 ways Remaining 2 balls can be arranged in 2! Ways Hence, total number of ways = 3C1 × 5C3 × 2! = 3 × 10 × 2 = 60---(A)


Case B : 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in the remaining 1 box) The two boxes (in each of them, two balls are put) can be selected in 3C2 ways. Now, two balls for the first selected box can be selected in 5C2 ways Two balls for the second selected box can be selected in 3C2 ways Remaining 1 ball can be placed only in 1 way Hence, total number of ways = 3C2 × 5C2 × 3C2 × 1 = 3 × 10 ×3 = 90 ---(B) From (A) and (B), The required number of ways = 60 + 90 = 150

104. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if no box can be empty , all balls are identical but all boxes are different? A. 8 B. 6


C. 4 D. 2


Answer : Option B

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 7] Here n = 3, k = 5. Hence, as per the above formula, the required number of ways = (k-1)C(n-1) = 4C2 = 6 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 2 boxes)


A box (in which 3 balls are put) can be selected in 3C1 ways Now, the three balls can be selected only in 1 way (as all the balls are identical) Remaining 2 balls can be arranged in only 1 way (as all the balls are identical) Hence, total number of ways = 3C1 = 3---(A) Case B : 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in the remaining 1 box) The two boxes (in each of them, two balls are put) can be selected in 3C2 ways. Now, two balls for the first selected box can be selected only in 1 way(as all the balls are identical) Two balls for the second selected box can be selected in only in 1 way(as all the balls are identical) Remaining 1 ball can be placed only in 1 way Hence, total number of ways = 3C2 = 3---(A) From (A) and (B), The required number of ways = 3 + 3 = 6


105. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if all balls are identical and all boxes are different? A. 32 B. 21 C. 18 D. 11


Answer : Option B

Explanation :

[Reference : Distribution of k balls into n boxes: formula 5] Here n = 3, k = 5. Hence, as per the above formula, the required number of ways = (n+k-1)Ck = 7C5 = 21

106. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes so that no box can be empty if all balls are different but all boxes are identical? A. 25 B. 23 C. 15 D. 6


Answer : Option A


Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference: Distribution of k balls into n boxes: formula 11] Here n = 3, k =5. Hence, as per the above formula, the required number of ways = S(k,n)

=1 n! ∑ i=0 n−1 (−1) i n C i (n−i) k =1 3! ∑ i=0 2 (−1) i 3 C i (3−i) 5 =1 6 [3 C 0 (3) 5 −3 C 1 (2) 5 +3 C 2 (1) 5 ]=1 6 [(3) 5 −3(2) 5 +3(1) 5 ] =1 6 [243−96+3]=1 6 ×150=25

---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 2 boxes) A box (in which 3 balls are put) can be selected only in 1 way (As all boxes


are identical) Now, the three balls can be selected in 5C3 ways Remaining 2 balls can be arranged in the remaining 2 boxes (1 ball in each box) only in 1 way (As all boxes are identical) Hence, total number of ways = 5C3 = 10---(A) Case B : 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in the remaining 1 box) The two boxes (in each of them, two balls are put) can be selected only in 1 way(As all boxes are identical) Now, two balls for the first selected box can be selected in 5C2 ways Two balls for the second selected box can be selected in 3C2 ways Remaining 1 ball can be placed in the remaining 1 box only in 1 way Hence, total number of ways = 5C2 × 3C2 = 5 × 3 = 15 ---(B) From (A) and (B),


The required number of ways = 10 + 15 = 25

107. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes so that no box remains empty if all balls and boxes are identical? A. 1 B. 4 C. 2 D. 6


Answer : Option C

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference: Distribution of k balls into n boxes: formula 15] Here n = 3, k =5. Hence, as per the above formula, the required number of ways = P(k,n) = P(5,3) The partitions of 5 into 3 parts are 1 + 1 + 3


1 + 2 + 2 Hence the number of partitions of 5 into 3 parts are = 2 => P(5,3) = 2 =>The required number of ways = 2 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 2 boxes) Total number of ways in which this can be done = 1 (As both boxes and balls are identical) Case B : 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in the remaining 1 box) Total number of ways in which this can be done = 1 (As both boxes and balls are identical) From (A) and (B), The required number of ways = 1 + 1 = 2


108. Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes so that no box remains empty If all balls and boxes are identical but the boxes are placed in a row. A. 2 B. 4 C. 6 D. 1


Answer : Option C

Explanation :

Here, the balls and boxes are identical. But the boxes are placed in a row. Hence, we need to consider the boxes as distinct i.e., This should be treated as a problem where balls are identical and boxes are distinct. Now it can be solved in any of the following ways. ---------------------------------------------------------------------------------- Solution 1 : Using Formula ---------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes: formula 7] Here n = 3, k = 5. Hence, as per the above formula, the required number of ways


= (k-1)C(n-1) = 4C2 = 6 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ---------------------------------------------------------------------------------- Since no box can be left empty, there can be only two cases Case A : 1 , 1, 3 (i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 2 boxes) A box (in which 3 balls are put) can be selected in 3C1 ways Now, the three balls can be selected only in 1 way (as all the balls are identical) Remaining 2 balls can be arranged in only 1 way (as all the balls are identical) Hence, total number of ways = 3C1 = 3---(A) Case B : 1, 2, 2 (i.e., two balls are put in each of the two boxes and 1 ball is put in the remaining 1 box) The two boxes (in each of them, two balls are put) can be selected in 3C2 ways.


Now, two balls for the first selected box can be selected only in 1 way(as all the balls are identical) Two balls for the second selected box can be selected in only in 1 way(as all the balls are identical) Remaining 1 ball can be placed only in 1 way Hence, total number of ways = 3C2 = 3---(A) From (A) and (B), the required number of ways = 3 + 3 = 6

109. Eight balls of different colours need to be placed in three boxes of different sizes. Each box can hold all the eight balls. In how many ways can the balls be placed in the boxes so that no box remains empty ? A. 5796 B. 8212 C. 6016 D. 16800


Answer : Option A

Explanation :

[Reference : Distribution of k balls into n boxes: formula 3] Here n = 3, k = 8. Hence, as per the above formula, the required number of ways = S(k,n) × n!


=∑ i=0 n−1 (−1) i n C i (n−i) k =∑ i=0 2 (−1) i 3 C i (3−i) 8 =3 C 0 (3) 8 −3 C 1 (2) 8 +3 C 2 (1) 8 =(3) 8 −3(2) 8 +3 =6561−768+3=5796

110. The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least two gold coins is A. 24 B. 36 C. 48 D. 0


Answer : Option B

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula : Distributing Balls into Boxes ---------------------------------------------------------------------------------- [Reference : Distribution of k balls into n boxes : formula 5] Here coins are identical and persons are distinct. The number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least two gold coins = The number of ways in which 13 identical balls can be


distributed into three distinct boxes such that each box gets at least two balls Each box gets at least two balls. Hence, initially distribute 2 balls to each of the 3 boxes. Since balls are identical, there is only 1 way of doing this The number of balls left = 13 – 6 = 7 Now distribute these 7 identical balls into 3 distinct boxes

As per the formula mentioned at the beginning, we can solve the problem now. Here n = 3, k = 7. Hence,number of ways in which this can be done = (n+k-1)Ck = 9C7 = 36 i.e, required number of ways = 36 ---------------------------------------------------------------------------------- Solution 2 : Using Formula : Counting Integral Solutions ----------------------------------------------------------------------------------

The number of non-negative integer solutions of equation x1 + x2 + ... + xn = r is (n+r-1)Cr Read More …

Here coins are identical and persons are distinct.


Hence,the number of ways in which 13 gold coins can be distributed among three persons such that each one gets at least two gold coins = The number of integer solutions of equation x1 + x2 + x3 = 13 where xi ≥ 2 for all 1 ≤ i ≤ 3 Give 2 to x1, 2 to x2 and 2 to x3 so that the required number of solutions is equal to the number of solutions of x1 + x2 + x3 = 13-6 = 7 where xi ≥ 0 for all 1 ≤ i ≤ 3 Now this is in general form with n = 3 and r = 7 Total number of solutions = (n+r-1)Cr = 9C7 = 36

111. In how many ways can 30 identical apples be divided among 10 boys? A. 39C9 B. 30! 10! C. 30C10 D. 40C10


Answer : Option A

Explanation :

[Reference : Distribution of k balls into n boxes: formula 5] Here apples are identical and boys are distinct


Number of ways in which 30 identical apples can be divided among 10 boys = Number of ways in which 30 identical balls can be distributed into 10 distinct boxes This problem can be solved using the formula given at the beginning. Here n = 10, k = 30. Hence, the required number of ways = (n+k-1)Ck = 39C30 = 39C9 [∵ nCr = nC(n - r)]

112. In how many ways can 30 different toys be equally divided among 10 boys? A. 1030 B. 30! (3!) 10 C. 30C10 D. 3010


Answer : Option B

Explanation :

---------------------------------------------------------------------------------- Solution 1 : Using Formula ----------------------------------------------------------------------------------


Number of ways in which m × n distinct things can be distributed equally among n persons (each person gets m number of things) = Number of ways in which m × n distinct things can be divided equally into n groups (each group will have m things and the groups are numbered, i.e., distinct) =(mn)! (m!) n

[Reference : Division and Distribution of Distinct Objects - Case 4]

30 different toys need to be equally divided among 10 different boys Number of toys that each boy should get = 30/10 = 3 Here toys are different. Similarly boys are also different. Hence, all groups are distinct. i.e., we need to divide 30 different toys into 10 distinct groups containing 3 toys each. Hence, as per the above formulas, total number of ways

=30! (3!) 10 ---------------------------------------------------------------------------------- Solution 2 : Using Concepts ----------------------------------------------------------------------------------


30 different toys need to be equally divided among 10 different boys i.e., number of toys that each boy should get = 30/10 = 3 Number of ways of selecting 3 toys from 30 toys = 30C3 Number of ways of selecting 3 toys from remaining 27 toys = 27C3 Number of ways of selecting 3 toys from remaining 24 toys = 24C3 Number of ways of selecting 3 toys from remaining 21 toys = 21C3 Number of ways of selecting 3 toys from remaining 18 toys = 18C3 Number of ways of selecting 3 toys from remaining 15 toys = 15C3 Number of ways of selecting 3 toys from remaining 12 toys = 12C3 Number of ways of selecting 3 toys from remaining 9 toys = 9C3 Number of ways of selecting 3 toys from remaining 6 toys = 6C3 Number of ways of selecting 3 toys from remaining 3 toys = 3C3 Total number of ways = 30C3 × 27C3 × 24C3 × 21C3 × 18C3 × 15C3 × 12C3 × 9C3 × 6C3 × 3C3

=(30! 27!×3! )×(27! 24!×3! )×(24! 21!×3! )×(21! 18!×3! )×(18! 15!×3! ) ×(15! 12!×3! )×(12! 9!×3! )×(9! 6!×3! )×(6! 3!×3! )×(3! 0!×3! ) =(30! 3! )(1 3! )(1 3! )(1 3! )(1 3! )(1 3! )(1 3! )(1 3! )(1 3! )(1 3! ) =30! (3!) 10


113. In how many ways can 30 different toys be equally divided into 10 packets? A. 1030 B. 30! (3!) 10 C. 30! 10!×(3!) 10 D. 3010


Answer : Option C

Explanation :

Number of ways in which m × n distinct things can be divided equally into n groups (each group will have m things and the groups are unmarked, i.e., not distinct) =(mn)! (m!) n n!

[Reference : Division and Distribution of Distinct Objects - Case 3]

30 different toys need to be equally divided into 10 packets Number of toys in each packet = 30/10 = 3 Since packets do not have distinct identity, we can consider that all groups are identical (not distinct) i.e., we need to divide 30 different toys into 10 identical groups containing 3 toys each. Hence, as per the above formula, total number of ways


=30! 10!×(3!) 10

114. In how many ways can 30 identical toys be divided among 10 boys if each boy must get at least one toy? A. 30C10 B. 3010 C. 1030 D. 29C9


Answer : Option D

Explanation :

[Reference : Distribution of k balls into n boxes: formula 7] Here toys are identical and boys are different The number of ways in which 30 identical toys can be divided among 10 boys if each boy must get at least one toy = The number of ways in which 30 identical balls can be distributed into 10 boxes if each box must contain at least one ball Hence, this problem can be solved using the formula given at the top. n = 10, k = 30.


Hence, the required number of ways = (k-1)C(n-1) = 29C9

115. In how many ways can seven '#' symbol and five '*' symbol be arranged in a line so that no two '*' symbols occur together? A. 44 B. 56 C. 62 D. 28


Answer : Option B

Explanation :

There are 7 identical '#' symbols and 5 identical '*' symbols. We need to arrange these 12 symbols in a line so that no two '*' symbols occur together. The seven '#' symbols can be arranged in 1 way ---(A) (Because all these symbols are identical and order is not important) Now there are 8 positions to arrange the five '*' symbols so that no two '*' symbols occur together as indicated in the diagram below


The five '*' symbols can be placed in these 8 positions in 8C5 ways---(B) (Because all these symbols are identical and order is not important) From(A) and (B), the required number of ways = 1 × 8C5 = 8C5 = 8C3 [∵ nCr = nC(n - r)]

=8×7×6 3×2×1 =8×7=56

116. Naresh has 10 friends and he wants to invite 6 of them to a party. How many times will 3 particular fiends always attend the party? A. 720 B. 120 C. 126 D. 35


Answer : Option D

Explanation :

Initially invite the 3 particular friends. This can only be done in 1 way ---(A) Now he needs to invite 3 (=6-3) more friends from the remaining 7 (=10-3) friends. This can be done in 7C3 ways ---(B)


From(A) and (B), required number of ways = 1 × 7C3 = 7C3

=7×6×5 3×2×1 =7×5=35

117. Naresh has 10 friends and he wants to invite 6 of them to a party. How many times will 3 particular fiends never attend the party? A. 8 B. 720 C. 35 D. 7


Answer : Option D

Explanation :

Remove the 3 particular friends. This can only be done in 1 way ---(A) Now he needs to invite 6 friends from the remaining 7 (=10-3) friends. This can be done in 7C6 ways ---(B) From(A) and (B), required number of ways = 1 × 7C6 = 7C6 = 7C1[∵ nCr = nC(n - r)] = 7


118. In how many ways can 10 engineers and 4 doctors be seated at a round table without any restriction? A. 14C10 B. 14! C. 13! D. None of these


Answer : Option C

Explanation :

Number of circular permutations (arrangements) of n different things is (n-1)!

[Reference : Circular Permutations: Case 1]

Here, n = 10 + 4 = 14 Hence, the number of arrangements possible = (14-1)! = 13!

119. In how many ways can 10 engineers and 4 doctors be seated at a round table if all the 4 doctors sit together? A. 13! × 4! B. 14! C. 10! D. 10! × 4!


Answer : Option D

Explanation :



[ Circular Permutations: Case 1]

Since all the 4 doctors sit together, group them together and consider as a single doctor. Hence, n = total number of persons = 10 + 1 = 11 These 11 persons can be seated at a round table in (11-1)! = 10! ways ---(A) However these 4 doctors can be arranged among themselves in 4! Ways ---(B) From (A) and (B), required number of ways = 10! × 4!

120. In how many ways can 10 engineers and 4 doctors be seated at a round table if no two doctors sit together? A. 10! × 4! B. 9! × 10P4 C. 10! × 10P4 D. 13!


Answer : Option B

Explanation :



[Reference : Circular Permutations: Case 1]

No two doctors sit together. Hence, let's initially arrange the 10 engineers at a round table. Total number of ways in which this can be done = (10-1)! = 9! ---(A) Now there are 10 positions left (marked as *) to place the four doctors as shown below so that no two doctors can sit together.

Total number of ways in which this can be done = 10P4---(A) From (A) and (B), required number of ways = 9! × 10P4


451 Question & Answer 1: There is a toy train that can make 10 musical sounds. It makes 2 musical sounds after being defective. What is the probability that same musical sound would be produced 5 times consecutively? ( 1 of _____) ? Answer: 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 32 will be the answer. 2: Peter and Paul are two friends. The sum of their ages is 35 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter? Answer: 20 years. 3: The ages of two friends is in the ratio 6:5. The sum of their ages is 66.After how many years will the ages be in the ratio 8:7? Answer: 12 years. 4: (There was a long story, I'll cut short it). There are 5 materials to make a perfume: Lilac, Balsalmic, Lemon, Woody and Mimosaic. To make a perfume that is in demand the following conditions are to be followed: Lilac and Balsalmic go together. Woody and Mimosaic go together, Woody and Balsalmic never go together. Lemon can be added with any material. (Actually they had also mentioned how much amount of one can be added with how much quantity of the other; but that's not needed for the question.) All of the following combinations are possible to make a perfume EXCEPT: 1) Balsalmic and Lilac 2) Woody and Lemon 3) Mimosaic and Woody


4) Mimosaic and Lilac Answer: Mimosaic and Lilac. 5: A girl has to make pizza with different toppings. There are 8 different toppings. In how many ways can she make pizzas with 2 different toppings. Answer: 8 * 7 = 56 6: A triangle is made from a rope. The sides of the triangle are 25 cm, 11 cm and 31 cm. What will be the area of the square made from the same rope? Answer:280.5625 7: What is the distance between the z-intercept from the x-intercept in the equation ax+by+cz+d=0. (I do not remember the values of a,b,c,d) 8: An athlete decides to run the same distance in 1/4th less time that she usually took. By how much percent will she have to increase her average speed? Answer: 33.33% 9: A horse chases a pony 3 hours after the pony runs. Horse takes 4 hours to reach the pony. If the average speed of the horse is 35 kmph, what s the average speed of the pony? 10: There is 7 friends (A1,A2,A3....A7).If A1 have to have shake with all with out repeat. How many hand shakes possible? 11: There are two pipes A and B. If A filled 10 liters in a hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160….if B filled in (1/16) th of a tank in 3 hours,


how much time will it take to fill completely? Answer:7 hours 12: (Keywords): Sports readers,10 tables,4chairs per table, each table has different number of people then how many tables will left without at least one person? Ans : 6

13: The ages of two friends is in the ratio 5:6. After how many years will the ages be in the ratio 7:8? Answer: 10 years.

14: What is the distance of the z-intercept from the x-intercept in the equation ax+by+cz+d=0. (I do not remember the values of a,b,c,d)

15: An athlete decides to run the same distance in 1/4th less time that she usually took. By how much percent will she have to increase her average speed? Answer: 33.33%

16. A man whose age is 45 yrs has 3 sons named

John,jill,jack. He went to a park weekly twice.he loves his sons very much. On a certain day he find # shopkippers sailing different things. An apple

cost 1penny, 2chocalate costs 1penny.& 3 bananas cost 1 penny. He has bought equal no. of apple,

chocolate & banana for each son. If the total

amount he invest is 7 penny then how many he has bought from each piece for his son?

a)1app,1cho,1 banana

b)1 app,2cho,3 banana


c)1app,2cho,1banana

17. A scientist was researching on animal behavior in his lab. He was very interested in analyzing the behavior of bear. For some reason he travelled 1mile in north direction & reached at north pole.there he saw a bear .he then followed the bear around 1 hr with a speed of 2km/hr in east direction.After that he travelled in south direction & reached at his lab in2 hrs. Then what is the colour of the bear? I think ans is white

a)white b)black c)gray d)brown

18. In a particular city there are 100 homes numbered from 1,2,3………..100. Thecity was build

by a builder from chennei. There was 45 shop in the town which was build by a builder from Mumbai. THE 2nd builder can build the in ½ time as compared to 1st builder. If the 2nd builder builds in

15 days,then how many 2’s are used by the builder from Chennai in numbering the 100 homes?

a)17 b)18 c)19 ans d) 20 c)19

19.MR dash has 3 sons whose ages are respectively a,b,c. The grandfather has bought a

cycle for the eldest son, mother has bought a bag for the youngest one which cost Rs150/. The sum of two age of the elder son & one son is 15.The

difference of age of sons is 3 & 2.Then what is the

age of the eldest son?

a)10, b)11, c) 12,d)13

20. We all know that Arya bhatta is the greatest mathematics belongs to india . When his daughter


Mayabati was in her teen age he discovered a problem. At that time the age of mayabati is a

prime number,let that age is a. After some years her age becomes b. then Arya Bhatta was able to

solve that problem wit the help of he daughter mayabati. If a-b=5 & product of a& b is 26 then

what is the sum of two squares?

A)77 b) 45 c)89 d)67

21.how many 13 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if repetition of digits is allowed? Ans:5 to the power

12 22. (40*40* 40-

31*31*31)/(40*40+40*31+31*31)=?a simle calcutation 23. x/2y=2a,then 2x/x-2ay=?(some thing like this .very easy ) 24. A big Question describing a story.After that a

number is given eg 2880.by what if we divide the number it ll become a perfect square?Ans:5 25. 1st a story. Then a simple ratio problem. The question was if the ratio of age of two persons is 5:6,sum of present age is 33,then in how many

years the ratio of their age becomes 7:8?

a)3 b)4 c)5 d)6


26. Mr behera wants to build A house for his wife. In this there are 5 rooms each having equal area.

The length of each room is 4m,,breadth is 5 m. the height of the rooms are 2m. if to make a sq meter

we need 17 bricks ,then how many bricks are needed to make the floor of a particular room? 27. A very big story.on Tuesday college parking

place have only 4wheelers & bicycles,total no of

wheels was 182,yhen what is the possible no of bicycles?

a)20 b 19 c 18 d 17

28. Simple question bt big one on average age.sth like a,b,c weigheted separately 1st a,b,c ,then a&

b,then b&c ,then c&a at last abc,the last weight was 167,then what will be the avg weight of the 7 weight? 29. Arrange the jumbled letters to make a perfect word RGTEI(sth like this). Find to which category it

belong?(not so easy,I was bt able 2 solve the problem .the number of the question was 34)

A)town b)vegetable c)animal d) bird

30. 3 persons a,b,c were there A always says truth,B lies on Monday,tusday,& Wednesday.but C

lies on thrusday,Friday & saturday .one day A said”that B & C said to A that” B said “yesterday way one of the days when I lies”,C said that”yesterday way one of the days when I lies

too”.then which day was that? RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

Ans: a Sunday b thrusday c saterday d.Tuesday 31.a long story & with in it a mathematical series present like

8 6 17 14 35 30 71 _ 143.

32. One man want to build a wall the length and breadth of the wall are 20, 30 respectively. He need 35 bricks for one square centimeter then how many bricks he need?

Ans:l*b*35(no of bricks needed for sqcm)

33. one person had three children.he has 7 pennis.then how he can distribute the fruits among his child by folloing conditions.

a)he can get one water millon for 1 penny.

b)he can get 2 oranges for 1 penny.

c)he can get 3 grapes for 1 penny.

Ans:2 water millon 1 orange 1 grape

34. 1/3 rd of a number is more 3 than the 1/6th of a number then find the number?

Ans:18 35. In Tnagar many buildings were under residential category.for buildings they number as 1 to 100. For shops, corporation numbered between 150 and 200 only prime numbers. how many time 6 will appear in building numbering? Ans: For 1 to 10 - 1 six 2 to 20 - 1 six Similarly upto 59 we utilise six, 5 times


from 60 to 69 (including 66) - 11 times from 70 to 100 - 3, hence ans = 5+11+3 = 19 Ans:19.

36. one grand father has 3 grand child.eldest one age is 3 times of the youngest child age.sum of two youngest child age is more than two of eldest one age.find the eldest one age?

Ans: 15(we can easily predict from options, as we take y as 15)

37.difrence b/w two nubers is 4.and their product is 17.then find the sum of their squers?

Ans: 70 (By using (x-y)2=x2+y2-2xy)

38. I dont remember exactly the question, one logical problem stating the colour of beer? Ans: white.

39. find category from following Jumbled letters, parakeet(answer) Ans: bird(category)

40. which is the smallest digit when devides the 2880 gives perfect squre.?

Ans:5(we can easily predict from options, as we devides them with 2880)

41. I don’t have any brothers and sisters.by pointing a picture that man said that his father is my fathers .son then who is he? Ans:his son. 42. 6 persons standing in queue with different age group, after two years their average age will be 43 and seventh person joined with them. hence the current average age has become 45. find the age of seventh person?


Solution: Here the question appear as an easy one, but carried a lot of unwanted sentences and unwanted datas(i dint mention above) in exam which may confuse u on solving technique. So now we can compute x from above equation. (x = 41, 6x = 246) Let now we compute y, ((6x+y)/7) = 45, as we have value of x, compute y. Ans: 69

43.The ratio b/w the ages of two pwrsons is 6:5.and sum of there ages is 77 then how many years later there ratio becomes 8:7?

Ans: we can easily predict from options

44. Horse started to chase dog as it relieved stable two hrs ago. And horse started to ran with average speed 22km/hr, horse crossed 10 mts road and two small pounds with depth 3m, and it crossed two small street with 200 mts length. After traveling 6 hrs, 2hrs after sunset it got dog. compute the speed of dog? Ans: As we have speed and travel time of horse, we can get distance travelled by it... Hence d = 22*6 = 132km, Exactly this 132km was travelled by dog in 8 hours (as it started two hours earlier). Hence speed of dog = 132/8 = 16.5km/hr Ans:16.5km/hr.

45..six friends go to pizza corner there r 2 types of pizzas.and six different flavors r there they has to select 2 flavors from 6 flavors what’s chanses to select?

Ans:6C2

46.3, 22 , 7, 45, 15, ? , 31 Solution: Here it appear simple, because it arranged in arranged


in sequence manner, but the actual question was some what twist mentioning fibonacci series and more over question was in statements (no numbers).. hence first try to understand the question well. here let group alternate terms 3,7,15,31 (3+4 =7, 7+8 =15, 15+16=31) Similarly for second group (22,45,? (22+23 = 45, 45+46 = 91) hence ans is 91.

47.cycles and 4 wheelors problem?

Ans: We can easily predict from options

48.some irrivvelent data.in last two lines problem will be there.

One man walks certain distance with 5 kmph.and walk back the same

Ans: A

49.A and B tanks r there.1/8th of the tank B is filled in 22Hrs.what is time to fill the tank full?

50.5 friends went for week end party to Mc donalds restrurent and there they measure there weights .some irrrrrrrrrrrrrrrrilevent data.finel measure is 155 kg then find the average weight of 5 people?

Ans: 155/5=31

51.2 pots r there.1st pot is filled with ink and 2nd pot is filled with water.take 1 spoon of ink from 1st pot and pore it in 2nd pot.and take 1 spoon of mixture from 2nd pot and pore it in 2nd pot then which one of following is true?

Ans: Water in 1st pot is less than the ink in 2nd pot.

52:One electronic problem?Ohm’s law RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

Ans: V=IR

53.There r ten spots in library and each spot has 4 tables and ten readers ar there . sorry I don’t remember complete question?

Ans: None

54:lion and tiger r there.lion lies on Monday,tues,wends and tiger lies on thurs,frid,sat.

Lion said that today is one of those days when I lies.

Tiger said that today is one of those days when I lie too.Then find today?

Ans: Thursday

55. 6 persons standing in queue with different age group, after two years their average age will be 43

and seventh person joined with them. hence the current average age has become 45. find the age of seventh person?

Solution: it is given as after 2 yr average age wiil

be 43 so now the average is 2 yr.

After addition of 7th person avg is 45 so 7th person

wiil be 45+(6*(45-4))

Ans: 69

56. Horse started to chase dog as it relieved stable

two hrs ago. And horse started to ran with average RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

speed 22km/hr, horse crossed 10 mts road and two small pounds with depth 3m, and it crossed

two small street with 200 mts length. After traveling 6 hrs, 2hrs after sunset it got dog.

compute the speed of dog?

Ans: As we have speed and travel time of horse, we can get distance travelled by it...

Hence d = 22*6 = 132km,

Exactly this 132km was travelled by dog in 8 hours

(as it started two hours earlier). Hence speed of dog = 132/8 = 16.5km/hr

Ans:16.5km/hr.

57. 3, 22 , 7, 45, 15, ? , 31 Solution: Here it appear simple, because it

arranged in arranged in sequence manner, but the actual question was some what twist mentioning fibonacci series and more over question was in statements (no numbers).. hence first try to

understand the question well. here let group alternate terms 3,7,15,31 (3+4 =7,

7+8 =15, 15+16=31) Similarly for second group (22,45,? (22+23 = 45,


45+46 = 91) hence ans is 91.

58. In Tnagar many buildings were under

residential category.for buildings they number as 1 to 100. For shops, corporation numbered between

150 and 200 only prime numbers. how many time 6 will appear in building numbering?

Ans:

this type of question if it is asked how many

2,3,4,5,6,7,8,9 then you bindly write the answer as 20.but for 1 answer wiil be 21 as 100 is included

59. ((4x+3y)+(5x+9y))/(5x+5y) = ? as (x/2y) =

2

Ans: as x=2y put the value and get the answer. 60: If we subract a number with y, we get 4

increase of number, once it got divided by y itself.. Find that number??

Ans: 12 (we can easily predict from options, as we take y as 6)

61. I dont remember exactly the question, one


logical problem stating the colour of beer? Ans: white.what ever the question about the color

of beer means you wrote the answer as white because polar beer.this question is very lengthy

don’t burther about that.

62. Jumbled letters, parakeet(answer)

Ans: bird(category)

63. Im only son for my parents. (some irrelevant statements in the middle to distract u).The man in picture is my father's son.(some irrelevant

statements).who is he?

64.A toy train can make 10 sounds sound changes

aftr every 4

min.................

.....................................

.......................................... now train is defective and can make only 2

sounds................. .................................................................................

find probability that same sound is repeated 3

times consecutively(1 OUT OF __)? 1.16

2.8 3.12


4.4 ANS:

(1/2)*(1/2)*(1/2)=(1/8) thus 1 out of 8

65.

................................................

......................................................

.......................................................................

.........

.......................................................................

........... resistance is X ohm voltage Y then wat is current

1.

2. 3. 4. ans:V=IR

66. I have 3

grandsons........................................... age diff btw 2 of grandsons X yrs

1st grandson is twice elder than younger one

adiition off ages of all the three is y thn what is age of eldest grandson??(there is some value in X and Y)

67. Ferrari is leading car manufacturer.*Ferrari S.p.A.* is an Italian sports car ........................................


It has enjoyed great success. If Mohan's Ferrari is 3 times faster than his old

MERCEDES wich gave him 35 kmph

if Mohan travelled 490 km in his ferrari the hw much time(hours) he took??

1.8 2.4 3.7

4.7.33 (options may be different)

68.lion rat stayin in jungle

happily...........................................

.......................................................................

............. Lion lies on : MON TUE WED RAT lies on :WED THURS SAT

if lion says : I didnt lie yesterday RAT says : e1 i didnt lie yesterday so what day is today??

69. The ratio of current age of X and Y is 5:7,after hw many years der age ratio will b 7:9?

70. Inspired by fibonacci series sanket decided to create is own series which is 1,2,3,7,7,22,15,67,.... lik dis,then what no come immediately before 63?

ans= 202

xpalnaton ;check altenate

no.1,3,7,15=====>n*2+1 RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

similarly 2,7,22,67=======>n*3+1 so series is 1,2,3,7,7,22,15,67,31,202,63........

71. By using 1,2,3,4,5,how many 5 digit no. can

be formed which is divisible by 4,repetation of no. is allowed??

Ans-last 2 place should be divisible by 4

So possible values at last place are 12,24,32,44,52 this can be arranged in 5 ways

The rest 3 places can be filled in 5*5*5 way so total is 5^4

72. The cost 1 plum is 1 cent ,2 apples is 1 cent,3 banana is 1 cent........ if rahul buys same amount of fruits for his 3 sons spending 7 cent den what amount of fruit each child will get??

ans: 1 plum ,2 apple,1 banana

xplanation:7/3=2.333 cents for each child according to ans given for d sum each child will get 1 plum ,2 apple,1 banana

73. 2880 is divided by which smallest no. so we

get no. which is perfect square??? ans= 5 xplanation 2880/5=576 sure sort question.


74. There are to prime no........(with some nonsence stuff)..............

den addition of two prime no is 13,n multiplication is 21,den wat r some of der squares?

Explanation : XY=21 and X+Y=13...solve using

calci..ans of X & Y will b in points..den x2+y2=??

75. Smita was makin 1 design .....(again some

nonsence)....size of larger cube to be made is 5*5*5........ using smaller cubes of 1*1*1....she created solid

larger cube ..den she decided to make hollow cube...

den hw many 1*1*1 cubes rqd to make hollow

larger cube

Ans= 104

Explanation (25+25)+(15+15)+(12+12)=104

76. 2X/5Y=5X/3Y...den wat is x/y

77. A pizza parlor provides pizzas...there wer 2

toppings available initially peperoni and salami.

but now they,ve introduces 8 new toppings (some names) to select from...... a person wishes to buy two DIFFERENT pizzas of

NEW topping....in how many ways he can do that??

ans : 8 X 7=56 RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

78. Person travels to....(some nonsence stuff).....if

he goes from A to B with speed of 4kmph and returns back to B with speed of 5 kmph....what is his avg. speed of journey??(values may b different)

Ans: 4.44(its NOT 4.5) Explaination : 2PQ/(P+Q)===2*4*5/(4+5)=4.44kmph

79. There is a dice having value frm 1 ..6 on each face......and a pack of cards having face card aces .....

(hugh chunk of nonsense)......when 2 dies are thrown and their scores are added then which sum will come max number of times??

1.8 2.9 3.10 4.11

Ans: 8 Explanation : 8----2,6 3,5 4,4 9----4,5 3,6

10----5,5 4,6

11---- 5,6 thus 8's probability is more

80. ''susha brought terilon cloth and rope to (some nonsence nw jst go to last 2 lines)....''.... if rope is 153 mtr long and it is to be cut into

pieces of 1 mtr long then how many times will she RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

have to cut it?? Ans : 152 times

81. There are some 2 wheelers and 4 wheelers

parked ........(some nonsense)...........total number of wheels present is 240

then how many 4 wheelers wer there Ans-

This can be done by looking at the option first check the no of bicycles and then multiply it by 2.

And then substract the multiplication value from 240 if the value is divided by 4 then that is the answer

82. 1/3 of a number is 6 more than 1/6 of that number then what is the number

Ans=x/3=x/6+6=36

83. The cost of making a robot consists of material

cost,repairing cost,coloring cost and is in the ratio 3:4:5.if the material cost is 1200 then find out the cost of the robot.

Ans- simple 3 part is 1200 so 3+4+5=12 part=?

84. There are pepsi 1 litre & oil 1 litre .it is given as 1 spoon of pepsi is aken and is mixed with Oil. Then 1 spoon oil&pepsi is taken and is mixed with pepsi then which of the condition holds true.


Ans-the amount of pepsi in oil is mre than amount of oil in pepsi.

85. An tank is filled with water .in first hour 10 lit

,in second hour 20 lit and in 3 rd hour time 40 lit.if time taken to fill ¼ of the tank is 5 hr.what is the time required to fill up the tank.

Ans-as the water is filled as twice speed.and in 5 the hour ¼.so in 6 hour ½.so answer 7 th hour.

86. Which is the smallest no divides 2880 and gives a perfect square?

a.1 b.2 c.5 d.6 Ans: c

87. Two bowls are taken, one contains water and another contains tea.one spoon of water is added to second bowl and mixed well, and a spoon of mixture is taken from second bowl and added to the second bowl. Which statement will hold good for the above? (Ans: second liquid in first bowl is smaller than the first mixture in second bowl) 88. Form 8 digit numbers from by using 1, 2,3,4,5 with repetition is allowed and must be divisible by4?

a.31250 b.97656 c.78125 d.97657

Ans: c


89. Rearrange and categorize the word ‘RAPETEKA’? Ans: bird 87. In school there are some bicycles and 4wheeler wagons.one Tuesday there are 190 wheels in the campus. How many bicycles are there? Ans: 15 88. Key words in question (Fibonacci series, infinite series, in the middle of the question one number series is there….I got the series 3 12 7 26 15 ? Ans:54

(Logic: 3*2+1=7 12*2+2=26

7*2+1=15 26*2+2=54)

89. A lies on mon, tues, wed and speak truths on other days, B lies on thur, fri, sat and speaks truths on other days …one day a said I lied today and B said I too lied today. What is the day? 90. Man, Bear, North, South, walks. Ans: White 91. A father has 7 penny’s with him and 1 water melon is for 1p, 2chickoos for 1p, 3 grapes foe 1p.he has three sons. How can he share the fruits equally? Ans: 1 watermelon,2chickoos,1grape


92. (1/2) of a number is 3 times more than the (1/6) of the same number? Ans: 9 93. There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160….if B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely? Ans:7 hours 94. KEYWORDS:T.Nagar,Chennai,1-100,prime numbers b/n 140-180,How many 2’s are there? Ans: 20 (Not only 2’s ,1’s,3’s,4’s,5’s,6’s,7’s,8’s,9’s,0’s also 20) 95. A man is standing before a painting of a man and he says I have no bro and sis and his father is my father’s son? Ans: he himself

96. One question has last part like difference between two terms is 9 and product of two numbers is 14, what is the squares of sum of numbers?

Ans:109

97. What is the value of [(3x+8Y)/(x-2Y)]; if x/2y=2?

Ans:10 {the numerical may change)

98. A pizza shop made pizzas with to flavours.in home there are ‘N’ different flavors, in that ‘M’ flavors are taken to made pizza.in how many ways they can arrange?


(Logic: NcM )

99. One grandfather has three grandchildren, two of their age difference is 3, eldest child age is 3 times youngest child’s age and eldest child’s age is two times of sum of other two children. What is the age of eldest child?

Ans:15

100. KEYWORDS: one organization ,material labor and maintenance are in the ratio of 4:6:7,the material cost is:100,what is the total cost?

Ans: 425

101. KEYWORDS: density, reluctance, sensitivity, voltage ,current, what is the resistance Formula is “R=V/I”

102. In a market 4 man are standing .the average age of the four before 4years is 45,aftyer some days one man is added and his age is 49.what is the average weight of all?

Ans: 49

103. KEYWORDS: Sports readers,10 tables,4chairs per table, each table has different number of people then how many tables will left without at least one person?

Ans : 6

107. KEYWORDS: Die, card, coin, b/n 2 to 12

Ans: All are equal

108. In a school for a student out of a 100 he got 74 of average for 7 subjects and he got 79 marks in 8th subject. what is the average of all the subjects?

Ans: 74.625


109. In a question ,last part has ,the ages of two people has the ratio of 6:6 and by adding the numbers we get 44,after how many years the ratio would be 8:7?

Ans: 8

110.One train travels 200m from A to B with 70 km/ph. and returns to A with 80kmph, what is the average of their speed?

111. Two years before Paul’s age is 2times the Alice age and the present age of Paul is 6times the Alice. what is the presents Paul’s age???( 3years) “u try to solve this question once”

112. There is ferarri and benz car, benz speed is

say 10kmph and it cover 10 km.And if ferarri goes

with 3 times faster than benz.So in how much time

ferarri could take to cover same distance.

sol: as speed of ferarri is 3*10=30 so time will be

10/30

113. If one lady have 3 daughter and any of out 3

have diff, of ages is 3.And oldest is 3 times of

more than 2 than yougest after 2 years then tell

the age of oldest daughter.

Solution: let x is youngest ,y middle ,z oldest. so

y-x=3, z-y=3, and z=2=2(x+2) and put the

option answer try to get condition.(sorry i forgot

option but pattern ll be same)

114.One question like that ,there is fabonaci series

and you have to find one number ..clue-it based on RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

series

116.if a person moves 15km straight and turns

45km right and moves 15km straight then how

much distance he needs to walk to reach starting

point?

117.if there are 30 cans out of them one is

poisoned if a person tastes very little he will die

within 14 hours so if there are mice to test and 24

hours ,how many mices are required to find the

poisoned can?

.if atlantic is found in atlantic ocean ,india is found

in

indian ocean then which of the following cases are

true

118.if a and b are mixed in 3:5 ration and b,c are

mixed in 8:5 ration if the final mixture is 35

litres,find the amount { a/b=3*8/5*8 and

b/c=8*5/5*5 a/b/c=24:40:25

ans=40*35/(24+40+35)=1400/89

Ans=15.79}

of b in the final mixture


119.1!+2!+....50!=3*10^64?

120. 6 persons standing in queue with different

age group, after two years their average age will

be 43 and seventh person joined with them. hence

the current average age has become 45. find the

age of seventh person?

Solution: Here the question appear as an easy one,

but carried a lot of unwanted sentences and

unwanted datas(i dint mention above) in exam

which may confuse u on solving technique.

Let x be current average age of first 6 persons in

queue and current age of seventh person be y.

Then 6x will become the sum of those 6 persons

age.

Now, let compute the sum of those 6 persons after

two years, 6x+12 (as each and individual increase

their age by 2). hence its average become

(6x+12)/6 = 43 (give in question itself).

So now we can compute x from above equation. (x

= 41, 6x = 246)

Let now we compute y, ((6x+y)/7) = 45, as we RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

have value of x, compute y.

Ans: 69

121. Horse started to chase dog as it relieved

stable two hrs ago. And horse started to ran with

average speed 22km/hr, horse crossed 10 mts

road and two small pounds with depth 3m, and it

crossed two small street with 200 mts length. After

traveling 6 hrs, 2hrs after sunset it got dog.

compute the speed of dog?

Ans: As we have speed and travel time of horse,

we can get distance travelled by it...

Hence d = 22*6 = 132km,

Exactly this 132km was travelled by dog in 8 hours

(as it started two hours earlier).

Hence speed of dog = 132/8 = 16.5km/hr

Ans:16.5km/hr.

122. 3, 22 , 7, 45, 15, ? , 31


Solution: Here it appear simple, because it

arranged in arranged in sequence manner, but the

actual question was some what twist mentioning

fibonacci series and more over question was in

statements (no numbers).. hence first try to

understand the question well.

here let group alternate terms 3,7,15,31 (3+4 =7,

7+8 =15, 15+16=31)

Similarly for second group (22,45,? (22+23 = 45,

45+46 = 91) hence ans is 91.

123. In Tnagar many buildings were under

residential category.for buildings they number as 1

to 100. For shops, corporation numbered between

150 and 200 only prime numbers. how many time

6 will appear in building numbering?

Ans:

For 1 to 10 - 1 six

2 to 20 - 1 six

Similarly upto 59 we utilise six, 5 times

from 60 to 69 (including 66) - 11 times


from 70 to 100 - 3, hence ans = 5+11+3 = 19

Ans:19.

124. If we subract a number with y, we get 4

increase of number, once it got divided by y itself..

Find that number??

Ans: 12 (we can easily predict from options, as we

take y as 6)

125. Im only son for my parents. (some irrelevant

statements in the middle to distract u).The man in

picture is my father's son.(some irrelevant

statements).who is he?

Ans: he himself(blood relation type of question)..

126. It is the class with the seating arrangement

in 4 rows and 8 columns. When the teacher says

'start' the girl who is sitting in first row and first

column will say 1, then the next girl sitting behind

her will say 4, the next girl sitting behind that girl

will say 7, in a particular order each girl is telling a


number, the following girls told 10, 13 next turn is

yours what u will say?16

127. It is dark in my bedroom and I want to get two socks of the same color from my drawer, which contains 24 red and 24 blue socks. How

many socks do I have to take from the drawer to get at least two socks of the same color?

a) 2

b) 3

c) 48

d) 25 ; Solution: 3

128.Inspired by fibonacci series sanket decided to create is own series which is 1,2,3,7,7,22,15,67,.... lik dis,then what no come immediately before 63?

ans= 202

xpalnaton ;check altenate no.1,3,7,15=====>n*2+1 similarly 2,7,22,67=======>n*3+1 so series is 1,2,3,7,7,22,15,67,31,202,63........

129..valentine day 14 feb 2005,was celebrated by n and u on monday,he was very happy,he n she........ .........den day on 14 feb 2010???(similar to dis some date qn was der)


130.the cost 1 plum is 1 cent ,2 apples is 1 cent,3 banana is 1 cent........ if rahul buys same amount of fruits for his 3 sons spending 7 cent den what amount of fruit each child will get??

ans: 1 plum ,2 apple,1 banana xplanation:7/3=2.333 cents for each child according to ans given for d sum each child will get 1 plum ,2 apple,1 banana

131.there is a dice having value frm 1 ..6 on each face......and a pack of cards having face card aces ..... (hugh chunk of nonsense)......when 2 dies are thrown and their scores are added then which sum will come max number of times?? 1.8 2.9 3.10 4.11 Ans: 8 explanation : 8----2,6 3,5 4,4 9----4,5 3,6 10----5,5 4,6 11---- 5,6 thus 8's probability is more

132.''susha brought terilon cloth and rope to (some nonsence nw jst go to last 2 lines)....''.... if rope is 153 mtr long and it is to be cut into pieces of 1 mtr long then how many times will she have to cut it?? ans : 152 times


133.(dnt remembr the xact q but procedure was somethn lyk this) .........................8th year--1/1024,, 9th year--1/512,, 10th year--1/256 then aftr hw many years 1/32??? ans: 13

134.there are 2 cans A and B one of MILK and other of Water resp. , both of same qty...... first one teaspoon of milk from A can was added to B can... then one teaspoon from B can was added to A can then wich of the folloe\wing is true.. 1.Can A contain more milk than water in can B 2.Can A contain less milk than water in can B 3.both contain same qty of milk and water 4. Ans : option--2

135.If a pipe can fill the tank within 6 hrs but due to leak it took 30 min more now if the tank was full hw much tym will it take to get emptyed through the leak??(i dont remembr whole sum exactly)(lil bit tricky sum)

136.Avg wt of class is X kg(some number) after adding wt of the teacger avg wt of class becomes Y kg then wat is the wt of the teacher??


137.20 men shake hand with each other. Maximum no of handshakes w without cyclic handshakes.

138. 100 men & women dance with each other. Probability that a man cannot dance with more than two women. 139. Horse chasing a pony. Horse leaves stable aftr 2hrs from ponys departure. 4 hrs 2 catch pony. Find speed of pony. Given-speed of horse. 140. A man goes north 37km.turns left goes 2km.turns right goes 17km.turns right goes 2km. find distance b/w starting&ending point. 141. Lady hav 2 select gloves&hat from a basket I the dark.she can distinguish hat&gloves.14red,20blue,18green r there. Find probability that any selected glove pair has same colour. 142. Alice in wonderland meets a character goblet whose age is 2times alice.aftr 2 years age problem 143. Peter is 2times paul’s age was when peter’s age is same as paul’s present age.find pauls age. 144. From a rope a triangle is made of sides 21cm,24cm,28cm. from this a square is made. Find area of square. 145. In a supermarket average of 4people standin in queue taken 2yrs before is 55yrs. Now a person of 45 yrs is added current age. 146. A toy can produce 10diiff sounds. Nw toy is defective to produce 2 sounds in 3min. find probability that it produces 6 consecutive is 1 in(_)


147. 1/6th of a no is 4 times more than 2/3 of a no.find no 148. Age of 2 in d ratio 4:5. Total of 2 ages is 55. Aftr 2 yrs age in ratio 5:7 ages 149. A jogger jogs@1/6th of his usual speed. How much % she has 2 increase 2 reach normal pace of walking.

150: X is 3 years yunger to Y. X's father is a businessman who invested 10000/- at 8% rate of interest n obtained his amount after 10 years. Y's father is a job holder who invested around 20000 at 2% rate n obtained his amount after 20 years.Now Compunded both of dem get around ABC rs/-(dnt remenbr).After 5 years the ratio of ages of X n Y is 1:2. Now X's father is 20 years older to Y n Y' father is 30 years more than X. After 20 years again X's mother asks X's father to purchase a LCD TV whch costs around 45000/-.

what is the age of X n Y together?

Ans: answer lies in considering two statements 2gether i.e "X is 3 years younger to Y" n "After 5 years the ratio of ages of X n Y is 1:2"

151. 3, 22 , 7, 45, 15, ? , 31 Solution: Here it appear simple, because it arranged in arranged in sequence manner, but the actual question was some what twist mentioning fibonacci series and more over question was in statements (no numbers).. hence first try to understand the


question well. here let group alternate terms 3,7,15,31 (3+4 =7, 7+8 =15, 15+16=31) Similarly for second group (22,45,? (22+23 = 45, 45+46 = 91) hence ans is 91. 152. In Tnagar many buildings were under residential category.for buildings they number as 1 to 100. For shops, corporation numbered between 150 and 200 only prime numbers. how many time 6 will appear in building numbering? Solution: For 1 to 10 - 1 six 2 to 20 - 2 six Similarly upto 59 we utilise six, 6 times from 60 to 69 (including 66) - 11 times from 70 to 100 - 3, hence ans = 5+11+3 = 19 Ans:19. 153.The bacteria had a probability of splitting into three and a probability to die is one third of total bacteria..Let the probability be P. Some of them survived wit probability 1/5. then wic among the following relation is true? a)P=1/3+1/5*3 b)P=1/5*(1/8-3) c) 154. if a tank a can be filled within 10 hrs and tank B is 1/4th filled in 19 hrs.. then wat is the duration of the tank to fill completely?


155.A man looks at a painting and tells “Neither I hv brothers nor sisters, but the person in the painting is my father’s son”. Then who is in the painting? 156.A lady had fine gloves and hats. 25 blue,7 red and 9 grey.she had to select a pair among them. But there was no light so she had to select in darkness the correct pair wit a glove and a hat. Therefore how many combinations of same color she can select? 157.A old lady had three grandchildren,the difference between two children was 3 years. Her eldest grandchild was 3 times elder than the youngest one and the elder one 2 years more than the sum of the other two. Then wat is the age of the eldest child? 158.There was a grandmother in a village who had a grandchild. Upon asking her grandchild’s age she told dat she is as older as many days old as her daughter’s age in weeks and as many days as her own age in years. The sum of the three is 130. then how old is the child? 159.(98*98*98-73*73*73)/(98*98*98+73*73*73)=? 8.which is the smallest number wic on dividing 2880 to make it a perfect square? a.6 b.5 c.4 d.3 Ans :5 160.Leena cut small cubes of 10 cubic cms each. Which she joined to form a cube wit 10 cubes length, 5 cubes in depth and 5 cubes wide. How many more small cubes does she require to form a perfect cube?


10.the first two numbers are 1 and 2. The numbers in series are 3,6,7,14, , 32? Which number comes before 32 161.The age of two people is in the ratio 6:8. the sum of their ages is 77. after 2 years the ratio of their ages becomes 5:7. wat is their present age? 162.if a and b are mixed in 3:5 ration and b,c are mixed in 8:5 ration if the final mixture is 35 litres,find the amount of b in the final mixture

163. A vendor sells 1 apple for 1 penny, 2 grapes for 1 penny, 3 bananas for 1 penny. A man spends 7 penny and gives equal amount of fruits to each of his three daughters. What is the possible number of fruits each daughter gets? Ans: 1 apple,2 grape,2 banana

164. 5 persons standing in queue with different age group, two years ago their average age will be X(I couldn’t remember) and 6th person joined with them. hence the current average age has become Y(I couldn’t remember). find the age of seventh person?

165. Horse started to chase dog as it relieved stable three hrs ago. Avg speed of the horse was given and the time horse chased dog was also given. What is the speed of the dog?

166. 5,9,12,18,26,36,47,72,--? Here odd terms have differences as multiples of 7 and even terms adds with themselves to form the next number. So answer is 75.

167. In Tnagar many buildings were under residential category. for buildings they number as 1 to 100. For shops, corporation numbered between 150 and 200 only prime numbers. how many time 4 will appear in building numbering? Ans: 19


168. Jumbled letters, choices were given whether the word is bird or city or sweet……parakeet(answer) Ans: bird(category)

169. Lion tells lie on Monday, Tuesday, and Wednesday. Rat tells lie on Thursday, Friday and Saturday. Both of them speak truth on other days. Lion tells, “Yesterday was one of the days which I tell lying”. Rat also tells, “Yesterday was one of the days which I tell lying”. what day was yesterday?

170. There were three different gloves. 13 red, 27 black and 40 green. How many gloves one has to take so as to ensure that there is at least one pair in each color?

171. Probability of occurrence of some events was given. Have to find total probability of specified group of events.

172. One person has no siblings and says,” the guy in the photo

is the only son of my father ‘s son”. What is the relation of the guy to the person?

173. Difference of two numbers is 6. Product of them is 13. What is the sum of their squares?

174. Voltage and current are given, resistance was asked. V=IR

175. Speed and distance were given and time taken was

asked.T=D/S 176. A problem on finding the age of the grand mother. 177. A lady builds 9cm length,10cm width,3cm height box

using 3cubic cm cubes. What is the minimum number of cubes required to build the box?

178. When a pair of dice is thrown, what number has the higher probability to occur…the sum of 8 or 9 or 10?

179. A person has to make 146 pieces of a long bar. He takes 4 seconds to cut a piece. What is the total time taken by him in seconds to make 146 pieces?


180) 6 persons standing in queue with different age group, after two years their average age will be 43 and seventh person joined with them. hence the current average age has become 45. find the age of seventh person? Solution: Here the question appear as an easy one, but carried a lot of unwanted sentences and unwanted datas(i dint mention above) in exam which may confuse u on solving technique. Let x be current average age of first 6 persons in queue and current age of seventh person be y. Then 6x will become the sum of those 6 persons age. Now, let compute the sum of those 6 persons after two years, 6x+12 (as each and individual increase their age by 2). hence its average become (6x+12)/6 = 43 (give in question itself). So now we can compute x from above equation. (x = 41, 6x = 246) Let now we compute y, ((6x+y)/7) = 45, as we have value of x, compute y. Ans: 69 181) Horse started to chase dog as it relieved stable two hrs ago. And horse started to ran with average speed 22km/hr, horse crossed 10 mts road and two small pounds with depth 3m, and it crossed two small street with 200 mts length. After traveling 6 hrs, 2hrs after sunset it got dog. compute the speed of dog? Ans: As we have speed and travel time of horse, we can get distance travelled by it...


Hence d = 22*6 = 132km, Exactly this 132km was travelled by dog in 8 hours (as it started two hours earlier). Hence speed of dog = 132/8 = 16.5km/hr Ans:16.5km/hr. 182) 3, 22 , 7, 45, 15, ? , 31 Solution: Here it appear simple, because it arranged in arranged in sequence manner, but the actual question was some what twist mentioning fibonacci series and more over question was in statements (no numbers).. hence first try to understand the question well. here let group alternate terms 3,7,15,31 (3+4 =7, 7+8 =15, 15+16=31) Similarly for second group (22,45,? (22+23 = 45, 45+46 = 91) hence ans is 91. 183) In Tnagar many buildings were under residential category.for buildings they number as 1 to 100. For shops, corporation numbered between 150 and 200 only prime numbers. how many time 6 will appear in building numbering? Solution: For 1 to 10 - 1 six


2 to 20 - 1 six Similarly upto 59 we utilise six, 5 times from 60 to 69 (including 66) - 11 times from 70 to 100 - 3, hence ans = 5+11+3 = 19 Ans:19. 184) ((4x+3y)+(5x+9y))/(5x+5y) = ? as (x/2y) = 2 Ans: 2(simple algebra, i think u no need of explanation) 185) If we subract a number with y, we get 4 increase of number, once it got divided by y itself.. Find that number?? Ans: 12 (we can easily predict from options, as we take y as 6) 186) I dont remember exactly the question, one logical problem stating the colour of beer? Ans: white. 187) Jumbled letters, parakeet(answer) Ans: bird (category) 188) ratio proportional problem with age. Sorry, dint remember exact question. 189) one question like. (209*144)^2 + (209*209)+(209*144)+(144*144) = ? Ans: here you can use calc, many(4 to 5) questions were depend upon calc alone.(no need problem solving technique). 190) Im only son for my parents. (some irrelevant statements in


the middle to distract u).The man in picture is my father's son.(some irrelevant statements).who is he? Ans: he himself(blood relation type of question).

191 By which number should we divide the number 2880 to make it perfect square?

Ans: 5

192. There is a problem to find out the color of beer. It is full of unwanted data in problem. for that the

Ans: white

193. 1/3 ofsome number is 5 more than 1/6 th of that number. Find the number.

Ans: 30

194. Difference of two numbers is 4 and their product is 13. Find the sum of squares of that numbers.

Note: this problem starts with the story of aryabhatta , ignore all and read the problem from last line.

195. How many of 14 digit numbers we can make with 1,2,3,4,5 that are divisible by 4. Repetitions allowed.

Ans: we have calculate the value (5power12)*4 (like 5*5*5*5*5*5*5*5*5*5*5*5*4= value will be given in answers check it out)

196. There is a lengthy problem with details of Chennai city. At last they ask how many 6’s came when we give numbering to 100 buildings from 1 to 100.

Ans: 20


197. Rearrange the alphabets REGHFTYD(SOMETHING LIKE THAT) . find the type of rearranged word belongs to:

a. Animal

b. Tree

c. Bird

d. Thing

Ans: c (bird)

198. There is a factory which is producing the bicycles and four wheelers. One day the total production of wheels is 158. Find out the possible no. of bicycles produced

a. 6

b. 7

c. 8

d. 9

Ans: 7 (note: there is a probability of the answer of 19 also for this question . take care of it. )Also they change the total number of wheels to 198.

199. Four years hence the average of 6 members is 45. Now a person is added and the avg becomes 48. What is the age of added person?

Ans: 42

There is another question with diff. details. For that ans is :69 (problem is not remembered but ans is 69 only)


200. A dog two hours early before the horse started. The horse reached the dog after 6 hours with the speed of 16kmph . find the speed of dog. I cant remember the exact figures.

Ans may be 16.5kmph

201. There is a problem with blood relations. A man is saying while pointing to a person who is painting like this:” I am the only son to my father. His father is the son my father”. Find the relation to him with the painting person.

ans: his son

202. There is problem on probabilities. There are gloves and hats with three different colors with some totals are given for each type. Then asked to find out the probability of taking the pair of glove and hat of same color in dark.

203. There is a bacteria which has the probability of die 1/3 of its total number or it may tripled. Find out the probability

A. P=1/3+(2/3*p^3)

B. P=2/3+(2/3*p^3)

C. P=2/3+(1/3*p^3)

D. P=2/3+(2/3*p^3)

I marked it as A. check it out

204. There are two tanks A,B. A will be fill up 1ltr in one hour. B tank will fill up double in every hour (lik e 10, 20, 40, 80, 160…..). if the tank B is filled 1/16 in 13 hours how much time it will take to fill up totally.


Ans: 17 hours (note: here no need with A tank details. but they gave to confuse, all problems are like this ,avoid unnecessary data)

205. In a hotel we can order two types of varieties, but we can make 6 more varieties in home. One can want the four varieties with two from hotel must. Find how many ways one can order.

Ans: 12 ways

206. There is a series 13,14,27,30,55,62,?, 126. Find the missing.

This is combination of two series; 13,27 55, ? and 14, 30, 62, 126 (14, 14*2+2(30),30*2+2(62),….)

13, 13*2+1 (27), 27*2+1(55), ?=55*2+1=111 .For this numbers may be changed but the logic is same.

207. There are three frnds x, y , z gone to excursion with their girl frnds. there they wanted to find their weights but their GF’s are not accept to check their weight( all unnecessary data) . Then they check weights as x, y, z individually and then x and y, y and z, x and z ,then all(x,y,z). the last measure is 171. Then find the avg of all these seven measures.

208) Two tanks A and B. A fills 1 ltr/1 hr... b fills 10, 20, 30..... per hour. if this is (passage unnecessary). if 1/4th tnk of b takes 15 hrs to fill how much it time will t take to fill complete tank?

209) Out of 7 children the youngest is boy then find the probability that all the remaining children are boys

Ans: 1/2^6 = 1/64


210) The three sides of a triangle are given.16, 14, 21 cms and this triangle is conveted into a square .so what will be the area of the square generated?

Ans: - (14+16+21)/4. Then you will get the 1 side of a square and now find the area of a square.ie, side^2

211) An equation of the form 4x+6y-2z=32 . Find the difference between x intercept and z intercept?

Ans: x/a+y/b+z/c=1

212) A toy train can make 10 sounds sound changes after every 4 min................. ..................................... .......................................... now train is defective and can make only 2 sounds................. ................................................................................. Find probability that same sound is repeated 5 times consecutively (1 OUT OF __)? Ans: 1/32

213) ) 20men and 20 women are there, they dance with each other, is there possibilty that 2 men are dancing with same women and vice versa. Ans-never

214) 10 people are there, they are shaking hands together, how many hand shakes possible, if they are in no pair of cyclic sequence. Ans-9

215) In school there are some bicycles and 4wheeler wagons. One Tuesday there are 234 wheels in the campus. How many


bicycles are there? Ans: go wid options. muliply each option wid 2 and subtact the obtained no from 234. if it is exactly divisible by 4, that is the ans....

216) A father has 7 penny’s with him and 1 water melon is for 1p, 2chickoos for 1p, 3 grapes foe 1p.he has three sons. How can he share the friuts equally? Ans: 1 watermelon,2chickoos,1grape

217) A piza shop made pizzas with to flavours.in home there are ‘9’ different flavors, in that ‘2’ flavors are taken to made piza in how many ways they can arrange? (Logic: NcM, N= 9, M=2 )

218) one organization ,material ,labor and maintenance are in the ratio of 4:6:7, if the material cost is: 272 ,what is the total cost?

Ans: 4x=272 ==> x=68 ; now total cost = 272 + 6(68)+7(68).

219) 4 years before Paul’s age is 3times the Alice age and the present age of Paul is 6times the Alice. what is the presents Paul’s age???

Ans: x-4 = 3(y-4) ; x=6y : solve u will get it..

220) In a question ,last part has ,the ages of two people has the ratio of 6:5 and by adding the numbers we get 55,after how many years the ratio would be 8:7? Ans: easy u can do it... simple eqtns

221) In a room ..................... ( unwanted stuff)............................. Sports readers,10 tables,4chairs per


table, each table has different number of people then how many tables will left without at least one person? Ans : 6

222) passage................... joe is taller than jerry and 3 pillers. kistern is shorter than joe and 2 pillers is jerry shorter/taller than kistern ?

Ans:

223) a volume of A are having in a container of sphere... how many semi hemispheres of B volume each will be rqred to transfer all the A in to semi hemispheres.....?

Ans: A= x B

224) Question based on V=I*R but in dis question most of data given are ridiculas like volume ,density,length,height similar long story are given

Ans:

225) Peter and Paul are two friends. The sum of their ages is 42 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter?

Ans:

226) A horse chases a pony 2 hours after the pony runs. Horse takes 4 hours to reach the pony. If the average speed of the horse is 81 kmph, what s the average speed of the pony? (This question was really long with loads of irrelevant statement)

Ans:

227) difrence b/w two nubers is 4.and their product is 17.then find the sum of their squers? Ans:


228) A, B, C, D, E are there among A, B, C are boys and D, E are girls =====> D is to the left of A and no girl sits at the middle and at the extemes. Then what is the order of their sittings..

Ans:

229.) some ages problem............ then asked the answer in binary

Ans:

230) .................(some chetta chikkati sollu)..unwanted data............ folowed by a formula diameter d = 10*(t-14), t>14... then what id diameter after t=40..?

Ans:

231) Some denominations question... like u have 31 ps, and tckt cost is between 1 to 31, u have to give the exact denominations for the ticket. find all the no of possible denominations u may prdict and u must be left wid atleast few paise...?

232) Direction problems..... A man goes 50 km NORTH, then turned left walked 40 km, then turned RIGHT...? in which direction is he in?

Ans: NORTH

Questions 233) Out of 6 children the youngest is boy then find the probability that all the remaining children are boys. Ans: 1/2^5 = 1/32

234) A man went 1mile to east den 1mile to north (un wanted stuff) and killed a bear what is the color of the bear?


Ans: White

235) Some age, average related problems (practice R. S.Aggarwal )

236) (1/2) of a number is 3 times more than the (1/6) of the same number..?

237) There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160….if B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely?

Ans: 7 hours

238) In a market 4 man are standing .the average age of the four before 4years is 45, after some days one man is added and his age is 49.what is the average weight of all?

Ans: 49

239) One train travels 200m from A to B with 70 km/ph. and returns to A with 80kmph, what is the average of their speed?

Ans: apply 2xy/x+y

240) The three sides of a triangle are given.18, 18, 28 cms and this triangle is conveted into a square .so what will be the area of the square generated? Ans: - (18+18+28)/4. Then you will get the 1 side of a square and now find the area of a square.ie, side^2

241) An equation of the form 7x+17y+ 3z=54. Find the difference between x intercept and z intercept? Hint: x/a+y/b+z/c=1 Ans: convert the above equation to this form and see the difference between a and c then you will get the answer RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

242) Permutation problem don’t remember exactly, but it was almost like there are n people sitting. Find the number of handshakes

243) Average wt of class is (some number) kg after adding wt of the teacher average wt of class becomes some number kg then what is the wt of the teacher?? 244) A pizza shop, there were 2 kinds of pizzas available. But now they have introduces 8 new types, a person buy two different type pizzas of new type in how many ways he can select? Ans: 8 X 7=56

245) Series Problem like 4 12 x 44 46 132 134. Find x? (I could not solve out this in exam).

246) There are 1000 pillars for a temple. 3 friends Linda, Chelsey, Juli visited that temple. (Som unrelated stuff) Linda is taller than Chelsey and taller than 2 of 1000 pillars. Juli is shorter than Linda. Find the correct sentence?

a) Linda is shorter among them

b) Chelsey is taller than Juli

c) Chelsey is shorter than Juli

d) Cannot determine who is taller among Chelsey and Juli

Ans:: d

247) A toy train can make 10 sounds sound changes after every 4 min................. ..................................... .......................................... now train is defective and can make only 2 sounds................. RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

................................................................................. Find probability that same sound is repeated 3 times consecutively (1 OUT OF __)?

1.16 2.8 3.12 4.4 Ans: (1/2)*(1/2)*(1/2) = (1/8)

248) Probability problem (little tricky)

249) Some Statement and Conclusion type problems

250) Entry ticket to an exhibition ranges from 1p to 7p. You need to provide exact change at the counter. You have 7p coin. In how many parts will u divide 7p so that u will provide the exact change required and carry as less coins as possible?

a) 8 b) 7 c) 5 d) 3 (I cud not solve out the answer in exam)

251) Dhoni and Ponting are waiting for the toss to happen. Umpire found that the coin to be tossed is missing. Ponting then takes a dice (1-6) from his pocket and asks the umpire to toss with it. Umpire feels both the captains may not get fair chance with dice. Dhoni den suggests a solution to umpire which den wud give fair chance to both captains. What would be the idea of Dhoni?

a) Toss the dice if even no comes captain wil win the toss and if it is odd he loses.

(It’s the only option I remember and I think its da answer)

252) Now pet's age is to times when paul was once. But at that time paul's age=pet's current age ,how old is pet?


253) Block has 10,9,5 size, how many unit cube is needed to make a block of that size? 254) 23 people are there, they are shaking hands together, how many hand shakes possible, if they are in pair of cyclic sequence. Ans-22 255) 10men and 10 women are there, they dance with each other, is there possibilty that 2 men are dancing with same women and vice versa. Ans-never 256) A lady took out jacket and gloves, which are avialable in blue 26, yellow 30 and red 56. Power goes off, she can distinguish between gloves and jacket but not in colors. What's the possibilty their she will pick up pair of gloves of each color 257) B is taller than j and 3 pillers. P is shorter than B and 2 pillers is j shorter/taller than P? Ans-irrelevant question 258) Sangakara and ponting selects batting by using a dice, but dice is biased so to resolve ponty takes out a coin, what is the probability that dice shows correct option. 259) In school there are some bicycles and 4wheeler wagons. One Tuesday there are 58 wheels in the campus. How many


bicycles are there? Ans: 7 260. Two bowls are taken, one contains water and another contains tea.one spoon of water is added to second bowl and mixed well, and a spoon of mixture is taken from second bowl and added to the second bowl. Which statement will hold good for the above? (Ans: second liquid in first bowl is smaller than the first mixture in second bowl) 261. Which is the smallest no divides 2880 and gives a perfect square? a.1 b.2 c.5 d.6 Ans: c 262. Form 8 digit numbers from by using 1, 2,3,4,5 with repetition is allowed and must be divisible by4? a.31250 b.97656 c.78125 d.97657 Ans: c 263. One problem on (a3-b3)/(a2+ab+b2) Ans: ‘a-b’ 264. Rearrange and categorize the word ‘RAPETEKA’? Ans: bird(parakeet) 265. Key words in question (Fibonacci series, infinite series, in


the middle of the question one number series is there….I got the series 3 12 7 26 15 b ? Ans:54 (Logic: 3*2+1=7 12*2+2=26 7*2+1=15 26*2+2=54) 266. A father has 7 penny’s with him and 1 water melon is for 1p, 2chickoos for 1p, 3 grapes foe 1p.he has three sons. How can he share the fruits equally? Ans: 1 watermelon,2chickoos,1grape 267. A lies on mon, tues, wed and speak truths on other days, B lies on thur, fri, sat and speaks truths on other days.. one day a said I lied today and B said I too lied today. What is the day? 268. Man, Bear, North, South, walks. Ans: White 269. (1/2) of a number is 3 times more than the (1/6) of the same number? Ans: 9(for any no it can be true) 270. There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160….if B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely? Ans:7 hours


271. KEYWORDS:T.Nagar,Chennai,1-100,prime numbers b/n 140-180,How many 2’s are there? Ans: 20 (Not only 2’s ,1’s,3’s,4’s,5’s,6’s,7’s,8’s,9’s,0’s also 20) 272. One question has last part like difference between two terms is 9 and product of two numbers is 14, what is the squares of sum of numbers? Ans:109 273. A man is standing before a painting of a man and he says I have no bro and sis and his father is my father’s son? Ans: His son 274. What is the value of [(3x+8Y)/(x-2Y)]; if x/2y=2? Ans:10 {the numerical may change) 275. A pizza shop made pizzas with to flavours.in home there are ‘N’ different flavors, in that ‘M’ flavors are taken to made pizza.in how many ways they can arrange? (Logic: NcM ) 276. One grandfather has three grandchildren, two of their age difference is 3, eldest child age is 3 times youngest child’s age and eldest child’s age is two times of sum of other two children. What is the age of eldest child? Ans:18 277. In a market 4 man are standing .the average age of the four before 4years is 45,aftyer some days one man is added and his age is 49.what is the average weight of all?


Ans: 49 278. KEYWORDS: one organization ,material ,labor and maintenance are in the ratio of 4:6:7,the material cost is:100,what is the total cost? Ans: 425 279. KEYWORDS: density, reluctance, sensitivity, voltage ,current, what is the resistance Formula is “R=V/I” 280. KEYWORDS: Sports readers,10 tables,4chairs per table, each table has different number of people then how many tables will left without at least one person? Ans : 6 281. In a school for a student out of a 100 he got 74 of average for 7 subjects and he got 79 marks in 8th subject. what is the average of all the subjects? Ans: the xtra 5 marks will be distributed in 8 semester, 5/8=.625 74+.625=74.625 282. In a question ,last part has ,the ages of two people has the ratio of 6:5 and by adding the numbers we get 44,after how many years the ratio would be 8:7? Ans: 8 283. Two years before Paul’s age is 2times the Alice age and the present age of Paul is 6times the Alice. what is the presents Paul’s age???( 3years) “u try to solve this question once”


284.One train travels 200m from A to B with 70 km/ph. and returns to A with 80kmph, what is the average of their speed? Aptitude test 285. A man whose age is 45 yrs has 3 sons named John,jill,jack. He went to a park weekly twice.he

loves his sons very much. On a certain day he find

# shopkippers sailing different things. An apple cost 1penny, 2chocalate costs 1penny.& 3 bananas cost 1 penny. He has bought equal no. of apple,

chocolate & banana for each son. If the total amount he invest is 7 penny then how many he has bought from each piece for his son?

a)1app,1cho,1 banana

b)1 app,2cho,3 banana

c)1app,2cho,1banana

D)2 app,2cho,2 banana

286. A scientist was researching on animal behavior in his lab. He was very interested in

analyzing the behavior of bear. For some reason he travelled 1mile in north direction & reached at north pole.there he saw a bear .he then followed the bear around 1 hr with a speed of 2km/hr in

east direction.After that he travelled in south

direction & reached at his lab in2 hrs. Then what is the colour of the bear? I think ans is white a)white b)black c)gray d)brown


287. How many 9 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if

repetition of digits is allowed?

288. Calcutation based on A^3-B^3 formula 289. A long story based ALICE AND WONDER LAND after that simple age question .

290. Question based on V=I*R but in dis question most of data given are ridiculas like volume

,density,length,height similar long story are given 291. A direct question on blood relation.

292. A big Question describing a story ARYABHATT AND HIS DAUGHTER LILAVATI

After that a number is given eg 2088.by what if we

divide the number it ll become a perfect square? 2293 1st a story. Then a simple ratio problem. The question was if the ratio of age of two persons is 5:6,sum of present age is 33,then in how many

years the ratio of their age becomes 7:8? a)3 b)4 c)5 d)6

294. A very big story.on Tuesday college parking

place have only 4wheelers & bicycles,total no of wheels was 182,yhen what is the possible no of bicycles?


a)20 b 19 c 18 d 17 295. Simple question bt big one on average age.sth like a,b,c weigheted separately 1st a,b,c ,then a& b,then b&c ,then c&a at last abc,the last

weight was 167,then what will be the avg weight of the 7 weight? 296. Arrange the jumbled letters to make a perfect word RGTEI(sth like this). Find to which category it belong? (plz do not these type q becoz its time

consuming) A)town b)vegetable c)animal d) bird 297) Simple puzzle based on IQ 3 persons a,b,c were there A always says truth,B lies on Monday,tusday,& Wednesday.but C lies on

thrusday,Friday & saturday .one day A said”that B & C said to A that” B said “yesterday way one of the days when I lies”,C said that”yesterday way

one of the days when I lies too”.then which day was that? Ans: a Sunday b thrusday c saterday d…. 298) A girl has to make pizza with different toppings. There are 8 different toppings. In how many ways can she make pizzas with 2 different toppings. Answer: 8 * 7 = 56 299) Peter and Paul are two friends. The sum of their ages is 35 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter?


300) Question based on pipe & ciston (geometrical series) A and B tanks r there.1/8th of the tank B is filled in 22Hrs.what is time to fill the tank full? 301) 5 friends went for week end party to Mc donalds restrurent and there they measure there weights .some irrrrrrrrrrrrrrrrilevent data.finel measure is 155 kg then find the average weight of 5 people? ans:155/5=31 302) 2 pots are there.1st pot is filled with ink and 2nd pot is filled with water.take 1 spoon of ink from 1st pot and pore it in 2nd pot.and take 1 spoon of mixture from 2nd pot and pore it in 2nd pot then which one of following is true? ans: Water in 1st pot is less than the ink in 2nd pot. 303) There r ten spots in library and each spot has 4 tables and ten readers ar there . 10 student come into liberary and want 2 study in how many ways that they sit in d liberary so that no chair would be blank? ans :1 304.Question 1: There is a toy train that can make 10 musical sounds. It makes 2 musical sounds after being defective. What is the probability that same musical sound would be produced 5 times consecutively? ( 1 of _____) ? Answer: 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 32 will be the answer. (My friends got similar question with "3 times consecutively"; for that the answer would be 8) 305.Question 2: Peter and Paul are two friends. The sum of their ages is 35 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter?


Luckily, I had this question twice. Initially I left it because of the confusing statement but when I got the same question second time with different names of the friends (Pooja and Prasad) then I gave a little extra time and solved it. Answer: 20 years. I simply substituted the answers in the statement to find the best fit, because the statement in the question was pretty confusing. 306.Question 3: The ages of two friends is in the ratio 6:5. The sum of their ages is 66.After how many years will the ages be in the ratio 8:7? Answer: 12 years. 307.Question 4: (There was a long story, I'll cut short it). There are 5 materials to make a perfume: Lilac, Balsalmic, Lemon, Woody and Mimosaic. To make a perfume that is in demand the following conditions are to be followed: Lilac and Balsalmic go together. Woody and Mimosaic go together, Woody and Balsalmic never go together. Lemon can be added with any material. (Actually they had also mentioned how much amount of one can be added with how much quantity of the other; but that's not needed for the question.) All of the following combinations are possible to make a perfume EXCEPT: 1) Balsalmic and Lilac 2) Woody and Lemon 3) Mimosaic and Woody 4) Mimosaic and Lilac Answer: Mimosaic and Lilac. I have made the question here really easy to understand. But the actual question was in a twisted language and it was difficult to find the answer. It took me some time to get to the answer. 308.Question 5: A girl has to make pizza with different toppings. There are 8 different toppings. In how many ways can


she make pizzas with 2 different toppings. Answer: 8 * 7 = 56 309.Question 6: A triangle is made from a rope. The sides of the triangle are 25 cm, 11 cm and 31 cm. What will be the area of the square made from the same rope? Answer:280.5625 310.Question 7: What is the distance between the z-intercept from the x-intercept in the equation ax+by+cz+d=0. (I do not remember the values of a,b,c,d) 311.Question 8: An athlete decides to run the same distance in 1/4th less time that she usually took. By how much percent will she have to increase her average speed? Answer: 33.33% 312.Question 9: A horse chases a pony 3 hours after the pony runs. Horse takes 4 hours to reach the pony. If the average speed of the horse is 35 kmph, what s the average speed of the pony? (This question was really long with loads of irrelevant statement) 313.Question 10: There is 7 friends (A1,A2,A3....A7).If A1 have to have shake with all with out repeat. How many hand shakes possible?(I dont know the exact question but like this only) 314.Question 11: There are two pipes A and B. If A filled 10 liters in a hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160….if B filled in (1/16) th of a tank in 3 hours, how much time will it take to fill completely? Answer:7 hours 315.Question 12: KEYWORDS: Sports readers,10 tables,4chairs per table, each table has different number of people then how many tables will left without at least one


person? Ans : 6 316.) Two pipes A and B fill at A certain rate B is filled at 10,20,40,80,. If 1/16 of B if filled in 17

hours what time it will take to get completely filled Ans 21 317.) In a shopping mall with a staff of 5 members

the average age is 45 years. After 5 years a person

joined them and the average age is again 45 years. What’s the age of 6th person?

318) Find (4x+2y)/ (4x-2y) if x/2y=2

319) Find average speed if a man travels at speed of 24kmph up and 36kmph down at an altitude of

200m. Formula is 2xy/(x+y)

320.) Same model as 4th question. But it is on flat surface. Formula is same 2xy/(x+y).

321) Six friends go to pizza corner there are 2 types of pizzas. And six different flavors are there

they have to select 2 flavors from 6 flavors. In how

many ways we can select? Ans: 6C2 322) 3, 15, x, 51, 53,159,161. Find X

Ans: 17


323) 3 friends A, B, C went for week end party to McDonald’s restaurant and there they measure

there weights in some order IN 7 rounds. A;B;C;AB;BC;AC;ABC. Final round measure is 155

kg then find the average weight of all the 7 rounds? Ans: 4(155)/7=31 324) There is a toy train that can make 10 musical

sounds. It makes 2 musical sounds after being defective. What is the probability that same

musical sound would be produced 5 times

consecutively? (1 of )? Ans: 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32 325) (There was a long story, I'll cut short it).

There are 5 materials to make a perfume: Lilac, Balsamic, Lemon, and Woody and MI mosaic. To

make a perfume that is in demand the following

conditions are to be followed: Lilac and Balsamic go together. Woody and MI mosaic go together; Woody and Balsamic never go together. Lemon

can be added with any material. (Actually they had also mentioned how much amount of one can be added with how much quantity of the other; but that's not needed for the question.) All of the

following combinations are possible to make a perfume except: 1) Balsamic and Lilac

2) Woody and Lemon 3) MI mosaic and Woody 4) MI mosaic and Lilac


326) A triangle is made from a rope. The sides of

the triangle are A cm, B cm and C cm (I do not remember the numerical value). What will be the

area of the square made from the same rope? Ans: ((A+B+C)/4)2 327) What is the distance of the z-intercept from the x-intercept in the equation ax+by+cz=d (I do

not remember the values of a, b, c, d). Ans: sqrt ((d/a) 2+ (d/c) 2) 328) A scientist in Antarctic region conducts research on bears came to know that bears changes according to the location .once he moves 1 mile towards north, then he moves 2 miles

towards east, then 1 mile towards south. Now the color of bear he found will be in: Ans: white 329) (1/3) of a number is 3 times more than the (1/6) of the same number?

Ans is 18 330) There are 11 boys in a family. Youngest child is a boy. What is the probability of all are boys? a) 2 b) 2! C) 2048 d) 1024 331) A boy bought a roll A of 56 inches wide and

141 yards long. He also bought B of 77 inches wide of length 333yards. We don’t want any details of B. Some irrelevant matter. Final question is Time


taken for cutting A into 1 yard piece is 2 seconds. Time taken to cut into 141 pieces of 1 yard each

is? Ans is 2(141) =242 332) A Person buys a horse for 15 ponds, after one

year he sells it for 20 pounds. After one year, again he buys the same horse at 30 pounds and sells it for 40 pounds. What is the profit for that

person? Ans is 15 pounds 333) John buys a cycle for 31 dollars and given a cheque of amount 35 dollars. Shop Keeper exchanged the cheque with his neighbor and gave change to John. After 2 days, it is known that

cheque is bounced. Shop keeper paid the amount to his neighbor. The cost price of cycle is 19

dollars. What is the profit/loss for shop keeper? Ans is 23(cost price + change given). 334 In a family there are some boys and girls. All

boys told that they are having equal no of brothers and sisters and girls told that they are having twice the no. of brothers than sisters. How many boys and girls present in a family? Ans is 4 boys and 3 girls 335) There are certain number of hats and gloves

in a box. They are of 41 red, 23 green, 11 orange. Power gone. But a woman can differentiate


between hats and gloves.How many draws are required to obtain a pair of each color.

336) There is a die with 10 faces. It is not known

that fair or not. 2 captains want to toss die for batting selection. What is the possible solution

among the following? a) If no. is odd it is head, if no. is even it is tail b) If no. is odd it is tail, if no. is even it is head c) Toss a die until all the 10 digits appear on top face. And if first no. in the sequence is odd then

consider it as tail. If it is even consider it as head. I didn’t remembered last option and I don’t know answer. 337) 2 years ago of A is x times that of B. 3 Years

hence the age of A is 4/3 times of B. What is the present age of B in binary form? I didn’t remember the exact values of x and y. You

can solve easily. 338) metal strip of width ‘x’ cm. 2 metal strips are

placed one over the other, then the combine length of 2 strips is ‘y’. If ‘z’ strips are placed in that manner. What is the final width of that arrangement? Ans is (z-1) (y-x) +x. 339) There are 100 men and 100 women on the

dance floor. They want to dance with each other. Then which of the following statements is always true:


a) There are 2 men who danced with equal no. of women’s b) There are 2 women who danced with equal no. of men 340) A game is played between 2 players and one

player is declared as winner. All the winners from first round are played in second round. All the winners from second round are played in third

round and so on. If 8 rounds are played to declare only one player as winner, how many players are

played in first round Ans is 28. 341) There are 3 boys A, B, C and 2 Girls D, E. D always sit right to A. Girls never sit in extreme

positions and in the middle position. C always sits in the extreme positions. Who is sitting immediate

right to E? Ans is B or C 342) 49 members attended the party. In that 22

are males, 17 are females. The shake hands between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible? Ans is 12C2 343) There are 1000 pillars for a temple. 3 friends

Linda, Chelsey, Juli visited that temple. (Some unrelated stuff) Linda is taller than Chelsea and


taller than 2 of 1000 pillars. Julia is shorter than Linda. Find the correct sentence? a) Linda is shorter among them b) Chelsea is taller than Julia c) Chelsea is shorter than Julia d) Cannot determine who is taller among Chelsea

and Julia Ans: d

344) Entry ticket to an exhibition ranges from 1p to 31p. You need to provide exact change at the

counter. You have 31p coin. In how many parts

will u divide 31p so that u will provide the exact change required and carry as less coins as possible? a) 22 b) 31 c) 6 d) 32 Ans is 6

345) There are 2 friends Peter and Paul. Peter age

is twice as old as Paul when peter was as old as Paul is now. Sum of the present ages of Peter and Paul is 35.What is the present age of Peter?

Ans is 20 346) A lady took out jacket and gloves, which are avialable in blue 26, yellow 30 and red 56. Power

goes off, she can distinguish between gloves and jacket but not in colors. What's the possibilty their she will pick up pair of gloves of each color. 347) Two bowls are taken, one contains water and another contains tea.one spoon of water is added


to second bowl and mixed well, and a spoon of mixture is taken from second bowl and added to

the second bowl. Which statement will hold good for the above?

(Ans: second liquid in first bowl is smaller than the first mixture in second bowl)

348) Rearrange and categorize the word ‘RAPETEKA’? Ans: bird(parakeet)

349) A lies on mon, tues, wed and speak truths on other days, B lies on Thurs, Fri, Sat and speaks truths on other days. One day a said I lied today

and B said I too lied today. What is the day?

350) One grandfather has three grandchildren, two of their age difference is 3, eldest child age is 3 times youngest child’s age and eldest child’s age is

two times of sum of other two children. What is the age of eldest child?

Ans: 18 351) Now pet's age is to times when paul was once. But at that time paul's age=pet's current age

,how old is pet? 352) Block has 10,9,5 size, how many unit cube is

needed to make a block of that size? 353) 23 people are there, they are shaking hands

together, how many hand shakes possible, if they RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

are in pair of cyclic sequence. Ans-22

354) 10men and 10 women are there, they dance

with each other, is there possibilty that 2 men are dancing with same women and vice versa.

Ans-never 355) A lady took out jacket and gloves, which are

avialable in blue 26, yellow 30 and red 56. Power goes off, she can distinguish between gloves and

jacket but not in colors. What's the possibilty their

she will pick up pair of gloves of each color 356) B is taller than j and 3 pillers. P is shorter than B and 2 pillers is j shorter/taller than P?

Ans-irrelevant question

357) Sangakara and ponting selects batting by

using a dice, but dice is biased so to resolve ponty takes out a coin, what is the probability that dice shows correct option.

358) In school there are some bicycles and 4wheeler wagons. One Tuesday there are 58 wheels in the campus. How many bicycles are

there? Ans: 7

359. Two bowls are taken, one contains water and another contains tea.one spoon of water is added to second bowl and mixed well, and a spoon of


mixture is taken from second bowl and added to the second bowl. Which statement will hold good

for the above? (Ans: second liquid in first bowl is smaller than the

first mixture in second bowl)

360. Which is the smallest no divides 2880 and gives a perfect square? a. 1 b.2 c.5 d.6

Ans: c

361. Form 8 digit numbers from by using 1,

2,3,4,5 with repetition is allowed and must be divisible by4? a. 31250 b.97656 c.78125 d.97657 Ans: c

362. One problem on (a3-b3)/(a2+ab+b2)

Ans: ‘a-b’

363. Rearrange and categorize the word ‘RAPETEKA’?

Ans: bird(parakeet) 364. Key words in question (Fibonacci series, infinite series, in the middle of the question one

number series is there. I got the series 3 12 7 26 15 b? Ans:54 (Logic: 3*2+1=7 12*2+2=26 7*2+1=15 26*2+2=54)


365. A father has 7 penny’s with him and 1 water melon is for 1p, 2chickoos for 1p, 3 grapes foe 1p.

He has three sons. How can he share the fruits equally?

Ans: 1 watermelon, 2chickoos, 1grape

366. A lies on mon, tues, wed and speak truths on other days, B lies on thur, fri, sat and speaks truths on other days.. one day a said I lied today

and B said I too lied today. What is the day?

367. Man, Bear, North, South, Walks.

Ans: White 368. (1/2) of a number is 3 times more than the (1/6) of the same number?

Ans: 9(for any no it can be true)

369. There are two pipes A and B. If A filled 10

liters in hour B can fills 20 liters in same time. Likewise B can fill 10, 20, 40, 80,160. If B filled in (1/16) th of a tank in 3 hours, how much time will

it take to fill completely? Ans: 7 hours 370. Keywords: T.Nagar, Chennai, 1-100, prime

numbers b/n 140-180, How many 2’s are there? Ans: 20 (Not only 2’s ,1’s,3’s,4’s,5’s,6’s,7’s,8’s,9’s,0’s also 20)

371. One question has last part like difference between two terms is 9 and product of two


numbers is 14, what is the squares of sum of numbers?

Ans: 109

372. A man is standing before a painting of a man and he says I have no bro and sis and his father is

my father’s son? Ans: His son

373. What is the value of [(3x+8Y)/(x-2Y)]; if x/2y=2?

Ans: 10 {the numerical may change)

374. A pizza shop made pizzas with to flavours.in home there are ‘N’ different flavors, in that ‘M’ flavors are taken to made pizza.in how many ways

they can arrange? (Logic: NcM )

375. One grandfather has three grandchildren, two of their age difference is 3, eldest child age is 3 times youngest child’s age and eldest child’s age is

two times of sum of other two children. What is the age of eldest child? Ans: 18

376. In a market 4 man are standing. The average age of the four before 4 years is 45, aftyer some days one man is added and his age is 49. What is

the average weight of all? Ans: 49


377. Keywords: One organization, material, labor and maintenance are in the ratio of 4:6:7, the

material cost is: 100, what is the total cost? Ans: 425

378. Keywords: Density, Reluctance, Sensitivity,

Voltage, Current, what is the Resistance Formula is “R=V/I”

379. Keywords: Sports readers, 10 tables, 4chairs per table, each table has different number of

people then how many tables will left without at

least one person? Ans : 6 380. Keywords: Die, card, coin, b/n 2 to 12

Ans: All are equal

381. In a school for a student out of a 100 he got

74 of average for 7 subjects and he got 79 marks in 8th subject. what is the average of all the subjects?

The extra 5 marks will be distributed in 8 semester, 5/8=.625 74+.625=74.625

382. In a question ,last part has ,the ages of two people has the ratio of 6:5 and by adding the numbers we get 44,after how many years the ratio

would be 8:7? Ans: 8


383. Two years before Paul’s age is 2times the Alice age and the present age of Paul is 6times the

Alice. What is the presents Paul’s age?( 3years) “You try to solve this question once”

384. One train travels 200m from A to B with 70

km/ph. and returns to A with 80kmph, what is the average of their speed? 385. Which is the smallest no divides 2880 and gives a perfect square? a.1 b.2 c.5 d.6 Ans: c

386. In school there are some bicycles and

4wheeler wagons.one Tuesday there are 190

wheels in the campus. How many bicycles are there? Ans: 15

387. Man, Bear, North, South, walks. Colour of

bear(Hint: North pole ) Ans: White

388. A father has 7 penny�s with him and 1 water

melon is for 1p, 2chickoos for 1p, 3 grapes foe

1p.he has three sons. How can he share the fruits

equally? Ans: 1 watermelon, 2chickoos, 1grape

389. (1/2) of a number is 3 times more than the (1/6) of the same number?


Ans: 9

390. There are two pipes A and B. If A filled 10 liters in hour B can fills 20 liters in same time.

Likewise B can fill 10, 20, 40, 80,160�.if B filled in

(1/16) th of a tank in 3 hours, how much time will it take to fill completely? Ans: 7 hours 391. In a market 4 man are standing .the average

age of the four before 4years is 45,aftyer some days one man is added and his age is 49.what is the average weight of all? Ans: 49 392. There are 10 reading spots in a room. Each

reading spot has a round table. Each round table has 4 chair. If different no of persons are sitting at

each reading spot. And if there are 10 persons

inside the room then how many reading spots donot have atleast a single reader. (1) 5 (2) 6 (3) 7 (4) None Ans 6. because different no of persons are sitting on round table. So possible differnt combinations for 10 people will be 1 2 3 4. because max 4 people can sit on round table. so round tables left

are 6. 393. A person do rock climbing at an altitude of

800 m.He go up by 7 mph. and come down by 9 mph. what was his av speed. Ans (7+9)/2=8.


394. A boy want to make a cuboid of dimension 5m, 6m, 7m. from small cubes of .03 m3. later he

realized he can make same cuboid by making it hollow. Then it take some cubes less. What is the

no. of these cube. Ans. Vol of solid cuboid= 5*6*7=210 m3. Vol of its

inner cuboid by removal of which the cuboid will be hollow= (5-2)*(6-2)*(7-2)=60 m3, then ans will be 60/.03 395. Two years ago A was 6 times older than B.

Now he is 2 times older than B. What is the age of

A. Ans. Age of A=5, Age of B= Two and half. 396. What is thew value of (78*78*78-

45*45*45)/(78*78+78*45+45*45) Ans. 78-45=33. a3-b3=(a-b)(a2+ab+b2)

397) Two pipes A and B fill at A certain rate B is

filled at 10,20,40,80,. If 1/16 of B if filled in 17 hours what time it will take to get completely filled Ans 21

398) In a shopping mall with a staff of 5 members

the average age is 45 years. After 5 years a person

joined them and the average age is again 45 years. What’s the age of 6th person?

399) Find (4x+2y)/ (4x-2y) if x/2y=2


400] Find average speed if a man travels at speed

of 24kmph up and 36kmph down at an altitude of

200m.formula is 2xy/(x+y)

401) Same model as 4th question. But it is on flat

surface. Formula is same 2xy/(x+y).

402) Six friends go to pizza corner there r 2 types

of pizzas. And six different flavors r there they have to select 2 flavors from 6 flavors. In how

many ways we can select? Ans: 6C2

403) 3, 15, x, 51, 53,159,161. Find X Ans: 17 404) 3 friends A, B, C went for week end party to

McDonald’s restaurant and there they measure there weights in some order IN 7 rounds. A;B;C;AB;BC;AC;ABC. Final round measure is 155

kg then find the average weight of all the 7

rounds? Ans: 4(155)/7=31

405) There is a toy train that can make 10 musical

sounds. It makes 2 musical sounds after being

defective. What is the probability that same musical sound would be produced 5 times consecutively? (1 of )? Answer: 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

406) (There was a long story, I'll cut short it).

There are 5 materials to make a perfume: Lilac, RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

Balsamic, Lemon, and Woody and MI mosaic. To make a perfume that is in demand the following

conditions are to be followed: Lilac and Balsamic go together. Woody and MI mosaic go together;

Woody and Balsamic never go together. Lemon can be added with any material. (Actually they had

also mentioned how much amount of one can be added with how much quantity of the other; but that's not needed for the question.) All of the

following combinations are possible to make a perfume except:

1) Balsamic and Lilac

2) Woody and Lemon 3) MI mosaic and Woody 4) MI mosaic and Lilac

407) A triangle is made from a rope. The sides of the triangle are A cm, B cm and C cm (I do not

remember the numerical value). What will be the

area of the square made from the same rope? Ans: ((A+B+C)/4)2 408) What is the distance of the z-intercept from the x-intercept in the equation ax+by+cz=d (I do not remember the values of a, b, c, d). Ans: sqrt ((d/a) 2+ (d/c) 2) 409) A scientist in Antarctic region conducts research on bears came to know that bears

changes according to the location .once he moves 1 mile towards north, then he moves 2 miles


towards east, then 1 mile towards south. Now the color of bear he found will be in: Ans: white

410) (1/3) of a number is 3 times more than the

(1/6) of the same number? Ans is 18 411) There are 11 boys in a family. Youngest child is a boy. What is the probability of all are boys? a) 2 b) 2! C) 2048 d) 1024 412) A boy bought a roll A of 56 inches wide and

141 yards long. He also bought B of 77 inches wide

of length 333yards. We don’t want any details of B. Some irrelevant matter. Final question is Time taken for cutting A into 1 yard piece is 2 seconds. Time taken to cut into 141 pieces of 1 yard each

is? Ans is 2(141) =242 413) A Person buys a horse for 15 ponds, after one year he sells it for 20 pounds. After one year, again he buys the same horse at 30 pounds and

sells it for 40 pounds. What is the profit for that person? Ans is 15 pounds 414) John buys a cycle for 31 dollars and given a cheque of amount 35 dollars. Shop Keeper exchanged the cheque with his neighbor and gave

change to John. After 2 days, it is known that cheque is bounced. Shop keeper paid the amount


to his neighbor. The cost price of cycle is 19 dollars. What is the profit/loss for shop keeper? Ans is 23(cost price + change given). 415) In a family there are some boys and girls. All boys told that they are having equal no of brothers

and sisters and girls told that they are having twice the no. of brothers than sisters. How many boys and girls present in a family? Ans is 4 boys and 3 girls

416) There are certain number of hats and gloves

in a box. They are of 41 red, 23 green, 11 orange.

Power gone. But a woman can differentiate

between hats and gloves.How many draws are required to obtain a pair of each color.

417) There is a die with 10 faces. It is not known that fair or not. 2 captains want to toss die for batting selection. What is the possible solution among the following?

a) If no. is odd it is head, if no. is even it is tail b) If no. is odd it is tail, if no. is even it is head c) Toss a die until all the 10 digits appear on top

face. And if first no. in the sequence is odd then consider it as tail. If it is even consider it as head. I didn’t remembered last option and I don’t know answer. 418) 2 years ago of A is x times that of B. 3 Years

hence the age of A is 4/3 times of B. What is the

present age of B in binary form? RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

I didn’t remember the exact values of x and y. You can solve easily. 419) metal strip of width ‘x’ cm. 2 metal strips are

placed one over the other, then the combine length of 2 strips is ‘y’. If ‘z’ strips are placed in

that manner. What is the final width of that arrangement? Ans is (z-1) (y-x) +x. 420) There are 100 men and 100 women on the

dance floor. They want to dance with each other.

Then which of the following statements is always true: a) There are 2 men who danced with equal no. of women’s b) There are 2 women who danced with equal no. of men 421) A game is played between 2 players and one player is declared as winner. All the winners from first round are played in second round. All the

winners from second round are played in third round and so on. If 8 rounds are played to declare only one player as winner, how many players are played in first round Ans is 28. 422) There are 3 boys A, B, C and 2 Girls D, E. D

always sit right to A. Girls never sit in extreme positions and in the middle position. C always sits


in the extreme positions. Who is sitting immediate right to E? Ans is B or C 423) 49 members attended the party. In that 22 are males, 17 are females. The shake hands

between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible? Ans is 12C2 424) There are 1000 pillars for a temple. 3 friends

Linda, Chelsey, Juli visited that temple. (Some unrelated stuff) Linda is taller than Chelsea and taller than 2 of 1000 pillars. Julia is shorter than Linda. Find the correct sentence? a) Linda is shorter among them b) Chelsea is taller than Julia c) Chelsea is shorter than Julia d) Cannot determine who is taller among Chelsea and Julia

Ans: d

425) Entry ticket to an exhibition ranges from 1p to 31p. You need to provide exact change at the

counter. You have 31p coin. In how many parts will u divide 31p so that u will provide the exact change required and carry as less coins as

possible? a) 22 b) 31 c) 6 d) 32

Ans is 6 RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

426) There are 2 friends Peter and Paul. Peter age is twice as old as Paul when peter was as old as

Paul is now. Sum of the present ages of Peter and Paul is 35.What is the present age of Peter? Ans is 20

427) Two tanks A and B. A fills 1 ltr/1 hour B fills

10, 20, 30 per hour. If this is (passage unnecessary). If 1/4th tank of B takes 15 hours to fill how much it time will to take to fill complete tank?

428) Out of 7 children the youngest is boy then find the probability that all the remaining children

are boys Ans: 1/2^6 = 1/64 429) The three sides of a triangle are given. 16, 14, 21 cms and this triangle is conveted into a

square. So what will be the area of the square generated?

Ans:(14+16+21)/4. Then you will get the 1 side of

a square and now find the area of a square.ie, side^2

430) An equation of the form 4x+6y-2z=32 . Find the difference between x intercept and z intercept?Ans: x/a+y/b+z/c=

431) A Toy train can make 10 sounds sound changes after every 4 mins, now train is defective

and can make only 2 sounds. Find probability that RangaRakesRangaRakes tamilnavarasam.comtamilnavarasam.com

same sound is repeated 5 times consecutively (1 out of)?

Ans: 1/32 432) 20men and 20 women are there, they dance with each other, is there possibilty that 2 men are

dancing with same women and vice versa. Ans: Never 433) 10 people are there, they are shaking hands together, how many hand shakes possible, if they

are in no pair of cyclic sequence.

Ans-9 434) In school there are some bicycles and 4wheeler wagons. One Tuesday there are 234

wheels in the campus. How many bicycles are there?

Ans: Go with options. Multiply each option with 2 and subtact the obtained no from 234. If it is exactly divisible by 4, that is the answer. 435) A father has 7 penny’s with him and 1 water melon is for 1p, 2 chickoos for 1p, 3 grapes foe 1p. He has three sons. How can he share the friuts

equally? Ans: 1 watermelon, 2chickoos, 1grape 436) A piza shop made pizzas with to flavours. In home there are ‘9’ different flavors, in that ‘2’ flavors are taken to made piza in how many


ways they can arrange? (Logic: NcM, N= 9, M=2 ) 437) One organization, material, labor and

maintenance are in the ratio of 4:6:7, if the material cost is: 272, what is the total cost?

Ans: 4x=272 => x=68; now total cost = 272 + 6(68)+7(68).

438) 4 years before Paul’s age is 3times the Alice age and the present age of Paul is 6times the Alice. What is the presents Paul’s age?

Ans: x-4 = 3(y-4); x=6y: Solve you will get it.

439) In a question, last part has, the ages of two

people has the ratio of 6:5 and by adding the numbers we get 55, after how many years the

ratio would be 8:7? Ans: Easy you can do it, simple equtions

440) In a room (unwanted stuff) Sports readers,

10 tables, 4 chairs per table, each table has different number of people then how many tables will left without at least one person? Ans : 6

441) Passage joe is taller than jerry and 3

pillers. kistern is shorter than joe and 2 pillers is jerry shorter/taller than kistern ?

442) a volume of A are having in a container of sphere. how many semi hemispheres of B volume


each will be required to transfer all the A in to semi hemispheres? Ans: A= x B 443) Question based on V=I*R but in dis question most of data given are ridiculas like volume,

density, length, height similar long story are given

444) Peter and Paul are two friends. The sum of

their ages is 42 years. Peter is twice as old as Paul was when Peter was as old as Paul is now. What is the present age of Peter?

445) A horse chases a pony 2 hours after the pony runs. Horse takes 4 hours to reach the pony. If the

average speed of the horse is 81 kmph, what s the

average speed of the pony? (This question was really long with loads of irrelevant statement)

446) Difference between two numbers is 4 and their product is 17. Then find the sum of their squers?

447) A, B, C, D, E are there among A, B, C are

boys and D, E are girls D is to the left of A and no girl sits at the middle and at the extemes. Then what is the order of their sittings.

448) Some ages problem, then asked the answer in binary

449) (Some chetta chikkati solutions) unwanted

data folowed by a formula diameter d = 10*(t-14), t>14 then what id diameter after t=40?


450) Some denominations question like you have 31 paise, and ticket cost is between 1 to 31, you

have to give the exact denominations for the ticket. Find all the no. of possible denominations

you may prdict and you must be left wid atleast few paise?

451) Direction problems. A man goes 50 km North,

then turned left walked 40 km, then turned right? In which direction is he in?

Ans: North


Age calculation 1. Ten years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages? A. 45 B. 40 C. 35 D. 30


Answer : Option C

Explanation :

Let the present age of P and Q be 3x and 4x respectively. Ten years ago, P was half of Q in age => (3x – 10) = ½ (4x – 10) => 6x – 20 = 4x – 10 => 2x = 10 => x = 5 total of their present ages = 3x + 4x = 7x = 7 × 5 = 35

2. Father is aged three times more than his son Sunil. After 8 years, he would be two and a half times of Sunil's age. After further 8 years, how many times would he be of Sunil's age? A. 4 times B. 4 times C. 2 times D. 3 times



Answer : Option C

Explanation :

Assume that Sunil's present age = x. Then father's present age = 3x + x = 4x After 8 years, fathers age = 2 ½ times of Sunils' age => (4x+8) = 2 ½ × (x+8) => 4x + 8 = (5/2) ×(x + 8) => 8x + 16 = 5x+ 40 => 3x = 40-16 = 24 => x = 24/3 = 8 After further 8 years, Sunil's age = x + 8+ 8 = 8+8+8 = 24 Father's age = 4x + 8 + 8 = 4 × 8 + 8 + 8 = 48 Father's age/Sunil's age = 48/24 = 2

3. A man's age is 125% of what it was 10 years ago, but 83 1/3 % of what it will be after ten 10 years. What is his present age? A. 70 B. 60 C. 50 D. 40



Answer : Option C

Explanation :

Let the age before 10 years = x Then 125x/100 = x + 10 =>125x = 100x + 1000 => x = 1000/25 = 40 Present age = x + 10 = 40 +10 = 50

4. A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son? A. 23 years B. 22 years C. 21 years D. 20 years


Answer : Option B

Explanation :

Let the present age of the son = x years Then present age the man = (x+24) years Given that in 2 years, man's age will be twice the age of his son => (x+24) +2 = 2(x+2)


=> x = 22

5. Present ages of Kiran and Syam are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Syam's present age in years? A. 28 B. 27 C. 26 D. 24


Answer : Option D

Explanation :

Ratio of the present age of Kiran and Syam = 5 : 4 => Let the present age of Kiran = 5x Present age of Syam = 4x After 3 years, ratio of their ages = 11:9 => (5x + 3) : (4x + 3) = 11 : 9 => (5x+3) / (4x+3) = 11/9 => 9(5x + 3) = 11(4x + 3) => 45x + 27 = 44x + 33 => x = 33-27 =6 Syam's present age = 4x = 4×6 = 24


6. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. Find out the age of the youngest child? A. 6 years B. 5 years C. 4 years D. 3 years


Answer : Option C

Explanation :

Let the age of the youngest child = x Then the ages of 5 children can be written as x, (x+3), (x+6),(x+9) and (x+12) X + (x+3) + (x+6) + (x+9) + (x+12) = 50 => 5x + 30 =50 => 5x = 20 => x = 20/5 = 4

7. A is two years older than B who is twice as old as C. The total of the ages of A, B and C is 27. How old is B? A. 10 B. 9 C. 8 D. 7


Answer : Option A

Explanation :


Let the age of C = x. Then Age of B = 2x Age of A = 2 + 2x The total age of A,B and C =27 => (2+2x) + 2x + x = 27 => 5x = 25 => 25/5 = 5 B's age = 2x = 2×5 = 10

8. The Average age of a class of 22 students is 21 years. The average increased by 1 when the teacher's age also included. What is the age of the teacher? A. 48 B. 45 C. 43 D. 44


Answer : Option D

Explanation :

Total age of all students = 22×21 Total age of all students + age of the teacher = 23 × 22 Age of the teacher = 23×22 – 22×21 = 22(23-21) = 22×2 = 44


9. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, what was the son's age five years back? A. 20 years B. 18 years C. 14 years D. 22 years


Answer : Option C

Explanation :

Let the son's present age be x years. Then, (38 - x) = x => 2x = 38 => x = 38/2 = 19 Son's age 5 years back = 19-5 = 14

10. Ayisha's age is 1/6th of her father's age. Ayisha 's father's age will be twice the age of Shankar's age after 10 years. If Shankar's eight birthdays was celebrated two years before, then what is Ayisha 's present age. A. 10 years B. 12 years C. 8 years D. 5 years


Answer : Option D

Explanation :

Consider Ayisha's present age = x


Then her father's age = 6x Given that Ayisha 's father's age will be twice the age of Shankar's age after 10 years => Shankar's age after 10 years = ½(6x + 10) = 3x + 5 Also given that Shankar's eight birthdays was celebrated two years before => Shankar's age after 10 years = 8 + 12 = 20 => 3x + 5 = 20 => x = 15/3 = 5 => Ayisha 's present age = 5 years

11. The sum of the present ages of a son and his father is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, what will be son's age? A. 23 years B. 22 years C. 21 years D. 20 years


Answer : Option D

Explanation :

Let the present age of the son = x, then Present age of the father = 60-x


Six years ago father's age was 5 times the age of the son => (60-x)-6 = 5(x-6) => 84 = 6x => x = 84/6 = 14 Son's age after 6 years = x + 6 = 14 + 6 = 20

12. Kiarn is younger than Bineesh by 7 years and their ages are in the respective ratio of 7 : 9, how old is Kiran? A. 25 B. 24.5 C. 24 D. 23.5


Answer : Option B

Explanation :

Let the ages of Kiran and Bineesh are 7x and 9x respectively 7x = 9x-7 => x = 7/2 = 3.5 Kiran's age = 7x = 7 × 3.5 = 24.5

13. Six years ago, the ratio of the ages of Vimal and Saroj was 6 : 5. Four years


hence, the ratio of their ages will be 11 : 10. What is Saroj's age at present? A. 18 B. 17 C. 16 D. 15


Answer : Option C

Explanation :

Given that , six years ago, the ratio of the ages of Vimal and Saroj = 6 : 5 Hence we can assume that The age of Vimal six years ago = 6x The age of Saroj six years ago = 5x After 4 years, the ratio of their ages = 11 : 10 => (6x+10) / (5x + 10) = 11/10 => 10(6x + 10) = 11(5x + 10) => 5x = 10 => x = 10/5 = 2 Saroj's present age = 5x + 6 = 5×2 +6 = 16

14. At present, the ratio between the ages of Shekhar and Shobha is 4 : 3. After 6 years, Shekhar's age will be 26 years. Find out the age of Shobha at present? A. 15 years B. 14 years


C. 13 years D. 12 years


Answer : Option A

Explanation :

After 6 years, Shekhar's age will be 26years => Present age of Shekhar = 26 – 6 = 20 Let present age of Shobha = x Then 20/x = 4/3 => x = 20×3/4 = 15

15. My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born? A. 35 years B. 34 years C. 33 years D. 32 years


Answer : Option D

Explanation :

Let my age = x Then


My brother's age = x + 3 My mother's age = x + 26 My sister's age = (x + 3) + 4 = x + 7 My Father's age = (x + 7) + 28 = x + 35 => age my father when my brother was born = x + 35 – (x + 3) = 32

16. The present ages of A,B and C are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. What are their present ages (in years)? A. Insufficient data B. 16, 30, 40 C. 16, 28 40 D. 16, 28, 36


Answer : Option D

Explanation :

Let's take the present age of A,B and C as 4x, 7x and 9x respectively Then (4x - 8) + (7x – 8) + (9x – 8) = 56 => 20x = 80 => x = 4 Hence the present age of A, B and C are 4×4, 7×4 and 9×4 respectively ie, 16,28 and 36 respectively.


17. A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. What is the present age of the mother? A. 60 B. 50 C. 40 D. 30


Answer : Option C

Explanation :

Let the present age of the person = x . Then present age of the mother = 5x/2 Given that , after 8 years, the person will be one-half of the age of his mother. => (x + 8) = (1/2)(5x/2 + 8) => 2x + 16 = 5x/2 + 8 => x/2 = 8 => x = 16 Present age of the mother = 5x/2 = 5×16/2 = 40

18. A is as much younger than B and he is older than C. If the sum of the ages of B and C is 50 years, what is definitely the difference between B and A's age? A. Data inadequate B. 3 years C. 2 years D. 5 years



Answer : Option A

Explanation :

Age of C < Age of A < Age of B Given that sum of the ages of B and C is 50 years. => Let's take B's age = x and C's age = 50-x Now we need to find out B's age – A's age. But we cannot find out this with the given data.

19. Sobha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents? A. 6 years B. 5 years C. 4 years D. 3 years


Answer : Option A

Explanation :

Let Sobha's age = x and her brother's age = x-4 Then Sobha's father's age = x + 38


Sobha's mother's age = (x-4) + 36 => Sobha's father's age - Sobha's mother's age = (x + 38) – [(x-4) + 36] = x + 38 –x +4 – 36 = 6

20. The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. What is the ratio of their present ages? A. 7 : 3 B. 3 : 7 C. 9 : 4 D. 4 : 9


Answer : Option A

Explanation :

Let the age of the son before 10 years = x and age of the father before 10 years = 3x Now we can write as (3x + 20) = 2(x + 20) => x = 20 => Age of Father the son at present = x + 10 = 20 + 10 = 30 Age of the father at present = 3x + 10 = 3×20 + 10 = 70


Required ratio = 70 : 30 = 7 : 3

21. The ages of two persons differ by 16 years. 6 years ago, the elder one was 3 times as old as the younger one. What are their present ages of the elder person? A. 10 B. 20 C. 30 D. 40


Answer : Option C

Explanation :

Let's take the present age of the elder person = x and the present age of the younger person = x – 16 (x – 6) = 3 (x-16-6) => x – 6 = 3x – 66 => 2x = 60 => x = 60/2 = 30

22. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. What is father's present age? A. 30 years B. 31 years C. 32 yeas D. 33 years



Answer : Option D

Explanation :

Let the present age the son = x. Then present age of the father = 3x + 3 Given that ,three years hence, father's age will be 10 years more than twice the age of the son => (3x+3+3) = 2(x + 3) +10 => x = 10 Father's present age = 3x + 3 = 3×10+ 3 = 33

23. Kamal was 4 times as old as his son 8 years ago. After 8 years, Kamal will be twice as old as his son. Find out the present age of Kamal. A. 40 years B. 38 years C. 42 years D. 36 years


Answer : Option A

Explanation :

Let the age of the son before 8 years = x. Then age of Kamal before 8 years ago = 4x


After 8 years, Kamal will be twice as old as his son => 4x + 16 = 2(x + 16) => x = 8 Present age of Kamal = 4x + 8 = 4×8 +8 = 40

24. If 6 years are subtracted from the present age of Ajay and the remainder is divided by 18, then the present age of Rahul is obtained. If Rahul is 2 years younger to Denis whose age is 5 years, then what is Ajay 's present age? A. 50 years B. 60 years C. 55 years D. 62 years


Answer : Option B

Explanation :

Present age of Denis = 5 years Present age of Rahul = 5-2 = 3 Let the present age of Ajay = x Then (x-6)/18 = present age of Rahul = 3 => x- 6 = 3×18 = 54 => x = 54 + 6= 60

25. The ratio of the age of a man and his wife is 4:3. At the time of marriage the


ratio was 5:3 and After 4 years this ratio will become 9:7. How many years ago were they married? A. 8 years B. 10 years C. 11 years D. 12 years


Answer : Option D

Explanation :

Let the present age of the man and his wife be 4x and 3x respectively. After 4 years this ratio will become 9:7 => (4x + 4)/ (3x + 4) = 9/7 => 28x + 28 = 27x + 36 => x = 8 => Present age of the man = 4x = 4×8 = 32 Present age of his wife = 3x = 3×8 = 24 Assume that they got married before t years. Then (32 – t) / (24 – t) = 5/3 => 96 – 3t = 120 – 5t => 2t = 24


=> t = 24/2 = 12

26. The product of the ages of Syam and Sunil is 240. If twice the age of Sunil is more than Syam's age by 4 years, what is Sunil's age? A. 16 B. 14 C. 12 D. 10


Answer : Option C

Explanation :

Let the age of Sunil = x and age of Syam = y. Then xy = 240 ---(1) 2x = y + 4 => y = 2x – 4 => y = 2(x – 2) ---(2) Substituting equation (2) in equation (1). We get x × 2(x-2) = 240 => x (x-2) = 240/2 => x (x -2) = 120 ---(3) We got a quadratic equation to solve.


Always time is precious and objective tests measures not only how accurate you are but also how fast you are. We can either solve this quadratic equation in the traditional way. But more easy way is just substitute the values given in the choices in the quadratic equation (equation 3 ) and see which choice satisfy the equation. Here the option A is 10. If we substitute that value in the quadratic equation, x(x-2) = 10 × 8 which is not equal to 120 Now try option B which is 12. If we substitute that value in the quadratic equation, x(x-2) = 12 × 10 = 120. See, we got that x = 12. Hence Sunil's age = 12 (Or else, we cal solve the quadratic equation by factorization as, x(x- 2) = 120 => x2 – 2x - 120 = 0 =>(x-12)(x+10) = 0 => x = 12 or -10. Since x is age and can not be negative, x = 12 Or by using quadratic formula as x=−b±b 2 −4ac − − − − − − − √ 2a =2±(−2) 2 −4×1×(−120) − − − − − − − − − − − − − − − − − − − √ 2×1 =2±4+480 − − − − − − √ 2 =2±484 − − − √ 2 =2±22 2 =12 or −10 Since age is positive, x = 12 )

27. One year ago, the ratio of Sooraj's and Vimal's age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Vimal? A. 32 B. 34 C. 36 D. 38



Answer : Option C

Explanation :

Let take the age of Sooraj and Vimal , 1 year ago as 6x and 7x respectively. Given that, four years hence, this ratio would become 7: 8. => (6x + 5)/(7x + 5) = 7/8 => 48x + 40 = 49x + 35 => x = 5 Vimal's present age = 7x + 1 = 7×5 + 1 = 36

28. The total age of A and B is 12 years more than the total age of B and C. C is how many year younger than A? A. 10 B. 11 C. 12 D. 13


Answer : Option C

Explanation :

Given that A+B = 12 + B + C => A – C = 12 + B – B = 12 => C is younger than A by 12 years


29. Sachin's age after 15 years will be 5 times his age 5 years back. Find out the present age of Sachin? A. 10 years B. 11 years C. 12 years D. 13 years


Answer : Option A

Explanation :

Let the present age of Sachin = x Then (x+15) = 5(x-5) => 4x = 40 => x = 10


Important Formulas - Area

• Pythagorean Theorem (Pythagoras' theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c2 = a2 + b2 where c is the length of the hypotenuse and a and b are the lengths of the other two sides

• Pi is a mathematical constant which is the ratio of a circle's circumference to its diameter. It is denoted by π

π≈3.14≈22 7

• Geometric Shapes and solids and Important Formulas

Geometric

Shapes

Descrip

tion Formulas


Rectangle

l = Length

b = Breadth

d= Length of diagonal

Area = lb

Perimeter = 2(l + b)

d = l 2 +b 2 − − − − − √

Square

a = Length of a side

d= Length of diagonal

Area = a 2 =1 2 d 2

Perimeter = 4a

d = 2 √ a

Parallelogram

b and c are sides

Area = bh

Perimeter = 2(b + c)


b = base

h = height Rhombus

a = length of each side

b = base

h = height

d1, d2 are the diagonal

Area = bh (Formula 1 for area)

Area = 1 2 d 1 d 2 (Formula 2 for area)

Perimeter = 4a

Triangle

a , b and c are sides

b =

Area = 1 2 bh (Formula 1 for area)

Area = S(S−a)(S−b)(S−c) − − − − − − − − − − − − − − − − − − √ where S is the semiperimeter = a+b+c 2 (Formula 2 for area - Heron's


base

h = height

formula)

Perimeter = a + b + c

Radius of incircle of a triangle of area A = A S where S is the semiperimeter = a+b+c 2

Equilateral Triangle

a = side

Area = 3 √ 4 a 2

Perimeter = 3a

Radius of incircle of an equilateral triangle of side a = a 23 √

Radius of circumcircle of an equilateral triangle of side a = a 3 √

Base a is parallele to base b

Trapezium (Trapezoid in American English)

h = height

Area = 1 2 (a+b)h


Circle

r = radius

d = diameter

d = 2r

Area = πr 2 =1 4 πd 2

Circumference = 2πr=πd

Circumference d =π

Sector of Circle

r = radius

θ = central angle

Area, A =⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ θ 360 πr 2 (if angle measure is in degrees) 1 2 r 2 θ (if angle measure is in radians)

Arc Length, s = ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ θ 180 πr (if angle measure is in degrees) rθ (if angle measure is in radians)

Plese note that in the radian system for angular measurement, 2π radians = 360° ⇒1 radian = 180° π ⇒1° = π 180 radians Hence, Angle in Degrees = Angle in Radians ×180° π Angle in Radians = Angle in Degrees ×π 180°


Ellipse

Major axis length = 2a

Minor axis length = 2b

Area = πab

Perimeter ≈2πa 2 +b 2 2 − − − − − − √

Rectangular Solid

l = length

w = width

h = height

Total Surface Area = 2lw + 2wh + 2hl = 2(lw + wh + hl)

Volume = lwh

Cube

s = edge

Total Surface Area = 6s2

Volume = s3


Right Circular Cylinder

h = height

r = radius of base

Lateral Surface Area = (2 π r)h

Total Surface Area = (2 π r)h + 2 (π r2)

Voulme = (π r2)h

Pyramid

h = height

B = area of the base

Total Surface Area = B + Sum of the areas of the trianguar sides

Volume = 1 3 Bh

Right Circular Cone

h = height

r = radius of base

Lateral Surface Area =πrr 2 +h 2 − − − − − − √ =π rs where s is the slant height = r 2 +h 2 − − − − − − √

Total Surface Area = πrr 2 +h 2 − − − − − − √ +πr 2 =πrs+πr 2


Sphere

r = radius

d = diameter

d = 2r

Surface Area = 4πr 2 =πd 2

Volume = 4 3 πr 3 =1 6 πd 3

• Important properties of Geometric Shapes I. Properties of Triangle

i. Sum of the angles of a triangle = 180° ii. Sum of any two sides of a triangle is greater than

the third side. iii. The line joining the midpoint of a side of a

triangle to the positive vertex is called the median iv. The median of a triangle divides the triangle into

two triangles with equal areas v. Centroid is the point where the three medians of a

triangle meet. vi. Centroid divides each median into segments with

a 2:1 ratio vii. Area of a triangle formed by joining the

midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.

viii. An equilateral triangle is a triangle in which all three sides are equal

ix. In an equilateral triangle, all three internal angles are congruent to each other


x. In an equilateral triangle, all three internal angles are each 60°

xi. An isosceles triangle is a triangle with (at least) two equal sides

xii. In isosceles triangle, altitude from vertex bisects the base.

II. Properties of Quadrilaterals

A. Rectangle

i. The diagonals of a rectangle are equal and bisect each other

ii. opposite sides of a rectangle are parallel iii. opposite sides of a rectangle are congruent iv. opposite angles of a rectangle are congruent v. All four angles of a rectangle are right angles

vi. The diagonals of a rectangle are congruent

B. Square

vii. All four sides of a square are congruent viii. Opposite sides of a square are parallel

ix. The diagonals of a square are equal x. The diagonals of a square bisect each other at

right angles xi. All angles of a square are 90 degrees.

C. Parallelogram

xii. The opposite sides of a parallelogram are equal in length.


xiii. The opposite angles of a parallelogram are congruent (equal measure).

xiv. The diagonals of a parallelogram bisect each other.

xv. Each diagonal of a parallelogram divides it into two triangles of the same area

D. Rhombus

xvi. All the sides of a rhombus are congruent xvii. Opposite sides of a rhombus are parallel.

xviii. The diagonals of a rhombus are unequal and bisect each other at right angles

xix. Opposite internal angles of a rhombus are congruent (equal in size)

xx. Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)

Other properties of quadrilaterals

xxi. The sum of the interior angles of a quadrilateral is 360 degrees

xxii. A square and a rhombus on the same base will have equal areas.

xxiii. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

xxiv. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

xxv. Each diagonal of a parallelogram divides it into two triangles of the same area


III. Sum of Interior Angles of a polygon

i. The sum of the interior angles of a polygon = 180(n - 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 - 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 - 2) = 180 × 2 = 360 °

1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? A. 4.04 % B. 2.02 % C. 4 % D. 2 %


Answer : Option A

Explanation :


Error = 2% while measuring the side of a square. Let the correct value of the side of the square = 100

Then the measured value = 100×(100+2) 100 =102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000 Calculated Value of the area of the square = 102 × 102 = 10404 Error = 10404 - 10000 = 404

Percentage Error = Error Actual Value ×100=404 10000 ×100=4.04%

2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of the lawn is 2109 sq. m. what is the width of the road? A. 5 m B. 4 m C. 2 m D. 3 m


Answer : Option D


Explanation :

Please refer the diagram given above. Area of the park = 60 × 40 = 2400 m2 Given that area of the lawn = 2109 m2 ∴ Area of the cross roads = 2400 - 2109 = 291 m2 Assume that the width of the cross roads = x Then total area of the cross roads


= Area of road 1 + area of road 2 - (Common Area of the cross roads) = 60x + 40x - x2 (Let's look in detail how we got the total area of the cross roads as 60x + 40x - x2 As shown in the diagram, area of the road 1 = 60x. This has the areas of the parts 1,2 and 3 given in the diagram Area of the road 2 = 40x. This has the parts 4, 5 and 6 You can see that there is an area which is intersecting (i.e. part 2 and part 5) and the intersection area = x2. Since 60x + 40x covers the intersecting area (x2) two times ( part 2 and part 5) ,we need to subtract the intersecting area of (x2) once time to get the total area. . Hence total area of the cross roads = 60x + 40x - x2) Now, we have Total areas of cross roads = 60x + 40x - x2 But area of the cross roads = 291 m2 Hence 60x + 40x - x2 = 291 => 100x - x2 = 291 => x2 - 100x + 291 = 0 => (x - 97)(x - 3) = 0 => x = 3 (x can not be 97 as the park is only 60 m long and 40 m wide)


3. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area? A. 30 % B. 28 % C. 32 % D. 26 %


Answer : Option B

Explanation :

--------------------------------------------------------- Solution 1 --------------------------------------------------------- Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000

Lost 20% of length ⇒New length = Original length×(100−20) 100 =100×80 100 =80 Lost 10% of breadth ⇒New breadth= Original breadth×(100−10) 100 =100×90 100 =90

New area = 80 × 90 = 7200 Decrease in area = Original Area - New Area = 10000 - 7200 = 2800


Percentage of decrease in area = Decrease in Area Original Area ×100=2800 10000 ×100=28%

--------------------------------------------------------- Solution 2 --------------------------------------------------------- Let original length = l and original breadth = b Then original area = lb

Lost 20% of length ⇒New length = Original length×(100−20) 100 =l×80 100 =80l 100 Lost 10% of breadth ⇒New breadth= Original breadth×(100−10) 100 =b×90 100 =90b 100 New area =80l 100 ×90b 100 =7200lb 10000 =72lb 100 Decrease in area = Original Area - New Area = lb−72lb 100 =28lb 100 Percentage of decrease in area = Decrease in Area Original Area ×100 =(28lb 100 ) lb ×100=28lb×100 100lb =28%

4. If the length of a rectangle is halved and its breadth is tripled,


what is the percentage change in its area? A. 25 % Increase B. 25 % Decrease C. 50 % Decrease D. 50 % Increase


Answer : Option D

Explanation :

--------------------------------------------------------- Solution 1 --------------------------------------------------------- Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000

Length of the rectangle is halved ⇒New length = Original length 2 =100 2 =50 breadth is tripled ⇒New breadth= Original breadth×3=100×3=300

New area = 50 × 300 = 15000 Increase in area = New Area - Original Area = 15000 - 10000= 5000

Percentage of Increase in area = Increase in Area Original Area ×100=5000 10000 ×100=50%



Length of the rectangle is halved ⇒New length = Original length 2 =l 2 breadth is tripled ⇒New breadth = Original breadth×3=3b New area = l 2 ×3b=3lb 2 Increase in area = New Area - Original Area = 3lb 2 −lb=lb 2 Percentage of Increase in area = Increase in Area Original Area ×100 =(lb 2 ) lb ×100=lb×100 2lb =50%

5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges? A. 35% B. 30 % C. 20 % D. 25%


Answer : Option B


Explanation :

--------------------------------------------------------- Solution 1 ---------------------------------------------------------

Consider a square plot as shown above and let the length of each side = 1

Then length of the diagonal = (1+1) − − − − − − √ =2 √

Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2

Distance travelled if walked diagonally = BD = 2 √ =1.41

Distance Saved = 2 - 1.41 = .59


Percent distance saved = .59 2 ×100=.59×50≈30%

--------------------------------------------------------- Solution 2 ---------------------------------------------------------

Consider a square plot as shown above and let the length of each side = x

Then length of the diagonal = (x+x) − − − − − − √ =2x − − √


Distance travelled if walked along the edges = BC + CD = x + x = 2x

Distance travelled if walked diagonally = BD = 2x − − √ =1.41x

Distance Saved = 2x - 1.41x = .59x

Percent distance saved = .59x 2x ×100=.59×50≈30%

6. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? A. 95 B. 92 C. 88 D. 82


Answer : Option C

Explanation :

Given that area of the field = 680 sq. feet => lb = 680 sq. feet Length(l) = 20 feet


=> 20 × b = 680

⇒b=680 20 =34 feet

Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? A. 126 sq. ft. B. 64 sq. ft. C. 100 sq. ft. D. 102 sq. ft.


Answer : Option A

Explanation :

Let l = 9 ft. Then l + 2b = 37 => 2b = 37 - l = 37 - 9 = 28 => b = 28/2 = 14 ft. Area = lb = 9 × 14 = 126 sq. ft.


8. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot? A. 14 metres B. 20 metres C. 18 metres D. 12 metres


Answer : Option B

Explanation :

lb = 460 m2 ------(Equation 1) Let the breadth = b

Then length, l = b×(100+15) 100 =115b 100 ------(Equation 2)

From Equation 1 and Equation 2,

115b 100 ×b=460 b 2 =46000 115 =400 ⇒b=400 − − − √ =20 m

9. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares? A. 400 B. 365 C. 385 D. 315



Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 - x) hectares.

Difference of the areas of the two parts = x - (700 - x) = 2x - 700 one-fifth of the Average of the two areas = 1 5 [x+(700−x)] 2 =1 5 ×700 2 =350 5 =70

Given that difference of the areas of the two parts = one-fifth of the Average of the two areas => 2x - 700 = 70 => 2x = 770

⇒x=770 2 =385

Hence, Area of smaller part = (700 - x) = (700 – 385) = 315 hectares.

10. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq.


metre. A. Rs.12000 B. Rs.19500 C. Rs.18000 D. Rs.16500.


Answer : Option D

Explanation :

Area = 5.5 × 3.75 sq. metre. Cost for 1 sq. metre. = Rs. 800 Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle? A. 18 cm B. 16 cm C. 40 cm D. 20 cm


Answer : Option C

Explanation :

Let breadth = x cm Then length = 2x cm Area = lb = x × 2x = 2x2


New length = (2x - 5) New breadth = (x + 5) New Area = lb = (2x - 5)(x + 5) But given that new area = initial area + 75 sq.cm. => (2x - 5)(x + 5) = 2x2 + 75 => 2x2 + 10x - 5x - 25 = 2x2 + 75 => 5x - 25 = 75 => 5x = 75 + 25 = 100 => x = 100/5 = 20 cm

Length = 2x = 2 × 20 = 40cm

12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus? A. equal to ½ B. equal to ¾ C. greater than 1 D. equal to 1


Answer : Option D

Explanation :

A square and a rhombus on the same base will have equal areas.

Hence ratio of the areas of the square and the rhombus will be equal to 1 since they stand on the same base


================================================================ Note : Please find the proof of the formula given below which you may like to go through

Let ABCD be the square and ABEF be the rhombus Consider the right-angled triangles ADF and BCE We know that AD = BC (∵ sides of a square) AF = BE (∵ sides of a rhombus) ∴ DF = CE [∵ DF2 = AF2 - AD2 and CE2 = BE2 - BC2] Hence ∆ ADF = ∆ BCE => ∆ ADF + Trapezium ABCF= ∆ BCE + Trapezium ABCF => Area of square ABCD = Area of rhombus ABEF


13. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field. A. 37500 m2 B. 30500 m2 C. 32500 m2 D. 40000 m2


Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length

⇒b=60l 100 =3l 5

perimeter of the field = 800 m => 2 (l + b) = 800

⇒2(l+3l 5 )=800 ⇒l+3l 5 =400 ⇒8l 5 =400 ⇒l 5 =50 ⇒l=5×50=250 m b = 3l 5 =3×250 5 =2×50=150 m Area = lb = 250×150=37500 m 2

14. A room 5m 44cm long and 3m 74cm broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor?


A. 176 B. 124 C. 224 D. 186


Answer : Option A

Explanation :

l = 5 m 44 cm = 544 cm b = 3 m 74 cm = 374 cm Area = 544 × 374 cm2 Now we need to find out HCF(Highest Common Factor) of 544 and 374. Let's find out the HCF using long division method for quicker results)

374) 544 (1 374 170) 374 (2 340 34) 170 (5 170 0


Hence, HCF of 544 and 374 = 34 Hence, side length of largest square tile we can take = 34 cm Area of each square tile = 34 × 34 cm 2

Number of tiles required = 544×374 34×34 =16×11=176

15. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres? A. 60 m B. 100 m C. 75 m D. 50 m


Answer : Option A

Explanation :

Length of the plot is 20 metres more than its breadth. Hence, let's take the length as l metres and breadth as (l - 20) metres


Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l - 20) ] = 2(2l - 20) metres Cost per meter = Rs. 26.50 Total cost = 2(2l - 20) × 26.50 Total cost is given as Rs. 5300 => 2(2l - 20) × 26.50 = 5300 => (2l - 20) × 26.50 = 2650 => (l - 10) × 26.50 = 1325 => (l - 10) = 1325/26.50 = 50 => l = 50 + 10 = 60 metres

16. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)? A. 142000 B. 112800 C. 142500 D. 153600


Answer : Option D

Explanation :

l : b = 3 : 2 ------------------------------------------(Equation 1) Perimeter of the rectangular park = Distance travelled by the man at the speed of 12 km/hr in 8


minutes = speed × time = 12 × 8/60 (∵ 8 minute = 8/60 hour) = 8/5 km = 8/5 ×1000 m = 1600 m Perimeter = 2(l + b) => 2(l + b) = 1600 => l + b = 1600/2 = 800 m ---------------------------(Equation 2) From (Equation 1) and (Equation 2) l = 800 × 3/5 = 480 m b = 800 × 2/5 = 320 m (Or b = 800 - 480 = 320m) Area = lb = 480 × 320 = 153600 m2

17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%? A. 45% B. 44% C. 40% D. 42%


Answer : Option B

Explanation :

--------------------------------------------------------- Solution 1 ---------------------------------------------------------


Let original length = 100 and original breadth = 100 Then original area = 100 × 100 = 10000

Increase in 20% of length ⇒New length = Original length×(100+20) 100 =100×120 100 =120 Increase in 20% of breadth ⇒New breadth= Original breadth×(100+20) 100 =100×120 100 =120

New area = 120 × 120 = 14400 Increase in area = New Area - Original Area = 14400 - 10000 = 4400

Percentage increase in area = Increase in Area Original Area ×100=4400 10000 ×100=44%



Increase in 20% of length ⇒New length = Original length×(100+20) 100 =l×120 100 =120l 100 Increase in 20% of breadth ⇒New breadth= Original breadth×(100+20) 100 =b×120 100 =120b 100 New area =120l 100 ×120b 100 =14400lb 10000 =144lb 100 Increase in area = New Area - Original Area = 144lb 100 −lb=44lb 100 Percentage of increase in area = Increase in Area Original Area ×100 =(44lb 100 ) lb ×100=44lb×100 100lb =44%

18. If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area? A. 2800 m2 B. 2740 m2 C. 2520 m2 D. 2200 m2


Answer : Option C

Explanation :

l - b = 23 ...................(Equation 1) perimeter = 2(l + b) = 206 => l + b = 103.............(Equation 2) (Equation 1) + (Equation 2) => 2l = 23 + 103 = 126


=> l = 126/2 = 63 metre Substituting this value of l in (Equation 1), we get 63 - b = 23 => b = 63 - 23 = 40 metre Area = lb = 63 × 40 = 2520 m2

19. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? A. 16 cm B. 18 cm C. 14 cm D. 20 cm


Answer : Option B

Explanation :

Given that 2(l+b) b =5

=> 2l + 2b = 5b => 2l = 3b

=>b=2l 3


Also given that area = 216 cm 2 => lb = 216 cm 2

Substituting the value of b, we get, l×2l 3 =216 ⇒l 2 =3×216 2 =3×108=(3×3)×36 ⇒l=3×6=18 cm

20. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? A. 814 B. 802 C. 836 D. 900


Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm b = 9 m 2 cm = 902 cm Area = 1517 × 902 cm2 Now we need to find out HCF(Highest Common Factor) of 1517 and 902. Let's find out the HCF using long division method for quicker results)


902) 1517 (1 902 615) 902 (1 615 287) 615 (2 574 41) 287 (7 287 0 Hence, HCF of 1517 and 902 = 41 Hence, side length of largest square tile we can take = 41 cm Area of each square tile = 41 × 41 cm 2


Number of tiles required = 1517×902 41×41 =37×22=407×2=814

21. The diagonal of the floor of a rectangular room is 71 2 feet. The shorter side of the room is 4 1 2 feet. What is the area of the room? A. 27 square feet B. 22 square feet C. 24 square feet D. 20 square feet


Answer : Option A

Explanation :

Diagonal, d = 71 2 feet =15 2 feet Breadth, b = 41 2 feet=9 2 feet In the right-angled triangle PQR, l 2 =(15 2 ) 2 −(9 2 ) 2 =225 4 −81 4 =144 4 l=144 4 − − − − √ =12 2 feet = 6 feet Area = lb = 6×9 2 =27 feet 2

22. The diagonal of a rectangle is 41 − − √ cm and its area is


20 sq. cm. What is the perimeter of the rectangle? A. 16 cm B. 10 cm C. 12 cm D. 18 cm


Answer : Option D

Explanation :

For a rectangle, d 2 =l 2 +b 2 where l = length , b = breadth and d = diagonal of the of the rectangle

d=41 − − √ cm d 2 =l 2 +b 2 ⇒l 2 +b 2 =(41 − − √ ) 2 =41........(Equation 1) Area = lb = 20 cm 2 ............(Equation 2) Solving (Equation 1) and (Equation 2)

(a+b) 2 =a 2 +2ab+b 2

using the above formula, we have (l+b) 2 =l 2 +2lb+b 2 =(l 2 +b 2 )+2lb=41+(2×20)=81 ⇒(l+b)=81 − − √ =9 cm perimeter = 2(l+b)=2×9=18 cm


23. A tank is 25 m long, 12 m wide and 6 m deep. What is the cost of plastering of its walls and bottom at the rate of 75 paise per sq. m? A. Rs. 558 B. Rs. 502 C. Rs. 516 D. Rs. 612


Answer : Option A

Explanation :

Consider a rectangular solid of length l, width w and height h. Then 1. Total Surface area of a rectangular solid, S = 2lw + 2lh +


2wh = 2(lw + lh + wh) 2. Volume of a rectangular solid, V = lwh

In this case, l = 25 m, w = 12 m, h = 6 m and all surface needs to be plastered except the top Hence total area needs to be plastered = Total Surface Area - Area of the Top face = (2lw + 2lh + 2wh) - lw = lw + 2lh + 2wh = (25 × 12) + (2 × 25 × 6) + (2 × 12 × 6) = 300 + 300 + 144 = 744 m2 Cost of plastering = 744 × 75 = 55800 paise = Rs.558

24. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction? A. Rs.3500 B. Rs. 4200 C. Insufficient Data D. Rs. 4400


Answer : Option C

Explanation :


Let length and width of the rectangular plot be l and b respectively Total Area of the rectangular plot = 96 sq.m. Width of the pathway = 2 m Length of the remaining area in the plot = (l - 4) breadth of the remaining area in the plot = (b - 4) Area of the remaining area in the plot = (l - 4)(b - 4) Area of the pathway = Total Area of the rectangular plot - remaining area in the plot = 96 - [(l - 4)(b - 4)] = 96 - [lb - 4l - 4b + 16] = 96 - [96 - 4l - 4b + 16] = 96 - 96 + 4l + 4b - 16] = 4l + 4b - 16 = 4(l + b) - 16


We do not know the values of l and b and hence total area of the rectangular plot can not be found out. So we can not find out total cost of the construction.

25. The area of a parallelogram is 72 cm2 and its altitude is twice the corresponding base. What is the length of the base? A. 6 cm B. 7 cm C. 8 cm D. 12 cm


Answer : Option A

Explanation :

Area of a parallelogram , A = bh where b is the base and h is the height of the parallelogram

Let the base = x cm. Then the height = 2x cm (∵ altitude is twice the base)


Area = x × 2x = 2x2 But the area is given as 72 cm2 => 2x2 = 72 => x2 = 36 => x = 6 cm

26. Two diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter? A. 136 cm B. 156 cm C. 144 cm D. 121 cm


Answer : Option B

Explanation :

Remember the following two properties of a rhombus which will be useful in solving this question


1. The sides of a rhombus are congruent. 2. The diagonals of a rhombus are unequal and bisect each other at right angles. Let the diagonals be PR and SQ such that PR = 72 cm and SQ = 30 cm

PO = OR = 72 2 =36 cm SO = OQ = 30 2 =15 cm PQ = QR = RS = SP = 36 2 +15 2 − − − − − − − − √ =1296+225 − − − − − − − − − √ =1521 − − − − √ = 39 cm perimeter = 4 × 39 =156 cm

27. The base of a parallelogram is (p + 4), altitude to the base is (p - 3) and the area is (p2 - 4), find out its actual area. A. 40 sq. units B. 54 sq. units C. 36 sq. units D. 60 sq. units


Answer : Option D

Explanation :


Area of a parallelogram , A = bh where b is the base and h is the height of the parallelogram

Hence, we have p2 - 4 = (p + 4)(p - 3) => p2 - 4 = p2 - 3p + 4p - 12 => -4 = p - 12 => p = 12 - 4 = 8 Hence, actual area = (p2 - 4) = 82 - 4 = 64 - 4 = 60 sq. units

28. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle? A. 1443 √ −48π cm2 B. 1213 √ −36π cm2 C. 1443 √ −36π cm2 D. 1213 √ −48π cm2


Answer : Option A

Explanation :


Area of an equilateral triangle = 3 √ 4 a 2 where a is length of one side of the equilateral triangle

Area of the equilateral ∆ ABC = 3 √ 4 a 2 =3 √ 4 24 2 =1443 √ cm 2 .............(1)

Area of a triangle = 1 2 bh where b is the base and h is the height of the triangle

Let r = radius of the inscribed circle. Then Area of ∆ ABC


= Area of ∆ OBC + Area of ∆ OCA + area of ∆ OAB = (½ × r × BC) + (½ × r × CA) + (½ × r × AB) = ½ × r × (BC + CA + AB) = ½ x r x (24 + 24 + 24) = ½ x r x 72 = 36r cm2 ------------------------------------------ (2) From (1) and (2),

1443 √ =36r ⇒r=144 36 3 √ =43 √ −−−−−−−−−−−−(3)

Area of a circle = πr 2 where = radius of the circle

From (3), the area of the inscribed circle = πr 2 =π(43 √ ) 2 =48π−−−−−−−−−−−−(4) Hence , Area of the remaining portion of the triangle Area of ∆ ABC – Area of inscribed circle 1443 √ −48π cm 2


29. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed? A. 30 B. 44 C. 56 D. 60


Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b) where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important. Hence number of poles required = 280⁄5 = 56

30. If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter


A. 42 cm B. 64 cm C. 56 cm D. 52 cm


Answer : Option D

Explanation :

Let the diagonals be PR and SQ such that PR = 24 cm and SQ = 10 cm

PO = OR = 24 2 =12 cm SO = OQ = 10 2 =5 cm PQ = QR = RS = SP = 12 2 +5 2 − − − − − − − √ =144+25 − − − − − − − √ =169 − − − √ = 13 cm perimeter = 4 × 13 = 52 cm

31. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height? A. 11600 − − − − − √ cm B. 14400 − − − − − √ cm C. 10000 − − − − − √ cm D. 12040 − − − − − √ cm



Answer : Option A

Explanation :

The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal) Hence the length of the longest rod = AG Initially let's find out AC. Consider the right angled triangle ABC


AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000

⇒AC = 8000 − − − − √ cm

Consider the right angled triangle ACG

AG2 = AC 2 + CG 2

=(8000 − − − − √ ) 2 +60 2 =8000+3600=11600 ⇒AG = 11600 − − − − − √ cm ⇒The length of the longest rod = 11600 − − − − − √ cm


Important Formulas - Average • Average

Average = Sum of observations Number of observations

• Average Speed If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,the average

1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? A. 6.25 B. 5.5 C. 7.4 D. 5


Answer : Option A

Explanation :

Runs scored in the first 10 overs = 10×3.2=32 Total runs = 282 remaining runs to be scored = 282 - 32 = 250


remaining overs = 40 Run rate needed = 250 40 =6.25

2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? A. 4800 B. 4991 C. 5004 D. 5000


Answer : Option B

Explanation :

Let the sale in the sixth month = x Then 6435+6927+6855+7230+6562+x 6 =6500 => 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 × 6500 = 39000 => 34009 + x = 39000 => x = 39000 - 34009 = 4991

3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero , at the most?


A. 1 B. 20 C. 0 D. 19


Answer : Option D

Explanation :

Average of 20 numbers = 0 =>Sum of 20 numbers 20 =0 => Sum of 20 numbers = 0 Hence at the most, there can be 19 positive numbers. (Such that if the sum of these 19 positive numbers is x, 20th number will be -x)

4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team. A. 23 years B. 20 years C. 24 years D. 21 years



Answer : Option A

Explanation :

Number of members in the team = 11 Let the average age of of the team = x =>Sum of the ages of all the 11 members of the team 11 =x => Sum of the ages of all the 11 members of the team = 11x Age of the captain = 26 Age of the wicket keeper = 26 + 3 = 29 Sum of the ages of 9 members of the team excluding captain and wicket keeper = 11x - 26 - 29 = 11x - 55 Average age of 9 members of the team excluding captain and wicket keeper =11x−55 9 Given that 11x−55 9 =(x−1) => 11x - 55 = 9(x - 1)


=> 11x - 55 = 9x - 9 => 2x = 46 =>x=46 2 =23 years

5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A? A. 2000 B. 3000 C. 4000 D. 5000


Answer : Option C

Explanation :

Let the monthly income of A = a monthly income of B = b monthly income of C = a a+b=2×5050−−−−−−−−−(Equation1) b+c=2×6250−−−−−−−−−(Equation2) a+c=2×5200−−−−−−−−−(Equation3) (Equation 1) + (Equation 3) - (Equation 2)


=>a+b+a+c−(b+c)=(2×5050)+(2×5200)−(2×6250) => 2a = 2(5050 + 5200 - 6250) => a = 4000 => Monthly income of A = 4000

6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year? A. Rs. 8 B. Rs. 7.98 C. Rs. 6.2 D. Rs. 8.1


Answer : Option B

Explanation :

Total Cost = 4000×3 Total diesel used = 4000 7.5 +4000 8 +4000 8.5 average cost per litre of diesel = 4000×3 (4000 7.5 +4000 8 +4000 8.5 ) =3 (1 7.5 +1 8 +1 8.5 ) It is important how you proceed from this stage. Remember time is very important here and if we solve this completely in the traditional way, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices In this case


answer = 3 (1 7.5 +1 8 +1 8.5 ) ≈3 (1 8 +1 8 +1 8 ) ≈3 (3 8 ) ≈8 Means we got that answer is approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these? It will be easy to figure out. Just see here the denominator was 1 7.5 +1 8 +1 8.5 and we approximated it as 3 8 . However 1 7.5 +1 8.5 =1 8−.5 +1 8+.5 =8+.5+8−.5 (8−.5)(8+.5) =16 (8 2 −.5 2 ) [because a 2 −b 2 =(a−b)(a+b)] =16 (64−.25) ie, 1 7.5 +1 8.5 =16 (64−.25) We know that 1 8 +1 8 =1 4 =16 64 => 1 7.5 +1 8.5 >1 8 +1 8 Early we had approximated the denominator as 3 8 However from the above mentioned equations, now you know that actually denominator is slightly greater than 3 8 It means that answer is slightly lower that 8. Hence we can pick the choice 7.98 as the answer Try to remember the relations between numbers and which can help you to save a lot of time which can be very precious in competitive exams

7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran? A. 70 kg B. 69 kg C. 61 kg D. 67 kg



Answer : Option D

Explanation :

Let Kiran's weight = x. Then According to Kiran, 65 < x < 72 ------------(equation 1) According to brother, 60 < x < 70 ------------(equation 2) According to mother, x≤68 ------------(equation 3) Given that equation 1,equation 2 and equation 3 are correct. By combining these equations, we can write as 65<x≤68 That is x = 66 or 67 or 68 average of different probable weights of Kiran = 66+67+68 3 =67

8. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. A. 48.55 B. 42.25 C. 50 D. 51.25



Answer : Option A

Explanation :

Average weight of 16 boys = 50.25 Total Weight of 16 boys = 50.25×16 Average weight of remaining 8 boys = 45.15 Total Weight of remaining 8 boys = 45.15×8 Total weight of all boys in the class = (50.25×16)+(45.15×8) Total boys = 16+8=24 Average weight of all the boys = (50.25×16)+(45.15×8) 24 =(50.25×2)+(45.15×1) 3 =(16.75×2)+15.05=33.5+15.05 =48.55

9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday? A. 290 B. 304 C. 285 D. 270


Answer : Option C

Explanation :

in a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday


Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday Average visitors on Sundays = 510 Total visitors of 5 Sundays = 510×5 Average visitors on other days = 240 Total visitors of other 25 days = 240×25 Total visitors = (510×5)+(240×25) Total days= 30 Average number of visitors per day = (510×5)+(240×25) 30 =(51×5)+(24×25) 3 =(17×5)+(8×25)=85+200=285

10. A student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half 1/2. What is the number of students in the class? A. 45 B. 40 C. 35 D. 30


Answer : Option B

Explanation :

Let the total number of students = x The average marks increased by 1 2 due to an increase of 83-63=20 marks. But total increase in the marks = 1 2 ×x=x 2 Hence we can write as x 2 =20 ⇒x=20×2=40


11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is A. 322 7 years B. 315 7 years C. 281 7 years D. 305 7 years


Answer : Option B

Explanation :

Total age of the grandparents = 67×2 Total age of the parents = 35×2 Total age of the grandchildren = 6×3 Average age of the family = (67×2)+(35×2)+(6×3) 7 =134+70+18 7 =222 7 =315 7

12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B? A. 31 kg B. 281 2 kg C. 32 kg D. 301 2 kg


Answer : Option A

Explanation :


Let the weight of A, B and C are a,b and c respectively. Average weight of A,B and C = 45 a+b+c=45×3=135---equation(1) average weight of A and B = 40 a+b=40×2=80---equation(2) average weight of B and C = 43 b+c=43×2=86 ---equation(3) equation(2)+equation(3)- equation(1) =>a+b+b+c−(a+b+c)=80+86−135 =>b=80+86−135=166−135=31 => weight of B = 31

13. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students? A. 53.23 B. 54.68 C. 51.33 D. 50


Answer : Option B

Explanation :

Average marks of batch1 = 50 Students in batch1 = 55 Total marks of batch1=55×50 Average marks of batch2 = 55 Students in batch2 = 60


Total marks of batch2=60×55 Average marks of batch3 = 60 Students in batch3 = 45 Total marks of batch3=45×60 Total students = 55 + 60 + 45 = 160 Average marks of all the students = (55×50)+(60×55)+(45×60) 160 =275+330+270 16 =875 16 =54.68

14. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband? A. 40 B. 32 C. 28 D. 30


Answer : Option A

Explanation :

let the present age of the husband = h present age of the wife = w


present age of the child = c 3 years ago, average age of husband, wife and their child = 27 =>Sum of age of husband, wife and their child before 3 years = 3×27=81 => (h-3) + (w-3) + (c-3) = 81 => h + w + c = 81+9 = 90 -------------------equation(1) 5 years ago, average age of wife and child = 20 =>Sum of age of wife and child before 5 years = 2×20=40 => (w-5) + (c-5) = 40 => w + c = 40+10 = 50 -------------------equation(2) Substituting equation(2) in equation(1) => h + 50 = 90 => h = 90 - 50 = 40 =>present age of the husband = 40


15. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person? A. 75 Kg B. 50 Kg C. 85 Kg D. 80 Kg


Answer : Option C

Explanation :

Total increase in weight = 8×2.5=20 If x is the weight of the new person, total increase in weight = x - 65 => 20 = x - 65 => x = 20 + 65 = 85

16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class? A. 38.25 B. 37.25 C. 38.5 D. 37



Answer : Option B

Explanation :

Total weight of students in division A = 36×40 Total weight of students in division B = 44×35 Total students = 36+44=80 Average weight of the whole class = (36×40)+(44×35) 80 =(9×40)+(11×35) 20 =(9×8)+(11×7) 4 =72+77 4 =149 4 =37.25

17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning? A. 39 B. 35 C. 42 D. 40.5


Answer : Option A

Explanation :

Let the average after 17 innings = x Total runs scored in 17 innings = 17x then average after 16 innings = (x-3) Total runs scored in 16 innings = 16(x-3) We know that Total runs scored in 16 innings + 87 = Total runs


scored in 17 innings => 16(x-3) + 87 = 17x => 16x - 48 + 87 = 17x => x = 39

18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x? A. 12 B. 5 C. 7 D. 9


Answer : Option C

Explanation :

3+11+7+9+15+13+8+19+17+21+14+x 12 =12

=>137+x 12 =12 => 137 + x = 144 => x = 144 - 137 = 7

19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in


English, Mathematics, Chemistry, Biology and Physics. What is his average mark? A. 53 B. 54 C. 72 D. 75


Answer : Option D

Explanation :

Average mark = 76+65+82+67+85 5 =375 5 =75

20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey? A. 69.0 km /hr B. 69.2 km /hr C. 67.2 km /hr D. 67.0 km /hr


Answer : Option C

Explanation :

------------------------------------------- Solution 1 (Quick) --------------------------------------------


If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed of the whole journey = 2xy x+y kmph. By using the same formula, we can find out the average speed quickly average speed = 2×84×56 84+56 =2×84×56 140 =2×21×56 35 =2×3×56 5 =336 5 =67.2 ------------------------------------------- Solution 2 (Fundamentals) -------------------------------------------- Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics Train travels from A to B at 84 km per hour Let the distance between A and B = x Total time taken for traveling from A to B = distance speed =x 84 Train travels from B to A at 56 km per hour Total time taken for traveling from B to A = distance speed =x 56 Total distance travailed


= x+x=2x Total time taken = x 84 +x 56 Average speed = Total distance traveled Total time taken =2x x 84 +x 56 =2 1 84 +1 56 =2×84×56 56+84 =2×84×56 140 =2×21×56 35 =2×3×56 5 =336 5 =67.2

21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? A. 15 B. 16 C. 15.5 D. Insufficient Data


Answer : Option D

Explanation :

We do not have the number of boys and girls. Hence we can not find out the answer.

22. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. Find out the teacher's age in years? A. 51 years B. 49 years C. 53 years D. 50 years


Answer : Option A


Explanation :

average age of 36 students in a group is 14 => Sum of the ages of 36 students = 36×14 When teacher's age is included to it, the average increases by one => average = 15 => Sum of the ages of 36 students and the teacher =37×15 Hence teachers age = 37×15−36×14=37×15−14(37−1)=37×15−37×14+14 =37(15−14)+14=37+14=51

23. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number? A. 30 B. 40 C. 32.5 D. 35


Answer : Option D

Explanation :

Sum of 5 numbers = 5×27 Sum of 4 numbers after excluding one number = 4×25 Excluded number = 5×27−4×25=135−100=35


24. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player. A. 150 B. 174 C. 180 D. 166


Answer : Option B

Explanation :

Total runs scored by the player in 40 innings = 40×50 Total runs scored by the player in 38 innings after excluding two innings 38×48 Sum of the scores of the excluded innings = 40×50−38×48=2000−1824=176 Given that the scores of the excluded innings differ by 172. Hence let's take the highest score as x + 172 and lowest score as x Now x + 172 + x = 176 => 2x = 4 =>x=4 2 =2 highest score as x + 172 = 2 + 172 = 174


25. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches? A. 34.25 B. 36.4 C. 40.2 D. 32.25


Answer : Option A

Explanation :

Total runs scored in 10 matches = 10×38.9 Total runs scored in first 6 matches = 6×42 Total runs scored in the last 4 matches = 10×38.9−6×42 Average of the runs scored in the last 4 matches = 10×38.9−6×42 4 =389−252 4 =137 4 =34.25

26. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then A. None of these B. x = y + z C. 2x = y + z D. x = 2y + 2z


Answer : Option C

Explanation :

Average of 6 numbers = x => Sum of 6 numbers = 6x


Average of the 3 numbers = y => Sum of these 3 numbers = 3y Average of the remaining 3 numbers = z => Sum of the remaining 3 numbers = 3z Now we know that 6x = 3y + 3z => 2x = y + z

27. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ? A. 32.5 km/hr. B. 35 km/hr. C. 37.5 km/hr D. 40 km/hr


Answer : Option C

Explanation :

------------------------------------------- Solution 1 (Quick) --------------------------------------------


If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed of the whole journey = 2xy x+y kmph. By using the same formula, we can find out the average speed quickly average speed = 2×50×30 50+30 =2×50×30 80 =2×50×3 8 =50×3 4 =25×3 2 =75 2 =37.5 ------------------------------------------- Solution 2 (Fundamentals) -------------------------------------------- Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics. Total time taken for traveling one side = distance speed =150 50 Total time taken for return journey = distance speed =150 30 Total distance travailed = 150+150=2×150 Total time taken = 150 50 +150 30 Average speed = Total distance traveled Total time taken =2×150 150 50 +150 30 =2 1 50 +1 30 =2×50×30 30+50 =2×50×30 80 =2×50×3 8 =50×3 4 =25×3 2 =75 2 =37.5


28. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family ? A. 21 years B. 20 years C. 18 years D. 19 years


Answer : Option D

Explanation :

Total of the age of husband and wife = 2×23=46 Total of the age of husband and wife after 5 years + Age of the 1 year old child =46+5+5+1=57 Average age of the family=57 3 =19

29. In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination? A. 12 B. 11 C. 13 D. 14


Answer : Option B

Explanation :

Let the number of subjects = x


Then, Total marks he scored for all subjects = 63x If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65 => Total marks he would have scored for all subjects = 65x Now we can form the equation as 65x - 63x = the additional marks of the student = 20 + 2 = 22 => 2x = 22 =>x=22 2 =11


Important Concepts - Banker's Discount Assume that a merchant A purchases goods worth, say Rs.1000 from another merchant B at a credit of say 4 months.

Then B prepares a bill called bill of exchange (also called Hundi). On receipts of goods, A gives an agreement by signing on the bill allowing B to withdraw the money from A’s bank exactly after 4 months of the date of the bill.

The date exactly after 4 months is known as nominally due date. Three more days (called grace days) are added to this date to get a date known as legally due date.

The amount given on the bill is called the Face Value (F) which is Rs.1000 in this case.

Assume that B needs this money before the legally due date. He can approach a banker or broker who pays him the money against the bill, but somewhat less than the face value. The banker deducts the simple interest on the face value for the unexpired time. This deduction is known as Bankers Discount (BD). In another words, Bank Discount (BD) is the simple interest on the face value for the period from the date on which the bill was discounted and the legally due date.

The present value is the amount which, if placed at a particular rate for a specified period will amount to that sum of money at the end of the specified period. The interest on the present value is called the True Discount (TD). If the banker deducts the true discount on the face value for the unexpired time, he will not gain anything.


Banker’s Gain (BG) is the difference between banker’s discount and the true discount for the unexpired time.

Note: When the date of bill is not given, grace days are not to be added.

B. Important Formulas - Banker's Discount

Let F = Face Value of the Bill, TD = True Discount, BD = Bankers Discount, BG = Banker’s Gain, R = Rate of Interest, PW = True Present Worth and T = Time in Years

BD = Simple Interest on the face value of the bill for unexpired time = FTR 100

PW = F 1+T(R 100 )

TD = Simple Interest on the present value for unexpired time = PW × TR 100 =FTR 100+(TR)

TD = BD ×100 100+ TR

PW = F - TD


F = BD × TD (BD – TD)

BG = BD – TD = Simple Interest on TD = (TD) 2 PW

TD = PW × BG − − − − − − − − − √

TD = BG ×100 TR

C. A Simple Example to understand the Basic Concepts of Banker's Discount

What is the present value, true discount, banker's discount and banker's gain on a bill of Rs.104500 due in 9 months at 6% per annum?

F = Rs. 104500

T = 9 months = 9 12 years = 3 4 years

R = 6%


Banker's Discount, BD = FTR 100 =104500×3 4 ×6 100 =1045×3 4 ×6= Rs. 4702.50

Present value, PW = F 1+T(R 100 ) =104500 1+(3 4 )(6 100 ) = Rs. 100000

True Discount, TD = PW × TR 100 =100000×3 4 ×6 100 =1000×3 4 ×6= Rs. 4500

Banker's Gain, BG = BD – TD = 4702.50 - 4500 = Rs.202.50

1. The banker's discount on a bill due 4 months hence at 15% is Rs. 420. What is the true discount? A. Rs. 410 B. Rs. 400 C. Rs. 390 D. Rs. 380


Answer : Option B

Explanation :

TD = BD ×100 100+ TR =420×100 100+(4 12 ×15) =420×100 10


0+(1 3 ×15) =420×100 100+5 =420×100 105 =84×100 21 =4×100=400

2. The banker's discount on a certain amount due 2 years hence is 11⁄10 of the true discount. What is the rate percent? A. 1% B. 5% C. 10% D. 12%


Answer : Option B

Explanation :

Let TD = Rs. 1

Then BD = 11 10 ×1= Rs. 11 10

T = 2 R = ?

F = BD × TD (BD – TD) =(11 10 ×1) (11 10 −1) =11 10 1 10 = Rs. 11

BD =FTR 100 ⇒11 10 =11×2×R 100 ⇒110=22R ⇒R=110 22 =5%


3. The present worth of a sum due sometimes hence is Rs.5760 and the baker's gain is Rs.10. What is the true discount? A. Rs. 480 B. Rs. 420 C. Rs. 120 D. Rs. 240


Answer : Option D

Explanation :

TD = PW × BG − − − − − − − − − √ =5760×10 − − − − − − − − √ =57600 − − − − − √ = Rs. 240

4. What is the banker's discount if the true discount on a bill of Rs.540 is Rs.90 ? A. Rs. 108 B. Rs. 120 C. Rs. 102 D. Rs. 106


Answer : Option A

Explanation :

Present Worth, PW = F - TD = 540 - 90 = Rs. 450 Simple Interest on the Present Worth = True Discount Hence Simple Interest on 450 = 90 ------(Equation 1)


Simple Interest on the face value = Bankers Discount => Simple Interest on 540 = Bankers Discount From Equation 1, Simple Interest on 450 = 90

Hence, Simple Interest on 540 = 90 450 ×540=540 5 = Rs. 108

=> Bankers Discount = Rs. 108

5. A bill for Rs. 3000 is drawn on 14th July at 5 months. It is discounted on 5th October at 10%. What is the Banker's Discount? A. Rs. 60 B. Rs. 82 C. Rs. 90 D. Rs. 120


Answer : Option A

Explanation :

F = Rs. 3000 R = 10% Date on which the bill is drawn = 14th July at 5 months


Nominally Due Date = 14th December Legally Due Date = 14th December + 3 days = 17th December Date on which the bill is discounted = 5th October Unexpired Time = [6th to 31st of October] + [30 Days in November] + [1st to 17th of December] = 26 + 30 + 17 = 73 Days

=73 365 year=1 5 year

BD = Simple Interest on the face value of the bill for unexpired time = FTR 100 =3000×1 5 ×10 100 =30×1 5 ×10 = Rs. 60

6. The bankers discount and the true discount of a sum at 10% per annum simple interest for the same time are Rs.100 and Rs.80 respectively. What is the sum and the time? A. Sum = Rs.400 and Time = 5 years

B. Sum = Rs.200 and Time = 2.5 years

C. Sum = Rs.400 and Time = 2.5 years

D. Sum = Rs.200 and Time = 5 years



Answer : Option C

Explanation :

BD = Rs.100 TD = Rs.80 R = 10%

F = BD × TD (BD – TD) =100×80 (100–80) =100×80 20 =Rs.400

BD = Simple Interest on the face value of the bill for unexpired time = FTR 100 ⇒100=400×T×10 100 ⇒100=4×T×10 ⇒10=4×T ⇒T=10 4 =2.5 years

7. The banker's gain on a sum due 6 years hence at 12% per annum is Rs. 540. What is the banker's discount? A. 1240 B. 1120 C. 1190 D. 1290


Answer : Option D

Explanation :


TD = BG ×100 TR =540×100 6×12 =90×100 12 =15×100 2 = Rs. 750

BG = BD – TD => 540 = BD - 750 => BD = 540 + 750 = 1290

8. The present worth of a certain bill due sometime hence is Rs. 1296 and the true discount is Rs. 72. What is the banker's discount? A. Rs. 76 B. Rs. 72 C. Rs. 74 D. Rs. 4


Answer : Option A

Explanation :

BG = (TD) 2 PW =72 2 1296 =72×72 1296 =12×12 36 =12 3 = Rs. 4

BG = BD – TD


=> 4 = BD - 72 => BD = 72 + 4 = Rs. 76

9. The banker's discount of a certain sum of money is Rs. 36 and the true discount on the same sum for the same time is Rs. 30. What is the sum due? A. Rs. 180 B. Rs. 120 C. Rs. 220 D. Rs. 200


Answer : Option A

Explanation :

F = BD × TD (BD – TD) =36×30 (36–30) =36×30 6 =36×5 = Rs. 180

10. The banker's gain on a bill due 1 year hence at 10% per annum is Rs. 20. What is the true discount? A. Rs. 200 B. Rs. 100 C. Rs. 150 D. Rs. 250


Answer : Option A

Explanation :


TD = BG ×100 TR =20×100 1×10 = Rs. 200

11. The banker's gain of a certain sum due 3 years hence at 10% per annum is Rs. 36. What is the present worth ? A. Rs. 400 B. Rs. 300 C. Rs. 500 D. Rs. 350


Answer : Option A

Explanation :

T = 3 years R = 10%

TD = BG ×100 TR =36×100 3×10 =12×10=Rs. 120

TD = PW × TR 100 ⇒120=PW ×3×10 100 ⇒1200=PW ×3 PW =1200 3 = Rs. 400

12. The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is: A. Rs. 21 B. Rs. 17 C. Rs. 18 D. Rs. 34



Answer : Option D

Explanation :

BG = (TD) 2 PW =(340) 2 3400 =340×340 3400 =340 10 =Rs. 34

13. The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. What is the time? A. 3 months B. 4 months C. 5 months D. 6 months


Answer : Option B

Explanation :

Bankers Discount, BD = Simple Interest on the face value of the bill for unexpired time. True Discount, TD = Simple Interest on the present value for unexpired time. Simple Interest on Rs. 1600 = True Discount on Rs.1680 => Rs. 1600 is the Present Worth (PW) of Rs. 1680


=> Rs. 80 is the Simple Interest of Rs.1600 at 15%

⇒80=1600×T×15 100 ⇒80=16×T×15 ⇒5=T×15 ⇒1=T×3 => T = 1 3 year=12 3 months=4 months

14. The banker's gain on a certain sum due 21 2 years hence is 9 25 of the banker's discount. What is the rate percent? A. 181 3 % B. 181 2 % C. 241 3 % D. 221 2 %


Answer : Option D

Explanation :

T = 21 2 years=5 2 years

Let the banker's discount, BD = Rs. 1

Then, banker's gain, BG = 9 25 ×1 = Rs. 9 25

BG = BD – TD ⇒9 25 =1 - TD ⇒TD=1−9 25 =16 25

F = BD × TD (BD – TD) =1×16 25 (1–16 25 ) =16 25 9 25 = Rs. 16 9

BD = Simple Interest on the face value of the bill for unexpired time


= FTR 100 ⇒1=16 9 ×5 2 ×R 100 ⇒100=16 9 ×5 2 × R ⇒100=16×5×R 9×2 ⇒100=8×5×R 9 ⇒R = 100×9 8×5 =100×9 40 =5×9 2 =45 2 =221 2 %

15. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 360. The banker's discount is: A. Rs. 1360 B. Rs. 1000 C. Rs. 360 D. Rs. 640


Answer : Option A

Explanation :

BG = Rs. 360 T = 3 years R = 12%

TD = BG ×100 TR =360×100 3×12 =Rs. 1000

BG = BD – TD => BD = BG + TD = 360 + 1000 = Rs. 1360

16. The true discount on a certain sum due 6 months hence at 15% is Rs. 240. What is the banker's discount on the same sum


for the same time at the same rate? A. None of these B. Rs. 278 C. Rs. 228 D. Rs. 258


Answer : Option D

Explanation :

TD = Rs. 240

T = 6 months = 1 2 year

R = 15%

TD = BG ×100 TR ⇒240=BG×100 (1 2 ×15) BG =240×15 100×2 =120×15 100 =Rs. 18

BG = BD – TD => 18 = BD - 240 => BD = 18 + 240 = Rs. 258

17. A bill is discounted at 10% per annum. If banker's discount is allowed, at what rate percent should the proceeds be invested so that nothing will be lost? A. 101 9 % B. 111 9 %


C. 11 D. 102 9 %


Answer : Option B

Explanation :

Let the amount = Rs.100 Then BD = Rs.10 (∵ banker's discount, BD is the simple Interest on the face value of the bill for unexpired time and bill is discounted at 10% per annum) Proceeds = Rs.100 – Rs.10 = Rs.90 Hence we should get Rs.10 as the interest of Rs.90 for 1 year so that nothing will be lost

⇒10=90×1×R 100 ⇒R=10×100 90 =100 9 =111 9 %

18. A banker paid Rs.5767.20 for a bill of Rs.5840, drawn of Apr 4 at 6 months. If the rate of interest was 7%, what was the day on which the bill was discounted? A. 3rd March B. 3rd September C. 3rd October D. 3rd August



Answer : Option D

Explanation :

F = Rs.5840 R = 7% BD = 5840 - 5767.20 = Rs.72.8

BD = FTR 100 ⇒72.8=5840×T×7 100 ⇒ T=72.8×100 7×5840 =10.4×100 5840 =1040 5840 =104 584 =13 73 years =13×365 73 Days=65 Days ⇒Unexpired Time = 65 Days

Given that Date of Draw of the bill = 4 th April at 6 months => Nominally Due Date = 4 th October => Legally Due Date = (4 th October + 3 days) = 7 th October Hence, The date on which the bill was discounted


= (7 th October - 65 days) = (7 th October - 7 days in October - 30 days in September - 28 days in August) = 3 rd August

19. The banker's discount on a sum of money for 3 years is Rs. 1116. The true discount on the same sum for 4 years is Rs. 1200. What is the rate per cent? A. 8% B. 12% C. 10% D. 6%


Answer : Option D

Explanation :

BD for 3 years = Rs. 1116

BD for 4 years = 1116 3 ×4 = Rs. 1488

TD for 4 years = Rs. 1200

F = BD × TD (BD – TD) =1488×1200 (1488–1200) =1488×1200 288 =124×1200 24 =124×100 2 =62×100=Rs. 6200


=> Rs.1488 is the simple interest on Rs. 6200 for 4 years

⇒1488=6200×4×R 100 ⇒R=1488×100 6200×4 =372×100 6200 =372 62 =6%

20. The true discount on a bill of Rs. 2160 is Rs. 360. What is the banker's discount? A. Rs. 432 B. Rs. 422 C. Rs. 412 D. Rs. 442


Answer : Option A

Explanation :

F = Rs. 2160 TD = Rs. 360 PW = F - TD = 2160 - 360 = Rs. 1800 True Discount is the Simple Interest on the present value for unexpired time =>Simple Interest on Rs. 1800 for unexpired time = Rs. 360 Banker's Discount is the Simple Interest on the face value of the bill for unexpired time


= Simple Interest on Rs. 2160 for unexpired time

=360 1800 ×2160=1 5 ×2160=Rs. 432

21. The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. What is the present worth? A. Rs. 600 B. Rs. 500 C. Rs. 400 D. Rs. 300


Answer : Option A

Explanation :

T = 2 years R = 10%

TD = BG ×100 TR =24×100 2×10 =12×10=Rs. 120

TD = PW × TR 100 ⇒120=PW ×2×10 100 ⇒1200=PW ×2 PW =1200 2 = Rs. 600

22. The true discount on a bill for Rs. 2520 due 6 months hence at 10% per annum is A. Rs. 180 B. Rs. 140


C. Rs. 80 D. Rs. 120


Answer : Option D

Explanation :

F= Rs. 2520

T=6 months = 1 2 year

R = 10%

TD = FTR 100+(TR) =2520×1 2 ×10 100+(1 2 ×10) =2520×1 2 ×10 100+(1 2 ×10) =1260×10 100+5 =12600 105 =2520 21 =Rs. 120

23. What is the present worth of a bill of Rs.1764 due 2 years hence at 5% compound interest is A. Rs. 1600 B. Rs. 1200 C. Rs. 1800 D. Rs. 1400


Answer : Option A

Explanation :

Since the compound interest is taken here,


PW (1+5 100 ) 2 =1764 PW (1+1 20 ) 2 =1764 PW (21 20 ) 2 =1764 PW ×441 400 =1764 ⇒PW =1764×400 441 =4×400=Rs. 1600

24. If the discount on Rs. 498 at 5% simple interest is Rs.18, when is the sum due? A. 8 months B. 11 months C. 10 months D. 9 months


Answer : Option D

Explanation :

F = Rs. 498 TD = Rs. 18 PW = F - TD = 498 - 18 = Rs. 480 R = 5%

TD = PW × TR 100 ⇒18=480× T ×5 100 ⇒18=24× T ⇒ T =18 24 =3 4 years=12×3 4 months = 9 months

25. What is the difference between the banker's discount and the true discount on Rs.8100 for 3 months at 5%


A. Rs. 2 B. Rs. 1.25 C. Rs. 2.25 D. Rs. 0.5


Answer : Option B

Explanation :

F = Rs. 8100 R = 5%

T = 3 months = 1 4 years

BD = FTR 100 =8100×1 4 ×5 100 =2025 20 =405 4 = Rs. 101.25

TD = FTR 100+(TR) =8100×1 4 ×5 100+(1 4 ×5) =2025×5 100+(5 4 ) =2025×5×4 400+5 =2025×5×4 405 =405×5×4 81 =45×5×4 9 =5×5×4= Rs. 100

BD - TD = Rs. 101.25 - Rs. 100 = Rs. 1.25

26. The B.G. on a certain sum 4 years hence at 5% is Rs. 200. What is the present worth? A. Rs. 4500 B. Rs. 6000 C. Rs. 5000 D. Rs. 4000



Answer : Option C

Explanation :

T = 4 years R = 5% Banker's Gain, BG = Rs.200

TD = BG ×100 TR =200×100 4×5 = Rs. 1000

TD = PW × BG − − − − − − − − − √ 1000=PW ×200 − − − − − − − − − √ 1000000=PW ×200 PW =1000000 200 = Rs. 5000

27. The B.D. and T.D. on a certain sum is Rs.200 and Rs.100 respectively. Find out the sum. A. Rs. 400 B. Rs. 300 C. Rs. 100 D. Rs. 200


Answer : Option D

Explanation :

F = BD × TD (BD – TD) =200×100 200−100 =200×100 100 = Rs. 200


28. The banker's discount on a bill due 6 months hence at 6% is Rs. 18.54. What is the true discount? A. Rs. 24 B. Rs. 12 C. Rs. 36 D. Rs. 18


Answer : Option D

Explanation :

T = 6 months = 1 2 year

R = 6% TD = BD ×100 100+ TR =18.54×100 100+(1 2 ×6) =18.54×100 103 =1854 103 =Rs. 18

29. The banker's discount on a sum of money for 11 2 years is Rs. 120. The true discount on the same sum for 2 years is Rs.150. What is the rate per cent? A. 31 3 % B. 41 3 % C. 32 3 % D. 42 3 %


Answer : Option A

Explanation :

BD for 11 2 years = Rs.120

BD for 2 years = 120×2 3 ×2= Rs.160


TD for 2 years = Rs. 150

F = BD × TD (BD – TD) =160×150 160–150 =160×150 10 =Rs. 2400

=> Rs.160 is the simple interest on Rs. 2400 for 2 years

⇒160=2400×2×R 100 ⇒R=160×100 2400×2 =160 48 =20 6 =10 3 =31 3 %

30. The present worth of a certain bill due sometime hence is Rs. 400 and the true discount is Rs. 20. What is the banker's discount? A. Rs. 19 B. Rs. 22 C. Rs. 20 D. Rs. 21


Answer : Option D

Explanation :

BG = (TD) 2 PW =20 2 400 = Rs. 1

BG = BD – TD => 1 = BD - 20 => BD = 1 + 20 = Rs. 21


Important Formulas - Boats and Streams • Downstream

In water, the direction along the stream is called downstream.

• Upstream In water, the direction against the stream is called upstream.

• Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then Speed downstream = (u + v) km/hr Speed upstream = (u - v) km/hr.

• Let the speed downstream be a km/hr and the speed upstream be b km/hr, then Speed in still water =1 2 (a+b) km/hr Rate of stream =1 2 (a−b) km/hr

Some more short-cut methods


• Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey = Speed downstream × Speed downstream Speed in still water =(x+y)(x−y) x km/hr

• Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance =(x 2 −y 2 )t 2y km

• A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2 hours. If the speed of the stream is y km/hr, then the speed of the man in still water =y(t 2 +t 1 t 2 −t 1 ) km/hr

• A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places =t(x 2 −y 2 ) 2x km


• A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then x=y(n+1 n−1 )

1. A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is: A. 8.5 km/hr B. 10 km/hr. C. 12.5 km/hr D. 9 km/hr


Answer : Option B

Explanation :

Man's speed with the current = 15 km/hr =>speed of the man + speed of the current = 15 km/hr speed of the current is 2.5 km/hr Hence, speed of the man = 15 - 2.5 = 12.5 km/hr


man's speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km/hr

2. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is: A. 10 B. 6 C. 5 D. 4


Answer : Option C

Explanation :

Speed of the motor boat = 15 km/hr Let speed of the stream = v Speed downstream = (15+v) km/hr Speed upstream = (15-v) km/hr Time taken downstream = 30 (15+v) Time taken upstream = 30 (15−v) total time = 30 (15+v) +30 (15−v) It is given that total time is 4 hours 30 minutes = 41 2 hour = 9 2 hour i.e., 30 (15+v) +30 (15−v) =9 2 ⇒1 (15+v) +1 (1


5−v) =9 2×30 =3 20 ⇒15−v+15+v (15+v)(15−v) =3 20 ⇒30 15 2 −v 2 =3 20 ⇒30 225−v 2 =3 20 ⇒10 225−v 2 =1 20 ⇒225−v 2 =200 ⇒v 2 =225−200=25 ⇒v=5 km/hr

3. In one hour, a boat goes 14 km/hr along the stream and 8 km/hr against the stream. The speed of the boat in still water (in km/hr) is: A. 12 km/hr B. 11 km/hr C. 10 km/hr D. 8 km/hr


Answer : Option B

Explanation :

------------------------------------------------------------------- Solution 1 : Using Formula -------------------------------------------------------------------

Let the speed downstream be a km/hr and the speed upstream be b km/hr, then Speed in still water =1 2 (a+b) km/hr Rate of stream =1 2 (a−b) km/hr [Read more ...]

Speed in still water = 1 2 (14+8) kmph = 11 kmph.


------------------------------------------------------------------- Solution 2 : Using Principles ------------------------------------------------------------------- Let speed of the boat in still water = a and speed of the stream = b Then a + b = 14 a - b = 8 Adding these two equations, we get 2a = 22 => a = 11 ie, speed of boat in still water = 11 km/hr

4. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is: A. 1 km/hr. B. 2 km/hr. C. 1.5 km/hr. D. 2.5 km/hr.


Answer : Option A

Explanation :

Assume that he moves 4 km downstream in x hours Then, speed downstream = distance time =4 x km/hr


Given that he can row 4 km with the stream in the same time as 3 km against the stream i.e., speed upstream = 3 4 of speed downstream => speed upstream = 3 x km/hr He rows to a place 48 km distant and come back in 14 hours =>48 (4 x ) +48 (3 x ) =14 ==>12x+16x=14 =>6x+8x=7 =>14x=7 =>x=1 2 Hence, speed downstream = 4 x =4 (1 2 ) = 8 km/hr speed upstream = 3 x =3 (1 2 ) = 6 km/hr Now we can use the below formula to find the rate of the stream


Hence, rate of the stream = 1 2 (8−6)=1 km/hr

5. A boatman goes 2 km against the current of the stream in 2 hour and goes 1 km along the current in 20 minutes. How long will it take to go 5 km in stationary water? A. 2 hr 30 min B. 2 hr C. 4 hr D. 1 hr 15 min



Answer : Option A

Explanation :

Speed upstream = 2 2 =1 km/hr Speed downstream = 1 (20 60 ) =3 km/hr Speed in still water = 1 2 (3+1)=2 km/hr Time taken to travel 5 km in still water = 5 2 =21 2 hours = 2 hour 30 minutes

6. Speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph. A man rows to a place at a distance of 4864 km and comes back to the starting point. The total time taken by him is: A. 700 hours B. 350 hours C. 1400 hours D. 1010 hours


Answer : Option A

Explanation :

Speed downstream = (14 + 1.2) = 15.2 kmph Speed upstream = (14 - 1.2) = 12.8 kmph Total time taken = 4864 15.2 +4864 12.8 = 320 + 380 = 700 hours


7. The speed of a boat in still water in 22 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 24 minutes is: A. 9.4 km B. 10.2 km C. 10.4 km D. 9.2 km


Answer : Option C

Explanation :

Speed downstream = (22 + 4) = 26 kmph Time = 24 minutes = 24 60 hour = 2 5 hour distance travelled = Time × speed = 2 5 ×26 = 10.4 km

8. A boat covers a certain distance downstream in 1 hour, while it comes back in 11⁄2 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water? A. 14 kmph B. 15 kmph C. 13 kmph D. 12 kmph


Answer : Option B

Explanation :


Let the speed of the water in still water = x Given that speed of the stream = 3 kmph Speed downstream = (x+3) kmph Speed upstream = (x-3) kmph He travels a certain distance downstream in 1 hour and come back in 11⁄2 hour. ie, distance travelled downstream in 1 hour = distance travelled upstream in 11⁄2 hour since distance = speed × time, we have (x+3)×1=(x−3)3 2 => 2(x + 3) = 3(x-3) => 2x + 6 = 3x - 9 => x = 6+9 = 15 kmph

9. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively? A. 5 : 6 B. 6 : 5


C. 8 : 3 D. 3 : 8


Answer : Option C

Explanation :

Let the rate upstream of the boat = x kmph and the rate downstream of the boat = y kmph Distance travelled upstream in 8 hrs 48 min = Distance travelled downstream in 4 hrs. Since distance = speed × time, we have x×84 5 =y×4 x×44 5 =y×4 x×11 5 =y--- (equation 1) Now consider the formula given below


Hence, speed of the boat = y+x 2 speed of the water = y−x 2 Required Ratio


= (y+x 2 ):(y−x 2 ) =(y+x):(y−x) =(11x 5 +x):(11x 5 −x)∵ (Substituted the value of y from equation 1) =(11x+5x):(11x−5x) =16x:6x =8:3

10. A boat can travel with a speed of 22 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 54 km downstream A. 5 hours B. 4 hours C. 3 hours D. 2 hours


Answer : Option D

Explanation :

Speed of the boat in still water = 22 km/hr speed of the stream = 5 km/hr Speed downstream = (22+5) = 27 km/hr Distance travelled downstream = 54 km Time taken = distance speed =54 27 = 2 hours

11. A boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water?


A. 5 kmph B. 4.95 kmph C. 4.75 kmph D. 4.65


Answer : Option B

Explanation :

Speed downstream = 22 4 = 5.5 kmph Speed upstream = 22 5 = 4.4 kmph Speed of the boat in still water = 5.5+4.4 2 = 4.95 kmph

12. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is: A. 3 : 1 B. 1 : 3 C. 1 : 2 D. 2 : 1


Answer : Option A

Explanation :

Let speed upstream = x Then, speed downstream = 2x Speed in still water = 2x+x 2 =3x 2 Speed of the stream


= 2x−x 2 =x 2 Speed in still water : Speed of the stream = 3x 2 :x 2 = 3 : 1

13. A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place? A. 3.2 km B. 3 km C. 2.4 km D. 3.6 km


Answer : Option C

Explanation :

Speed in still water = 5 kmph Speed of the current = 1 kmph Speed downstream = (5+1) = 6 kmph Speed upstream = (5-1) = 4 kmph Let the requited distance be x km Total time taken = 1 hour =>x 6 +x 4 =1 => 2x + 3x = 12 => 5x = 12


=> x = 2.4 km

14. A man can row three-quarters of a kilometre against the stream in 111⁄4 minutes and down the stream in 71

⁄2minutes. The speed (in km/hr) of the man in still water is: A. 4 kmph B. 5 kmph C. 6 kmph D. 8 kmph


Answer : Option B

Explanation :

Distance = 3 4 km Time taken to travel upstream = 111 4 minutes = 45 4 minutes = 45 4×60 hours = 3 16 hours Speed upstream = Distance Time =(3 4 ) (3 16 ) = 4 km/hr Time taken to travel downstream = 71 2 minutes = 15 2 minutes = 15 2×60 hours = 1 8 hours Speed downstream = Distance Time =(3 4 ) (1 8 ) = 6 km/hr Rate in still water = 6+4 2 =10 2 =5 kmph

15. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is: A. 4 mph B. 2.5 mph


C. 3 mph D. 2 mph


Answer : Option D

Explanation :

Speed of the boat in still water = 10 mph Let speed of the stream be x mph Then, speed downstream = (10+x) mph speed upstream = (10-x) mph Time taken to travel 36 miles upstream - Time taken to travel 36 miles downstream = 90 60 hours =>36 10−x −36 10+x =3 2 =>12 10−x −12 10+x =1 2 =>24(10+x)−24(10−x)=(10+x)(10−x) =>240+24x−240+24x=(100−x 2 ) =>48x=100−x 2 =>x 2 +48x−100=0 =>(x+50)(x−2)=0 =>x = -50 or 2 Since x can not be negative, x = 2 mph

16. Tap 'A' can fill the tank completely in 6 hrs while tap 'B' can empty it by 12 hrs. By mistake, the person forgot to close the tap 'B', As a result, both the taps, remained open. After 4 hrs, the person realized the mistake and immediately closed the tap 'B'. In how much time now onwards, would the tank be full?


A. 2 hours B. 4 hours C. 5 hours D. 1 hour


Answer : Option B

Explanation :

Tap A can fill the tank completely in 6 hours => In 1 hour, Tap A can fill 1⁄6 of the tank Tap B can empty the tank completely in 12 hours => In 1 hour, Tap B can empty 1⁄12 of the tank i.e., In one hour, Tank A and B together can effectively fill 1

⁄6 - 1⁄12 = 1⁄12 of the tank

=> In 4 hours, Tank A and B can effectively fill 1

⁄12 × 4 = 1⁄3 of the tank. Time taken to fill the remaining 1−1 3 =2 3 of the tank =(2 3 ) (1 6 ) = 4 hours

17. A Cistern is filled by pipe A in 8 hrs and the full Cistern can be leaked out by an exhaust pipe B in 12 hrs. If both the pipes are opened in what time the Cistern is full? A. 12 hrs B. 24 hrs C. 16 hrs D. 32 hrs



Answer : Option B

Explanation :

Pipe A can fill 1⁄8 of the cistern in 1 hour. Pipe B can empty 1

⁄12 of the cistern in 1 hour Both Pipe A and B together can effectively fill 1

⁄8-1⁄12= 1⁄24 of the

cistern in 1 hour i.e, the cistern will be full in 24 hrs.

18. In a river flowing at 2 km/hr, a boat travels 32 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr, find the total journey time. A. 10 hours B. 12 hours C. 14 hours D. 16 hours


Answer : Option B

Explanation :

------------------------------------------------------------ Solution 1 ------------------------------------------------------------ speed of the boat = 6 km/hr


Speed downstream = (6+2) = 8 km/hr Speed upstream = (6-2) = 4 km/hr Distance travelled downstream = Distance travelled upstream = 32 km Total time taken = Time taken downstream + Time taken upstream =32 8 +32 4 =32 8 +64 8 =96 8 = 12 hr ------------------------------------------------------------ Solution 2 ------------------------------------------------------------

A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places =t(x 2 −y 2 ) 2x km [Read more ...]

x = 6 km/hr y = 2 km/hr distance = 32 km


As per the formula, we have 32=t(6 2 −2 2 ) 2×6 =>32=32t 12 =>t=12 hr

19. Two pipes A and B can fill a tank in 10 hrs and 40 hrs respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank? A. 8 hours B. 6 hours C. 4 hours D. 2 hours


Answer : Option A

Explanation :

Pipe A can fill 1⁄10 of the tank in 1 hr Pipe B can fill 1⁄40 of the tank in 1 hr Pipe A and B together can fill 1

⁄10 + 1⁄40 = 1⁄8 of the tank in 1 hr i.e., Pipe A and B together can fill the tank in 8 hours

20. A boat covers a certain distance downstream in 4 hours but takes 6 hours to return upstream to the starting point. If the speed of the stream be 3 km/hr, find the speed of the boat in still water


A. 15 km/hr B. 12 km/hr C. 13 km/hr D. 14 km/hr


Answer : Option A

Explanation :

------------------------------------------------------------------------- Solution 1 ------------------------------------------------------------------------- Let the speed of the water in still water = x Given that speed of the stream = 3 kmph Speed downstream = (x+3) kmph Speed upstream = (x-3) kmph He travels a certain distance downstream in 4 hour and come back in 6 hour. ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour since distance = speed × time, we have (x+3)4=(x−3)6 => (x + 3)2 = (x - 3)3 => 2x + 6 = 3x - 9 => x = 6+9 = 15 kmph


------------------------------------------------------------------------- Solution 2 -------------------------------------------------------------------------

A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2 hours. If the speed of the stream is y km/hr, then the speed of the man in still water =y(t 2 +t 1 t 2 −t 1 ) km/hr [Read more ...]

t1 = 4 hour t2 = 6 hour y = 3 km/hr By using the the abov formula, Speed of the boat in still water =y(t 2 +t 1 t 2 −t 1 )=3(6+4 6−4 )= 15 km/hr

21. If a man rows at the rate of 5 kmph in still water and his rate against the current is 3 kmph, then the man's rate along the current is: A. 5 kmph B. 7 kmph C. 12 kmph D. 8 kmph



Answer : Option B

Explanation :

Let the rate along with the current is x km/hr x+3 2 =5 => x + 3 = 10 => x = 7 kmph

22. A man can row 8 km/hr in still water. If the river is running at 3 km/hr, it takes 3 hours more in upstream than to go downstream for the same distance. How far is the place? A. 32.5 km B. 25 km C. 27.5 km D. 22.5 km


Answer : Option C

Explanation :

-------------------------------------------------------- Solution 1 -------------------------------------------------------- Let the speed downstream = x and speed updstream = y


Then x+y 2 =8 => x + y = 16 ---(Equation 1) x−y 2 =3 => x - y = 6 ---(Equation 2) (Equation 1 + Equation 2) = > 2x = 22 => x = 11 km/hr (Equation 1 - Equation 2) = > 2y = 10 =>y = 5 km/hr Time taken to travel upstream = Time taken to travel downstream + 3 Let distance be x km Then x 5 =x 11 +3 => 11x = 5x + 165 => 6x = 165


=> 2x = 55 => x = 27.5 -------------------------------------------------------- Solution 2 ---------------------------------------------------------

Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance =(x 2 −y 2 )t 2y km [Read more ...]

x = 8 km/hr y = 3 km/hr t = 3 hours As per the formula, we have distance=(8 2 −3 2 )3 2×3 =55×3 2×3 =55 2 = = 27.5 km

23. A man can row 4 kmph is still water. If the river is running at 2 kmph it takes 90 min to row to a place and back. How far is the place?


A. 2 km B. 4 km C. 5 km D. 2.25 km


Answer : Option D

Explanation :

Speed in still water = 4 kmph Speed of the stream = 2 kmph Speed upstream = (4-2)= 2 kmph Speed downstream = (4+2)= 6 kmph Total time = 90 minutes = 90

⁄60 hour = 3⁄2 hour Let L be the distance. Then L 6 +L 2 =3 2 => L + 3L = 9 => 4L = 9 => L = 9⁄4= 2.25 km

24. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in 6 hours less than it takes him to travel the same distance upstream. But if he could double his


usual rowing rate for his 24-mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour? A. 21 3 mph B. 11 3 mph C. 12 3 mph D. 22 3 mph


Answer : Option D

Explanation :

Let the speed of Rahul in still water be x mph and the speed of the current be y mph Then, Speed upstream = (x - y) mph Speed downstream = (x + y) mph Distance = 12 miles Time taken to travel upstream - Time taken to travel downstream = 6 hours ⇒12 x−y −12 x+y =6 ⇒12(x+y)−12(x−y)=6(x 2 −y 2 ) ⇒24y=6(x 2 −y 2 ) ⇒4y=x 2 −y 2 ⇒x 2 =(y 2 +4y)⋯(Equation 1) Now he doubles his speed. i.e., his new speed = 2x Now, Speed upstream = (2x - y) mph


Speed downstream = (2x + y) mph In this case, Time taken to travel upstream - Time taken to travel downstream = 1 hour ⇒12 2x−y −12 2x+y =1 ⇒12(2x+y)−12(2x−y)=4x 2 −y 2 ⇒24y=4x 2 −y 2 ⇒4x 2 =y 2 +24y⋯(Equation 2) (Equation 1 × 4)⇒4x 2 =4(y 2 +4y)⋯(Equation 3) (From Equation 2 and 3, we have)y 2 +24y=4(y 2 +4y) ⇒y 2 +24y=4y 2 +16y ⇒3y 2 =8y ⇒3y=8 y=8 3 mph i.e., speed of the current = 8 3 mph=22 3 mph

25. A man can row 40 kmph in still water and the river is running at 10 kmph. If the man takes 1 hr to row to a place and back, how far is the place? A. 16.5 kmph B. 12.15 kmph C. 2.25 kmph D. 18.75 kmph


Answer : Option D

Explanation :

Let the distance be x Speed upstream = (40 - 10) = 30 kmph Speed downstream = (40 + 10) = 50 kmph


Total time taken = 1 hr x 50 +x 30 =1 ⇒8x 150 =1 ⇒x=150 8 = 18.75 kmph

26. A boatman can row 96 km downstream in 8 hr. If the speed of the current is 4 km/hr, then find in what time will be able to cover 8 km upstream? A. 6 hr B. 2 hr C. 4 hr D. 1 hr


Answer : Option B

Explanation :

Speed downstream = 96⁄8 = 12 kmph

Speed of current = 4 km/hr Speed of the boatman in still water = 12-4 = 8 kmph Speed upstream = 8-4 = 4 kmph Time taken to cover 8 km upstream = 8

⁄4 = 2 hours

27. The speed of a boat in still water is 10 km/hr. If it can travel 78 km downstream and 42 km upstream in the same time, the


speed of the stream is A. 3 km/hr B. 12 km/hr C. 1.5 km/hr D. 4.4 km/hr


Answer : Option A

Explanation :

Let the speed of the stream be x km/hr. Then Speed upstream = (10-x) km/hr Speed downstream = (10+x) km/hr Time taken to travel 78 km downstream = Time taken to travel 42 km upstream =>78 10+x =42 10−x =>26 10+x =14 10−x =>13 10+x =7 10−x =>130−13x=70+7x =>20x=60 =>x=3 km/hr

28. A man can row at a speed of 12 km/hr in still water to a certain upstream point and back to the starting point in a river which flows at 3 km/hr. Find his average speed for total journey. A. 123 4 km/hr B. 113 4 km/hr C. 121 4 km/hr D. 111 4 km/hr



Answer : Option D

Explanation :

Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey = Speed downstream × Speed downstream Speed in still water =(x+y)(x−y) x km/hr [Read more ...]

Speed of the man in still water = 12 km/hr Speed of the stream = 3 km/hr Speed downstream = (12+3) = 15 km/hr Speed upstream = (12-3) = 9 km/hr Average Speed = Speed downstream × Speed downstream Speed in still water =15×9 12 =15×3 4 =45 4 =111 4 km/hr

29. A boatman can row 3 km against the stream in 20 minutes and return in 18 minutes. Find the rate of current A. 1⁄2 km/hr B. 1 km/hr


C. 1⁄3 km/hr D. 2⁄3 km/hr


Answer : Option A

Explanation :

Speed upstream = 3 (20 60 ) = 9 km/hr Speed downstream = 3 (18 60 ) = 10 km/hr Rate of current = 10−9 2 =1 2 km/hr

30. A boat takes 38 hours for travelling downstream from point A to point B and coming back to point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B? A. 240 km B. 120 km C. 360 km D. 180 km


Answer : Option C

Explanation :

velocity of the stream = 4 kmph Speed of the boat in still water is 14 kmph Speed downstream = (14+4) = 18 kmph


Speed upstream = (14-4) = 10 kmph Let the distance between A and B be x km Time taken to travel downstream from A to B + Time taken to travel upstream from B to C(mid of A and B) = 38 hours ⇒x 18 +(x 2 ) 10 =38 ⇒x 18 +x 20 =38 ⇒19x 180 =38 ⇒x 180 =2 ⇒x=360 i.e., the distance between A and B = 360 km

31. If a man's rate with the current is 15 km/hr and the rate of the current is 11⁄2 km/hr, then his rate against the current is A. 12 km/hr B. 10 km/hr C. 10.5 km/hr D. 12.5 km/hr


Answer : Option A

Explanation :

Speed downstream = 15 km/hr Rate of the current= 11

⁄2 km/hr


Speed in still water = 15 - 11⁄2 = 131

⁄2 km/hr Rate against the current = 131

⁄2 km/hr - 11⁄2 = 12 km/hr

32. The speed of the boat in still water in 12 kmph. It can travel downstream through 45 kms in 3 hrs. In what time would it cover the same distance upstream? A. 8 hours B. 6 hours C. 4 hours D. 5 hours


Answer : Option D

Explanation :

Speed of the boat in still water = 12 km/hr Speed downstream = 45

⁄3 = 15 km/hr Speed of the stream = 15-12 = 3 km/hr Speed upstream = 12-3 = 9 km/hr Time taken to cover 45 km upstream = 45

⁄9 = 5 hours

33. A boat goes 8 km upstream in 24 minutes. The speed of stream is 4 km/hr. The speed of boat in still water is: A. 25 km/hr B. 26 km/hr


C. 22 km/hr D. 24 km/hr


Answer : Option D

Explanation :

Speed upstream = 8 (24 60 ) = 20 km/hr Speed of the stream = 4 km/hr speed of boat in still water = (20+4) = 24 km/hr

34. The Speed of a boat in still water is 25 kmph. If it can travel 10 km upstream in 1 hr, What time it would take to travel the same distance downstream? A. 22 minutes B. 30 minutes C. 40 minutes D. 15 minutes


Answer : Option D

Explanation :

Speed of boat in still water = 25 km/hr Speed upstream = 10

⁄1 = 10 km/hr Speed of the stream = (25-10) = 15 km/hr


Speed downstream = (25+15) = 40 km/hr Time taken to travel 10 km downstream = 10 40 hours=10×60 40 = 15 minutes

35. A boat can travel with a speed of 12 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream. A. 4 hr B. 4.25 hr C. 4.5 hr D. 6 hr


Answer : Option B

Explanation :

speed of boat in still water = 12 km/hr speed of the stream = 4 km/hr Speed downstream = (12+4) = 16 km/hr Time taken to travel 68 km downstream = 68

⁄16 = 17⁄4 = 4.25 hours

36. A man can row 7.5 kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream.


A. 10 km/hr B. 2.5 km/hr C. 5 km/hr D. 7.5 km/hr


Answer : Option B

Explanation :

Given that, time taken to travel upstream = 2 × time taken to travel downstream When distance is constant, speed is inversely proportional to the time Hence, 2 × speed upstream = speed downstream Let speed upstream = x Then speed downstream = 2x we have, 1 2 (x+2x) = speed in still water ⇒1 2 (3x)=7.5 3x = 15 x = 5 i.e., speed upstream = 5 km/hr Rate of stream = 1 2 (2x−x)=x 2 =5 2 =2.5 km/hr


37. A boat moves downstream at the rate of one km in 5 minutes and upstream at the rate of 4 km an hour. What is the velocity of the current? A. 4 km/hr B. 2 km/hr C. 3 km/hr D. 1 km/hr


Answer : Option A

Explanation :

Speed downstream = 1 (5 60 ) =12 km/hr Speed upstream = 4 1 = 4 km/hr velocity of the current = 1 2 (12−4)=4 km/hr

38. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 24 minutes is A. 3.6 km B. 2.4 km C. 3.2 km D. 7.2 km


Answer : Option D

Explanation :

speed of a boat in still water = 15 km/hr Speed of the current = 3 km/hr


Speed downstream = (15+3) = 18 km/hr Distance travelled downstream in 24 minutes=24 60 ×18=2×18 5 = 7.2 km

39. The current of a stream runs at the rate of 2 km per hr. A motor boat goes 10 km upstream and back again to the starting point in 55 min. Find the speed of the motor boat in still water? A. 22 km/hr B. 12 km/hr C. 20 km/hr D. 16 km/hr


Answer : Option A

Explanation :

Let the speed of the boat in still water = x km/hr Speed of the current = 2 km/hr Then, speed downstream = (x+2) km/hr speed upstream = (x-2) km/hr Total time taken to travel 10 km upstream and back = 55 minutes = 55

⁄60 hour = 11⁄12 hour

⇒10 x−2 +10 x+2 =11 12 120(x+2)+120(x−2)=11(x 2 −4) 240x=11x 2 −44 11x 2 −240x−44=0 11x 2 −242x+2x−4


4=0 11x(x−22)+2(x−22)=0 (x−22)(11x+2)=0 x=22 or −2 11 Since x can not be negative, x = 22 i.e., speed of the boat in still water = 22 km/hr

40. The speed of a boat in still water is 8 kmph. If it can travel 1 km upstream in 1 hr, What time it would take to travel the same distance downstream? A. 1 minute B. 2 minutes C. 3 minutes D. 4 minutes


Answer : Option D

Explanation :

Speed of the boat in still water = 8 km/hr Speed upstream = 1

⁄1 = 1 km/hr Speed of the stream = 8-1 = 7 km/hr Speed downstream = (8+7) = 15 km/hr Time taken to travel 1 km downstream = 1 15 hr = 1×60 15 = 4 minutes

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Important Formulas - Calendar • Odd Days

Number of days more than the complete weeks are called odd days in a given period.

. Leap Year A leap year has 366 days. In a leap year, the month of February has 29 days

a. Every year divisible by 4 is a leap year, if it is not a century. Examples: 1952, 2008, 1680 etc. are leap years. 1991, 2003 etc. are not leap years

b. Every 4th century is a leap year and no other century is a leap year. Examples: 400, 800, 1200 etc. are leap years. 100, 200, 1900 etc. are not leap years

• Ordinary Year The year which is not a leap year is an ordinary year. An ordinary year has 365 days


• Counting odd days and Calculating the day of any particular date I. 1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)

Hence number of odd days in 1 ordinary year= 1.

II. 1 leap year ≡ 366 days ≡ (52 weeks + 2 days) Hence number of odd days in 1 leap year= 2.

III. 100 years ≡ (76 ordinary years + 24 leap years ) ≡ (76 x 1 + 24 x 2) odd days ≡ 124 odd days. ≡ (17 weeks + 5 days) ≡ 5 odd days. Hence number of odd days in 100 years = 5.

IV. Number of odd days in 200 years = (5 x 2) = 10 ≡ 3 odd days


V. Number of odd days in 300 years = (5 x 3) = 15 ≡ 1 odd days

VI. Number of odd days in 400 years = (5 x 4 + 1) = 21 ≡ 0 odd days Similarly, the number of odd days in all 4th centuries (400, 800, 1200 etc.) = 0

VII. Mapping of the number of odd day to the day of the week

Number of Odd Days

: 0 1 2 3 4 5 6

Day of the week

: Sunday

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Last day of a century cannot be Tuesday or Thursday or Saturday.


For the calendars of two different years to be the same, the following conditions must be satisfied.

. Both years must be of the same type. i.e., both years must be ordinary years or both years must be leap years.

i. 1st January of both the years must be the same day of the week.

1. What day of the week does May 28 2006 fall on A. Saturday B. Monday C. Sunday D. Thursday


Answer : Option C

Explanation :

28th May 2006 = (2005 years + period from 1-Jan-2006 to 28-May-2006) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)


Number of odd days in the period 2001-2005 = 4 normal years + 1 leap year = 4 x 1 + 1 x 2 = 6 Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may) = 148 148 days = 21 weeks + 1 day = 1 odd day Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day 0 odd day = Sunday Hence May 28 2006 is Sunday.

2. What will be the day of the week 15th August, 2010? A. Thursday B. Sunday C. Monday D. Saturday



Answer : Option B

Explanation :

15th Aug 2010 = (2009 years + period from 1-Jan-2010 to 15-Aug-2010) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Number of odd days in the period 2001-2009 = 7 normal years + 2 leap year = 7 x 1 + 2 x 2 = 11 = (11 - 7x1) odd day = 4 odd day Days from 1-Jan-2010 to 15-Aug-2010 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 30(Jun) + 31(Jul) + 15(Aug) = 227 227 days = 32 weeks + 3 day = 3 odd day


Total number of odd days = (0 + 4 + 3) = 7 odd days = 0 odd day 0 odd day = Sunday Hence 15th August, 2010 is Sunday.

3. Today is Monday. After 61 days, it will be A. Thursday B. Sunday C. Monday D. Saturday


Answer : Option D

Explanation :

61 days = 8 weeks 5 days = 5 odd days Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days) = Saturday

4. On what dates of April, 2001 did Wednesday fall? A. 2nd, 9th, 16th, 23rd B. 4th, 11th, 18th, 25th


C. 3rd, 10th, 17th, 24th D. 1st, 8th, 15th, 22nd, 29th


Answer : Option B

Explanation :

We need to find out the day of 01-Apr-2001 01-Apr-2001 = (2000 years + period from 1-Jan-2001 to 01-Apr-2001) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Days from 1-Jan-2001 to 01-Apr-2001 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 1(Apr) = 91 91 days = 13 weeks = 0 odd day Total number of odd days = (0 + 0) = 0 odd days 0 odd day = Sunday. Hence 01-Apr-2001 is Sunday.


Hence first Wednesday of Apr 2011 comes in 04th and successive Wednesdays come in 11th, 18th and 25th

5. How many days are there in x weeks x days A. 14x B. 8x C. 7x2 D. 7


Answer : Option B

Explanation :

x weeks x days = (7×x)+x=7x+x= 8x days

6. The calendar for the year 2007 will be the same for the year A. 2017 B. 2018 C. 2014 D. 2016


Answer : Option B

Explanation :


For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.

Year

: 2007

2008

2009

2010

2011

2012

2013

2014

2015

2016

2017

Odd Days

: 1 2 1 1 1 2 1 1 1 2 1

Take the year 2014 given in the choice. Total odd days in the period 2007-2013 = 5 normal years + 2 leap year = 5 x 1 + 2 x 2 = 9 odd days = 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything) Take the year 2016 given in the choice. Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year = 7 x 1 + 2 x 2 = 11 odd days


= 4 odd days (Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days) Take the year 2017 given in the choice. Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year = 7 x 1 + 3 x 2 = 13 odd days = 6 odd days Take the year 2018 given in the choice. Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year = 8 x 1 + 3 x 2 = 14 odd days = 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything) Also, both 2007 and 2018 are not leap years.


Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007

7. Which of the following is not a leap year? A. 1200 B. 800 C. 700 D. 2000


Answer : Option C

Explanation :

Remember the leap year rule (Given in the formulas) 1. Every year divisible by 4 is a leap year, if it is not a century. 2. Every 4th century is a leap year, but no other century is a leap year. 800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc). Hence 800,1200 and 2000 are leap years


700 is not a 4th century, but it is a century. Hence it is not a leap year

8. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008? A. Wednesday B. Sunday C. Friday D. Tuesday


Answer : Option D

Explanation :

Given that January 1, 2007 was Monday. Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007) Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday

9. 8th Dec 2007 was Saturday, what day of the week was it on 8th Dec, 2006? A. Sunday B. Tuesday


C. Friday D. Tuesday


Answer : Option C

Explanation :

Given that 8th Dec 2007 was Saturday Number of days from 8th Dec, 2006 to 7th Dec 2007 = 365 days 365 days = 1 odd day Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday

10. On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004? A. Sunday B. Friday C. Saturday D. Monday


Answer : Option A

Explanation :

Given that 8th Feb, 2005 was Tuesday


Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366 (Since Feb 2004 has 29 days as it is a leap year) 366 days = 2 odd days Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunday

11. The last day of a century cannot be A. Monday B. Wednesday C. Tuesday D. Friday


Answer : Option C

Explanation :

We know that number of odd days in 100 years = 5 Hence last day of the first century is Friday Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything)


Hence last day of the 2nd century is Wednesday Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce multiples of 7 from odd days which will not change anything) Hence last day of the 3rd century is Monday We know that umber of odd days in 400 years = 0. (? 5 x 4 + 1 = 21 = 0) Hence last day of the 4th century is Sunday Now this cycle will be repeated. Hence last day of a century will not be Tuesday or Thursday or Saturday. its better to learn this by heart which will be helpful to save time in objective type exams


12. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009? A. Saturday B. Wednesday C. Thursday D. Saturday


Answer : Option C

Explanation :

Number of odd days in 2008 = 2 (since it is a leap year) (we have taken the complete year 2008 because we need to find out the odd days from 01-Jan-2008 to 31-Dec-2008, that is the whole year 2008) Given that January 1, 2008 is Tuesday Hence January 1, 2009 = (Tuesday + 2 odd days) = Thursday

13. If Jan 1, 2006 was a Sunday, What was the day of the week Jan 1, 2010? A. Friday B. Thursday C. Tuesday D. Saturday



Answer : Option A

Explanation :

Given that Jan 1 2006 was a Sunday Number of odd days in the period 2006-2009 = 3 normal years + 1 leap year = 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved) Hence Jan 1 2010 = (Sunday + 5 odd days) = Friday

14. What was the day of the week on 17th June 1998? A. Monday B. Sunday C. Wednesday D. Friday



Answer : Option C

Explanation :

17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17-Jun-1998) We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400) Number of odd days in the period 1601-1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything) Number of odd days in the period 1901-1997 = 73 normal years + 24 leap year = 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days


Number of days from 1-Jan-1998 to 17-Jun-1998 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun) = 168 168 days = 0 odd day Total number of odd days = (0 + 1 + 2 + 0) = 3 3 odd days = Wednesday Hence 17th June 1998 is Wednesday.

15. 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004? A. Friday B. Saturday C. Wednesday D. Sunday


Answer : Option D

Explanation :

Number of days from 6th March, 2004 to 5th March 2005 = 365 days


(Though Feb 2004 has 29 days as it is a leap year, it will not come in the required period) 365 days = 1 odd day Given that 6th March, 2005 is Monday Hence 6th March, 2004 = (Monday - 1 odd day) = Sunday

16. What day of the week was 1 January 1901 A. Monday B. Tuesday C. Saturday D. Friday


Answer : Option B

Explanation :

1 Jan 1901 = (1900 years + 1-Jan-1901) We know that number of odd days in 400 years = 0 Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)


Number of odd days in the period 1601-1900 = Number of odd days in 300 years = 5 x 3 = 15 = 1 (As we can reduce perfect multiples of 7 from odd days without affecting anything) 1-Jan-1901 = 1 odd day Total number of odd days = (0 + 1 + 1) = 2 2 odd days = Tuesday Hence 1 January 1901 is Tuesday.

17. What day of the week will 22 Apr 2222 be? A. Monday B. Tuesday C. Sunday D. Thursday


Answer : Option A


Explanation :

22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222) We know that number of odd days in 400 years = 0 Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400) Number of odd days in the period 2001-2200 = Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce perfect multiples of 7 from odd days without affecting anything) Number of odd days in the period 2201-2221 = 16 normal years + 5 leap years = 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days Number of days from 1-Jan-2222 to 22 Apr 2222


= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112 112 days = 0 odd day Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day 1 odd days = Monday Hence 22 Apr 2222 is Monday.

18. Today is Thursday. The day after 59 days will be? A. Monday B. Tuesday C. Saturday D. Sunday


Answer : Option D

Explanation :

59 days = 8 weeks 3 days = 3 odd days Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days) = Sunday


19. What is the year next to 1990 which will have the same calendar as that of the year 1990? A. 1992 B. 2001 C. 1995 D. 1996


Answer : Option B

Explanation :

For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0. Take the year 1992 from the given choices. Total odd days in the period 1990-1991= 2 normal years ≡ 2 x 1 = 2 odd days Take the year 1995 from the given choices. Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year ≡ 4 x 1 + 1 x 2 = 6 odd days Take the year 1996 from the given choices. Number of odd days in the period 1990-1995= 5 normal years + 1 leap year ≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days (As we can reduce multiples of 7 from odd days which will not change anything)


Though number of odd days in the period 1990-1995 is 0, there is a catch here. 1990 is not a leap year whereas 1996 is a leap year. Hence calendar for 1990 and 1996 will never be the same. Take the year 2001 from the given choices. Number of odd days in the period 1990-2000= 8 normal years + 3 leap years ≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days Also, both 1990 and 2001 are normal years. Hence 1990 will have the same calendar as that of 2001

20. January 1, 2004 was a Thursday, what day of the week lies on January 1 2005. A. Saturday B. Monday C. Saturday D. Tuesday


Answer : Option C

Explanation :

Given that January 1, 2004 was Thursday. Odd days in 2004 = 2 (because 2004 is a leap year) (Also note that we have taken the complete year 2004 because we need to find out


the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004) Hence January 1, 2005 = (Thursday + 2 Odd days) = Saturday

21. If the first day of a year (other than leap year) was Friday, then which was the last day of that year? A. Saturday B. Friday C. Tuesday D. Monday


Answer : Option B

Explanation :

Given that first day of a normal year was Friday Odd days of the mentioned year = 1 (Since it is an ordinary year) Hence First day of the next year = (Friday + 1 Odd day) = Saturday ? Last day of the mentioned year = Friday


22. If 1st October is Sunday, then 1st November will be A. Saturday B. Thursday C. Wednesday D. Tuesday


Answer : Option C

Explanation :

Given that 1st October is Sunday Number of days in October = 31 31 days = 3 odd days (As we can reduce multiples of 7 from odd days which will not change anything) Hence 1st November = (Sunday + 3 odd days) = Wednesday

23. Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today? A. Wednesday B. Saturday C. Friday D. Sunday



Answer : Option B

Explanation :

Clearly it can be understood from the question that 9 days ago was a Thursday Number of odd days in 9 days = 2 (As 9-7 = 2, reduced perfect multiple of 7 from total days) Hence today = (Thursday + 2 odd days) = Saturday

24. 1.12.91 is the first Sunday. Which is the fourth Tuesday of December 91? A. 20.12.91 B. 22.12.91 C. 24.12.91 D. 25.12.91


Answer : Option C

Explanation :

Given that 1.12.91 is the first Sunday Hence we can assume that 3.12.91 is the first Tuesday


If we add 7 days to 3.12.91, we will get second Tuesday If we add 14 days to 3.12.91, we will get third Tuesday If we add 21 days to 3.12.91, we will get fourth Tuesday => fourth Tuesday = (3.12.91 + 21 days) = 24.12.91

25. If the day before yesterday was Thursday, when will Sunday be? A. Day after tomorrow B. Tomorow C. Two days after today D. Today


Answer : Option B

Explanation :

Day before yesterday was Thursday =>Yesterday was a Friday => Today is a Saturday => Tomorrow is a Sunday


26. The second day of a month is Friday, What will be the last day of the next month which has 31 days? A. Friday B. Saturday C. Wednesday D. Data inadequate


Answer : Option D

Explanation :

We cannot find out the answer because the number of days of the current month is not given.

27. How many days will there be from 26th January,1996 to 15th May,1996(both days included)? A. 102 B. 103 C. 111 D. 120


Answer : Option C

Explanation :

Number of days from 26-Jan-1996 to 15-May-1996 (both days included) = 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May) = 111


28. If 25th of August in a year is Thursday, the number of Mondays in that month is A. 4 B. 5 C. 2 D. 3


Answer : Option B

Explanation :

Given that 25th August = Thursday Hence 29th August = Monday So 22nd,15th and 8th and 1st of August will be Mondays Number of Mondays in August = 5

29. If the seventh day of a month is three days earlier than Friday, What day will it be on the nineteenth day of the month? A. Saturday B. Monday C. Sunday D. Wednesday


Answer : Option C

Explanation :


Given that seventh day of a month is three days earlier than Friday => Seventh day is Tuesday => 14th is Tuesday => 19th is Sunday

30. Every second Saturday and all Sundays are holidays. How many working days will be there in a month of 30 days beginning on a Saturday? A. 24 B. 23 C. 18 D. 21


Answer : Option B

Explanation :

Mentioned month has begins on a Saturday and has 30 days Sundays = 2nd, 9th, 16th, 23rd, 30th => Total Sundays = 5 Second Saturdays = 8th and 22nd


Total Second Saturdays = 2 Total Holidays = Total Sundays + Total Second Saturdays = 5 + 2 = 7 Total days in the month = 30 Total working days = 30 - 7 = 23


Important Formulas - Chain Rule • Direct Proportion

Two quantities are said to be directly proportional, if on the increase or decrease of the one, the other increases or decreases the same extent. Examples i. Cost of the goods is directly proportional to the

number of goods. (More goods, More cost)

ii. Amount of work done is directly proportional to the number of persons who did the work. (More persons, More Work)

• Indirect Proportion (inverse proportion) Two quantities are said to be indirectly proportional (inversely proportional) if on the increase of the one, the other decreases to the same extent and vice-versa. Examples i. Number of days needed to complete a work is

indirectly proportional (inversely proportional) with the number of persons who does the work (More Persons, Less Days needed)


ii. The time taken to travel a distance is indirectly proportional (inversely proportional) with the speed in which one is travelling (More Speed, Less Time)

1. If the cost of x metres of wire is d rupees, then what is the cost of y metres of wire at the same rate? A. Rs. (xd y ) B. Rs. xd C. Rs. (yd x ) D. Rs. yd


Answer : Option C

Explanation :

cost of x metres of wire = Rs. d

cost of 1 metre of wire = Rs.(d x ) cost of y metre of wire = Rs.(y×d x )=Rs. (yd x )

2. In a dairy farm, 40 cows eat 40 bags of husk in 40 days. In how many days one cow will eat one bag of husk? A. 1 B. 40 C. 20 D. 26


Answer : Option B


Explanation :

Assume that in x days, one cow will eat one bag of husk. More cows, less days (Indirect proportion) More bags, more days (direct proportion) Hence we can write as

Cows Bags 40:1 1:40 ⎫ ⎭ ⎬ ::x:40

⇒40×1×40=1×40×x ⇒x=40

3. If 7 spiders make 7 webs in 7 days, then how many days are needed for 1 spider to make 1 web? A. 1 B. 7 C. 3 D. 14


Answer : Option B

Explanation :

Let, 1 spider make 1 web in x days. More spiders, Less days (Indirect proportion) More webs, more days (Direct proportion)


Hence we can write as

Spiders Webs 7:1 1:7 ⎫ ⎭ ⎬ ::x:7

⇒7×1×7=1×7×x ⇒x=7

4. 4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days? A. 4 B. 16 C. 8 D. 1


Answer : Option B

Explanation :

Let the required number of mats be x More mat-weavers, more mats (direct proportion) More days, more mats (direct proportion) Hence we can write as

Mat-weavers Days 4:8 4:8 ⎫ ⎭ ⎬ ::4:x


⇒4×4×x=8×8×4 ⇒x=2×2×4=16

5. If a quarter kg of potato costs 60 paise, how many paise does 200 gm cost? A. 65 paise B. 70 paise C. 52 paise D. 48 paise


Answer : Option D

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let 200 gm potato costs x paise Cost of ¼ Kg potato = 60 Paise => Cost of 250 gm potato = 60 Paise (∵ 1 Kg = 1000 gm => ¼ Kg = 1000/4 gm = 250 gm) More quantity, More Paise (direct proportion) Hence we can write as

Quantity 200:250 }::x:60


⇒200×60=250×x ⇒4×60=5×x ⇒4×12=x ⇒x=48

------------------------------------------- Solution 2 ------------------------------------------- Cost of ¼ Kg potato = 60 Paise => Cost of 250 gm potato = 60 Paise ( ∵ 1 Kg = 1000 gm => ¼ Kg = 1000 / 4 gm = 250 gm)

=> Cost of 200 gm potato = 60×200 250 =60×4 5 =48paise

6. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal? A. 50 B. 30 C. 40 D. 10


Answer : Option B

Explanation :

Meal for 200 children = Meal for 120 men


Meal for 1 child = Meal for 120 200 men Meal for 150 children = Meal for 120×150 200 men=Meal for 90 men

Total mean available = Meal for 120 men Renaming meal = Meal for 120 men - Meal for 90 men = Meal for 30 men

7. 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work? A. 26 B. 22 C. 12 D. 24


Answer : Option D

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let the required number of days be x More men, less days (indirect proportion) Hence we can write as

Men 36:27 }::x:18


⇒36×18=27×x ⇒12×18=9×x ⇒12×2=x ⇒x=24

------------------------------------------- Solution 2 (Using Time and Work) ------------------------------------------- Amount of work 36 men can do in 1 day = 1/18

Amount of work 1 man can do in 1 day = 1 18×36 Amount of work 27 men can do in 1 day = 27 18×36 =3 18×4 =1 24 ⇒ 27 men can complete the work in 24 days

------------------------------------------- Solution 3 (Using Time and Work) -----------------------------------------

If M 1 men can do W 1 work in D 1 days working H 1 hours per day and M 2 men can do W 2 work in D 2 days working H 2 hours per day (where all men work at the same rate), then M 1 D 1 H 1 W 1 =M 2 D 2 H 2 W 2

In this case,


M1 = 36, M 2 = 27 D1 = 18, D 2 = x W1 = W 2 H1 = H 2 ( ∵ We can assume like this as these vales are not explicitly given)

Hence, M 1 D 1 =M 2 D 2 ⇒36×18=27x ⇒x=36×18 27 =4×18 3 =4×6=24

8. A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 21 revolutions, what will be the number of revolutions made by the larger wheel? A. 15 B. 12 C. 21 D. 9


Answer : Option D

Explanation :

Let the number of revolutions made by the larger wheel be x More cogs, less revolutions (Indirect proportion) Hence we can write as


Cogs 6:14 }::x:21

⇒6×21=14×x ⇒6×3=2×x ⇒3×3=x ⇒x=9

9. 3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day should 4 pumps work in order to empty the tank in 1 day? A. 10 B. 12 C. 8 D. 15


Answer : Option B

Explanation :

Let the required hours needed be x More pumps, less hours (Indirect proportion) More Days, less hours (Indirect proportion) Hence we can write as

Pumps Days 3:4 2:1 ⎫ ⎭ ⎬ ::x:8

⇒3×2×8=4×1×x ⇒3×2×2=x ⇒x=12


10. 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? A. 9 B. 12 C. 10 D. 13


Answer : Option D

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let the required number of days be x More persons, less days (indirect proportion) More hours, less days (indirect proportion) Hence we can write as

Persons Hours 39:30 5:6 }::x:12

⇒39×5×12=30×6×x ⇒39×5×2=30×x ⇒39=3×x ⇒x=13


------------------------------------------- Solution 2 (Using Time and Work) ------------------------------------------- Amount of work 39 persons can do in 1 day, working 5 hours a day = 1/12

Amount of work 1 person can do in 1 day, working 5 hours a day = 1 12×39 Amount of work 1 person can do in 1 day, working 1 hours a day = 1 12×39×5 Amount of work 30 person can do in 1 day, working 1 hours a day = 30 12×39×5 Amount of work 30 person can do in 1 day, working 6 hours a day = 30×6 12×39×5 =30 2×39×5 =3 39 =1 13 ⇒ 30 persons can complete the work ,working 6 hours a day in 13 days

------------------------------------------- Solution 3 (Using Time and Work) -------------------------------------------



In this case, M1 = 39, M 2 = 30 D1 = 12, D 2 = x W1 = W 2 H1 = 5, H 2 = 6

Hence, M 1 D 1 H 1 =M 2 D 2 H 2 ⇒39×12×5=30×x×6 ⇒39×2×5=30×x ⇒39=3×x ⇒x=39 3 =13

11. A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth? A. 205 B. 200 C. 180 D. 195


Answer : Option D

Explanation :

Let the required number of seconds be x More cloth, More time, (direct proportion) Hence we can write as


Cloth 0.128:25 }::1:x

⇒0.128x=25 ⇒x=25 .128 =25000 128 =3125 16 ≈195

12. A contract is to be completed in 56 days if 104 persons work, each working at 8 hours a day. After 30 days, 2/5 of the work is completed. How many additional persons should be deployed so that the work will be completed in the scheduled time,each persons now working 9 hours a day. A. 160 B. 150 C. 24 D. 56


Answer : Option D

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Persons worked = 104 Number of hours each person worked per day = 8 Number of days they worked = 30 Work completed = 2/5 Remaining days = 56 - 30 = 26 Remaining Work to be completed = 1 - 2/5 = 3/5


Let the total number of persons who do the remaining work = x Number of hours each person needs to be work per day = 9 More days, less persons(indirect proportion) More hours, less persons(indirect proportion) More work, more persons(direct proportion) Hence we can write as

Days Hours Work 30:26 8:9 3 5 :2 5 ::x:104

⇒30×8×3 5 ×104=26×9×2 5 ×x ⇒x=30×8×3 5 ×104 26×9×2 5 =30×8×3×104 26×9×2 =30×8×104 26×3×2 =30×8×4 3×2 =5×8×4=160 Number of additional persons required = 160 - 104 = 56

------------------------------------------- Solution 2 (Using Time and Work) ------------------------------------------- Persons worked = 104 Number of hours each person worked per day = 8 Number of days they worked = 30


Work completed = 2/ 5 Remaining days = 56 - 30 = 26 Remaining Work to be completed = 1 - 2/ 5 = 3/ 5 Let the total number of persons who do the remaining work = x Number of hours each person needs to be work per day = 9

Amount of work 1 person did in 1 day, working 1 hours a day = (2 5 ) 104×30×8 Now, the amount of work x person should do in 1 day, working 1 hours a day = (3 5 ) 26×9 ⇒x=(3 5 ) 26×9 (2 5 ) 104×30×8 =30×8×3 5 ×104 26×9×2 5 =30×8×3×104 26×9×2 =30×8×104 26×3×2 =30×8×4 3×2 =5×8×4=160 Number of additional persons required = 160 - 104 = 56

------------------------------------------- Solution 3 (Using Time and Work) ------------------------------------------



Persons worked (M 1) = 104 Number of hours each person worked per day (H 1) = 8 Number of days they worked (D 1) = 30 Work completed (W 1)= 2/5 Remaining days (D 2)= 56 - 30 = 26 Remaining Work to be completed (W 2)= 1 - 2/5 = 3/5 Let the total number of persons who do the remaining work (M 2) = x Number of hours each person needs to be work per day (H 2) = 9


M 1 D 1 H 1 W 1 =M 2 D 2 H 2 W 2 ⇒104×30×8 (2 5 ) =x×26×9 (3 5 ) ⇒104×30×8 2 =x×26×9 3 ⇒52×30×8=x×26×3 ⇒2×30×8=3x ⇒x=2×10×8=160 Number of additional persons required = 160 - 104 = 56

13. x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days? A. x 2 y 2 units B. y 3 x 2 units C. x 3 y 2 units D. y 2 x 2 units


Answer : Option B

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let amount of work completed by y men working y hours per in y days = w units More men, more work(direct proportion) More hours, more work(direct proportion) More days, more work(direct proportion) Hence we can write as

Men Hours Days x:y x:y x:y ::x:w


⇒x 3 w=y 3 x ⇒w=y 3 x x 3 =y 3 x 2


Amount of work completed by 1 man in 1 day, working 1 hours a day = x x 3 =1 x 2 Amount of work y men in y days, working y hours a day = y 3 ×1 x 2 =y 3 x 2



M1 = x


H1 = x D1 = x W1 = x M2 = y D2= y H2 = y Let W 2= w

M 1 D 1 H 1 W 1 =M 2 D 2 H 2 W 2 ⇒x 3 x =y 3 w ⇒x 2 =y 3 w ⇒w=y 3 x 2

14. 21 goats eat as much as 15 cows. How many goats each as much as 35 cows? A. 49 B. 32 C. 36 D. 41


Answer : Option A

Explanation :


15 cows ≡ 21 goats 1 cow ≡21 15 goats 35 cows ≡ 21×35 15 goats≡21×7 3 goats≡7×7 goats ≡ 49 goats

15. A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions? A. 12.5 m B. 10.5 m C. 14 D. 12


Answer : Option A

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let the required height of the building be x meter More shadow length, More height(direct proportion) Hence we can write as

shadow length 40.25:28.75 }::17.5:x

⇒40.25×x=28.75×17.5 ⇒x=28.75×17.5 40.25 =2875×175 40250 =2875×7 1610 =2875 230 =575 46 =12.5


------------------------------------------- Solution 2 -------------------------------------------

shadow of length 40.25 m ≡ 17.5 m high shadow of length 1 m ≡ 17.5 40.25 m high shadow of length 28.75 m ≡ 28.75×17.5 40.25 m high = 12.5 m high

16. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes? A. 1800 B. 900 C. 2500 D. 2700


Answer : Option A

Explanation :

Let required number of bottles be x More machines, more bottles(direct proportion) More minutes, more bottles(direct proportion) Hence we can write as

machines minutes 6:10 1:4 ⎫ ⎭ ⎬ ::270:x


⇒6×1×x=10×4×270 ⇒x=10×4×270 6 =10×4×90 2 =10×4×45=1800

17. A person works on a project and completes 5/8 of the job in 10 days. At this rate, how many more days will he it take to finish the job? A. 7 B. 6 C. 5 D. 4


Answer : Option B

Explanation :

------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Number of days he worked = 10 Work completed = 5/8 Let the required number of days = x Remaining Work to be completed = 1 - 5/8 = 3/8 More work, more days(direct proportion) Hence we can write as


Work 5 8 :3 8 }::10:x

⇒5 8 ×x=3 8 ×10 ⇒5×x=3×10 ⇒x=3×2=6

------------------------------------------- Solution 2 (Using Time and Work) ------------------------------------------- Number of days he worked = 10 Work completed = 5/ 8 Let the required number of days = x Remaining Work to be completed = 1 - 5/ 8 = 3/ 8

Amount of work 1 person did in 1 day = (5 8 ) 10 Now, the amount of work 1 person should do in x days = 3 8 ⇒x=(3 8 ) ⎛ ⎝ ⎜ ⎜ ⎜ (5 8 ) 10 ⎞ ⎠ ⎟ ⎟ ⎟ =3 (5 10 ) =3×10 5 =6




Here, M1 = M 2 H1 = H 2 D1 = 10 W1 = 5/8 Let D 2 = x W2= 1 - 5/8 = 3/8


Hence, the equation can be written as D 1 W 1 =D 2 W 2 ⇒10 (5 8 ) =x (3 8 ) ⇒10 5 =x 3 ⇒2=x 3 ⇒x=2×3=6

18. A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. Find out the number of days for which the remaining food will last. A. 44 B. 42 C. 40 D. 38


Answer : Option B

Explanation :

Given that fort had provision of food for 150 men for 45 days Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days Remaining men after 10 days = 150 - 25 = 125 Assume that after 10 days,the remaining food is sufficient for 125 men for x days More men, Less days (Indirect Proportion)


⇒Men 150:125 }::x:35

⇒150×35=125x ⇒6×35=5x ⇒x=6×7=42 ⇒The remaining food is sufficient for 125 men for 42 days

19. If the price of 357 apples is Rs.1517.25, what will be the approximate price of 49 dozens of such apples? A. Rs. 2500 B. Rs. 2300 C. Rs. 2200 D. Rs. 1400


Answer : Option A

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let the required price be x More apples, More price(direct proportion) Hence we can write as

apples 357:(49×12) }::1517.25:x


⇒357x=(49×12)×1517.25 ⇒x=49×12×1517.25 357 =7×12×1517.25 51 =7×4×1517.25 17 =7×4×89.25≈2500

------------------------------------------- Solution 2 -------------------------------------------

price of 357 apples = Rs.1517.25 price of 1 apple = Rs. 1517.25 357 price of 49 dozens apples = Rs.(49×12×1517.25 357 )≈Rs. 2500

20. On a scale of a map 0.6 cm represents 6.6km. If the distance between two points on the map is 80.5 cm , what is the the actual distance between these points? A. 885.5 km B. 860 km C. 892.5 km D. 825 km


Answer : Option A

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule)


-------------------------------------------------------------------- Let the required actual distance be x km More scale distance, More actual distance(direct proportion) Hence we can write as

scale distance .6:80.5 }::6.6:x

⇒.6x=80.5×6.6 ⇒.1x=80.5×1.1 ⇒x=80.5×11=885.5

------------------------------------------ Solution 2 -------------------------------------------

.6 cm in map ≡ actual distance of 6.6 km 1 cm in map ≡6.6 .6 km 80.5 cm in map ≡80.5×6.6 .6 km = 885.5 km

21. A rope can make 70 rounds of the circumference of a cylinder whose radius of the base is 14cm. how many times can it go round a cylinder having radius 20 cm? A. 49 rounds B. 42 rounds C. 54 rounds D. 52 rounds



Answer : Option A

Explanation :

Let the required number of rounds be x More radius, less rounds(Indirect proportion) Hence we can write as

radius 14:20 }::x:70

⇒14×70=20x ⇒14×7=2x ⇒x=7×7=49

22. 8 persons can build a wall 140m long in 42 days. In how many days can 30 persons complete a similar wall 100 m long? A. 12 B. 10 C. 8 D. 6


Answer : Option C

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- More persons, less days(indirect proportion)


More length of the wall, more days(direct proportion) Hence we can write as

persons Length of the wall 8:30 100:140 ⎫ ⎭ ⎬ ::x:42

⇒8×100×42=30×140×x ⇒x=8×100×42 30×140 =8×100×14 10×140 =8×100 10×10 =8


Work done by 8 persons working 42 days = 140 Work done by 1 person working 42 days =140 8 Work done by 1 person working 1 day =140 8×42 Work done by 30 persons working 1 day =30×140 8×42 =100 8 Assume that 30 persons working x days complete a similar wall 100 m ⇒Work done by 30 persons working x days =100 Hence x=100 (100 8 ) =8




M1 = 8 D1 = 42 W1= 140 M2 = 30 Let D 2 = x W2= 100 Here H 1 = H 2

M 1 D 1 W 1 =M 2 D 2 W 2 ⇒8×42 140 =30×x 100 ⇒8×42 14 =30×x 10 ⇒8×3=3×x ⇒x=8

23. A certain number of persons can finish a piece of work in


100 days. If there were 10 persons less, it would take 10 more days finish the work. How many persons were there originally? A. 90 B. 100 C. 110 D. 120


Answer : Option C

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Assume that x persons can finish a piece of work in 100 days Also it is given that (x-10) persons can finish a piece of work in 110 days (∵ 100 + 10 = 110) More persons, less days(indirect proportion) Hence we can write as

persons x:(x−10) }::110:100

⇒100x=110(x−10) ⇒100x=110x−1100 ⇒10x=1100 ⇒x=1100 10 =110

-------------------------------------------


Solution 2 (Using Time and Work) -------------------------------------------

Assume that x persons can finish the work in 100 days ⇒Work done by 1 person in 1 day = 1 100x .......(Equation 1) Also it is given that (x-10) persons can finish the work in 110 days(∵ 100 + 10 = 110) ⇒Work done by 1 person in 1 day = 1 110(x−10) .......(Equation 2) But (Equation 1) = (Equation 2) ⇒1 100x =1 110(x−10) ⇒110(x−10)=100x ⇒110x−1100=100x ⇒10x=1100 x=1100 10 =110



Assume that x persons can finish a piece of work in 100 days Also it is given that (x-10) persons can finish a piece of work in 110 days ( ∵ 100 +


10 = 110) M1 = x D1 = 100 M2 = (x-10) D2 = 110 Here H 1 = H 2 and W 1 = W 2

Hence M 1 D 1 =M 2 D 2 ⇒100x=110(x−10) ⇒100x=110x−1100 ⇒10x=1100 x=1100 10 =110

24. 9 examiners can examine a certain number of answer books in 12 days by working 5 hours a day. How many hours in a day should 4 examiners work to examine twice the number of answer books in 30 days? A. 9 B. 10 C. 11 D. 12



Answer : Option A

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let required number of hours be x More examiners, less hours(indirect proportion) More days, less hours(indirect proportion) More answer books, more hours(direct proportion) Hence we can write as

examiners days answer books 9:4 12:30 2:1 ⎫ ⎭ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ::x:5

⇒9×12×2×5=4×30×1×x ⇒x=9×12×2×5 4×30×1 =9×3×2×5 30 =9×2×5 10 =9


Given that Work done by 9 examiners in 12 day in working 5 hours a day =1 ⇒Work done by 1 examiner in 1 day


in 1 hour =1 9×12×5 ......(Equation 1) ⇒Work needs to be done by 4 examiners in 30 days working x hours a day =2 (∵ twice work to be completed) ⇒Work needs to be done by 1 examiner in 1 day working x hours a day =2 4×30 ...(Equation 2) From (Equation 1) and (Equation 2), ⇒x=(2 4×30 ) (1 9×12×5 ) =2×9×12×5 4×30 =9×12×5 2×30 =9×6×5 30 =9



M1 = 9 D1 = 12 H1 = 5 W1 = 1


M2 = 4 D2 = 30 H2 = x W2 = 2

M 1 D 1 H 1 W 1 =M 2 D 2 H 2 W 2 ⇒9×12×5 1 =4×30×x 2 ⇒9×12×5=2×30×x ⇒3×12×5=2×10×x ⇒3×6×5=10×x ⇒3×3×5=5×x ⇒x=3×3=9

25. 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type? A. 20 metric tonnes B. 22 metric tonnes C. 24 metric tonnes D. 26 metric tonnes


Answer : Option D

Explanation :


-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Let required amount of coal be x metric tonnes More engines, more amount of coal (direct proportion) If 3 engines of first type consume 1 unit, then 1 engine will consume 1/3 unit which is its the rate of consumption. If 4 engines of second type consume 1 unit, then 1 engine will consume 1/4 unit which is its the rate of consumption More rate of consumption, more amount of coal (direct proportion) More hours, more amount of coal(direct proportion) Hence we can write as

engines rate of consumption hours 9:8 1 3 :1 4 8:13 ⎫ ⎭ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ::24:x

⇒9×1 3 ×8×x=8×1 4 ×13×24 ⇒3×8×x=8×6×13 ⇒3×x=6×13 ⇒x=2×13=26

26. A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the


provisions will now last just as long as before. How long was that? A. 50 days B. 30 days C. 40 days D. 60 days


Answer : Option A

Explanation :

------------------------------------------------------------- Solution 1 -------------------------------------------------------------

Assume that initially garrison had provisions for x men for y days So, after 10 days, garrison had provisions for x men for (y-10) days Also, garrison had provisions for 4x 5 men for y days (∵x−x 5 =4x 5 ) More men, Less days (Indirect Proportion) ⇒Men x:4x 5 }::y:(y−10)

⇒x(y−10)=4xy 5 ⇒(y−10)=4y 5 ⇒5(y−10)=4y ⇒5y−50=4y ⇒y=50

------------------------------------------- Solution 2 -------------------------------------------


Assume that amount of food eaten by 1 man in 1 day = x, total men = y and the garrison had provisions for z days Then total quantity of food = xyz amount of food eaten by y men in 10 days = 10xy Remaining food = xyz - 10xy = xy(z - 10) -------(Equation 1) After 10 days, total men = 4y/5 ( ∵ y - y/ 5 = 4y / 5) food taken by 4y / 5 men in 1 day = 4xy / 5 -------(Equation 2) From Equations 1 and 2,


Time taken for 4y / 5 men to complete xy(z - 10) food

=xy(z−10) (4xy 5 ) =5(z−10) 4

Given that number of days remain the same

⇒5(z−10) 4 =z ⇒5z−50=4z ⇒z=50

27. A garrison of 500 persons had provisions for 27 days. After 3 days a reinforcement of 300 persons arrived. For how many more days will the remaining food last now? A. 12 days B. 16 days C. 14 days D. 15 days


Answer : Option D

Explanation :

------------------------------------------------------------- Solution 1 ------------------------------------------------------------- Given that fort had provision for 500 persons for 27 days Hence, after 3 days, the remaining food is sufficient for 500 persons for 24 days


Remaining persons after 3 days = 500 + 300 = 800 Assume that after 10 days,the remaining food is sufficient for 800 persons for x days More men, Less days (Indirect Proportion)

⇒Men 500:800 }::x:24

⇒500×24=800x ⇒5×24=8x ⇒x=5×3=15

------------------------------------------- Solution 2 ------------------------------------------- Assume that amount of food taken by 1 man in 1 day = x Given that the garrison had provisions for 500 persons for 27 days

⇒Total quantity of food = 500×27×x=13500x Amount of food eaten by 500 persons in 3 days = 500×3×x=1500x Remaining food = 13500x−1500x=12000x........(Equation 1) After 10 days, total


persons = 500 + 300 = 800 Food eaten by 800 persons 1 day = 800x ...................(Equation 2)

From Equations 1 and 2, Time taken for 800 persons to complete 12000x food

=12000x 800x =120 8 =15

28. A hostel had provisions for 250 men for 40 days. If 50 men left the hostel, how long will the food last at the same rate? A. 48 days B. 50 days C. 45 days D. 60 days


Answer : Option B

Explanation :

------------------------------------------------------------- Solution 1 ------------------------------------------------------------- A hostel had provisions for 250 men for 40 days If 50 men leaves the hostel, remaining men = 250 - 50 = 200 We need to find out how long the food will last for these 200


men. Let the required numer of days = x days More men, Less days (Indirect Proportion)

⇒Men 250:200 }::x:40

⇒250×40=200x ⇒5×40=4x ⇒x=5×10=50

------------------------------------------- Solution 2 ------------------------------------------- Assume that amount of food taken by 1 man in 1 day = x Given that the hostel had provisions for 250 men for 40 days

⇒Total quantity of food = 250×40×x=10000x...(Equation 1)

If 50 men leaves the hostel, remaining men = 250 - 50 = 200


Food eaten by 200 men 1 day = 200x ...................(Equation 2)

From Equations 1 and 2, Time taken for 200 men to complete 10000x food

=10000x 200x =100 2 =50

29. in a camp, food was was sufficient for 2000 people for 54 days. After 15 days, more people came and the food last only for 20 more days. How many people came? A. 1900 B. 1800 C. 1940 D. 2000


Answer : Option A

Explanation :

------------------------------------------------------------- Solution 1 ------------------------------------------------------------- Given that food was sufficient for 2000 people for 54 days Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 - 15 = 39)


Let x number of people came after 15 days. Then, total number of people after 15 days = (2000 + x) Then, the remaining food was sufficient for (2000 + x) people for 20 days More men, Less days (Indirect Proportion)

⇒Men 2000:(2000+x) }::20:39

⇒2000×39=(2000+x)20 ⇒100×39=(2000+x) ⇒3900=2000+x x=3900−2000=1900

------------------------------------------ Solution 2 ------------------------------------------ Assume that amount of food eaten by 1 person in 1 day = k Given that food was sufficient for 2000 people for 54 days Hence,Total quantity of food = 2000 × 54 × k = 108000k


Amount of food eaten by 2000 persons in 15 days = 2000 × 15 × k = 30000k Remaining food = 108000k - 30000k = 78000k ............................(Equation 1) Let x number of people came after 15 days. Then, total number of people after 15 days = (2000 + x) Food eaten by (2000 + x) persons 1 day = (2000 + x)k ...................(Equation 2) From Equations 1 and 2, Time taken for (2000 + x) persons to complete 78000k food

=78000k (2000+x)k =78000 (2000+x)


Given that food lasted only for 20 more days

⇒78000 (2000+x) =20 ⇒78000=20(2000+x) ⇒3900=2000+x ⇒x=3900−2000=1900

30. If 40 men can make 30 boxes in 20 days, How many more men are needed to make 60 boxes in 25 days? A. 28 B. 24 C. 22 D. 26


Answer : Option B

Explanation :

-------------------------------------------------------------------- Solution 1 (Chain Rule) -------------------------------------------------------------------- Given that 40 men can make 30 boxes in 20 days Let x more men are needed to make 60 boxes in 25 days Then (40 + x) men can make 60 boxes in 25 days More boxes, more men(direct proportion) More days, less men(indirect proportion) Hence we can write as


boxes days 30:60 25:20 ⎫ ⎭ ⎬ ::40:(40+x)

⇒30×25×(40+x)=60×20×40 ⇒25×(40+x)=2×20×40 ⇒5×(40+x)=2×4×40 ⇒(40+x)=2×4×8=64 ⇒x=64−40=24

------------------------------------------- Solution 2 (Using Time and Work) ------------------------------------------- Let x more men are needed to make 60 boxes in 25 days Then (40 + x) men can make 60 boxes in 25 days

Work done by 40 men in 20 days = 30 Work done by 1 man in 1 day =30 40×20 Work done by (40 + x) men in 25 days =30(40+x)×25 40×20 ...(Equation 1) If (40 + x) men can make 60 boxes in 25 days, ⇒Work done by (40 + x) men in 25 days =60.....................(Equation 2) Hence, from equation 1 and equation 2, 30(40+x)×25 40×20 =60 ⇒30(40+x)×25=60×40×20 ⇒(40+x)×25=2×40×20 ⇒(40+x)×5=2×8×20 ⇒(40+x)=2×8×4=64 ⇒x=64−40=24




M1 = 40 D1 = 20 W1= 30 Let M 2 = x Let D 2 = 25 W2= 60 Here H 1 = H 2 Hence, M 1 D 1 W 1 =M 2 D 2 W 2 ⇒40×20 30 =x×25 60 ⇒2(40×20)=x×25 ⇒2(8×20)=x×5 ⇒x=2(8×4)=64 Hence, additional men required = 64 - 40 = 24


Important Formulas –

Mixtures and Allegations • Alligation

It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

• Mean Price Mean price is the cost price of a unit quantity of the mixture

• Suppose a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid = [x(1−y x ) n ] units.

• Rule of Alligation If two ingredients are mixed, then (Quantity of cheaper Quantity of dearer )=(C.P. of dearer - Mean Price Mean price - C.P. of cheaper )

Cost Price(CP) of a unit quantity of cheaper (c)

Cost Price(CP) of a unit quantity of dearer (d)

Mean Price


(m)

(d - m) (m - c)

• => (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)

1. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container? A. 26 litres B. 29.16 litres C. 28 litres D. 28.2 litres


Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid


=x(1−y x ) n Hence milk now contained by the container = 40(1−4 40 ) 3 =40(1−1 10 ) 3 40×9 10 ×9 10 ×9 10 =4×9×9×9 100 =29.16

2. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ? A. Rs.182.50 B. Rs.170.5 C. Rs.175.50 D. Rs.180


Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135) 2 =130.5 Hence let's consider that the mixture is formed by mixing two varieties of tea. one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. Now let's find out x. By the rule of alligation, we can write as


Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea

130.50 x

Mean Price

153

(x - 153) 22.50

(x - 153) : 22.5 = 1 : 1 =>x - 153 = 22.50 => x = 153 + 22.5 = 175.5

3. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? A. 5litres, 7 litres B. 7litres, 4 litres C. 6litres, 6 litres D. 4litres, 8 litres


Answer : Option C

Explanation :

---------------------------------------- Solution 1 ----------------------------------------


Let x and (12-x) litres of milk be mixed from the first and second container respectively Amount of milk in x litres of the the first container = .75x Amount of water in x litres of the the first container = .25x Amount of milk in (12-x) litres of the the second container = .5(12-x) Amount of water in (12-x) litres of the the second container = .5(12-x) Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3 : 5

⇒(.25x+6−.5x) (.75x+6−.5x) =3 5 ⇒(6−.25x) (.25x+6) =3 5 ⇒30−1.25x=.75x+18 ⇒2x=12 ⇒x=6 Since x = 6, 12-x = 12-6 = 6 Hence 6 and 6 litres of milk should mixed from the first and second container respectively ---------------------------------------- Solution 2 ----------------------------------------

Let cost of 1 litre milk be Rs.1 Milk in 1 litre mix in 1st can = 3 4 litre Cost Price(CP) of 1 litre mix in 1st can


= Rs.3 4 Milk in 1 litre mix in 2nd can = 1 2 litre Cost Price(CP) of 1 litre mix in 2nd can = Rs.1 2 Milk in 1 litre of the final mix=5 8 Cost Price(CP) of 1 litre final mix = Rs.5 8 => Mean price = 5 8 By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can CP of 1 litre mix in 1st can

1/2 3/4

Mean Price

5/8

3/4 - 5/8 = 1/8 5/8 - 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1 ie, from each can, 1 2 ×12=6 litre should be taken

4. Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ? A. 3: 4 B. 4 : 3 C. 9 : 7 D. 7 : 9


Answer : Option D

Explanation :


Let Cost Price(CP) of 1 litre spirit be Rs.1 Quantity of spirit in 1 litre mixture from vessel A = 5/7 Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7 Quantity of spirit in 1 litre mixture from vessel B = 7/13 Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13 Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13 Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price By the rule of alligation, we can write as

CP of 1 litre mixture from vessel A

CP of 1 litre mixture from vessel B

5/7 7/13

Mean Price

8/13

8/13 - 7/13 = 1/13 5/7 - 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

5. The cost of Type 1 material is Rs. 15 per kg and Type 2


material is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then what is the price per kg of the mixed variety of material? A. Rs. 19 B. Rs. 16 C. Rs. 18 D. Rs. 17


Answer : Option C

Explanation :

--------------------------------- Solution 1 --------------------------------- Cost Price(CP) of Type 1 material is Rs. 15 per kg Cost Price(CP) of Type 2 material is Rs. 20 per kg Type 1 and Type 2 are mixed in the ratio of 2 : 3

Hence Cost Price(CP) of the resultant mixture = (15×2)+(20×3) (2+3) =(30+60) 5 =90 5 =18 => rice per kg of the mixed variety of material = Rs.18 --------------------------------- Solution 2 --------------------------------- Cost Price(CP) of Type 1 material is Rs. 15 per kg Cost Price(CP) of Type 2 material is Rs. 20 per kg


Let Cost Price(CP) of resultant mixture be Rs.x per kg By the rule of alligation, we have

CP of Type 1 material CP of Type 2 material

15 20

Mean Price

x

(20-x) (x-15)

=> Type 1 material : Type 2 material = (20-x) : (x-15) Given that Type 1 material : Type 2 material = 2 : 3 => (20-x) : (x-15) = 2 : 3

⇒(20−x) (x−15) =2 3 ⇒3(20−x)=2(x−15) ⇒60−3x=2x−30 ⇒90=5x =>x=90 5 =18

=>price per kg of the mixed variety of material = Rs.18

6. Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg. A. 4 : 3 B. 3 : 4 C. 2 : 3 D. 3 : 2



Answer : Option C

Explanation :

CP of 1Kg 1st kind rice = Rs.7.20 CP of 1Kg 2nd kind rice = Rs.5.70 CP of 1Kg mixed rice = Rs.6.30 By the rule of alligation, we have

CP of 1Kg 1st kind rice CP of 1Kg 2nd kind rice

7.2 5.7

Mean Price

6.3

6.3 - 5.7 = .6 7.2 - 6.3 = .9

Required Ratio = .6 : .9 = 6:9 = 2:3

7. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine did the cask originally hold? A. 30 litres B. 26 litres C. 24 litres D. 32 litres



Answer : Option C

Explanation :

Let initial quantity of wine = x litre

After a total of 4 operations, quantity of wine = x(1−y x ) n =x(1−8 x ) 4

Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

Hence we can write as x(1−8 x ) 4 x =16 81 ⇒(1−8 x ) 4 =(2 3 ) 4 ⇒(1−8 x )=2 3 ⇒(x−8 x )=2 3 ⇒3x−24=2x ⇒x=24

8. A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is A. 4 3 B. 3 4 C. 3 2 D. 2 3


Answer : Option D

Explanation :


Concentration of alcohol in 1st Jar = 40% Concentration of alcohol in 2nd Jar = 19% After the mixing, Concentration of alcohol in the mixture = 26% By the rule of alligation,

Concentration of alcohol in 1st Jar

Concentration of alcohol in 2nd Jar

40% 19%

Mean

26%

7 14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

9. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ? A. 60 Kg B. 63 kg C. 58 Kg D. 56 Kg


Answer : Option B

Explanation :


Selling Price(SP) of 1 Kg mixture= Rs. 9.24 Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100 (100+Profit%) ×SP=100 (100+10) ×9.24 =100 110 ×9.24=92.4 11 =Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind CP of 1 kg sugar of 2nd kind

Rs. 9 Rs. 7

Mean Price

Rs.8.4

8.4 - 7 = 1.4 9 - 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the ratio 1.4 : .6 = 14 : 6 = 7 : 3 Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar then x : 27 = 7 : 3

⇒x 27 =7 3 ⇒x=27×7 3 =63

10. In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per


Kg ? A. 7 : 8 B. 8 : 7 C. 6 : 7 D. 7 ; 6


Answer : Option B

Explanation :


Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind

9.3 10.80

Mean Price

10

10.8-10 = .8 10 - 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

11. In what ratio must tea worth Rs. 60 per kg be mixed with tea worth Rs. 65 a Kg such that by selling the mixture at Rs. 68.20 a Kg ,there can be a gain 10%? A. 3 : 2 B. 2 : 3 C. 4 : 3 D. 3 : 4


Answer : Option A


Explanation :

Cost Price(CP) of 1 Kg mixture = Rs. 68.20 Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100 (100+Profit%) ×SP=100 (100+10) ×68.20 =100 110 ×68.20=682 11 =Rs.62

By the rule of alligation:

CP of 1 kg tea of 1st kind CP of 1 kg tea of 2nd kind

60 65

Mean Price

62

65 - 62 = 3 62 - 60 = 2

Hence required ratio = 3 : 2

12. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? A. 23 B. 21 C. 19 D. 17



Answer : Option B

Explanation :

Let's initial quantity of P in the container be 7x and initial quantity of Q in the container be 5x Now 9 litres of mixture are drawn off from the container Quantity of P in 9 litres of the mixtures drawn off = 9×7 12 =63 12 =21 4 Quantity of Q in 9 litres of the mixtures drawn off = 9×5 12 =45 12 =15 4 Hence Quantity of P remains in the mixtures after 9 litres is drawn off =7x−21 4 Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15 4 Since the container is filled with Q after 9 litres of mixture is drawn off, Quantity of Q in the mixtures=5x−15 4 +9=5x+21 4 Given that the ratio of P and Q becomes 7 : 9 ⇒(7x−21 4 ):(5x+21 4 )=7:9 ⇒(7x−21 4 ) (5x+21 4 ) =7 9 63x−(9×21 4 )=35x+(7×21 4 ) 28x=(16×21 4 ) x=(16×21 4×28 ) litres of P contained in the container initially = 7x=(7×16×21 4×28 )=16×21 4×4 =21

13. A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup? A. 1 3 B. 1 4 C. 1 5 D. 1 6



Answer : Option C

Explanation :

Let the quantity of the initial liquid in the vessel = 8 litre and quantity of water in the initial liquid = 3 litre and quantity of syrup in the initial liquid = 5 litre Let x litre of the mixture is drawn off and replaced with water Quantity of water in the new mixture = 3−3x 8 +x Quantity of syrup in the new mixture = 5−5x 8 Given that in the new mixture, quantity of water = quantity of syrup ⇒3−3x 8 +x=5−5x 8 ⇒10x 8 =2 ⇒5x 4 =2 ⇒x=8 5 ⇒8 5 litre

Initially we assumed that the quantity of the initial liquid in the vessel = 8 litre for the ease of calculations. For that 8/5 litre of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup Now, if the initial liquid in the vessel = 1 litre, quantity of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup

=8 5 ×1 8 =1 5

It means 1/5 of the mixture has to be drawn off and replaced with water so that the mixture


may be half water and half syrup

14. In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre? A. 1 : 3 B. 2 : 2 C. 1 : 2 D. 3 : 1


Answer : Option C

Explanation :


Cost Price of 1 litre of water Cost Price of 1 litre of milk

0 12

Mean Price

8

12-8=4 8-0=8

Required Ratio = 4 : 8 = 1 : 2

15. In what ratio must tea at Rs.62 per Kg be mixed with tea at Rs. 72 per Kg so that the mixture must be worth Rs. 64.50 per Kg?


A. 1 : 2 B. 2 : 1 C. 3 : 1 D. 1 : 3


Answer : Option C

Explanation :


Cost of 1 kg of 1st kind tea Cost of 1 kg tea of 2nd kind tea

62 72

Mean Price

64.5

72-64.5=7.5 64.5-62=2.5

Required Ratio = 7.5 : 2.5 = 3 : 1

16. In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs. 20 per kg respectively to obtain a mixture worth Rs.16.50 per Kg? A. 1 : 2 B. 2 : 1 C. 3 : 7 D. 7 : 3


Answer : Option D


Explanation :

By the rule of alligation,we have

CP of 1 kg of 1st variety pulse CP of 1 kg of 2nd variety pulse

15 20

Mean Price

16.5

20-16.5 = 3.5 16.5-15=1.5

Required Ratio = 3.5 : 1.5 = 35 : 15 = 7 : 3

17. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is A. 300 B. 400 C. 600 D. 500


Answer : Option C

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part Profit% by selling 2nd part

8 18


Net % profit

14

18-14=4 14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3 Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3 5 =600Kg

18. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture? A. 25% B. 20% C. 22% D. 24%


Answer : Option B

Explanation :

----------------------------------------------------------------------------

Solution 1 - Using the concepts learned in Profit and Loss

----------------------------------------------------------------------------- If a trader professes to sell his goods at cost price, but uses false weights, then Gain%=[Error (True Value−Error) ×100]% Here Gain= 25% Here error = quantity of water he mixes in the milk


= x Here the true value = true quantity of milk = T So the formula becomes 25 = x (T−x) ×100 ⇒1=x (T−x) ×4 ⇒T−x=4x ⇒T=5x percentage of water in the mixture = x T ×100=x 5x ×100=1 5 ×100=20%

--------------------------------------------------------------------------- Solution 2 –

Using the concepts learned in Mixtures and Alligations

---------------------------------------------------------------------------Let CP of 1 litre milk = Rs.1 Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1 Given than Gain = 25% Hence CP of 1 litre mixture = 100 (100+Gain%) ×SP =100 (100+25) ×1=100 125 =4 5


CP of 1 litre milk CP of 1 litre water

1 0

CP of 1 litre mixture

4/5

4/5 - 0 = 4/5 1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1 5 ×100=20%

19. In what ratio must water be mixed with milk to


gain 162 3 % on selling the mixture at cost price? A. 6 : 1 B. 1 : 6 C. 1 : 4 D. 4 : 1


Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1 SP of 1 litre mixture = CP of 1 litre milk = Rs.1 Gain=162 3 %=50 3 % CP of 1 litre mixture = 100 (100+Gain%) ×SP =100 (100+50 3 ) ×1=100 (350 3 ) =300 350 =6 7


CP of 1 litre water CP of 1 litre milk

0 1

CP of 1 litre mixture

6/7

1 - 6/7 = 1/7 6/7 - 0 = 6/7

Quantity of water : Quantity of milk = 1/7 : 6/7 = 1 : 6

20. In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth Rs.8 per Kg? A. 5 : 4 B. 2 : 1 C. 3 : 2 D. 4 : 3



Answer : Option D

Explanation :


CP of 1 kg Rice of 1st kind CP of 1 kg Rice of 2nd kind

7.1 9.2

Mean Price

8

9.2 - 8 = 1.2 8 - 7.1 = .9

Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3

21. How many Kg of rice at Rs.6.60 per Kg. be mixed with 56Kg of rice at Rs.9.60 per Kg to get a mixture worth Rs.8.20 per Kg A. 56 Kg B. 52 Kg C. 44 Kg D. 49 Kg


Answer : Option D

Explanation :



Cost of 1 kg of 1st kind rice Cost of 1 kg of 2nd kind rice

6.6 9.6

Price of 1Kg of the mixture

8.2

9.6 - 8.2 = 1.4 8.2 - 6.6 = 1.6

Quantity of 1st kind rice : Quantity of 2nd kind rice = 1.4 : 1.6 = 7 : 8 Quantity of 1st kind rice : 56 = 7 : 8

=>Quantity of 1st kind rice = 56×7 8 =49

22. How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it A. 4 litre B. 2 litre C. 1 litre D. 3 litre


Answer : Option B

Explanation :


% Concentration of water in pure water : 100

% Concentration of water in the given mixture : 10


Mean % concentration

20

20 - 10 = 10 100 - 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8 Here Quantity of the mixture = 16 litres => Quantity of water : 16 = 1 : 8

Quantity of water = 16×1 8 =2 litre

23. We have a 630 ml of mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3? A. 70 ml B. 60 ml C. 80 ml D. 50 ml


Answer : Option A

Explanation :

concentration of water in mixture1 = 2/9 (Since the ratio of milk and water = 7:2) ---item(1) concentration of water in pure water= 1 ---item(2) Now the above mentioned items are mixed to form a mixture2 where milk and water ratio = 7: 3


=>concentration of water in mixture2 = 3/10 By the rule of alligation, we have

concentration of water in mixture1 : 2/9

concentration of water in pure water : 1

Mean concentration

3/10

1 - 3/10 = 7/10 3/10 - 2/9 = 7/90

=> Quantity of mixture1 : Quantity of water = 7/10 : 7/90 = 1/10 : 1/90 = 1 : 1/9 Given that Quantity of mixture1 = 630 ml => 630 : Quantity of water = 1 : 1/9

⇒Quantity of water = 630×1 9 =70 ml

24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is A. 25% B. 20% C. 30% D. 33%



Answer : Option D

Explanation :

----------------------------------------------------------

Method-1 to solve the question

----------------------------------------------------------

We have a 11 litre solution containing 42% of alcohol in the water => Quantity of alcohol in the solution= 11×42 100 Now 3 litre of water is added to the solution => Total Quantity of the new solution = 11+3=14 Percentage of alcohol in the new solution=11×42 100 14 ×100 =11×3 100 =33%

----------------------------------------------------------

Method-2 to solve the question

---------------------------------------------------------- %Concentration of alcohol in pure water = 0 %Concentration of alcohol in mixture = 42 quantity of water : Quantity of mixture = 3 : 11 Let the %Concentration of alcohol in the new mixture = x


%Concentration of alcohol : 0 in pure water

%Concentration of alcohol in mixture : 42

Mean %Concentration

x


42 - x x - 0 = x

But (42 - x) : x = 3 : 11

⇒42−x x =3 11 ⇒42−x x =3 11 ⇒42×11−11x=3x ⇒14x=42×11 ⇒x=3×11=33 => The percentage of alcohol in the new mixture is 33%

25. Rs.460 was divided among 41 boys and girls such that each boy Rs.12 and each girl got Rs.8. What is the number of boys? A. 33 B. 30 C. 36 D. 28


Answer : Option A

Explanation :

------------------------------------------------------------------- Solution 1 ------------------------------------------------------------------- Assume that the number of boys = b and number of girls is g number of boys + number of girls = 41 => b + g = 41 ------------ (Equation 1)


Given that each boy got Rs.12 and each girl got Rs.8 and Total amount = Rs.460 => 12b + 8g = 460 -------- (Equation 2) Now we need solve Equation 1 and Equation 2 to get b and g

(Equation1)×8=>8b+8g=8×41=328−−−−−−−−(Equation3)

(Equation 2) - (Equation 3) = 4b = 460 - 328 = 132

=>b=132 4 =33 ------------------------------------------------------------------- Solution 2 ------------------------------------------------------------------- Given that Amount received by a boy = Rs.12 and Amount received by a girl =Rs.8 Total amount = 460 Given that number of boys + Number of girls = 41 Hence mean amount = 460/41 By the rule of alligation, we have

Amount received by a boy Amount received by a


girl 12 8

Mean amount

460/41

460/41 - 8 =132/41 12 - 460/41 = 32/41

Number of boys : Number of girls = 132/41 : 32/41 = 132 : 32 = 66 : 16 = 33 : 8 Given that number of boys + Number of girls = 41

Hence number of boys = 41×33 41 =33

26. A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%. A. 1200 Kg B. 1400 Kg C. 1600 Kg D. 800 Kg


Answer : Option A

Explanation :


% Profit by selling part1 % Profit by selling part2

8 12


Net % Profit

11

12 - 11 = 1 11 - 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3 Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3 4 =1200

27. In 40 litres of a mixture the ratio of milk to water is 7:1. In order to make the ratio of milk to water as 3:1, the quantity of water that should be added to the mixture will be A. 52 3 litre B. 41 3 litre C. 62 3 litre D. 6 litre


Answer : Option C

Explanation :


Concentration of water in pure water : 1

Concentration of water in mixture : 1/8

Concentration of water in the final mixture

1/4


1/4 - 1/8 = 1/8 1 - 1/4 = 3/4

Quantity of water : Quantity of mixture = 1/8 : 3/4 = 1 : 6 Given that quantity of mixture = 40 litre =>Quantity of water : 40 = 1 : 6

⇒Quantity of water = 40×1 6 =62 3 litre

28. Some amount out Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent of 6% per annum was A. Rs. 2400 B. Rs. 2200 C. Rs. 2000 D. Rs. 1800


Answer : Option C

Explanation :

Total simple interest received , I = Rs.1600 Principal , p = 7000 period, n = 5 years Rate of interest, r = ?


Simple Interest, I = pnr 100 =>1600=7000×5×r 100 =>r=1600×100 7000×5 =160 35 =32 7 % By the rule of alligation, we have

Rate of interest % from part1 Rate of interest % from part2

6 4

Net rate of interest %

32/7

32/7 - 4 =4/7 6 - 32/7 = 10/7

=> Part1 : part2 = 4/7 : 10/7 = 4 : 10 = 2 : 5 Given that total amount = Rs.7000

The amount lent of 6% per annum (part1 amount) = 7000×2 7 =Rs.2000

29. In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10% A. 1.5 Kg B. 2 Kg C. .5 Kg D. 1 Kg


Answer : Option D


Explanation :


Percentage concentration of manganese in the mixture : 20

Percentage concentration of manganese in pure iron : 0

Percentage concentration of manganese in the final mixture

10

10 - 0 = 10 20 - 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1 Given that Quantity of the mixture = 1 Kg Hence Quantity of iron to be added = 1 Kg


Important Formulas - Pipes and Cistern • Inlet of a tank or a cistern or reservoir

Inlet is a pipe connected with a tank or cistern or reservoir. It is used to fill the tank.

• Outlet of a tank or a cistern or reservoir Outlet is a pipe connected with a tank or cistern or reservoir. It is used to empty the tank.

• If a pipe can fill a tank in x hours, part filled in 1 hour = 1 x

• If a pipe can empty the tank in y hours, part emptied in 1 hour = 1 y

• Let a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours. We can examine two cases here. Case 1: x < y In this case, the net part filled in 1 hour = 1 x –1 y Case 2: y < x In this case, the net part emptied in 1 hour = 1 y –1 x


1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: A. 39 17 hours B. 113 17 hours C. 28 11 hours D. 41 2 hours


Answer : Option A

Explanation :

Pipes A and B can fill a tank in 5 and 6 hours respectively =>Part filled by pipe A in hour = 1

⁄5 and Part filled by pipe B in hour = 1

⁄6 Pipe C can empty it in 12 hours => Part emptied by pipe C in 1 hour = 1

⁄12

Net part filled by Pipes A,B and C together in 1 hour = 1 5 +1 6 −1 12 =17 60 i.e, the pipe can be filled in 60 17 =39 17 hours

2. Two pipes A and B can fill a cistern in 371⁄2 minutes and 45


minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if pipe B is turned off after: A. 5 min B. 9 min C. 10 min D. 15 min


Answer : Option B

Explanation :

Pipe A can fill a tank in 371⁄2 minutes = 75⁄2 minutes

=>Part filled by pipe A in 1 minute = 2

⁄75 Pipe B can fill a tank in 45 minutes => Part filled by pipe B in 1 minute = 1

⁄45

=> Part filled by Pipe A and B in 1 minute = =2 75 +1 45 =6+5 225 =11 225 Assume that B is turned off after x minutes. i.e., for x minutes, both pipe A and B were open.

Part filled in x minutes by Pipe A and B = x×11 225 =11x 225 Now, the cistern must be filled in (30-x) minutes by pipe A alone

Part filled in (30-x) minutes by pipe A = (30−x)×2 75 =2(30−x) 75


11x 225 +2(30−x) 75 =1 11x 225 +2(30−x) 75 =1 ⇒11x+6(30−x)=225 ⇒11x+180−6x=225 ⇒5x=45 x=9

3. A pump can fill a tank with water in 2 hours. Because of a leak, it took 22⁄3 hours to fill the tank. The leak can drain all the water of the tank in: A. 6 hours B. 8 hours C. 9 hours D. 10 hours


Answer : Option B

Explanation :

Let the leak can drain all the water of the tank in y hours Part of the tank filled by the pipe in 1 hr = 1/2 Part of the tank emptied by the leak in 1 hr = 1/y

1 2 −1 y =3 8 ⇒1 y =1 2 −3 8 =1 8 ⇒y=8 i.e., the leak can drain all the water of the tank in 8 hours

4. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is


the proportion of the solution R in the liquid in the tank after 3 minutes? A. 6 11 B. 5 11 C. 7 11 D. 8 11


Answer : Option A

Explanation :

Part of the tank filled by pipe A in 1 minute = 1⁄30

Part of the tank filled by pipe B in 1 minute = 1

⁄20 Part of the tank filled by pipe C in 1 minute = 1

⁄10 Here we have to find the proportion of the solution R. Pipe C discharges chemical solution R

Part of the tank filled by pipe C in 3 minutes = 3×1 10 =3 10

Part of the tank filled by pipe A,B and C together in 1 minute = 1 30 +1 20 +1 10 =11 60 Part of the tank filled by pipe A,B and C together in 3 minute = 3×11 60 =11 20

Required proportion = (3 10 ) (11 20 ) =(3×20) (10×11) =6 11

5. A tank is filled by three pipes with uniform flow. The first


two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is: A. 30 hours B. 15 hours C. 10 hours D. 6 hours


Answer : Option B

Explanation :

Let the first pipe alone can fill the tank in x hours Then the second pipe can fill the tank in (x-5) hours and the third pipe can fill the tank in (x-5)-4 = (x-9) hours part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr

1 x +1 x−5 =1 x−9 (From here, better to find the value of x from the choices which will be easier. Or we can solve it as given below)

(x−5)(x−9)+x(x−9)=x(x−5) x 2 −14x+45+x 2 −9x=x 2 −5x −14x+45+x 2 −9x=−5x x 2 −18x+45=0 (x−15)(x−3)=0 x=15 or 3


We can not take the value of x = 3 because, (x-9) becomes negative which is not possible because the third pipe can fill the tank in (x-9) hours Hence, x = 15

6. Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately? A. 6 hours B. 2 hours C. 4 hours D. 3 hours


Answer : Option A

Explanation :

Let pipe A alone can fill the tank in x hours Then pipe B can fill the tank in (x+6) hours

Part filled by pipe A in 1 hr = 1 x Part filled by pipe B in 1 hr = 1 x+6 Part filled by pipe A and pipe B in 1 hr = 1 x +1 x+6 It is given that pipes A and B together can fill the cistern in 4


hours. i.e., Part filled by pipes A and B in 1 hr = 1/4

1 x +1 x+6 =1 4 (From here, better to find the value of x from the choices which will be easier. Or we can solve it as given below)

4(x+6)+4x=x(x+6) 4x+24+4x=x 2 +6x x 2 −2x−24=0 (x−6)(x+4)=0 x=6 or -4 Since x can not be negative, x = 6 i.e.,pipe A alone can fill the tank in 6 hours

7. Two pipes A and B can fill a tank in 12 and 24 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank? A. 9 min B. 8 min C. 6 min D. 4 min


Answer : Option B

Explanation :

Part filled by pipe A in 1 minute= 1 12 Part filled by pipe B in 1 minute= 1 24 Part filled by pipe A and pipe B in 1


minute= 1 12 +1 24 =1 8 i.e., both the pipe together can fill the tank in 8 minutes

8. Two pipes A and B can fill a tank in 15 minutes and 40 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank? A. 10 min 10 sec B. 25 min 20 sec C. 14 min 40 sec D. 20 min 10 sec


Answer : Option B

Explanation :

Part filled by pipe A in 1 minute= 1 15 Part filled by pipe B in 1 minute= 1 40 Part filled by pipe A and pipe B in 1 minute= 1 15 +1 40 =11 120 pipe A and pipe B were open for 4 minutes.

Part filled by pipe A and pipe B in these 4 minutes = 4×11 120 =11 30 Remaining part to be filled = 1−11 30 =19 30 Time taken by pipe B to fill this remaining part = (19 30 ) (1 40 ) =19×40 30 =19×4 3 =76 3 =251 3 minutes = 25 minutes 20 seconds


9. Two pipes can fill a tank in 25 and 30 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is: A. 250 gallons B. 450 gallons C. 120 gallons D. 150 gallons


Answer : Option B

Explanation :

Part filled by first pipe in 1 minute= 1 25 Part filled by second pipe in 1 minute= 1 30 Let the waste pipe can empty the full tank in x minutes

Then, part emptied by waste pipe in 1 minute= 1 x All the three pipes can fill the tank in 15 minutes

i.e., part filled by all the three pipes in 1 minute= 1 15

⇒1 25 +1 30 −1 x =1 15 ⇒1 x =1 25 +1 30 −1 15 =6+5−10 150 =1 150 ⇒x=150 i.e, the waste pipe can empty the full tank in 150 minutes Given that waste pipe can empty 3 gallons per minute ie, in 150 minutes, it can empty 150 × 3 = 450 gallons


Hence, the volume of the tank = 450 gallons

10. A tank is filled in 10 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank? A. 70 hours B. 30 hours C. 35 hours D. 50 hours


Answer : Option A

Explanation :

Let the pipe A can fill the tank in x hours

Then pipe B can fill the tank in x 2 hours and pipe C can fill the tank in x 4 hours

Part filled by pipe A in 1 hour = 1 x Part filled by pipe B in 1 hour = 2 x Part filled by pipe C in 1 hour = 4 x Part filled by pipe A, pipe B and pipe C in 1 hour = 1 x +2 x +4 x =7 x i.e., pipe A, pipe B and pipe C can fill the tank in x 7 hours Given that pipe A, pipe B and pipe C can fill the tank in 10 hours

=>x 7 =10 ⇒x=10×7=70 hours


11. One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in: A. 180 min B. 144 min. C. 126 min D. 114 min


Answer : Option A

Explanation :

Let the slower pipe alone can fill the tank in x minutes Then the faster pipe can fill the tank in x 4 minutes Part filled by the slower pipe in 1 minute = 1 x Part filled by the faster pipe in 1 minute = 4 x Part filled by both the pipes in 1 minute = 1 x +4 x

It is given that both the pipes together can fill the tank in 36 minutes ⇒ Part filled by both the pipes in 1 minute = 1 36

1 x +4 x =1 36 5 x =1 36 x=5×36=180 i.e.,the slower pipe alone fill the tank in 180 minutes

12. A tap can fill a tank in 4 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely? A. 3 hr B. 1 hr 30 min C. 2 hr 30 min D. 2 hr



Answer : Option C

Explanation :

A tap can fill a tank in 4 hours => The tap can fill half the tank in 2 hours

Remaining part = 1 2 After half the tank is filled, three more similar taps are opened. Hence, total number of taps becomes 4.

Part filled by one tap in 1 hour = 1 4 Part filled by four taps in 1 hour = 4×1 4 =1 i.e., 4 taps can fill remaining half in 30 minutes Total time taken = 2 hour + 30 minute = 2 hour 30 minutes

13. A tap can fill a tank in 4 hours. After half the tank is filled, two more similar taps are opened. What is the total time taken to fill the tank completely? A. 1 hr 20 min B. 4 hr C. 3 hr D. 2 hr 40 min


Answer : Option D


Explanation :

A tap can fill a tank in 4 hours => The tap can fill half the tank in 2 hours

Remaining part = 1 2 After half the tank is filled, two more similar taps are opened. Hence, total number of pipes becomes 3.

Part filled by one tap in 1 hour = 1 4 Part filled by three taps in 1 hour = 3×1 4 =3 4 Time taken to fill1 2 the tank by 3 pipes = (1 2 ) (3 4 ) =4 6 ==2 3 hour = 40 minutes Total time taken = 2 hour + 40 minute = 2 hour 40 minutes

14. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: A. 10 B. 12 C. 14 D. 16


Answer : Option C

Explanation :


A, B and C can fill a tank in 6 hours.

⇒Part filled by pipes A,B and C in 1 hr = 1 6 All these pipes are open for only 2 hours and then C is closed.

Part filled by pipes A,B and C in these 2 hours = 2 6 =1 3 Remaining part = 1−1 3 =2 3

This remaining part of 2 3 is filled by pipes A and B in 7 hours ⇒Part filled by pipes A and B in 1 hr=(2 3 ) 7 −2 21

Part filled by pipe C in 1 hr = (1 6 −2 21 )=7−4 42 =3 42 =1 14 i.e., C alone can fill the tank in 14 hours

15. A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half? A. 15 min B. 20 min C. 27.5 min D. 30 min


Answer : Option D

Explanation :

Part filled by pipe A in 1 minute = 1 60 Part filled by pipe B in 1 minute = 1 40 Part filled by both pipes A and pipe B in 1


minute = 1 60 +1 40 =2+3 120 =5 120 =1 24 Suppose the tank is filled in x minutes Then, To fill the tanker from empty state, B is used for x/2 minutes and A and B is used for the rest x/2 minutes

x 2 ×1 40 +x 2 ×1 24 =1 ⇒x 2 [1 40 +1 24 ]=1 ⇒x 2 ×8 120 =1 ⇒x 2 ×1 15 =1 x=15×2=30 minutes

16. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in: A. 62 3 hours B. 6 hours C. 71 2 hours D. 7 hours


Answer : Option D

Explanation :

Part filled by pipe A in 1 hour = 1 12 Part filled by pipe B in 1 hour = 1 15 Part filled by pipe C in 1 hour = 1 20 In first hour, A and B is open In second hour, A and C is open


then this pattern goes on till the tank fills

Part filled by pipe A and pipe B in 1 hour = 1 12 +1 15 =9 60 =3 20 Part filled by pipe A and pipe C in 1 hour = 1 12 +1 20 =8 60 =2 15 Part filled in 2 hour = 3 20 +2 15 =17 60 Part filled in 6 hour =17 60 ×3=17 20 Remaining part = (1−17 20 )=3 20

Now, 6 hours are over and only 3 20 part needed to be filled. At this 7th hour, A and B is open Time taken by pipe A and B to fill this 3 20 part = (3 20 ) (3 20 ) = 1 hour Total time taken = 6 hour + 1 hour = 7 hour

17. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 24 hours. How many liters does the cistern hold? A. 4010 litre B. 2220 litre C. 1920 litre D. 2020 litre


Answer : Option C

Explanation :

Part emptied by the leak in 1 hour = 1 6 Net part emptied by the leak and the inlet pipe in 1 hour = 1 24 Part filled by the


inlet pipe in 1 hour = 1 6 −1 24 =1 8 i.e., inlet pipe fills the tank in 8 hours = (8 × 60) minutes = 480 minutes Given that the inlet pipe fills water at the rate of 4 liters a minute Hence, water filled in 480 minutes = 480 × 4 = 1920 litre i.e., The cistern can hold 1920 litre

18. A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled? A. 4.8 hr B. 2.4 hr C. 3.6 hr D. 1.8 hr


Answer : Option A

Explanation :

Part filled by the first tap in 1 hour = 1 3 Part emptied by the second tap 1 hour = 1 8 Net part filled by both these taps in 1 hour = 1 3 −1 8 =5 24 i.e, the cistern gets filled in 24 5 hours = 4.8 hours


19. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 40 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank? A. 28 hr B. 16 hr C. 22 hr D. 32 hr


Answer : Option A

Explanation :

Part filled by pipe A in 1 hour = 1 5 Part filled by pipe B in 1 hour = 1 20 Part filled by pipe A and B in 1 hour = 1 5 +1 20 =1 4 i.e., A and B together can fill the tank in 4 hours Given that due to the leakage, it took 40 minutes more to fill the tank.

i.e., due to the leakage, the tank got filled in 440 60 hour=42 3 hour=14 3 hour ⇒Net part filled by pipe A and B and the leak in 1 hour = 3 14 ⇒Part emptied by the leak in 1 hour =1 4 −3 14 =1 28 i.e., the leak can empty the tank in 28 hours

20. Bucket P has thrice the capacity as bucket Q. It takes 60


turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum? A. 30 B. 45 C. 60 D. 80


Answer : Option C

Explanation :

Let capacity of bucket P = x

Then capacity of bucket Q = x 3 Given that it takes 80 turns for bucket P to fill the empty drum => capacity of the drum = 80x

Number of turns required if both P and Q are used = 80x x+x 3 =240x 3x+x =240 4 =60

21. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying of the tank is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump? A. 20 m3 / min. B. 40 m3 / min. C. 50 m3 / min. D. 60 m3 / min.



Answer : Option C

Explanation :

Let the filling capacity of the pump = x m3 / min. Then the emptying capacity of the pump = (x + 10) m3 / min.

Time required for filling the tank = 2400 x minutes Time required for emptying the tank = 2400 x+10 minutes Pump needs 8 minutes lesser to empty the tank than it needs to fill it

⇒2400 x −2400 x+10 =8 ⇒300 x −300 x+10 =1 ⇒300(x+10)−300x=x(x+10) ⇒3000=x 2 +10x ⇒x 2 +10x−3000=0 (x+60)(x−50)=0 x = 50 or -60 Since x can not be negative, x=50 i.e.,filling capacity of the pump = 50 m3 / min.

22. Two pipes A and B can separately fill a cistern in 40 minutes and 30 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 20 minutes. In how much time, the third pipe alone can empty the cistern? A. 120 min B. 100 min C. 140 min D. 80 min



Answer : Option A

Explanation :

Part filled by pipe A in 1 minute = 1 40 Part filled by pipe B in 1 minute = 1 30 Net part filled by pipe A, pipe B and the third pipe in 1 hour = 1 20

⇒Part emptied by the third pipe in 1 hour = 1 40 +1 30 −1 20 =3+4−6 120 =1 120 i.e., third pipe alone can empty the cistern in 120 minutes

23. Two pipes A and B can fill a tank in 9 hours and 3 hours respectively. If they are opened on alternate hours and if pipe A is opened first, how many hours, the tank shall be full? A. 4 hr B. 5 hr C. 2 hr D. 6 hr


Answer : Option B

Explanation :

Part filled by pipe A in 1 hour = 1 9 Part filled by pipe B in 1 hour = 1 3 Pipe A and B are opened alternatively.


Part filled in every 2 hour = 1 9 +1 3 =1+3 9 =4 9 Part filled in 4 hour = 2×4 9 =8 9 remaining part =1−8 9 =1 9 Now it is pipe A's turn.

Time taken by pipe A to fill the remaining 1 9 part =(1 9 ) (1 9 ) =1 hour Total time taken = 4 hour + 1 hour = 5 hour

24. 13 buckets of water fill a tank when the capacity of each bucket is 51 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 17 litres? A. 33 B. 29 C. 39 D. 42


Answer : Option C

Explanation :

--------------------------------------------- solution 1 -------------------------------------------- Capacity of the tank = (13 × 51) litre

Number of buckets required of capacity of each bucket is 17 litre = 13×51 17 =13×3=39 ---------------------------------------------


solution 2 (using principles of chain rule) -------------------------------------------- Let x be the number of buckets needed if the capacity of each bucket is 17 litres More capacity, less buckets (Indirect proportion) Capacity 51 : 17 :: x : 13 => 51 × 13 = 17 × x => 3 × 13 = x => x = 39

25. Pipe A can fill a tank in 8 hours, pipe B in 4 hours and pipe C in 24 hours. If all the pipes are open, in how many hours will the tank be filled? A. 2.4 hr B. 3 hr C. 4 hr D. 4.2 hr


Answer : Option A

Explanation :

Part filled by pipe A in 1 hour = 1 8 Part filled by pipe B in 1 hour = 1 4 Part filled by pipe C in 1 hour = 1 24 Part filled by pipe A, pipe B and pipe C in 1 hour


= 1 8 +1 4 +1 24 =10 24 i.e, pipe A, pipe B and pipe C together can fill the tank in 24 10 hours = 2.4 hours

26. Two pipes A and B can fill a tank is 8 minutes and 14 minutes respectively. If both the taps are opened simultaneously, and the tap A is closed after 3 minutes, then how much more time will it take to fill the tank by tap B? A. 6 min 15 sec B. 5 min 45 sec C. 5 min 15 sec D. 6 min 30 sec


Answer : Option B

Explanation :

Part filled by pipe A in 1 minute = 1 8 Part filled by pipe B in 1 minute = 1 14 Part filled by pipe A and pipe B in 1 minute = 1 8 +1 14 =11 56 Pipe A and pipe B were open for 3 minutes

Part filled by pipe A and pipe B in 3 minutes = 3×11 56 =33 56 Remaining part = 1−33 56 =23 56 Time taken by pipe B to fill this remaining part = (23 56 ) (1 14 ) =23×14 56 =23 4 minutes=53 4 minutes = 5 minutes 45 seconds

27. A water tank is two-fifth full. Pipe A can fill a tank in 12


minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely? A. 2.8 min B. 4.2 min C. 4.8 min D. 5.6 min


Answer : Option C

Explanation :

Since pipe B is faster than pipe A, the tank will be emptied.

Part filled by pipe A in 1 minute = 1 12 Part emptied by pipe B in 1 minute = 1 6 Net part emptied by pipe A and pipe B in 1 minute = 1 6 −1 12 =1 12 Time taken to empty 2 5 of the tank = (2 5 ) (1 12 ) =2×12 5 =4.8 min

28. Pipes A and B can fill a tank in 8 and 24 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in: A. 18 hr B. 6 hr C. 24 hr D. 12 hr


Answer : Option D

Explanation :


Part filled by pipe A in 1 minute = 1 8 Part filled by pipe B in 1 minute = 1 24 Part emptied by pipe C in 1 minute = 1 12 Net part filled by pipe A, pipe B and pipe C in 1 minute = 1 8 +1 24 −1 12 =2 24 =1 12 i.e, the tank will be filled in 12minutes

29. One pipe can fill a tank 6 times as fast as another pipe. If together the two pipes can fill the tank in 22 minutes, then the slower pipe alone will be able to fill the tank in: A. 164 min B. 154 min C. 134 min D. 144 min


Answer : Option B

Explanation :

Let the slower pipe alone can fill the tank in x minutes Then the faster pipe can fill the tank in x 6 minutes Part filled by the slower pipe in 1 minute = 1 x Part filled by the faster pipe in 1 minute = 6 x Part filled by both the pipes in 1 minute = 1 x +6 x

It is given that both the pipes together can fill the tank in 22 minutes ⇒ Part filled by both the pipes in 1 minute = 1 22

1 x +6 x =1 22 7 x =1 22 x=22×7=154 i.e.,the slower pipe alone fill the tank in 154 minutes


30. Two pipes A and B can fill a tank in 2 and 6 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank? A. 3 min B. 2.5 min C. 2 min D. 1.5 min


Answer : Option D

Explanation :

Part filled by the first pipe in 1 minute = 1 2 Part filled by the second pipe in 1 minute = 1 6 Net part filled by pipe A and pipe B in 1 minute = 1 2 +1 6 =2 3 i.e, pipe A and B together can fill the tank in 3/2 minutes = 1.5 minutes


Important Formulas - Time and Distance • Speed=Distance Time • Distance=Speed×Time • Time=Distance Speed • To convert Kilometers per Hour(km/hr) to Meters per

Second(m/s) x km/hr=x×5 18 m/s

• To convert Meters per Second(m/s) to Kilometers per Hour(km/hr) x m/s=x×18 5 km/hr

• If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy x+y kmph

• Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝ 1 Time (when distance is constant)

• If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1 a :1 b or b : a

1. A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is


A. 11 hrs B. 8 hrs 45 min C. 7 hrs 45 min D. 9 hts 20 min


Answer : Option C

Explanation :

---------------------------------------------------------------- Solution 1 ---------------------------------------------------------------- Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way - 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot


of time which will be a real benefit for you. ---------------------------------------------------------------- Solution 2 -----------------------------------------------------------------

Let the distance be x km. Then (Time taken to walk x km) + (Time taken to ride x km) = 5 hour 45 min =545 60 hour = 53 4 hour = 23 4 hour .......(Equation1) (Time taken to ride 2x km) = 5 hour 45 min - 2 = 3 hour 45 min =345 60 hour = 33 4 hour = 15 4 hour .......(Equation2) Solving equations 1 and 2 (Equation 1 )×2⇒(Time taken to walk 2x km)+ (Time taken to ride 2x km) = 23 2 hour .......(Equation3) (Equation 3 - Equation 2)⇒ Time taken to walk 2x km=23 2 −15 4 =46 4 −15 4 =31 4 hours = 73 4 hours = 7 hours 45 minutes

2. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? A. 8.2 B. 4.2 C. 6.1 D. 7.2


Answer : Option D

Explanation :


distance = 600 meter time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance time =600 300 =2m/s =2×18 5 km/hr=36 5 km/hr=7.2 km/hr

3. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? A. 12 B. 11 C. 10 D. 9


Answer : Option C

Explanation :

speed of the bus excluding stoppages = 54 kmph speed of the bus including stoppages = 45 kmph Loss in speed when including stoppages = 54 - 45 = 9kmph => In 1 hour, bus covers 9 km less due to stoppages


Hence, time that the bus stop per hour = time taken to cover 9 km

=distance speed =9 54 hour=1 6 hour = 60 6 min=10 min

4. A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km. A. 121 km B. 242 km C. 224 km D. 112 km


Answer : Option C

Explanation :

distance = speed x time Let time taken to travel the first half = x hr then time taken to travel the second half = (10 - x) hr Distance covered in the the first half = 21x Distance covered in the the second half = 24(10 - x)


But distance covered in the the first half = Distance covered in the the second half => 21x = 24(10 - x) => 21x = 240 - 24x => 45x = 240 => 9x = 48 => 3x = 16

⇒x=16 3 Hence Distance covered in the the first half = 21x=21×16 3 =7×16=112 km Total distance = 2×112=224 km

5. A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car? A. 30 km/hr B. 35 km/hr C. 25 km/hr D. 40 km/hr


Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40 60 hr+48 3600 hr=1+2 3 +1 75 =126 75 hr distanc


e = 42 km speed=distance time =42 (126 75 ) =42×75 126 ⇒5 7 of the actual speed = 42×75 126 ⇒actual speed = 42×75 126 ×7 5 =42×15 18 =7×15 3 =7×5=35 km/hr

6. A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km? A. 36 B. 38 C. 40 D. 42


Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph Time taken when moving at normal speed - time taken when moving 3 kmph faster = 40 minutes

⇒x v −x v+3 =40 60 ⇒x[1 v −1 v+3 ]=2 3 ⇒x[v+3−v v(v+3) ]=2 3 ⇒2v(v+3)=9x................(Equation1)

Time taken when moving 2 kmph slower - Time taken when moving at normal speed = 40


minutes

⇒x v−2 −x v =40 60 ⇒x[1 v−2 −1 v ]=2 3 ⇒x[v−v+2 v(v−2) ]=2 3 ⇒x[2 v(v−2) ]=2 3 ⇒x[1 v(v−2) ]=1 3 ⇒v(v−2)=3x................(Equation2) Equation1 Equation2 ⇒2(v+3) (v−2) =3 ⇒2v+6=3v−6 ⇒v=12 Substituting this value of v in Equation 1⇒2×12×15=9x =>x=2×12×15 9 =2×4×15 3 =2×4×5=40 Hence distance = 40 km

7. A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.? A. 5 B. 6 C. 7 D. 8


Answer : Option C

Explanation :

Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions) = 2 + 3 = 5 rounds per hour


=> They cross each other 5 times in 1 hour and 2 times in 1/2 hour Time duration from 8 am to 9.30 am = 1.5 hour Hence they cross each other 7 times before 9.30 am

8. Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart? A. 17 hr B. 14 hr C. 12 hr D. 19 hr


Answer : Option A

Explanation :

Relative speed = 5.5 - 5 = .5 kmph (because they walk in the same direction) distance = 8.5 km

time = distance speed =8.5 .5 =17 hr

9. In covering a distance of 30 km, Arun takes 2 hours more


than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun's speed? A. 8 kmph B. 5 kmph C. 4 kmph D. 7 kmph


Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph distance = 30 km

We know that distance speed =time Hence 30 x −30 y =2...........(Equation1) 30 y −30 2x =1...........(Equation2) Equation1 + Equation2⇒30 x −30 2x =3 ⇒30 2x =3 ⇒15 x =3 ⇒5 x =1 ⇒x=5 Hence Arun's speed = 5 kmph

10. A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour? A. 70.24 km/hr B. 74. 24 km/hr C. 71.11 km/hr D. 72.21 km/hr



Answer : Option C

Explanation :

------------------------------------------- Solution 1 (Quick) --------------------------------------------

If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy x+y kmph. By using the same formula, we can find out the average speed quickly average speed = 2×64×80 64+80 =2×64×80 144 =2×32×40 36 =2×32×10 9 =64×10 9 =71.11 kmph

------------------------------------------- Solution 2 (Fundamentals) -------------------------------------------- Car travels first 160 km at 64 km/hr

Time taken to travel first 160 km = distance speed =160 64

Car travels next160 km at 80 km/hr

Time taken to travel next 160 km = distance speed =160 80 Total distance traveled = 160+160=2×160 Total time taken = 160 64 +160 80 Average speed = Total distance traveled Total time


taken =2×160 160 64 +160 80 =2 1 64 +1 80 =2×64×80 80+64 =2×64×80 144 =2×8×80 18 =640 9 =71.11 km/hr

11. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot? A. 12 km B. 14 km C. 16 km D. 18 km


Answer : Option C

Explanation :

----------------------------------------------------------------------- Solution 1 ----------------------------------------------------------------------- Let the time in which he travelled on foot = x hr Then the time in which he travelled on bicycle = (9 - x) hr distance = speed x time

⇒4x+9(9−x)=61 ⇒4x+81−9x=61 ⇒5x=20 ⇒x=4 ⇒ distance travelled on foot = 4x=4×4=16 km

------------------------------------------- Solution 2 -------------------------------------------


let the distance he travelled on foot = x km Then the distance he travelled on bicycle = (61-x) km

Time = Distance Speed ⇒x 4 +(61−x) 9 =9 ⇒9x+4×61−4x=36×9 ⇒5x+244=324 ⇒5x=324−244=80 ⇒x=80 5 =16 km

12. Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance? A. 1 hr 42 min B. 1 hr C. 2 hr D. 1 hr 12 min


Answer : Option D

Explanation :

New speed = 6/7 of usual speed Speed and time are inversely proportional. Hence new time = 7/6 of usual time


Hence, 7/6 of usual time - usual time = 12 minutes => 1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

13. A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office? A. 3 km B. 4 km C. 5 km D. 6 km


Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy x+y kmph Hence, average speed = 2×3×2 2+3 =12 5 km/hr Total time taken = 5 hours ⇒Distance travelled=12 5 ×5=12 km ⇒Distance between his house and office =12 2 =6 km

14. A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?


A. 11.2 kmph B. 10 kmph C. 10.2 kmph D. 10.8 kmph


Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km Time taken to travel 10 km at an average speed of 12 km/hr = distance speed =10 12 hr Time taken to travel 12 km at an average speed of 10 km/hr = distance speed =12 10 hr Total time taken =10 12 +12 10 hr Average speed = distance time =22 (10 12 +12 10 ) =22×120 (10×10)+(12×12) 22×120 244 =11×120 122 =11×60 61 =660 61 ≈10.8 kmph

15. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 12 3 hours, it must travel at a speed of: A. 660 km/hr B. 680 km/hr C. 700 km/hr D. 720 km/hr


Answer : Option D

Explanation :


------------------------------------------------------------------------------Solution 1 (Recommended) ------------------------------------------------------------------------------

Speed and time are inversely proportional ⇒Speed ∝ 1 Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional Speed ∝ 1 Time ⇒Speed1 Speed2 =Time2 Time1 ⇒240 Speed2 =(12 3 ) 5 ⇒240 Speed2 =(5 3 ) 5 ⇒240 Speed2 =1 3 ⇒Speed2=240×3=720 km/hr

------------------------------------------- Solution 2 ------------------------------------------- Distance = Speed x Time = 240 x 5 km

New time = 12 3 hr=5 3 hr Hence, new speed = Distance Time =240×5 5 3 =240×3=720 km/hr

16. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5


minutes while stopping at the stations. What is the speed of the car? A. 80 kmph B. 102 kmph C. 120 kmph D. 140 kmph


Answer : Option C

Explanation :

Let speed of the car = x kmph Then speed of the train = (100+50) 100 x=150 100 x=3 2 x kmph Time taken by the car to travel from A to B=75 x hours Time taken by the train to travel from A to B=75 (3 2 x) +12.5 60 hours Since Both start from A at the same time and reach point B at the same time 75 x =75 (3 2 x) +12.5 60 25 x =12.5 60 x=25×60 12.5 =2×60=120

17. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. What is the duration of the flight ? A. 2 hour B. 11 2 hour C. 1 2 hour D. 1 hour


Answer : Option D


Explanation :

Let the duration of the flight = x hours Given that distance = 600 km

Speed = Distance Time =600 x ................(Equation1) Duration of the flight due to the slow down = (x+30 60 ) hours=(x+1 2 ) hours New speed = 600 (x+1 2 ) ................(Equation2) From Equations 1 and 2, Reduction in Speed =600 x −600 (x+1 2 ) Given that Reduction in average speed =200 km/hr ⇒600 x −600 (x+1 2 ) =200 ⇒3 x −3 (x+1 2 ) =1 ⇒3 x −6 2x+1 =1 ⇒3(2x+1)−6x x(2x+1) =1 ⇒6x+3−6x x(2x+1) =1 ⇒3 x(2x+1) =1 ⇒2x 2 +x−3=0................(Equation3) From here, you can get the answer using Trial and error method. If you try with the values given as the choices, you can see the value of x = 1 satisfies the equation 3. Hence answer is 1 hour Or, we can solve the equation 3 to get the answer ⇒2x 2 +x−3=0 ⇒(2x+3)(x−1)=0 ⇒x=1(Removing the -ve value for x) Hence answer is 1 hour

18. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him? A. 80 km B. 70 km C. 60 km D. 50 km



Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance Time =x 10 .....(Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr

⇒Speed = Distance Time =(x+20) 14 .....(Equation2)

From Equations 1 and 2,

x 10 =(x+20) 14 14x=10x+200 4x=200 x=200 4 =50

19. The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, What is the the speed of the first train? A. 85 km/hr B. 87.5 km/hr C. 90 km/hr D. 92.5 km/hr


Answer : Option B

Explanation :


------------------------------------------------------------------------------ Solution 1 (Recommended) ------------------------------------------------------------------------------

Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝ 1 Time (when distance is constant)

Here distance is constant and hence speed and time are inversely proportional Speed ∝ 1 Time ⇒Speed1 Speed2 =Time2 Time1 ⇒7 8 =4 Time1 ⇒Time1 = 4×8 7 hr ⇒Speed of the first train = Distance Time1 =400 (4×8 7 ) =100×7 8 =12.5×7=87.5 km/hr

------------------------------------------- Solution 2 ------------------------------------------- Since the ratio of the speeds of trains is 7 : 8 , let's assume that speed of the trains are 7x and 8x respectively. Given that second train runs 400 km in 4 hours.

⇒Speed of the 2nd train = Distance Time =400 4 =100 km/hr ⇒8x=100 ⇒x=100 8 =12.5 ⇒Speed of the first train = 7x=7×12.5=87.5 km/hr


20. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. What is the ratio of the speed of the train to that of the car? A. 3 : 4 B. 2 : 3 C. 1 : 2 D. 1 : 3


Answer : Option A

Explanation :

Let speed of the train = x kmph and speed of the car = y kmph Time needed for traveling 600 km if 120 km by train and the rest by car = 8 hr

⇒120 x +(600−120) y =8 ⇒120 x +480 y =8.............(Equation 1) ⇒15 x +60 y =1.............(Equation 1)

Time needed for traveling 600 km if 200 km by train and the rest by car = 8 hr 20 min

⇒200 x +(600−200) y =820 60 =81 3 =25 3 ⇒200 x +400 y =25 3 ⇒8 x +16 y =1 3 ⇒24 x +48 y =1.............(Equation 2) Solving Equation1 and Equation2 Here


Equation1 = Equation2 = 1 ⇒15 x +60 y =24 x +48 y ⇒12 y =9 x ⇒4 y =3 x ⇒x y =3 4 ⇒x:y=3:4

21. Arun is traveling on his cycle and has calculated to reach point A at 2 pm if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 pm? A. 8 kmph B. 10 kmph C. 12 kmph D. 14 kmph


Answer : Option C

Explanation :

Let the distance be x km if travels at 10 kmph, Arun will reach point A at 2 pm if travels at 15 kmph, Arun will reach point 12 noon => Time taken when traveling at 10 km = Time taken when traveling at 15 km + 2 hours

⇒x 10 =x 15 +2 ⇒x 10 −x 15 =2 ⇒3x−2x=2×30 ⇒x=60 ⇒Distance = 60 km


Examine the any statement say, if travels at 10 kmph, Arun will reach point A at 2 pm

⇒Time taken = Distance Speed =60 10 =6 hours ⇒He must have started 6 hours back 2 pm, ie, at 8 am ⇒Now he wants to reach at 1 pm. ie; time to be taken = 5 hours ⇒Speed needed = Distance Time =60 5 =12 kmph

22. A car travels at an average of 50 miles per hour for 21 2 hours and then travels at a speed of 70 miles per hour for 11 2 hours. How far did the car travel in the entire 4 hours? A. 210 miles B. 230 miles C. 250 miles D. 260 miles


Answer : Option B

Explanation :

Speed1 = 50 miles/hour Time1 = 21 2 hour=5 2 hour Distance1 = Speed1 × Time1 = 50×5 2 =25×5=125 miles Speed2 = 70 miles/hour Time2 = 11 2 hour=3 2 hour Distance2 = Speed2 × Time2 = 70×3 2 =35×3=105 miles Total Distance = Distance1 + Distance2 =125+105=230 miles


23. The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours? A. 422 km B. 552 km C. 502 km D. 492 km


Answer : Option B

Explanation :

Given that distance travelled in 1st hour = 35 km and speed of the bus increases by 2 km after every one hour Hence distance travelled in 2nd hour = 37 km Hence distance travelled in 3rd hour = 39 km Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)] This is an Arithmetic Progression(AP) with first term, a=35, number of terms,n = 12 and common difference, d=2.


The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . . is called an Arithmetic Progression(AP) where a is the first term and d is the common difference of the AP

Sum of the first n terms of an Arithmetic Progression(AP), S n =n 2 [2a+(n−1)d] where n = number of terms

Hence, [35+37+39+... (12 terms)] =S 12 =12 2 [2×35+(12−1)2] =6[70+22]=6×92=552

Hence the total distance travelled = 552 km

24. Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper? A. 1800 ft B. 2810 ft C. 3020 ft D. 2420 ft


Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s Time = 11/5 second

Distance = Speed × Time = 1100 ×11 5 =220×11=2420 ft


25. An athlete runs 200 metres race in 24 seconds. What is his speed? A. 20 km/hr B. 25 km/hr C. 27.5 km/hr D. 30 km/hr


Answer : Option D

Explanation :

Speed=Distance Time =200 24 m/s=200 24 ×18 5 km/hr =40×3 4 km/hr=10×3 km/hr=30 km/hr

26. A train is moving at the speed of 80 km/hr. What is its speed in metres per second? A. 222 9 m/s B. 22 m/s C. 211 9 m/sec D. 21 m/s


Answer : Option A

Explanation :

Speed = 80 km/hr = 80×5 18 m/s = 40×5 9 m/s = 200 9 m/s = 222 9 m/s

27. The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr.


Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time will they meet? A. 10.30 a.m. B. 10 a.m. C. 12 noon D. 11 a.m.


Answer : Option D

Explanation :

Assume that they meet x hours after 8 a.m. Then, train1,starting from A , travelling towards B, travels x hours till the trains meet

⇒Distance travelled by train1 in x hours = Speed ×Time = 60x

Then, train2, starting from B , travelling towards A, travels (x-1) hours till the trains meet

⇒Distance travelled by train2 in (x-1) hours = Speed ×Time = 75(x-1)

Total distance travelled = Distance travelled by train1 + Distance travelled by


train2 => 330 = 60x + 75(x-1) => 12x + 15(x-1) = 66 => 12x + 15x - 15 = 66 => 27x = 66 + 15 = 81 => 3x = 9 => x = 3 Hence the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.

28. A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in metres)? A. 1250 B. 1280 C. 1320 D. 1340



Answer : Option A

Explanation :

Speed = 5 km/hr Time = 15 minutes = 1 4 hour Length of the bridge = Distance Travelled by the man = Speed × Time = 5×1 4 km =5×1 4 ×1000 metre=1250 metre

29. A train travelled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point? A. 6 hrs 21 min B. 7 hrs 14 min C. 7 hrs 22 min D. 6 hrs


Answer : Option A

Explanation :

Time taken to travel 600 km = Distance Speed =600 100 =6 hour

Now we need to find out the number of stops in the 600 km travel. Given that train stops after every 75 km.

600 75 =8


It means train stops 7 times before 600 km and 1 time just after 600 km. Hence we need to take only 7 stops into consideration for the 600 km travel. Hence, total stopping time in the 600 km travel = 7 x 3 = 21 minutes Total time needed to reach the destination = 6 hours + 21 minutes = 6 hrs 21 min

30. A person travels from A to B at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips? A. 60 km/hr B. 56 km/hr C. 52 km/hr D. 48 km/hr


Answer : Option D

Explanation :

------------------------------------------- Solution 1 (Quick) --------------------------------------------


If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy x+y kmph. By using the same formula, we can find out the average speed quickly Speed with which he travels from A to B = x = 40 km/hr Speed with which he travels from B to A = x = 40×(100+50) 100 =60 km/hr average speed = 2×40×60 40+60 =48 km/hr

------------------------------------------- Solution 2 (Fundamentals) ------------------------------------------- Assume that distance from A to B = x km Speed with which he travels from A to B = x = 40 km/hr

Time to travel from A to B = distance speed =x 40 hr Speed with which he travels from B to A = 40×(100+50) 100 =60 km/hr Time to travel from B to A = distance speed =x 60 Total distance traveled = x+x=2x Total time taken = x 40 +x 60 Average speed = Total distance traveled Total time taken =2x x 44 +x 60 =2 1 60 +1 40 =2×2400 40+60 =2×24=48 km/hr

31. A man in a train notices that he can count 21 telephone posts in one minute. If they are known to be 50 metres apart, at what speed is the train travelling?




Answer : Option D

Explanation :

The man in the train notices that he can count 21 telephone posts in one minute. Number of gaps between 21 posts are 20 and Two posts are 50 metres apart. It means 20 x 50 meters are covered in 1 minute.

Distance = 20×50 meter = 20×50 1000 km = 1 km Time = 1 minute = 1 60 hour Speed = Distance Time =1 (1 60 ) =60 km/hr

32. A truck covers a distance of 550 metres in 1 minute whereas a train covers a distance of 33 kms in 45 minutes. What is the ratio of their speed? A. 2 : 1 B. 1 : 2 C. 4 : 3 D. 3 : 4


Answer : Option D


Explanation :

Speed of the truck = Distance Time =550 1 =550 meters/minute Speed of the train = Distance Time =33 45 km/minute = 33000 45 meters/minute Speed of the truck Speed of the train =550 (33000 45 ) =550×45 33000 =55×45 3300 =11×45 660 =11×9 132 =9 12 =3 4

Hence, Speed of the truck : Speed of the train = 3 : 4

33. A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, what should be his speed in km/hr? A. 14 km/hr B. 12 km/hr C. 10 km/hr D. 8 km/hr


Answer : Option B

Explanation :

The person needs to cover 6 km in 45 minutes Given that he covers one-half of the distance in two-thirds of the total time => He covers half of 6 km in two-thirds of 45 minutes


=> He covers 3 km in 30 minutes Hence, now he need to cover the remaining 3 km in the remaining 15 minutes Distance = 3 km Time = 15 minutes = 1/4 hour

Required Speed=Distance Time =3 (1 4 ) =12 km/hr


Important Formulas –

Time and Work

1.If A can do a piece of work in n days, work done

by A in 1 day = 1/n

2.If A does 1/n work in a day, A can finish the work

in n days

3.If M1 men can do W1 work in D1 days working

H1 hours per day and M2 men can do W2 work in

D2 days working H2 hours per day (where all

men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2


4.If A can do a piece of work in p days and B can

do the same in q days, A and B together can

finish it in pq / (p+q) days

5.If A is thrice as good as B in work, then

Ratio of work done by A and B = 3 : 1

Ratio of time taken to finish a work by A and B =

1 : 3

1. P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left? A. 8/15 B. 7/15 C. 11/15 D. 2/11


Answer : Option A

Explanation :


Amount of work P can do in 1 day = 1/15 Amount of work Q can do in 1 day = 1/20 Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60 Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15 Fraction of work left = 1 – 7/15= 8/15

2. P can lay railway track between two stations in 16 days. Q can do the same job in 12 days. With the help of R, they completes the job in 4 days. How much days does it take for R alone to complete the work? A. 9(3/5) days B. 9(1/5) days C. 9(2/5) days D. 10 days


Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/16 Amount of work Q can do in 1 day = 1/12 Amount of work P, Q and R can together do in 1 day = 1/4


Amount of work R can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48 => Hence R can do the job on 48/5 days = 9 (3/5) days

3. P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day? A. 10 days B. 14 days C. 15 days D. 9 days


Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20 Amount of work Q can do in 1 day = 1/30 Amount of work R can do in 1 day = 1/60 P is working alone and every third day Q and R is helping him Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5 So work completed in 15 days = 5 × 1/5 = 1


Ie, the work will be done in 15 days

4. A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in - days if they work together. A. 18 days B. 22 ½ days C. 24 days D. 26 days


Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days Hence, if the difference is 2 days, B can complete the work in 3 days => if the difference is 60 days, B can complete the work in 90 days => Amount of work B can do in 1 day= 1/90 Amount of work A can do in 1 day = 3 × (1/90) = 1/30 Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45


=> A and B together can do the work in 45/2 days = 22 ½ days

5. A can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C? A. Rs. 380 B. Rs. 600 C. Rs. 420 D. Rs. 400


Answer : Option D

Explanation :

Amount of work A can do in 1 day = 1/6 Amount of work B can do in 1 day = 1/8 Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24 Amount of work A + B + C can do = 1/3 Amount of work C can do in 1 day = 1/3 - 7/24 = 1/24 work A can do in 1 day: work B can do in 1 day: work C can do in 1 day = 1/6 : 1/8 : 1/24 = 4 : 3 : 1


Amount to be paid to C = 3200 × (1/8) = 400

6. 6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days. A. 4 days B. 6 days C. 2 days D. 8 days


Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b Work done by 6 men and 8 women in 1 day = 1/10 => 6m + 8b = 1/10 => 60m + 80b = 1 --- (1) Work done by 26 men and 48 women in 1 day = 1/2 => 26m + 48b = ½ => 52m + 96b = 1--- (2)


Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200 Work done by 15 men and 20 women in 1 day = 15/100 + 20/200 =1/4 => Time taken by 15 men and 20 women in doing the work = 4 days

7. A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- days. A. 12 hours B. 6 hours C. 8 hours D. 10 hours


Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4 Work done by B and C in 1 hour = 1/3 Work done by A and C in 1 hour = 1/2 Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12


Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hour

8. P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in A. 30 days B. 25 days C. 20 days D. 15 days


Answer : Option B

Explanation :

Work done by P and Q in 1 day = 1/10 Work done by R in 1 day = 1/50 Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50 But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50 => Work done by P in 1 day = 3/50 => Work done by Q and R in 1 day = 3/50


Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25 So Q alone can do the work in 25 days

9. A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work? A. 37 ½ days B. 22 days C. 31 days D. 22 days


Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5 Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1) Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B) Work done by A and B in 1 day = 1/15 ---(2) Work done by B in 1 day = 1/15 – 1/25 = 2/75 => B can complete the work in 75/2 days = 37 ½ days


10. Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed? A. 3 pm B. 2 pm C. 1:00 pm D. 11 am


Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8 Work done by Q in 1 hour = 1/10 Work done by R in 1 hour = 1/12 Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120 Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60 From 9 am to 11 am, all the machines were operating.


Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60 Pending work = 1- 37/60 = 23/60 Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11 which is approximately equal to 2 Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

11. P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work? A. 8 B. 5 C. 4 D. 6


Answer : Option D

Explanation :


Work done by P in 1 day = 1/18 Work done by Q in 1 day = 1/15 Work done by Q in 10 days = 10/15 = 2/3 Remaining work = 1 – 2/3 = 1/3 Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

12. 3 men and 7 women can complete a work in 10 days . But 4 men and 6 women need 8 days to complete the same work . In how many days will 10 women complete the same work? A. 50 B. 40 C. 30 D. 20


Answer : Option B

Explanation :

Work done by 4 men and 6 women in 1 day = 1/8 Work done by 3 men and 7 women in 1 day = 1/10 Let 1 man does m work in 1 day and 1 woman does w work in 1


day. The above equations can be written as 4m + 6w = 1/8 ---(1) 3m + 7w = 1/10 ---(2) Solving equation (1) and (2) , we get m=11/400 and w=1/400 Amount of work 10 women can do in a day = 10 × (1/400) = 1/40 Ie, 10 women can complete the work in 40 days

13. A and B can finish a work 30 days if they work together. They worked together for 20 days and then B left. A finished the remaining work in another 20 days. In how many days A alone can finish the work? A. 60 B. 50 C. 40 D. 30


Answer : Option A

Explanation :

Amount of work done by A and B in 1 day = 1/30 Amount of work done by A and B in 20 days = 20 × (1/30) = 20/30 = 2/3


Remaining work – 1 – 2/3 = 1/3 A completes 1/3 work in 20 days Amount of work A can do in 1 day = (1/3)/20 = 1/60 => A can complete the work in 60 days

14. A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in --- days. A. 5 5⁄11 B. 4 5⁄11 C. 6 4⁄11 D. 6 5⁄11


Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day => Number of hours A can complete the work = 12×8 = 96 hours => Work done by A in 1 hour = 1/96


B can complete the work in 8 days working 10 hours a day => Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80 Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours A and B works 8 hours a day Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 5

⁄11 days

15. P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work? A. 9 B. 11 C. 13 D. 15



Answer : Option C

Explanation :

Work done by P in 1 day = 1/23 Let work done by Q in 1 day = q q × (130/100) = 1/23 => q = 100/(23×130) = 10/(23×13) Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13 => P and Q together can do the work in 13 days

16. P, Q and R can complete a work in 24, 6 and 12 days respectively. The work will be completed in --- days if all of them are working together. A. 2 B. 3 3⁄7 C. 4 ¼ D. 5


Answer : Option B

Explanation :


Work done by P in 1 day = 1/24 Work done by Q in 1 day = 1/6 Work done by R in 1 day = 1/12 Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24 => Working together, they will complete the work in 24/7 days = 3 3⁄7 days

17. 10 men can complete a work in 7 days. But 10 women need 14 days to complete the same work. How many days will 5 men and 10 women need to complete the work? A. 5 B. 6 C. 7 D. 8


Answer : Option C

Explanation :

Work done by 10 men in 1 day = 1/7 Work done by 1 man in 1 day = (1/7)/10 = 1/70 Work done by 10 women in 1 day = 1/14


Work done by 1 woman in 1 day = 1/140 Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140) = 5/70 + 10/140 = 1/7 => 5 men and 10 women can complete the work in 7 days

18. Kamal will complete work in 20 days. If Suresh is 25% more efficient than Kamal, he can complete the work in --- days. A. 14 B. 16 C. 18 D. 20


Answer : Option B

Explanation :

Work done by Kamal in 1 day = 1/20 Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16 => Suresh can complete the work in 16 days


19. Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages? A. 7 hour 15 minutes B. 7 hour 30 minutes C. 8 hour 15 minutes D. 8 hour 30 minutes


Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3 Pages typed by Suresh in 1 hour = 50/5 = 8 Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3 Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4 = 8 ¼ hours = 8 hour 15 minutes

20. P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work


last? A. 5 days B. 10 days C. 14 days D. 22 days


Answer : Option B

Explanation :

Work done by P in 1 day = 1/20 Work done by Q in 1 day = 1/12 Work done by P in 4 days = 4 × (1/20) = 1/5 Remaining work = 1 – 1/5 = 4/5 Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15 Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6 Total days = 4 + 6 = 10

21. P takes twice as much time as Q or thrice as much time as R


to finish a piece of work. They can finish the work in 2 days if work together. How much time will Q take to do the work alone? A. 4 B. 5 C. 6 D. 7


Answer : Option C

Explanation :

Let P takes x days to complete the work Then Q takes x/2 days and R takes x/3 days to finish the work Amount of work P does in 1 day = 1/x Amount of work Q does in 1 day = 2/x Amount of work R does in 1 day = 3/x Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x 6/x = 2 => x = 12 => Q takes 12/2 days = 6 days to complete the work


22. P and Q can complete a work in 15 days and 10 days respectively. They started the work together and then Q left after 2 days. P alone completed the remaining work. The work was finished in --- days. A. 12 B. 16 C. 20 D. 24


Answer : Option A

Explanation :

Work done by P in 1 day = 1/15 Work done by Q in 1 day = 1/10 Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6 Work done by P and Q in 2 days = 2 × (1/6) = 1/3 Remaining work = 1 – 1/3 = 2/3 Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days Total time taken = 2 + 10 = 12 days


23. P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 10 days. How many days more will P take to finish the work? A. 10 B. 15 C. 18 D. 22


Answer : Option C

Explanation :

Let work done by P in 1 day = p, Work done by Q in 1 day = q, Work done by R in 1 day = r p + q = 1/30 q + r = 1/24 r + p = 1/20 Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120


= 1/8 => p + q + r = 1/16 => Work done by P,Q and R in 1 day = 1/16 Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8 Remaining work = 1 = 5/8 = 3/8 Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day = 1/16 – 1/24 = 1/48 Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18

24. P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in --- days A. 1 B. 2 C. 3 D. 4



Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12 Work done by P in 1 day = 2 × (1/12) = 1/6 Work done by P and Q in 1 day = 1/12 + 1/6 = ¼ => P and Q can finish the work in 4 days

25. A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is A. 1:3 B. 4:3 C. 2:3 D. 2:1


Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16 Work done by 1 woman in 1 day = 1/(16×20)


Work done by 16 men in 1 day = 1/15 Work done by 1 man in 1 day = 1/(15×16) Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20 = 1/3 :1/4 = 4:3

26. P and Q need 8 days to complete a work. Q and R need 12 days to complete the same work. But P, Q and R together can finish it in 6 days. How many days will be needed if P and R together do it? A. 3 B. 8 C. 12 D. 4


Answer : Option B

Explanation :

Let work done by P in 1 day = p work done by Q in 1 day =q


Work done by R in 1 day = r p + q = 1/8 ---(1) q + r= 1/12 ---(2) p+ q+ r = 1/6 ---(3) (3) – (2) => p = 1/6 - 1/12 = 1/12 (3) – (1) => r = 1/6 – 1/8 = 1/24 p + r = 1/12 + 1/24 = 3/24 = 1/8 => P and R will finish the work in 8 days

27. P can do a work in 24 days. Q can do the same work in 9 days and R can do the same in 12 days. Q and R start the work and leave after 3 days. P finishes the remaining work in --- days. A. 7 B. 8 C. 9 D. 10


Answer : Option D

Explanation :


Work done by P in 1 day = 1/24 Work done by Q in 1 day = 1/9 Work done by R in 1 day = 1/12 Work done by Q and R in 1 day = 1/9 + 1/12 = 7/36 Work done by Q and R in 3 days = 3×7/36 = 7/12 Remaining work = 1 – 7/12 = 5/12 Number of days in which P can finish the remaining work = (5/12) / (1/24) = 10

28. If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600. A. 12 B. 14 C. 16 D. 18


Answer : Option C

Explanation :


Wages of 1 woman for 1 day = 21600 40×30 Wages of 1 man for 1 day = 21600×2 40×30 Wages of 1 man for 25 days = 21600×2×25 40×30 Number of men = 14400 (21600×2×25 40×30 ) =144 (216×50 40×30 ) =144 9 =16

29. P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day? A. Rs.40 B. Rs.70 C. Rs.90 D. Rs.100


Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 ---(1) Amount Earned by P and R in 1 day = 600/5 = 120 ---(2) Amount Earned by Q and R in 1 day = 910/7 = 130 ---(3)


(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day - Amount Earned by P,Q and R in 1 day = 120+130-180 = 70 =>Amount Earned by R in 1 day = 70

30. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days? A. Rs.1104 B. Rs.1000 C. Rs.934 D. Rs.1210


Answer : Option A

Explanation :

Assume that cost of keeping a cow for 1 day = c , cost of keeping a goat for 1 day = g Cost of keeping 20 cows and 40 goats for 10 days = 460 Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46


=> 20c + 40g = 46 => 10c + 20g = 23 ---(1) Given that 5g = c Hence equation (1) can be written as 10c + 4c = 23 => 14c =23 => c=23/14 cost of keeping 50 cows and 30 goats for 1 day = 50c + 30g = 50c + 6c (substituted 5g = c) = 56 c = 56×23/14 = 92 Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104


31. There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work? A. 3 1⁄4 days B. 4 1⁄3 days C. 5 1⁄6 days D. 6 1⁄5 days


Answer : Option C

Explanation :

Work completed in 1st day = 1/16 Work completed in 2nd day = (1/16) + (1/16) = 2/16 Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16 … An easy way to attack such problems is from the choices. You can see the choices are very close to each other. So just see one by one.


For instance, The first choice given in 3 1⁄4

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16 The work done in 4 days = (1+2+3+4)/16 = 10/16 The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn't it? The work done in 6 days = (1+2+3+4+5+6)/16 > 1 Hence the answer is less than 6, but greater than 5. Hence the answer is 5 1

⁄6 days. (Just for your reference, work done in 5 days = 15/16. Pending work in 6th day = 1 – 15/16 = 1/16. In 6th day, 6 people are working and work done = 6/16. To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days. Hence total time required = 5 + 1/6 = 5 1

⁄6 days


Important Formulas - Time • Minute Spaces

The face or dial of clock is a circle whose circumference is divided into 60 equal parts, named minute spaces

• Hour hand and minute hand

A clock has two hands. The smaller hand is called the hour hand or short hand and the larger one is called minute hand or long hand.

• 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min.

(In 60 minutes, hour hand will move 5 min spaces while the minute hand will move 60 min spaces. In effect the space gain of minute hand with respect to hour hand will be 60 - 5 = 55 minutes.)

• Both the hands of a clock coincide once in every hour. • The hands of a clock are in the same straight line when they

are coincident or opposite to each other. • When the two hands of a clock are at right angles, they are

15 minute spaces apart. • When the hands of a clock are in opposite directions, they

are 30 minute spaces apart. • Angle traced by hour hand in 12 hrs = 360° • Angle traced by minute hand in 60 min. = 360°. • If a watch or a clock indicates 9.15, when the correct time

is 9, it is said to be 15 minutes too fast.


• If a watch or a clock indicates 8.45, when the correct time is 9, it is said to be 15 minutes too slow.

• The hands of a clock will be in straight line but opposite in direction, 22 times in a day

• The hands of a clock coincide 22 times in a day • The hands of a clock are straight 44 times in a day • The hands of a clock are at right angles 44 times in a day • The two hands of a clock will be together between H and

(H+1) o' clock at (60H 11 )minutes past H o' clock. • The two hands of a clock will be in the same straight line

but not together between H and (H + 1) o' clock at ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (5H−30)12 11 minutes past H, when H > 6 (5H+30)12 11 minutes past H, when H <6

• Angle between Hands of a clock When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H'o clock =30(H−M 5 )+M 2 degree When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H'o clock =30(M 5 −H)−M 2 degree

• The two hands of the clock will be at right angles between H and (H + 1) o' clock at (5H±15)12 11 minutes past H 'o clock

• The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by =(720 11 −M)(60×24 M ) minutes

• Bbetween H and (H + 1) o' clock, the two hands of a clock are M minutes apart at (5H±M)12 11 minutes past H 'o clock


1. An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? A. 154° B. 180° C. 170° D. 160°


Answer : Option B

Explanation :

We know that Angle traced by hour hand in 12 hrs = 360° From 8 to 2, there are 6 hours

The angle traced by the hour hand in 6 hours = 6×360 12 =180°

2. A clock is started at noon. By 10 minutes past 5, the hour hand has turned through A. 155° B. 145° C. 152° D. 140°


Answer : Option A


Explanation :

We know that Angle traced by hour hand in 12 hrs = 360°

Time duration from noon to 10 minutes past 5 = 5 hours 10 minutes =510 60 hour=31 6 hours Hence the angle traced by hour hand from noon to 10 minutes past 5 = 31 6 ×360 12 =31 6 ×30=31×5=155°

3. At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together? A. 5 minutes past 7 B. 53 11 miinutes past 7 C. 51 11 minutes past 7 D. 55 11 minutes past 7


Answer : Option D

Explanation :

The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock at ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (5H−30)12 11 minutes past H, when H > 6 (5H+30)12 11 minutes past H, when H <6

Here H = 7. Hands of the clock will point in opposite directions at (5×7−30)12 11 minutes past 7 =5×12 11 minutes past 7=60 11 minutes past 7=55 11 minutes past 7

------------------------------------------- Solution 2


------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. If the hands of the clock are in the same straight line, but not together, they will be 30 minutes apart. At 7'o clock, the hands of the clock are 25 minutes apart.Hence the minute hand should gain 5 minutes so that the hands will be 30 minutes apart

We know that, in 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand. Hence to gain 5 minutes for the minute hand, time needed = 60 55 ×5 =60 11 minutes=55 11 minutes That means when the time is 55 11 minutes past 7, the hands of a clock will be in the same straight line but, not together


4. At what time between 5.30 and 6 will the hands of a clock be at right angles? A. 44 minutes past 5 B. 447 11 minutes past 5 C. 437 11 minutes past 5 D. 43 minutes past 5


Answer : Option C

Explanation :

The two hands of the clock will be at right angles between H and (H + 1) o' clock at (5H±15)12 11 minutes past H 'o clock

Let's see the times at which right angles are formed between 5 and 6

Let's take H=5. Hence the two hands will be at right angles between 5 and 6 at (5×5±15)12 11 minutes past 5 'o clock =(25±15)12 11 minutes past 5 'o clock =10×12 11 minutes past 5 'o clock and 40×12 11 minutes past 5 'o clock =120 11 minutes past 5 'o clock and 480 11 minutes past 5 'o clock =1010 11 minutes past 5 'o clock and 437 11 minutes past 5 'o clock 1010 11 minutes past 5 comes before 5.30. 437 11 minutes past 5 comes between 5.30 and 6. The question is to find out the time between 5.30 and 6 when the hands of a clock will be at right angles.Hence the time is 437 11 minutes past 5


------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. At 5, the hands are 25 minutes spaces apart To get a right angle when the time is between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 minutes We know that, in 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.

Hence to gain 40 minutes for the minute hand, time needed = 60 55 ×40 =12 11 ×40=480 11 =437 11 minutes That


means when the time is 437 11 minutes past 5, the hands of a clock will be at right angles

5. At what angle the hands of a clock are inclined at 15 minutes past 5? A. 671 2 ° B. 621 2 ° C. 70° D. 633 4 °


Answer : Option A

Explanation :

Angle between Hands of a clock When the minute hand is behind the hour hand, the angle between the two hands at M minutes past H'o clock =30(H−M 5 )+M 2 degree When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past H'o clock =30(M 5 −H)−M 2 degree

Here H = 5, M = 15 and the minute hand is behind the hour hand. Hence the angle

=30(H−M 5 )+M 2 =30(5−15 5 )+15 2 =30(5−3)+7.5 =30×2+7.5=67.5°

-------------------------------------------


Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.

15 minutes past 5 = 5 hour 15 minutes = 515 60 hour=51 4 hour = 21 4 hour

Angle traced by hour hand in 12 hours = 360°

Hence Angle traced by hour hand in 21 4 hour=360 12 ×21 4 =30×21 4 =30×5.25=157.5°

Angle traced by minute hand in 60 minutes = 360°

Angle traced by minute hand in 15 minutes = 360 60 ×15=90°

Required angle = 157.5 - 90 = 67.5°


6. At 3.40, the hour hand and the minute hand of a clock form an angle of A. 135° B. 130° C. 120° D. 125°


Answer : Option B

Explanation :


Here H = 3, M = 40 and minute hand is ahead of the hour hand. Hence the angle

=30(M 5 −H)−M 2 =30(40 5 −3)−40 2 =30(8−3)−20 =30×5−20=130°

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should


understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.

3.40 = 3 hour 40 minutes = 340 60 hour=32 3 hour = 11 3 hour


Hence Angle traced by hour hand in 11 3 hour=360 12 ×11 3 =30×11 3 =10×11=110°



Required angle = 240 - 110= 130°

7. The angle between the minute hand and the hour hand of a clock when the time is 8.30, is A. 75° B. 85° C. 80° D. 70°



Answer : Option A

Explanation :


Here H = 8, M = 30 and minute hand is behind the hour hand. Hence the angle

=30(H−M 5 )+M 2 =30(8−30 5 )+30 2 =30(8−6)+15 =30×2+15=75°

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in


the traditional way.

8.30 = 8 hour 30 minutes = 81 2 hour=17 2 hour


Hence Angle traced by hour hand in 17 7 hour=360 12 ×17 2 =30×17 2 =15×17=255°



Required angle = 255- 180= 75°

8. How many times in a day, are the hands of a clock in straight line but opposite in direction? A. 48 B. 22 C. 24 D. 12


Answer : Option B

Explanation :


The hands of a clock point in opposite directions (in the same straight line, making an angle 180° between them) 11 times in every 12 hours because between 5 and 7 they point in opposite directions at 6 'o clock only. Hence the hands point in the opposite directions 22 times in a day However this is already given as a formula and its is better to by heart the answer as 22 which can save time in competitive exams.(However if you should know the theory behind)

9. At what time between 3 o'clock and 4 o'clock, both the needles of a clock will coincide each other? A. 162 11 minutes past 3 B. 164 11 minutes past 3 C. 154 11 minutes past 3 D. 152 11 minutes past 3


Answer : Option B

Explanation :

The two hands of a clock will be together between H and (H+1) o' clock at (60H 11 )minutes past H o' clock.


Here H = 3. Hands will be together at 60H 11 minutes past 3 =60×3 11 minutes past 3 =180 11 minutes past 3=164 11 minutes past 3

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. At 3 o' clock, the hands are 15 minute spaces apart Hence minute hand needs to gain 15 more minute spaces so that the hands will coincide each other We know that 55 min spaces are gained by minute hand (with respect to hour hand)


in 60 min Hence time taken for gaining 15 minute spaces by minute hand

=60 55 ×15 minute=12 11 ×15 minute=180 11 minute=164 11 minute Hence hands will will coincide at 164 11 minute past 3

10. How many times will the hands of a clock coincide in a day? A. 24 B. 22 C. 20 D. 21


Answer : Option B

Explanation :

The hands of a clock coincide 11 times in every 12 hours (Between 11 and 1, they coincide only once, at 12 o'clock). 12:00 am 1:05 am 2:11 am


3:16 am 4:22 am 5:27 am 6:33 am 7:38 am 8:44 am 9:49 am 10:55 am 12:00 pm 1:05 pm 2:11 pm 3:16 pm 4:22 pm 5:27 pm 6:33 pm


7:38 pm 8:44 pm 9:49 pm 10:55 pm Hence the hands coincide 22 times in a day. However this is already given as a formula and its is better to by heart the answer as 22 which can save time in competitive exams.(However if you should know the theory behind)

11. How many times in a day, the hands of a clock are straight A. 22 B. 44 C. 48 D. 24


Answer : Option B

Explanation :

The hands of a clock point in opposite directions (in the same straight line, making an angle 180° between them) 11 times in every 12 hours because between 5 and 7 they


point in opposite directions at 6 'o clock only. Hence the hands point in the opposite directions 22 times in a day The hands of a clock coincide(0 ° between them) 11 times in every 12 hours (Between 11 and 1, they coincide only once, at 12 o'clock). Hence the hands coincide 22 times in a day. So In 24 hours, the hands come in opposite direction or coincide 44 times . However this is already given as a formula and its is better to by heart the answer as 44 which can save time in competitive exams.(However if you should know the theory behind)

12. How much does a watch lose per day, if its hands coincide every 64 minutes? A. 341 11 minute B. 328 11 minute C. 31 minute D. 332 11 minute


Answer : Option B

Explanation :


The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by =(720 11 −M)(60×24 M ) minutes

Here M = 64. The clock gains or losses in a day by

=(720 11 −M)(60×24 M )=(720 11 −64)(60×24 64 )=16 11 (60×3 8 ) =2 11 (60×3)=360 11 =328 11 minutes

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min


=> Time taken by the minute hand to gain 60 min spaces in a normal clock = 60 55 ×60=12 11 ×60=720 11 =655 11 min

In the given watch, hands coincide every 64 minutes. In another words, minute hand gains 60 min spaces in every 64 minutes for the given watch.

Loss in 64 min = 655 11 −64=15 11 =16 11 minute Loss in 24 hours = 16 11 ×1 64 ×24×60=1 11 ×1 4 ×24×60=360 11 =328 11 minute

13. At what time between 9 and 10 o' clock will the hands of a clock be together? A. 452 9 min past 9 B. 491 11 min past 9 C. 481 12 min past 9 D. 472 15 min past 9


Answer : Option B

Explanation :

The two hands of a clock will be together between H and (H+1) o' clock at (60H 11 ) minutes past H o' clock.


Here H = 9. Hands will be together at 60×9 11 minutes past 9 =540 11 minutes past 9=491 11 minutes past 9

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However its better to understand the basics. Please find the method given below to solve the same problem in the traditional way. At 9 o' clock, the hands are 45 minute spaces apart Hence minute hand needs to gain 45 more minute spaces so that the hands will coincide each other? We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 45 minute spaces by minute hand


=60 55 ×45 minute=12 11 ×45 minute=540 11 minute=491 11 minute Hence hands will coincide at 491 11 minute past 9

14. At what time between 4 and 5 o'clock will the hands of a watch point in opposite directions? A. 536 11 minutes past 4 B. 537 11 minutes past 4 C. 546 11 minutes past 4 D. 547 11 minutes past 4


Answer : Option C

Explanation :

The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock at ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (5H−30)12 11 minutes past H, when H > 6 (5H+30)12 11 minutes past H, when H <6

Here H = 4. Hands of the watch will point in opposite directions at (5×4+30)12 11 minutes past 4 =50×12 11 minutes past 4=600 11 minutes past 4 =546 11 minutes past 4

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However its better to


understand the basics. Please find the method given below to solve the same problem in the traditional way. At 4 o' clock, the hands are 20 minute spaces apart Hence minute hand needs to gain 50 more minute spaces so that the hands will point in opposite directions. We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 50 minute spaces by minute hand

=60 55 ×50 minute=12 11 ×50 minute=600 11 minute=546 11 minute Hence hands will point in opposite directions at 546 11 minute past 4


15. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is A. 3 pm B. 3.45 pm C. 3.30 pm D. 4 pm


Answer : Option D

Explanation :

Time from 7 am to 4.15 pm = 9 hours 15 minutes = 91 4 hours=37 4 hours 3 minute 5 seconds of the given clock = 3 minutes of a normal clock ⇒31 12 minutes of the given clock = 3 minutes of a normal clock ⇒37 12 minutes of the given clock = 3 minutes of a normal clock ⇒37 720 hours of the given clock = 3 60 hours of a normal clock ⇒37 720 hours of the given clock = 1 20 hours of a normal clock ⇒37 4 hours of the given clock = 1 20 ×720 37 ×37 4 hours of the given clock = 9 hours of the given clock Hence the correct time = 9 hours after 7 am = 4 pm

16. How many times are the hands of a clock at right angle in a day? A. 48 B. 44 C. 24 D. 22



Answer : Option B

Explanation :

In 12 hours, hands of a clock are at right angles at 22 times. In 24 hours, hands of a clock are at right angles at 44 times.

17. A watch which gains uniformly is 2 minutes low at noon on and is 4 min 48 sec fast at 2 pm on the following Monday. When was it correct? A. 2 pm on Tuesday B. 3 pm on Wednesday C. 2 pm on Wednesday D. 3 pm on Tuesday


Answer : Option C

Explanation :

Time from Monday noon (12 o' clock) to following Monday 2 pm = 7 day 2 hours

=7×24+2=168+2=170 hours

Total time gained Monday noon (12 o' clock) to following Monday 2 pm = 2 min + 4 min 48 sec

= 6 min 48 sec = 648 60 mins = 64 5 mins = 34 5 mins


=> The given watch gains 34 5 mins in 170 hours

Given that the said watch was 2 minutes low at Monday noon. Hence when it gained 2 minutes, the time was correct.

The given watch gains 34 5 mins in 170 hours => The given watch gains 2 mins in 170×5 34 ×2 hours =5×5×2 hours = 50 hours

Hence the time was correct after 50 hours from Monday noon = after 2 days 2 hours from Monday noon = 2 pm on Wednesday

18. What is the reflex angle between the hands of a clock at 10.25? A. 195° B. 1971 2 ° C. 180° D. 1931 2 °


Answer : Option B


Explanation :


Here H = 10 , M = 25 and the minute hand is behind the hour hand. Hence the angle

=30(H−M 5 )+M 2 =30(10−25 5 )+25 2 =30(10−5)+12.5 =30×5+12.5=150+12.5=162.5°

But the question is to find out the reflex angle. Reflex angle = 360 - 162.5 = 197.5 ° ------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should



10 hour 25 minutes = 1025 60 hour=105 12 hour=125 12 hour

Angle traced by hour hand in 12 hrs = 360°

Angle traced by hour hand in 125 12 hour = 360 12 ×125 12 =30×125 12 =10×125 4 =10×31.25=312.5°

Angle traced by minute hand in 60 min. = 360°.

Angle traced by minute hand in 25 min. = 360 60 ×25=6×25=150°.

Angle between the hands of the clock when the time is 10.25 = 312.5 ° -150 ° =162.5 °. Reflex angle = 360 - 162.5 = 197.5 °


19. The angle between the minute hand and the hour hand of a clock when the time is 4.20 is A. 10° B. 5° C. 0° D. 1°


Answer : Option A

Explanation :


Here H = 4, M = 20 the minute hand is slightly behind the hour hand. Hence the angle

=30(H−M 5 )+M 2 =30(4−20 5 )+20 2 =30(4−4)+10 =30×0+10=10°

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should



4 hour 20 minutes = 41 3 hour=13 3 hour


Angle traced by hour hand in 13 3 hour = 360 12 ×13 3 =30×13 3 =10×13=130°



Angle between the hands of the clock when the time is 4.20 = 130° - 120° =10°.

20. A clock is set at 5 am. If the clock loses 16 minutes in 24 hours, what will be the true time when the clock indicates 10 pm on 4th day? A. 9.30 pm B. 10 pm


C. 10.30 pm D. 11 pm


Answer : Option D

Explanation :

Time from 5 am to 10 pm on the 4th day = 3 days 17 hours = 3×24+17=89 hours Given that clock loses 16 minutes in 24 hours ⇒>23 hour 44 minutes of the given clock = 24 hours in a normal clock ⇒2344 60 hours of the given clock = 24 hours in a normal clock ⇒2311 15 hours of the given clock = 24 hours in a normal clock ⇒356 15 hours of the given clock = 24 hours in a normal clock ⇒89 hours of the given clock = 24×15 356 ×89 hours in a normal clock =24×15 4 =6×15=90 hours So the correct time is 90 hours after 5 am = 3 days 18 hours after 5 am = 11 pm on the 4th day

21. What is the angle between the hour and the minute hand of a clock when the time is 3.25? A. 47 B. 461 2 C. 46 D. 471 2


Answer : Option D

Explanation :



Here H = 3, M = 25 and the minute hand is ahead of the hour hand. Hence the angle

=30(M 5 −H)−M 2 =30(25 5 −3)−25 2 =30(5−3)−12.5 =30×2−12.5=60−12.5=47.5°

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way.

3 hour 25 minutes = 325 60 hour=35 12 hour=41 12 hour



Angle traced by hour hand in 41 12 hour = 360 12 ×41 12 =30×41 12 =10×41 4 =10×10.25=102.5°



Angle between the hands of the clock when the time is 10.25 = 150° - 102.5° = 47.5 °.

22. At what time between 8 and 9 o'clock will the hands of a clock are in the same straight line but not together? A. 118 11 minutes past 8 B. 108 11 minutes past 8 C. 1110 11 minutes past 8 D. 1010 11 minutes past 8


Answer : Option D

Explanation :

The two hands of a clock will be in the same straight line but not together between H and (H + 1) o' clock


at ⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (5H−30)12 11 minutes past H, when H > 6 (5H+30)12 11 minutes past H, when H <6

Here H = 8. Hands of the clock will point in opposite directions at (5×8−30)12 11 minutes past 8 =10×12 11 minutes past 8=120 11 minutes past 8=1010 11 minutes past 8

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. At 8 o' clock, the hands are 20 minute spaces apart Hence minute hand needs to gain 10 more minute spaces so that the hands will be in opposite direction


We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 10 minute spaces by minute hand

=60 55 ×10 minute=12 11 ×10 minute=120 11 minute=1010 11 minute Hence hands will be in opposite direction at 1010 11 minute past 8

23. At what time between 2 and 3 o'clock will the hands of a clock be together? A. 911 11 minutes past 2 B. 910 11 minutes past 2 C. 1011 11 minutes past 2 D. 1010 11 minutes past 2


Answer : Option D

Explanation :




------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in the traditional way. At 2 o' clock, the hands are 10 minute spaces apart Hence minute hand needs to gain 10 more minute spaces so that the hands will be in opposite direction We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 10 minute spaces by minute hand


=60 55 ×10 minute=12 11 ×10 minute=120 11 minute=1010 11 minute Hence hands of the clock will be together at 1010 11 minute past 2

24. At what time between 5 and 6 'o clock, will the hands of a clock be at right angle? A. 1010 11 minutes past 5 and 437 11 minutes past 5

B. 1010 11 minutes past 5 and 427 11 minutes past 5

C. 109 11 minutes past 5 and 427 11 minutes past 5

D. 109 11 minutes past 5 and 427 11 minutes past 5


Answer : Option A

Explanation :


Here H=5. Hence the two hands will be at right angles between 5 and 6 at (5×5±15)12 11 minutes past 5 'o clock =(25±15)12 11 minutes past 5 'o clock =10×12 11 minutes past 5 'o clock and 40×12 11 minutes past 5 'o clock =120 11 minutes past 5 'o clock and 480 11 minutes past 5 'o clock =1010 11 minutes past 5 'o clock and 437 11 minutes past 5 'o clock

-------------------------------------------


Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However its good to understand the basics. Please find the method given below to solve the same problem in the traditional way. At 5, the hands are 25 minutes spaces apart To get first right angle ,the minute hand has to gain 25 - 15 = 10 minutes To get second right angle ,the minute hand has to gain 25 + 15 = 40 minutes We know that, in 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.

To gain 10 minutes for the minute hand, time needed = 60 55 ×10 =12 11 ×10=120 11 =1010 11 minutes HTo gain 40 minutes for the minute hand, time needed = 60 55 ×40 =12 11 ×40=480 11 =437 11 minutes That means when the time is 1010 11 minutes past 5


and 437 11 minutes past 5, the hands of a clock will be at right angles

25. The minute hand of a clock overtakes the hour hand at intervals of 65 minutes. How much a day does the clock gain or loss? A. 109 143 minutes B. 119 143 minutes C. 1110 143 minutes D. 1010 143 minutes


Answer : Option D

Explanation :

The minute hand of a clock overtakes the hour hand at intervals of M minutes of correct time. The clock gains or loses in a day by =(720 11 −M)(60×24 M ) minutes

Here M = 65. The clock gains or losses in a day by

=(720 11 −M)(60×24 M )=(720 11 −65)(60×24 65 )=5 11 (12×24 13 ) =1440 143 =1010 143 minutes

26. Find the time between 4 and 5'o clock, when the two hands of a clock are 4 minutes apart?


A. 262 11 minutes past 4 and 175 11 minutes past 4

B. 261 11 minutes past 4 and 175 11 minutes past 4

C. 262 11 minutes past 4 and 174 11 minutes past 4



Answer : Option A

Explanation :

Bbetween H and (H + 1) o' clock, the two hands of a clock are M minutes apart at (5H±M)12 11 minutes past H 'o clock

Here H=4 and M=4. Hence the two hands are 4 minutes apart between 4 and 5 at (5×4±4)12 11 minutes past 4 'o clock =(20±4)12 11 minutes past 4 'o clock =16×12 11 minutes past 4 'o clock and 24×12 11 minutes past 4 'o clock =192 11 minutes past 4 'o clock and 288 11 minutes past 4 'o clock =175 11 minutes past 4 'o clock and 262 11 minutes past 4 'o clock

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However please find the method given below to solve the same problem in the traditional way.


At 4 o' clock, the hands are 20 minute spaces apart To get the first 4° between the hands ,the minute hand has to gain 20 - 4 = 16 minute spaces To get the second 4°between the hands ,the minute hand has to gain 20 + 4 = 24 minute spaces We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Time taken for gaining 16 minute spaces by minute hand

=60 55 ×16 minute=12 11 ×16 minute=192 11 minute=175 11 minutes

Time taken for gaining 24 minute spaces by minute hand


=60 55 ×24 minute=12 11 ×24 minute=288 11 minute=262 11 minutes Hence hands of the clock are 4 minutes apart at 175 11 minutes past 4 and 262 11 minute past 4

27. At what time between 5 and 6 will the hands of the clock coincide? A. 262 11 minutes past 5 B. 263 11 minutes past 5 C. 283 11 minutes past 5 D. 273 11 minutes past 5


Answer : Option D

Explanation :



------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However we should understand the basics for sure. Please find the method given below to solve the same problem in


the traditional way. At 5 o' clock, the hands are 25 minute spaces apart Hence minute hand needs to gain 25 more minute spaces so that the hands will be in opposite direction We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 25 minute spaces by minute hand

=60 55 ×25 minute=12 11 ×25 minute=300 11 minute=273 11 minute Hence hands of the clock will be together at 273 11 minute past 5

28. At what time between 6 and 7 will the hands be perpendicular A. 481 11 minutes past 6 and B. 48 minutes past 6 and


164 11 minutes past 6 163 11 minutes past 6 C. 491 11 minutes past 6 and 164 11 minutes past 6



Answer : Option C

Explanation :


Here H=6. Hence the two hands will be at right angles between 6 and 7 at (5×6±15)12 11 minutes past 6 'o clock =(30±15)12 11 minutes past 6 'o clock =15×12 11 minutes past 6 'o clock and 45×12 11 minutes past 6 'o clock =180 11 minutes past 6 'o clock and 540 11 minutes past 6 'o clock =164 11 minutes past 6 'o clock and 491 11 minutes past 6 'o clock

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However its better to understand the basics. Please find the method given below to solve the same


problem in the traditional way. At 5 o' clock, the hands are 30 minute spaces apart Hence minute hand needs to gain 15 more minute spaces or 45 more minute spaces so that the hands will be in right angles (90° between them) We know that 55 min spaces are gained by minute hand (with respect to hour hand) in 60 min Hence time taken for gaining 15 minute spaces by minute hand

=60 55 ×15=12 11 ×15=180 11 =164 11 minutes

Hence time taken for gaining 45 minute spaces by minute hand


=60 55 ×45=12 11 ×45=540 11 =491 11 minutes

Hence the hands will be perpendicular at 164 11 minutes past 6 and 491 11 minutes past 6

29. What is the angle between the hands at 4.40? A. 95° B. 100° C. 120° D. 110°


Answer : Option B

Explanation :


Here H = 4, M = 40 the minute hand is ahead of the hour hand. Hence the angle


=30(M 5 −H)−M 2 =30(40 5 −4)−40 2 =30(8−4)−20 =30×4−20=100°

------------------------------------------- Solution 2 ------------------------------------------- Its better to use formula as it can save lots of time in exams. However its better to understand the basics. Please find the method given below to solve the same problem in the traditional way.

4 hour 40 minutes = 42 3 hour=14 3 hour


Angle traced by hour hand in 14 3 hour = 360 12 ×14 3 =30×14 3 =10×14=140°




Angle between the hands of the clock when the time is 4.40 = 240° - 140° =100°. 30. A clock strikes 4 taking 9 seconds. In order to strike 12 at the same rate, the time taken is A. 33 seconds B. 30 seconds C. 36 seconds D. 27 seconds


Answer : Option A

Explanation :

There are 3 intervals when the clock strikes 4 Time taken at 3 intervals = 9 seconds

Time taken for 1 interval = 9 3 =3 seconds

In order to strike 12, there are 11 intervals. Hence time needed

=3×11=33 seconds


Impotant Formulas - Train Sum 1.km/hr to m/s conversion:

a km/hr =

a x5

m/s.18

1. m/s to km/hr conversion:

a m/s =

a x18

km/hr.5

2. Formulas for finding Speed, Time and Distance 3. Time taken by a train of length l metres to pass a pole or

standing man or a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length l metres to pass a stationery object of length b metres is the time taken by the train to cover (l + b) metres.

5. Suppose two trains or two objects bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relative speed is = (u - v) m/s.

6. Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:

The time taken by the trains to cross each other =(a + b)

sec.(u + v)

8. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then:


The time taken by the faster train to cross the slower train =

(a + b)

sec.(u - v)

9. If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A's speed) : (B's speed) = (b : a)

1. A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train? A. 190 metres B. 160 metres C. 200 metres D. 120 metres


Answer : Option C

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s Time taken to cross, t = 18 s Distance Covered, d = vt = (400/36)× 18 = 200 m


Distance covered is equal to the length of the train = 200 m

2. A train ,130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is A. 270 m B. 245 m C. 235 m D. 220 m


Answer : Option B

Explanation :

Assume the length of the bridge = x meter Total distance covered = 130+x meter total time taken = 30s speed = Total distance covered /total time taken = (130+x)/30 m/s => 45 × (10/36) = (130+x)/30 => 45 × 10 × 30 /36 = 130+x => 45 × 10 × 10 / 12 = 130+x => 15 × 10 × 10 / 4 = 130+x


=> 15 × 25 = 130+x = 375 => x = 375-130 =245

3. A train has a length of 150 meters . it is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train. A. 182 km/hr B. 180 km/hr C. 152 km/hr D. 169 km/hr


Answer : Option A

Explanation :

Length of the train, l = 150m Speed of the man , Vm= 2 km/hr Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr Relative Speed = Speed of train, Vt - Speed of man (As both are moving in the same direction) => 180 = Vt - 2


=> Vt = 180 + 2 = 182 km/hr

4. A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m? A. 120 sec B. 99 s C. 89 s D. 80 s


Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s t = (240+650)/10 = 89 seconds

5. A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 140 m long? A. 38 sec B. 35 s C. 44 sec D. 40 s



Answer : Option D

Explanation :

Speed = 45 km/hr = 45×(10/36) m/s = 150/12 = 50/4 = 25/2 m/s Total distance = length of the train + length of the platform = 360 + 140 = 500 meter Time taken to cross the platform = 500/(25/2) = 500×2/25 = 40 seconds

6. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . If they cross each other in 23 seconds, what is the ratio of their speeds? A. Insufficient data B. 3 : 1 C. 1 : 3 D. 3 : 2


Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively length of train1 = 27x


length of train2 = 17y Relative speed= x+ y Time taken to cross each other = 23 s => (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y) => 4x = 6y => x/y = 6/4 = 3/2

7. A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. how much time does it take for the train to pass the jogger? A. 46 B. 36 C. 18 D. 22


Answer : Option B

Explanation :


Distance to be covered = 240+ 120 = 360 m Relative speed = 36 km/hr = 36×10/36 = 10 m/s Time = distance/speed = 360/10 = 36 seconds

8. Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. If the faster train passes the slower train in 36 seconds,what is the length of each train? A. 88 B. 70 C. 62 D. 50


Answer : Option D

Explanation :

Assume the length of each train = x Total distance covered for overtaking the slower train = x+x = 2x Relative speed = 46-36 = 10km/hr = (10×10)/36 = 100/36 m/s Time = 36 seconds 2x/ (100/36) = 36


=> (2x × 36 )/100 = 36 => x = 50 meter

9. Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is A. 10.8 s B. 12 s C. 9.8 s D. 8 s


Answer : Option A

Explanation :

Distance = 140+160 = 300 m Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

10. Two trains are moving in opposite directions with speed of 60 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. the slower train cross the faster train in --- seconds A. 56 B. 48


C. 47 D. 26


Answer : Option B

Explanation :

Relative speed = 60+90 = 150 km/hr (Since both trains are moving in opposite directions) Total distance = 1.1+.9 = 2km Time = 2/150 hr = 1//75 hr = 3600/75 seconds = 1200/25 = 240/5 = 48 seconds

11. A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is A. None of these B. 280 meter C. 240 meter D. 200 meter


Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s Length of the train = speed × time taken to cross the man =


15×20 = 300 m Let the length of the platform = L Time taken to cross the platform = (300+L)/15 => (300+L)/15 = 36 => 300+L = 15×36 = 540 => L = 540-300 = 240 meter

12. A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train? A. 79.2 km/hr B. 69 km/hr C. 74 km/hr D. 61 km/hr


Answer : Option A

Explanation :

Let x is the length of the train and v is the speed Time taken to move the post = 8 s => x/v = 8


=> x = 8v --- (1) Time taken to cross the platform 264 m long = 20 s (x+264)/v = 20 => x + 264 = 20v ---(2) Substituting equation 1 in equation 2, we get 8v +264 = 20v => v = 264/12 = 22 m/s = 22×36/10 km/hr = 79.2 km/hr

13. Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction? A. 10 B. 25 C. 12 D. 20


Answer : Option C

Explanation :


speed of train1 = 120/10 = 12 m/s speed of train2 = 120/15 = 8 m/s if they travel in opposite direction, relative speed = 12+8 = 20 m/s distance covered = 120+120 = 240 m time = distance/speed = 240/20 = 12 s

14. Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is A. 2 : 3 B. 2 :1 C. 4 : 3 D. 3 : 2


Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train = 16 − − √ 9 √ = 4:3


15. A train having a length of 1/4 mile , is traveling at a speed of 75 mph. It enters a tunnel 3 ½ miles long. How long does it take the train to pass through the tunnel from the moment the front enters to the moment the rear emerges? A. 3 min B. 4.2 min C. 3.4 min D. 5.5 min


Answer : Option A

Explanation :

Total distance = 3 ½ + ¼ = 7/2 + ¼ = 15/4 miles Speed = 75 mph Time = distance/speed = (15/4) / 75 hr = 1/20 hr = 60/20 minutes = 3 minutes

16. A train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the train? A. 270 m B. 210 m C. 340 m D. 130 m


Answer : Option A


Explanation :

Speed= 72 kmph = 72×10/36 = 20 m/s Distance covered = 250+ x where x is the length of the train Time = 26 s (250+x)/26 = 20 250+x = 26×20 = 520 m x = 520-250 = 270 m

17. A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train? A. 62 m B. 54 m C. 50 m D. 55 m


Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph x/9 = ( v-2)(10/36) ---(1)


x/10 =( v-4) (10/36) --- (2) Dividing equation 1 with equation 2 10/9 = (v-2)/(v-4) => 10v - 40 = 9v - 18 => v = 22 Substituting in equation 1, x/9 = 200/36 => x

= 9×200/36 = 50 m

18. A train is traveling at 48 kmph . It crosses another train having half of its length , traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform? A. 500 m B. 360 m C. 480 m D. 400 m


Answer : Option D

Explanation :

Speed of train1 = 48 kmph


Let the length of train1 = 2x meter Speed of train2 = 42 kmph Length of train 2 = x meter (because it is half of train1's length) Distance = 2x + x = 3x Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s Time = 12 s Distance/time = speed => 3x/12 = 25 => x = 25×12/3 = 100 meter Length of the first train = 2x = 200 meter Time taken to cross the platform= 45 s Speed of train1 = 48 kmph = 480/36 = 40/3 m/s Distance = 200 + y where y is the length of the platform => 200 + y = 45×40/3 = 600 => y = 400 meter


19. A train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? A. 320 m B. 190 m C. 210 m D. 230 m


Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s time = 9s Total distance covered = 270 + x where x is the length of other train (270+x)/9 = 500/9 => 270+x = 500 => x = 500-270 = 230 meter

20. Two trains, each 100 m long are moving in opposite directions. They cross each other in 8 seconds. If one is moving twice as fast the other, the speed of the faster train is




Answer : Option B

Explanation :

Total distance covered = 100+100 = 200 m Time = 8 s let speed of slower train is v . Then the speed of the faster train is 2v (Since one is moving twice as fast the other) Relative speed = v + 2v = 3v 3v = 200/8 m/s = 25 m/s => v = 25/3 m/s Speed of the faster train = 2v = 50/3 m/s = (50/3)×(36/10) km/hr = 5×36/3 = 5×12 = 60 km/hr

21. Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards


P at a speed of 25 kmph. At what time will they meet? A. 10.30 a.m B. 10 a.m. C. 9.10 a.m. D. 11 a.m.


Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am Distance covered by train starting from P in x hours = 20x km Distance covered by train starting from Q in (x-1) hours = 25(x-1) Total distance = 110 => 20x + 25(x-1) = 110 => 45x = 135 => x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

22. A train overtakes two persons walking along a railway track. The first person walks at 4.5 km/hr and the other walks at 5.4


km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train? A. 81 km/hr B. 88 km/hr C. 62 km/hr D. 46 km/hr


Answer : Option A

Explanation :

Let x is the length of the train in meter and y is its speed in kmph x/8.4 = (y-4.5)(10/36) ---(1) x/8.5 = (y-5.4)(10/36) ---(2) Dividing 1 by 2 8.5/8.4 = (y-4.5)/ (y-5.4) => 8.4y - 8.4 × 4.5 = 8.5y - 8.5×5.4 .1y = 8.5×5.4 - 8.4×4.5 => .1y = 45.9-37.8 = 8.1 => y = 81 km/hr


23. A train , having a length of 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that of the train A. 10 sec B. 8 sec C. 6 sec D. 4 sec


Answer : Option C

Explanation :

Distance = 110 m Relative speed = 60+6 = 66 kmph (Since both the train and the man are in moving in opposite direction) = 66×10/36 mps = 110/6 mps Time = distance/speed = 110/(110/6) = 6 s

24. A 300 metre long train crosses a platform in 39 seconds while it crosses a post in 18 seconds. What is the length of the platform? A. 150 m B. 350 m C. 420 m D. 600 m



Answer : Option B

Explanation :

Length of the train = distance covered in crossing the post = speed × time = speed × 18 Speed of the train = 300/18 m/s = 50/3 m/s Time taken to cross the platform = 39 s (300+x)/(50/3) = 39 s where x is the length of the platform 300+x = (39 × 50) / 3 = 650 meter x = 650-300 = 350 meter

25. A train crosses a post in 15 seconds and a platform 100 m long in 25 seconds. Its length is A. 150 m B. 300 m C. 400 m D. 180 m


Answer : Option A

Explanation :

Assume x is the length of the train and v is the speed x/v = 15 => v = x/15


(x+100)/v = 25 => v = (x+100)/25 Ie, x/15 = (x+100)/25 => 5x = 3x+ 300 => x = 300/2 = 150

26. A train , 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)? A. 440 m B. 500 m C. 260 m D. 430 m


Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the train Time = 1 minute = 60 seconds Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s Distance/time = speed (800+x)/60 = 65/3


=> 800+x = 20×65 = 1300 => x = 1300 - 800 = 500 meter

27. Two train each 500 m long, are running in opposite directions on parallel tracks. If their speeds are 45 km/hr and 30 km/hr respectively, the time taken by the slower train to pass the driver of the faster one is A. 50 sec B. 58 sec C. 24 sec D. 22 sec


Answer : Option C

Explanation :

Relative speed = 45+30 = 75 km/hr = 750/36 m/s = 125/6 m/s We are calculating the time taken by the slower train to pass the driver of the faster one .Hence the distance = length of the smaller train = 500 m Time = distance/speed = 500/(125/6) = 24 s

28. Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely


passes a man sitting in the slower train in 5 seconds, the length of the fast train is : A. 19 m B. 277 9 m C. 132 9 m D. 33 m


Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s Time = 5 s Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 277 9 m = length of the fast train

29. Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is A. 42 B. 36 C. 28 D. 20


Answer : Option B

Explanation :


Distance covered = 120+120 = 240 m Time = 12 s Let the speed of each train = v. Then relative speed = v+v = 2v 2v = distance/time = 240/12 = 20 m/s Speed of each train = v = 20/2 = 10 m/s = 10×36/10 km/hr = 36 km/hr

30. A train 108 m long is moving at a speed of 50 km/hr . It crosses a train 112 m long coming from opposite direction in 6 seconds. What is the speed of the second train? A. 82 kmph B. 76 kmph C. 44 kmph D. 58 kmph


Answer : Option A

Explanation :

Total distance = 108+112 = 220 m Time = 6s Relative speed = distance/time = 220/6 m/s = 110/3 m/s


= (110/3) × (18/5) km/hr = 132 km/hr => 50 + speed of second train = 132 km/hr => Speed of second train = 132-50 = 82 km/hr

31. How many seconds will a 500 meter long train moving with a speed of 63 km/hr, take to cross a man walking with a speed of 3 km/hr in the direction of the train ? A. 42 B. 50 C. 30 D. 28


Answer : Option C

Explanation :

Distance = 500m Speed = 63 -3 km/hr = 60 km/hr = 600/36 m/s = 50/3 m/s Time taken = distance/speed = 500/(50/3) = 30 s


Maths solved problems.pdf

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Maths solved problems.pdf