Quantitative Aptitude - Vol 2

A. Krishnamurthy M.Sc. M.Phil Reproduction of this in any form without express permission is strictly prohibited. Notes are not for public distribution and are intended to supplement class room instruction. Mail-Id: [email protected]

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CHAPTER-1

Analytical Reasoning Analytical Reasoning forms an important part of generally all Campus Placement Test. The questions in this section can either test analytical or logical reasoning. This booklet will concentrate on the former. The thinking process associated with such problems resembles solving a case, wherein the clues here are in the form of certain conditions, which may or may not be mutually exclusive. The best way to get a hang of these problems is to work on them continuously. Initially, they may deter you but once the student follows the rules given below and solves many problems, it will appear simple.

Games1 in this section can both be verbal and numerical. The first part of this booklet will consist of the basics behind solving such problems, such as the elements forming the problem and the rules to follow to crack such problems. The second part will deal with verbal analytical reasoning wherein a few varieties of games / puzzles / problems will be provided, which will be followed by a discussion on solving the same. Questions may not be provided for a few of these games, as solving the conditions directly will give us the solution. The third part will deal with numerical reasoning, which basically requires a strong sense of arithmetic. The methodology for solving each game is only suggestive and the student can use his own discretion to solve it in an easier fashion.

Conditions As mentioned above, conditions are a set of clues, which when analysed, provide the solution for the problem. These conditions cannot be violated and all these conditions have to be analysed before attempting any question relating to the problem. These conditions are not necessarily in order and therefore, it becomes imperative for the student to go through all of them before identifying the important ones.

Rules of the Game There are a few rules of this game, which help in solving AR problems systematically. These rules help you frame the problem in the right perspective, organise your thoughts and finally, crack the problem and its associated set of questions. A few such rules have been given below.

1. Read, Read, and Read: Reading the problem and its conditions is a pre-requisite. The student has to spend sometime reading and analysing the statements. Read the conditions carefully. Please dont over-read into the conditions and statements. Its also important to study the implications of the conditions, i.e. try to analyse it from a what-if angle (though not the case in all the problems)

2. Identify the most important condition, if any. Since the conditions are not given in any specific order, the student has to peruse through all of them and identify the key condition, which will provide the platform from where the problem can be solved. If there are any restrictions in the problem, like for example, one of the elements has always to be at one place and then start by putting that element there. In case of ordering problems (wherein an order of the elements has to be ascertained), this becomes especially important.

3. Picturise the information. If, for example, the problem deals with 5 people standing at a bus stop, the student must be able to juxtapose it in his analysis. Since there wont be sufficient place to work out these kinds of problems, picturisation helps in arranging the information

4. Organise the information. Picturisation is the first step in organising information. Once the student has an idea of the problem, he has to organise the information in an easily comprehensible manner, i.e. he should try to associate it with whatever he feels comfortable. The student should put every bit of detail on paper, say, make notes. Some ways of organising information are


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Grids/Tables: Once the student has identified a particular structure of relationships, all the information can be arranged in the form of a multi-columnar grid. The different elements of the problem, as given by the conditions, must be entered onto the grid to facilitate a logical thinking process. It also helps in eliminating some of the answers provided in the solution.

Symbols/Notations: If the student is comfortable substituting symbols for the different conditions, he should do so, as, this, like tables, provides a simplistic view of the problem. These may differ according to the type of problem and the comfort level of the student.

5. Elimination of Answer choices. Some of the above rules do help one in eliminating answer choices. It is generally not advisable in case of AR to start off by eliminating choices. As you will see, many problems require a bottom-up approach, but, that is only after a careful perusal and analysis of the conditions. Eliminate choices when the conditions are insufficient to answer the questions.

6. Be careful of the language used. Certain words or phrases are oft-repeated and these can well form important clues in framing the solution.

7. Finally, answer all questions at the same time, if the game demands


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SOLVED EXAMPLES

The World This Week This game deals with scheduling a program over a time period, say a week or a year. It consists of a few conditions, from which the order of telecast has to be identified. These are problems, which require the identification of a particular order or a schedule. Other variants of this game include

Identifying people staying in an apartment complex, Ordering of books in a shelf etc. Example

A TV channel is going to telecast 6 soppy soaps over the week Sob, Cry, Weep, Wail, Lament and Moan. One of these will be telecast each day from Monday to Saturday, with Sunday being a FunDay. The conditions specified by the Boss are as follows:

Sob must be telecast earlier in the week than Lament. Weep must be telecast on Tuesday Wail must be telecast on the day immediately before or immediately after the day on which Cry

is telecast

Questions

Q1. If Cry is telecast on Thursday, the earliest day on which Lament can be featured is

a) Monday

b) Tuesday

c) Wednesday

d) Friday

e) Saturday

Q2. If Moan is to be telecast on Friday, Sob must be telecast on

a) Monday

b) Tuesday

c) Wednesday

d) Thursday

e) Saturday

Q3. If Wail is to be featured on Thursday, the latest day on which Sob can be telecast is

a) Monday

b) Tuesday

c) Wednesday

d) Friday

e) Saturday

Q4. Which of the following soaps can be telecast on Monday?

a) Cry

b) Moan

c) Lament

d) Weep

e) Wail


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Q5. If Moan is to be telecast on Thursday, which of the following is true?

a) Sob must be telecast on Wednesday

b) Cry must be telecast on Saturday

c) Wail must be telecast exactly two days after Lament is telecast

d) Lament must be telecast on Wednesday

e) Lament must be telecast later in the week than Moan

Q6. If Moan is to be telecast on Friday, what is the total number of acceptable schedules available to the TV channel?

a) 1

b) 2

c) 3

d) 4

e) 5

Solution/Discussion

We shall begin discussing the solution by following the rules that have been specified above. A thorough reading of the problem provides us with the following leads

It is imperative in these kinds of ordering problems that we begin with conditions that fix the exact position of one or more elements, and then work towards narrowing the possibilities for other elements. When the order cannot be determined by the condition, then, attack the questions and proceed with eliminating choices.

There are 3 conditions, of which, the second condition that Weep should be telecast on Tuesday can be considered to be the important condition since it provides an ideal base to work the problem.

The first condition suggests that Sob can be telecast on any day except Saturday, and Lament can be telecast on any day except Monday. Since Tuesday is already ruled out, the days of telecast for these two soaps are down to four.

From the third condition, we can ascertain that Cry cannot be telecast on Monday because Weep is to be telecast on Tuesday. Therefore, Cry can be telecast on any day from Wednesday to Saturday.

Let us now make a table to help organise this information

Days Monday Tuesday Wednesday Thursday Friday Saturday

Soaps Weep

We can proceed in the following manner

Calculations

Monday: No Lament and Cry.so its either Sob, Wail or Moan

Wednesday to Friday: All except Weep

Saturday: All except Weep and Sob

Since there can be many combinations of the telecast of these soaps, we can proceed with the problem by reading the questions. This approach will also help in eliminating answer choices.


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Solution 1: For every question, we can fill the table. As for this question, if Cry is telecast on Thursday, the table would look like this


Soaps Weep Cry

From our calculation earlier, we have seen that the earliest day that Lament can be telecast is on Wednesday. WE now check whether it is feasible with Cry being telecast on Thursday.


Soaps Sob Weep Lament Cry Wail Moan

We find that it is indeed feasible and therefore, the answer to this question is Wednesday

Solution 2: Moan has not figured in our calculations so far. So if we assume that Moan is telecast on Friday, then we have a table that looks like this


Soaps Weep Moan

Moans telecast on Friday provides us with only one combination of days for the telecast of Cry and Wail, Wednesday and Thursday, since they have to be telecast on consecutive days. And from the first condition, which specifies that Sob must be telecast earlier in the week than Lament, Sob must be telecast on Monday.


Soaps Sob Weep Cry Wail Moan Lament

The answer to the second question is Monday

Solution 3: We make a table again with Wail being telecast on Thursday.



This table is similar to the one used to arrive at Solution 2, but the question has been framed differently. Here we are supposed to find the latest day on which sob can be telecast and not the earliest day. From our earlier calculation, we find that Sob can be telecast on Monday and on any day from Wednesday to Friday. Since Wail is telecast on Thursday, Cry should be telecast on Wednesday or Friday. Since there is no condition relating to Moans telecast, Moan can be telecast on Monday and Sob on Friday. Do not forget that Lament has to be telecast after Sob.


Soaps Moan Weep Cry Wail Sob Lament

The answer to the third question is Friday


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Solution 4: From our earlier calculation, we can easily say that Moan is the only soap, among the answer choices, that can be telecast on Monday.

Solution 5: This question requires us to visualise different alternatives and ascertain their veracity.


Soaps Weep Moan

We can do this by analysing each of the answer choices. Never forget the calculations

Sob must be telecast on Wednesday: If Sob is telecast on Wednesday, Lament will have to be telecast on wither Friday or Saturday, which is not possible because Cry and Wail have to be telecast on consecutive days. Therefore, this is untrue.

Cry must be telecast on Saturday: Cry can be telecast on Friday or Saturday. Though this is not entirely untrue, it is an option we should keep in mind till we arrive at the exact answer.

Wail must be telecast exactly two days after Lament is telecast. Since, in this case, Lament has to be telecast on Wednesday, Wail can be telecast either on Friday or Saturday, i.e. either 2 or 3 days after lament is telecast. So, even this option is not entirely untrue and should be kept in abeyance till we arrive at the final answer.

Lament must be telecast on Wednesday: From the argument for the earlier answer choices, we find that Lament has to be telecast on Wednesday alone.

Lament must be telecast later in the week than Moan: this is not possible since Cry and Wail have to be telecast on consecutive days and only Friday and Saturday are available for them.

Therefore, after examining the answer choices, we find that the correct answer is (d).

Solution 6: We again make use of a table here, with Moan appearing on Friday


Soaps Weep Moan

The different schedules can be tabulated as under



Sob Weep Wail Cry Moan Lament

Since there are only 2 acceptable schedules, the answer is 2.


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Round Table In this game, there are a few people sitting around a table. The basic theme here is to identify, from the conditions, the exact placement of a person around the table.

Example:

There are six people, A, B, C, D, E and F sitting around a table.

1. The person sitting to the left of A faces D.

2. B and E sit on either side of C

3. A and B do not face each other.

Questions generally relate to the seating arrangement around the table and therefore are very simple. For example, Who sits in front of B? Who faces B if A and D swap places?

Solution/Discussion:

This is a very simplistic version of the problem. The best and the simplest method to solve round-table problems is to start by drawing a table and seating the students according to the conditions.

For example, from the first condition, by seating A at the top of the table, we can ascertain the exact seating position of D.

Since the identity of the person seated next to A cannot be ascertained with one condition alone, we move on to the second condition. After analysing the second condition along with the diagram above, we can ascertain the exact seating position of E, as illustrated by the following diagram.

The third condition that A and B do not face each other means that A is sitting to the left of A and facing D and E is facing A (from the second condition above). And therefore the final seating arrangement would like this

A

D

A

D C

A

D

F B

C

E


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Different Strokes This game is a variant of the above, with more number of elements. One thing to remember, the more the conditions, the easier it is to solve the problem. There are many variants of this game, like people standing in a bus-stop/railway station etc.

Example

There are 5 houses coloured Red, Green, Black, White and Blue in a row. Each of the houses is occupied by bachelors of different professions (Singer, Accountant, consultant, Software engineer and lawyer) and each of them has a different vehicle in which to commute to office. The colour of the vehicle is different from not only that of his house but also of his neighbours house.

1) A lives in a Blue House, which has the same number of houses on either side.

2) B is an accountant and the colour of his car is similar to the colour of the lawyers house.

3) E has a black car and his house is next to the Green house

4) The person living in the White House is a consultant

5) D is a software engineer living in the Black House

6) The lawyer drives a white car and he stays next to the Red House

1. Who is the lawyer?

a) A

b) E

c) D

d) C

2. Which of the following gives an orderly description of the houses?

a) GRBWB

b) BGWBR

c) GRBBW

d) WBBGR

3. What is the colour of the consultants car?

a) Black

b) Blue

c) Green

d) Red

4. Which of the following lives in the house next to B?

a) A

b) D

c) C

d) E

Solution/Discussion

The best method to solve these types of problems is to organise the information in the form of a multi-dimensional grid. The dimensions as provided in the case are: People, Profession, Houses and Cars.

We begin by filling in the information known to us in the correct boxes. First, we make individual grids connecting each of the dimensions and then consolidate them in one single grid based on our analysis.


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From condition 1, we know that A lives in the blue house and that it has two houses on either side. Grid 1 shows the relationship

Though there are two parts to the second condition, we take them one at a time. We match B with that of an accountant. Grid 3 shows the relationship

E has a black car and his house is next to the Green house, which means that the colour of Es house is neither black (the case specifically mentions that the house and a vehicle of a person cannot be of the same colour) nor green nor blue. Grid 2 shows this relationship

D is a software engineer living in the black house. From this, we ascertain that D is not the neighbour of E since E has a black car (refer to the case which specifically mentions that the colour of the vehicle is not similar to the colour of the neighbours house). We can now say that E and D live on either side of A.

Now for the remaining two conditions:

The lawyer drives a white car and stays next to the red house. Therefore the colour of the lawyers house is not white, black or red (which, in turn, means that the colour of Bs car is not white, black, or red)

The person living in the White House is a consultant. Therefore, the consultant is not the neighbour of the lawyer.

If we combine all the dimensions into one single grid, Grid 6, we can make the following guesses:

Either A or C should be a lawyer

Either C or E is a consultant.

E should be staying either in the white house or the red house.

The colour of Bs car is either Blue or Green.

The colour of the Lawyers house is Blue or Green. Since the lawyer stays next to the red house and drives a white car, his house can neither be red nor white in colour. This combined with the above mean that B is not the neighbour of the lawyer and A.

Since Es house is not green, black or blue in colour, he should be the neighbour of B. Since he stays next to the Green house, the colour of Bs house is Green, which means that B drives a blue car. This, in turn, implies that the colour of the lawyers house is blue, i.e. A is the lawyer. Therefore, the singer lives in the red house.

The lawyer stays next to the red house and therefore, the colour of Es house is red, which means that he is the singer and C is the consultant.

Grid 1

House Green Red Blue Black White

Name B E A D C

Grid 2

Car White Blue Green Red Black

Name A B C D E

Grid 3

Profession Lawyer Accountant consultant Engineer Singer

Name A B C D E

Grid 4


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Profession Singer Accountant Lawyer Consultant Engineer

House Red Green Blue White Black

Grid 5

Profession Singer Accountant Lawyer Consultant Engineer

Car Black Blue White Green Red

Grid 6

Name House Car Profession

A Blue Green Lawyer

B Green Blue Accountant

C White Green Consultant

D Black Red Engineer

E Red Black Singer

To show the same information in a multi-dimensional grid,

House GREEN RED BLUE BLACK WHITE

Person B E A D C

Profession ACCOUNTANT SINGER LAWYER ENGINEER CONSULTANT

Vehicle BLUE BLACK WHITE RED GREEN

With the above grid, all the questions can be answered.

This question can also be solved using symbols or notations. For example, the fact that the colour of Bs car is similar to the lawyers house can be recorded as

Bs car = Lawyers house, and so on. But, it is required that students be comfortable using notations in place of grids etc.


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Conditions Galore These games are slightly different from the others in that the problem does not specify the exact number of places to assign elements, rather, some elements have to be assigned for some questions and some others for other questions. These problems are of the selection variety wherein you are asked to distribute elements into groups. The strategy for this is similar to that of the earlier problems, namely ordering/scheduling problems, but because of the if-then relationship (or the probability relationship) prevalent, it requires a more organised solution process, especially with the use of symbols. Some of these relations are given below. Please note the implications of the a single statement.

Eg 1: If A is in the team, then B also has to be in the team.

If the team consists of A, then it will also have B, i.e. A, therefore B.

But, if B is in the team, it doesnt mean that A is in the team, i.e. B, therefore A is not a correct way of reasoning.

Similarly if B is not in the team, it also means that A is not in the team, i.e. not B, therefore not A. The converse, however, isnt true, i.e., if A is not in the team, it doesnt mean that A is not in the team, i.e. not A, not B is a wrong way of reasoning.

Eg 2: In a vote for a resolution, A and B never both aye.

This means that both never vote aye, but they can both vote nay. Please dont jump into conclusions that since they cannot both vote aye, they can also never both vote nay.

Example:

A track coach is deciding which and how many of her athletes L, M, N, O, P, R and S will compete in an upcoming track meet. She will decide according to the following guidelines:

1. If L competes, M must compete

2. If M and N both compete, O cannot compete

3. If N and O both compete, R cannot compete

4. If O competes, either P or S must compete

5. Either P or R must compete, but they cannot both compete

6. P and S cannot both compete.

Questions

1. If only 3 athletes can compete in the track meet, which of the following could be that group of athletes?

a) LMN

b) MPS

c) MPR

d) NOP

e) NOR

2. If O and S both compete in the track meet, which of the following must be true?

a) N competes

b) P competes

c) R competes

d) L does not compete

e) M does not compete


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3. If O and R both compete in the track meet, which of the following cannot be true?

a) M competes

b) N competes

c) S competes


e) P does not compete

4. If L and N both compete in the track meet, what is the maximum number of athletes who can compete

a) 3

b) 4

c) 5

d) 6

e) 7

5. If S competes in the track meet, which of the following combinations of 3 athletes can be among those who also compete?

a) LMP

b) LNO

c) LOP

d) MOR

e) NOR

Solution/Discussion

This problem requires the simplest of tables. The idea here is to identify who can compete together and who cannot and the best way to do that would be to draw a table which separates one from the other, i.e

Compete Not compete

Another simple method would be to symbolise the clues. But, as mentioned before, only people who are comfortable with symbolising should do so to avoid any confusion.

Some of the symbols used are:

Arrows: to indicate that if L competes, M must compete. L M Boxes: To indicate that if M and N both compete, O cannot compete. ~O Not equal: To indicate that P and S cannot both compete. P/S

A combination of symbols can also help in arriving at the solution.

As per the tabular method illustrated above, we can answer each question by deciding on whether the athletes compete together or not. This is another problem which can be solved using the questions, and thereby, the process of elimination. For example, if we consider the second question in the problem,

2. If O and S both compete in the track meet, which of the following must be true?

a) N competes

b) P competes

MN


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c) R competes


e) M does not compete

Using the table and the clues given above, we can arrive at the answer. The last clue clearly mentions that P and S cannot both compete. Therefore, put P in the Not compete column. Another clue specifies that either P or R must compete and not both. Since P is not competing, R can compete. So put R in the Compete column. Using these two clues alone, we can arrive at the answer choice, which is, R competes. We can also see that N can also compete in this case, but, if we read the question again it says which of the following must be true, and not can or may be true. Also, this is the logical answer, which appears on the face of the conditions. In such cases, it is advisable to take things at face value and leave the implications behind.

Compete Not compete

O S R P

Similarly, for the other questions.


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Liar Liar These are games which contain statements by a few people, which are either true or false. From these statements, the student has to answer a few questions depending on the kind of condition given in the problem. This problem requires considerable amount of time and the student has to be ready to invest it during the exam. It may seem confounding in the beginning but once the thinking process is in place and the approach methodical, it will appear simple.

Its time for the What-What island, where the inhabitants answer any question with two sentences; one of which is true and the other is false.

You are looking for Venkats house and you meet 3 people Anand, Ravi and Som. You ask them, Which is Venkats house?

Anand says: Venkats house is No.9. I am his neighbour.

Ravi says: Anand is not my neighbour. Anand and som live in the same house.

Som says: Venkats house is not No.9. Anand is Ravis neighbour.

There are only two houses and four people in What-what. Two people live in each house.

Q1. From the above, you can decide that

a) Venkat stays in house no.9

b) Venkat does not stay in house no. 9

c) Venkat does not stay in what-what

d) Ravi and Som stay together.

Q2. Who stays with Anand?

a) Ravi

b) Venkat

c) Som

d) Cant say

Solution/Discussion

This question is very different from the previous ones and requires some time on the part of the student to understand and solve it. A methodical process would be useful in this case.

To begin with, start with the first speaker and label one sentence each as true or false, similarly for the other speakers. Ascertain whether the labeling makes sense at the end of the third speaker.

For example, We label Anands first sentence as true (T) and the second as false (F) and similarly for Ravis and Soms sentences. The final output will look like:

Venkats house is no.9 and Anand is not his neighbour

Anand is not Ravis neighbour and Anand and som do not live in the same house.

Venkats house is not No.9 and Anand is not Ravis neighbour.

We can see the contradiction evident in the first and third sentences relating to Venkats house. Since this doesnt make much sense, we re-label the sentences accordingly, taking care of the initial contradiction.

If we label Anands first sentence as true, Soms first sentence has to be false, and thereby Soms second sentence is true. This means that Anand is Ravis neighbour and therefore, Ravis first sentence is false. But here lies another contradiction if Anand and Som live in the same house, then Ravi and Venkat live


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in the other house, but Anand;s second sentence specifically says that He is not Venkats neighbour. Therefore, even this is a wrong choice.

Now, we try the other way, by labeling Anands first sentence as false and Soms first sentence as true. By the chain of thoughts, it is evident that Anand is not Ravis neighbour and therefore Anand and Ravi live in the same house, which is also true in case of the second sentence of Anand. Therefore, Venkat does not stay in House no. 9 and Ravi stays with Anand.

Alpha-Numeric Here, numbers are coded as words or vice-versa and from the rules of this game, exact codes have to be identified. Variants of this game are

the usage of words only as codes substituting symbols for relations, i.e. if a*b means a is the father of b, and a@b means a is the sister of b, then how does one denote a paternal aunt

Two English words are codified as follows. Each number represents only one letter and each letter is represented by only one number.

Word 1: 8 3 7 6 3 2 9

Word 2: 3 6 7 5 8 4 1 6

The following rules are known to the person decoding them.

I. Letters T and R occur exactly three times

II. Letters S and A occur exactly two times.

III. Letters E, P, O and N occur exactly once.

IV. One of the words starts with T and the other with S.

V. E occurs only in word 1.

Solution/Discussion

We proceed methodically analysing one condition at a time.

From the first condition, it is clear that T or R should stand for 3 or 6.

From the second condition, it is clear that S or A should stand for 8 or 7

We will ignore the third condition for sometime because we cannot conclude anything from it.

From the fourth condition, we can conclude that T stands for 3 and S stands for 8. Therefore, R stands for 6 and A stands for 7. E stands for either 2 or 9. O,P and n can be substituted for 5,4 and 1 in Word 2.

Decoding the numbers individually, we find the following

Word 1: 8 3 7 6 3 2 9

S T ART

Word 2: 3 6 7 5 8 4 1 6

T RA S R

In this particular question, the words are not those that make some meaning. Its more open-ended than a word that has a meaning attached to it. I say so because it is not possible to decipher all the numbers correctly, and therefore, its left to the student as to what he makes of the numbers and words which cannot be correctly identified.


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Absolute Relatives In this game, again a variant of the first 2 games, from the data given, relationships have to be identified.

A, B, C, D, E, F, G and H are people who are related as below

I. A is the father of 2 children C (male) and D (female)

II. H is the mother of two children E (male) and F (female)

III. B is Es mother-in-law

IV. D is the daughter-in-law

V. Es wife is Fs sister-in-law

VI. Es son will also be As grandson and Cs daughter will also be Hs grand-daughter

Q1. Who is As wife?

a) H

b) F

c) B

d) E

e) NONE OF THESE

Q2. Who is Hs husband?

a) B

b) A

c) D

d) F

e) NONE OF THESE

Q3. Who is Ds mother-in-law?

a) B

b) H

c) F

d) E

e) NONE OF THESE

Q4. As daughter will also be

A. Bs daughter in law

B. Es wife

C. Hs daughter

a) A only

b) B only

c) C only

d) A and B only

e) B and C


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Q5. D is

a) Wife of A

b) Wife to E

c) Daughter of G

d) Fs sister

e) None of these

Q6. Es mother is also

a) Ds mother

b) Cs grandmother

c) Bs sister

d) Gs wife

e) As wife

Q7. If X is Cs son and Y is Es daughter then X and Y must be

a) Brother & Sister

b) Husband and Wife

c) Ds children

d) Hs grand children

e) None of these

Q8. Out of 8 people, how many must be male?

a) 4

b) 3

c) 5

d) 2

e) 6

Q9. Out of 8 people, how many must be female?

a) 2

b) 3

c) 4

d) 5

e) 6

Q10. Which of the following represent a correct pair of husband and wife?

a) A&H

b) B&G

c) E&D

d) A&F

e) C&D


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Solution/Discussion

How better to solve a family problem than by using a family tree. We can use symbols to denote the different relationships. Males and Females can be denoted using m and f respectively.

The following symbols can be used: (Indicative symbols)

Father: up-arrow

Mother: up-arrow

Mother-in-law: diagonal up arrow

Father-in-law: diagonal up arrow

Daughter-in-law: diagonal down arrow

Son: down-arrow

Daughter: down-arrow

Using the conditions given above, the relationships can be symbolised as follows:

Am Bf Hf

Cm Df Em Ff

Es wife Fs sister-in-law

Am H

Em Cm

Son daughter

From the symbols above, we can infer the following:

Since A has 2 children, C & D, E must be the son-in-law.

Since H has 2 children, E & F, C must be the daughter-in-law (which is also mentioned in the conditions)

Es wife therefore is D, who is the brother of C, who is Fs husband.

Since G is the only missing link, G must be the husband of H, just as B is the wife of A

The families are : AB, GH, ED and CF. There are an equal number of males and females in the family. Therefore, X and Y are the grand-children of AB and GH. The other questions can be solved similarly.


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Map Games These types of games describe connection between the elements specified in the game. Symbols are very much useful in this type of games.

In a message relay system there are exactly seven terminals F,G,H,J,K,L and M. A terminal can transmit any messages initiated by that terminal as well as any messages received from others, but only according to specific rules:

Messages can be transmitted in either direction between G and H, in either direction between J and M, and in either direction between K and L.

Messages can be transmitted from F to K, from H to J, from K to G, From M to F, and from M to H.

Q1. Which of the seven terminals can transmit messages directly to the greatest number of terminals?

a) F b) H c) J d) K

e) M

Q2. If a message initiated by G is to reach K, and is to be transmitted to no more terminals than necessary, it must be transmitted to a total of how many terminals, other than G and K?

a) 1 b) 2 c) 3 d) 4

e) 5

Q3. A message from H that eventually reaches L must have been transmitted to all of the following terminals except

a) F b) G c) J d) K

e) M

Q4. If J is removed from the message relay system for a day, it is still possible for a message to be transmitted on that day all the way along a route from

a) F to H b) G to K c) G to M d) H to K

e) L to M

Q5. If K is removed from the message relay system for a day, which of the following terminals cannot receive any messages from any other terminal on that day

a) F b) G c) H d) J

e) L

Q6. A message can travel along two alternative routes that have no terminal in common except the initiating terminal and the final recipient terminal if the initiating terminal and the final recipient terminal, respectively, are

a) G and J b) G and L c) H and L d) K and M

e) M and G

Q7. A message being transmitted along which of the following routes must reach each of the seven terminals at least once?

a) F to G and then to M b) J to H and then to L

c) L to H and then to M d) M to G and then to K

e) M to L and then to F

Solution/discussion

Arrows can be used to show the connection between the different elements. One-way or Two-way arrows can be used to differentiate between the connections.


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G H

I M

K L

F K

H J

K G

M F

M H

Using these connections, we make a link connecting these elements.

With this combined map, we can proceed to answering the questions. The process of elimination can also be used in this case. If there arises a need to redraw the connection chart for a particular question, go ahead and do it.

M J F H G K L


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Card Games Card games require a basic knowledge of playing cards, the suits in cards, etc. For the starter, there are 4 different suits, in 2 different colours Spades and Clubs, in black, and Hearts and Diamonds, in red. There are 13 cards in all, in each suit. The cards are Ace, 2,3,4,5,6,7,8,9, 10, Jack, Queen and King. The last three are picture cards. Though card and die games appear more in quantitative ability in the form of probability questions, they have known to appear in reasoning questions also.

From a normal pack of playing cards, 20 cards were used in a game. These were all the four suits of Ace, King, Queen, Jack and number ten. A, B, C and D are the players. Each player has all the five cards, in one or different suits. Cs five cards were in 3 different suits and consisted of 3 red and 2 black cards. Ds five cards were also in 3 different suits, his ace being in the same suit as his queen, and his king in the same suit as his jack. B held more than one black car. As five cards were all in the same suit. C held the king of Spades, and D, the ten of diamonds.

Q1. Between them, B,C and D held at least ..black cards

a) 6

b) 7

c) 4

d) 3

Q2. Who held the queen of hearts?

a) C

b) D

c) A

d) B

Q3. Who held the Ace of diamonds?

a) A

b) B

c) C

d) D

Q4. As five cards are all

a) Spades

b) Clubs

c) Hearts

d) Diamonds

Solution/Discussion

Firstly, we organise the given information. For our convenience, lets denote the different suits as S, D, H and C. Secondly, we start with identifying the important condition, if any. Lets begin by putting all the information together.

Cs cards were in 3 different suits and consisted of 3 red and 2 black cards. C held the King of spades.

Ds cards were also in 3 different suits, with ace and queen of the same suit and the king and jack of another suit. He has the ten of diamonds. Because 10 is of a different suit, the other cards may belong to spades, hearts or clubs.


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B held more than one black card.

All of As cards were in the same suit. Since C and D have one each of spades and diamonds, he must have either all hearts or all clubs.

Since we havent been able to glean more out of the conditions, we follow a bottom-up approach, and analyse the questions.

To answer the first question, we know that C has 2 black cards, B has more than one black card. 4 black cards are already accounted for. From what we have discussed above, we know that 2 of Ds cards must be black (either spades or clubs). Therefore, B,C and D account for at least 6 black cards (Please note the word at least, 6 may not be the ultimate number of black cards accounted for by them).

Since they account for 6 black cards, they must include at least 1 spade and 1 club. Therefore, all of As cards are hearts. So, the other 2 suits held by D are Spades and Clubs. Since C takes the King of spades, King and Jack must belong to clubs and Ace and Queen must belong to spades. Now we can make a small table with what we know.

Person Cards

A A K Q J 10 all hearts

B

C K s

D 10 d A Q s K J c

The cards that remain in each suit are:

Spades: J and 10

Clubs: A, Q and 10

Diamonds: A, K, Q and J

C has 3 red and 2 black colour cards. The only remaining cards in red are the ones in diamonds above.

Since C already has a K, he must have the A,Q and J of diamonds, which means that he also has the

10 of clubs. Now that C is over, all the remaining cards are assigned to B. The final table will look as

follows:

Person Cards

A A K Q J !0 h

B K d J 10 s A Q c

C K s A Q J d 10 c

D 10 d A Q s K J c


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Number Game This game requires a good knowledge of simple arithmetic, mainly LCM and factorisation. The problems will generally be in the form of a grid, wherein based on the conditions given, the grid has to be filled with the appropriate numbers.

One such game is illustrated below:

There are nine letters A to I, each represented by a different number from 1 to 9. The grid is positioned as below:

A B C

D

E F G

H

I

Each of the combinations of letters, A+B+C, C+D+E, E+F+G and G+H+I is equal to 13. Match the letters to the numbers.

Solution/discussion

We can begin by making a grid similar to that made by the letters above.

Here, distinction has to be made between A,B, D, F, H and I on one side and C,E, and G on the other, because the latter occur in two sums of numbers.

Since the numbers have to total 13, we can try different ways in which the number can be arrived at by adding three numbers.

1+3+9

A B C D E F G H I


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1+4+8

1+5+7

2+3+8

2+4+7

2+5+6

3+4+6

We can safely say that one of A,B,D,F,H and I can assume the number 9, and 9 can occur only with 1 and 3. So, if we assume A to be 9, then we can assume B as 3 and C as 1, which means that D and E should stand for either 8 and 4 or 5 and 7.

Using the first combination of 8 and 4, we find that E cannot be 8 because the only other combination possible with 8 is 2 and 3, and 3 is already present. So E has to be 4, which means that F and G should stand for 7 and 2. G, again, cannot stand for 7, because the only other possible combination with 7 is 1 and 5, and 1 is already present. If G stands for 2, then H and I stand for 5 and 6.

On the other hand, if D and E stand for 5 and 7, then, assuming E stands for 7, we find that the only other combination with 7 is F and G being 4 and 2. If G is 4, then H and I should be 3 and 6, but 3 has already been used, so, this combination is not possible. If G is 2, then H and I should be 3 and 8 or 4 and 7, and again, we find that in both cases, a number has already been used.

From these two alternatives, we find that the first one is a feasible alternative and therefore, one possible outcome of matching numbers can be,

A 9; B 3; C 1; D 8; E 4; F 7; G2; H 5; I 6

Please note that there can be other solutions based on the arrangement of numbers, say, for example, if a student assigns 9 to I, then the grid would look different. But it is preferable to begin with a number that can appear only once on the grid, in this case, 9.

Variations of this game include

Finding numbers for a similar grid, where products, instead of sums, are given. Identifying missing numbers in a sum/multiplication/division problem.


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Bulls Eye

This game basically involves scoring a certain number of points on the dartboard. This may involve more of trial and error than method. But, even in the former, a proper understanding of multiplication tables and factorisation is necessary. As the following problem will indicate, its never easy to solve such problems in a fraction of a minute unless one can calculate quickly.

The following numbers appear on a dartboard 46, 44, 42, 33, 31, 13 and 11, with 46 on the Bulls eye and the rest of the numbers appearing on either side of it. Ascertain the least number of attempts required for a person to score exactly 100 points.

Solution/Discussion

This problem requires the student to have quick calculation skills, especially multiplication. A good way to attack the problem is to use the bottom-up approach, i.e. start from 100 onwards and identify the numbers on the darts that will satisfy the given condition. From the board, we can easily identify that the score before the dart hits the final 11 points to score 100 should be 89. Similarly, it should be 87 for 13, and so on. Solving along these lines for all kinds of possible combinations, such as,

11+13=24, therefore 76 to be scored beforehand

11+31=44, therefore 56 to be scored beforehand

13*2 = 26, therefore 74 to be scored beforehand, and so on, till we can arrive at a proper match.

In this case, after a few combinations of numbers and their multiples, we identify that a person needs 8 attempts to score exactly 100 points. The 8 attempts can be broken down into 6 attempts at the target 13 and 2 at 11, which is equal to

13*6=78

11*2=22, totaling 100.

11

13

31

33

42

44

46


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An important point to note would be that the answer would definitely not be a very small number, because the analysis required for it would be very minimal.

Whos the winner??? This game is a variant of the number game illustrated above. These games generally consist of a table, wherein some numbers are missing or are substituted by letters. There are no explicit conditions in these games, and so require some trial and error to solve them. Its Copa America time and the South American football teams are playing each other for top honours. The table below gives the results at a certain point during the tournament. A team gets 2 points for a win, 1 point for a draw and no points for losing. Each letter below represents a distinct integer value. No two letters represent the same integer. Each team plays the other only once. Team Played Won Drawn Lost Points

A Y a Y

B Q

C X Y p a Q

D Q A

E Y A

The questions may be w.r.t. number of games played by a team, points taken, whom they still have to play against and values of the letters. A couple of them are given below: Q1. What is the value of q?

a) 0

b) 1

c) 2

d) 4 Q2. How many games has E played till now? a) 0

b) 1

c) 2

d) 3

Solution/Discussion

A team could have played its 4 matches, won all of them and scored 8 points, or a team would have lost all its matches. Therefore, the numbers can range from 0 to 8. All columns of the table are filled w.r.t team C, so, we use C as the starting point. At this point, it is advisable to look through the questions, and look at a question, which can help us solve the row pertaining to C. In this case, only the first question is relevant. When we analyse the first question, we find that q cannot take the value of 3. Also, Cs row consists of 5 different numbers, and he should have won at least one point. We can make a simple assumption that q is equal to 4 and start working the other numbers. If q is 4, then C must have played at least 2 games and won them both or played 3, won one, drawn two, or won two and lost one. C couldnt have played 4 games because the value of q is 4. Also, C couldnt have played 2 and won them both because 2 different letters, x and y, represent the columns of played and won respectively. So, he must have played 3 games. Now, we assume that C has won 2 matches and lost the other. Then, y represents 2 , p represents 0 and a represents 1.


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Once we have arrived at the numbers, well proceed to solving the whole table. A has played 2 matches, has 2 points from them, and has lost a match. Therefore, he must have won the other match. So, we fill in As row with the relevant numbers. Since only one column in team B and E has been filled, we shall ignore them for the time being and proceed with team D. We find that D has played 4 matches and has only one point. Therefore, he must have drawn a game and lost the rest. We fill in the columns with the respective numbers. At this stage, the table will look as follows:

Team Played Won Drawn Lost Points

A Y A p a Y

B Q

C X Y p a Q

D Q P a x A

E Y

Since team D has drawn a game and none of the other teams identified, A or C, has drawn a game, B or E should have drawn a game with D. B must have played at least 2 games since its score is 4. Now, if we analyse the Lost column, we find that 5 games have been lost but on comparison with the Won column, we find that only 3 games have been won. Therefore, we can be sure that B must have won at least one game. It can also happen that B would have drawn two of its games, one each with D and E. (For the time being, we keep the number of losses constant and work the problem). So, the other game must have been won by E. If we look at the table now,


A Y A p a Y

B X A y p Q

C X Y p a Q

D Q P a x A

E Y A a p X

Converting the table into its numerical form


A 2 1 0 1 2

B 3 1 2 0 4

C 3 2 0 1 4

D 4 0 1 3 1

E 2 1 1 0 3

From the table above, we can find that D has drawn with B and lost to the other 3 teams. B and C have to play one game each but they cant play themselves since the other 2 teams A and C have 2 games left to play. Therefore, since B has already played C, it must have beaten it, which, in turn, means that C must have beaten A for its only other win and is yet to play E. Similarly, B is yet to play A, and A and C have to play each other.


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Variants of this game include:

Identifying the correct answer choices given by students Ascertaining the number of runs scored, wickets taken etc. by players in a cricket match

We have tried to cover different problems that students can encounter during entrance examinations. Though it is not an exhaustive list of problems, students should find it easier to follow a methodical approach on solving analytical reasoning problems. The common mistakes made by students are-

Hurrying through the conditions and later, finding it difficult to answer the questions Following a set method of solving every game, irrespective of its type. Say, using symbols to solve every game and every question in the game.

Wasting time reading too much into the conditions, especially in the conditional problems.

Before signing off, some things to remember:

Read the conditions and the accompanying questions carefully Organise the information according to your convenience. If you are comfortable using symbols,

go ahead, but please take some time to identify the easiest approach. Its just a matter of a few seconds to read the conditions and identify the correct and easiest approach.

Use the bottom-up approach if the conditions are not sufficient to solve the questions. Conditional problems take some time. Understand the implications of the conditions while

solving them. In case of questions that may have two answers, go for the most logical one or the one that appears correct on the face of it, rather than working out its implications.

Last, but not the least, if nothing strikes you during the examination or if you feel that no approach will help, try Trial and Error. It should work.


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CHAPTER-2

Numerical Puzzles Recent papers of the Campus Placement Test have had a number of group problems in the quantitative sections. In this chapter, the basic approach to these group problems has been discussed with examples.

Each group of problems contains the directions that are nothing but the description of a numerical problem situation. The problem situations are from any of the already discussed areas in the quantitative ability fundamentals such as Interest Rates, Profit/Loss, Time, Speed, Functions, Geometry etc. We have called these situations as puzzles because all the unknown information has to be unveiled from one or two links provided generally at the end of the description of the situations.

As can be appreciated from the solved examples and exercise problems given under, two different approaches are required to solve these puzzles depending on the type of questions.

In the first type, all the unknown but related details have to be determined before finding the best answer choice for the questions (Please refer to solved example 1). The questions only pertain to the unknown variables in the problem situation.

In the second type each question contains additional data and the problem needs to be solved with this information separately for each question. In this type it is necessary that each question be treated independently. Also the information provided in one question pertains only to that question and should not be used in subsequent problems, unless otherwise mentioned.

Solved Examples

Example 1 : DIRECTIONS for questions 1 to 5.

Jeshwanth, Krishna and Lokesh invested in a certain business. As Lokesh was to manage the business alone, it was decided that Lokesh would take 20% of the profit earned by the business and the remaining amount would be shared by all the three in proportion to their investment.

Thus at the end of the first year each of them earned Rs.10,000/- from the business. While Krishna withdrew 50% of his investment in the second year, Jeshwanth invested his first years profit also into the business. The business earned the same amount of profit in the second year as in the first year. Jeshwanth got a share which was 2.5 times the share of Krishna.

1. What is Krishnas total profit for two years?

1. Rs.17,102 2. Rs.15,581.40 3. Rs.21,903.60 4. Rs.14,788

2. What is Krishnas investment in the second year?

1. Rs.20,000 2. Rs.18,000 3. Rs.28,000 4. Rs.15,000

3. What is Lokeshs investment in the business?

1. Rs.17,500 2. Rs.22,350 3. Rs.16,000 4. Rs.14,000


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4. What is the % profit on investment for Lokesh in the second year?

1. 38% 2. 65% 3. 29% 4. 47%

5. What is the profit earned by the business as a percentage of investment in the first year?

1. 42.75% 2. 31.25% 3. 52.22% 4. 35.35%

Ans: To answer all these questions, it is first necessary to organize the given information. From the data given it is clear that both Jeshwanth and Krishna get the same share of profit in the first year. Hence they must have invested the same amount in the first year. Let the investment made by Jeshwanth and Krishna in the first year be X. The total profit for the first year is Rs.30,000/- and 20% of this amount, that is Rs.6,000/- goes to Lokesh for managing the business alone. As Lokesh totally received a share of Rs.10,000/- in the first year, his share in proportion to his investment must be Rs.10,000 - Rs.6,000 = Rs.4,000. Clearly this amount is 40% of the share received by the other two partners. Therefore Lokeshs investment in the first year should be 40% of Jeshwanths or Krishnas investment i.e., 0.4X.

Now the data can be represented for two years as below.

Jeshwanth Krishna Lokesh

I Year

Investment X X 0.4X

Profit Rs.10,000 Rs.10,000 Rs.10,000

II Year

Investment X + 10,000 0.50X 0.40X

Profit 2.5Z Z 30,000 - 3.5Z

(assuming Z as Krishnas Profit)

As the ratio of the profit shared by Jeshwanth to that of Krishna is 2.5, the ratio of their investments must also be equal to 2.5.

Therefore

(X + 10,000)

0.5X = 2.5

On solving the above equation X = 40,000/-

The investments of Jeshwanth, Krishna and Lokesh for the second year are respectively Rs.50,000, Rs.20,000 and Rs.16,000.

The profit earned by the business in the second year is Rs.30,000. Out of this Rs.6,000 is paid to Lokesh for managing the business.

The balance amount of Rs.24,000/- is to be shared among the three in proportion to their investment. The final table looks as depicted below.


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Jeshwanth Krishna Lokesh

I Year

Investment Rs.40,000 Rs.40,000 Rs.16,000

Profit Rs.10,000 Rs.10,000 Rs.10,000

II Year

Investment Rs.50,000 Rs.20,000 Rs.16,000

Profit Rs.13,953.50 Rs.5,581.40 Rs.10,465.10

1. Krishnas total profit for two years is Rs.15,581.40.

2. Krishnas investment in the second year is Rs.20,000.

3. Lokeshs investment in the business is Rs.16,000.

4. Lokeshs profit to investment ratio is Rs. 16,00010,465.10

. This is equal to approx. 65%.

5. The profit earned by the business is Rs.30,000 and the total investment is Rs.96,000. The profit as a percentage of the investment is 30/96 = 31.25%. Answer is 2.

Example 2: DIRECTIONS for questions 6 to 10.

The PNV group is interested in investing in a High Technology Venture. The project details have been worked out and are as below.

Investment required = 250% of the annual sales desired.

Profit = 25%

Working capital requirement = 35% of cost of goods.

6. What should be the investment if PNV wants to earn a profit of Rs.2 crore in the first year?

1. Rs.25 cr. 2. Rs.35 cr. 3. Rs.28 cr. 4. Rs.18 cr.

Ans: % Profit = (Sales - Cost) / Cost x 100

25 = (S - C)/ C x 100 Or S = 1.25 C

Therefore profit = S - C = S - S/1.25 = 0.25S/1.25

Required profit is Rs.2 crores.


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Sales = Rs.10 crores.

It is given that the investment is 250% of annual sales desired. Hence the investment would be Rs.25 crores. Answer is 1.

7. If PNV would like to restrict its investments to Rs.10 crores, what will be the annual profit?

1. Rs.1 cr. 2. Rs.80 lakhs 3. Rs.1.2 cr. 4. Rs.2.2 cr.

Ans: The annual sales possible with an investment of Rs.10 crores is Rs.4 crores. The profit with this sales would be Rs.0.8 crores or Rs.80 lakhs. Answer is 2.

8. What is the working capital required for a sale of 15 crores?

1. Rs.3.7 cr. 2. Rs.2.9 cr. 3. Rs.5.3 cr. 4. Rs.4.2 cr.

Ans: The profit with a sales turnover of Rs.15 crores is Rs.3 crores. Hence the cost of goods will be Rs.12 crores. The working capital requirement is given to be 35% of the cost. Therefore the working capital required is 0.35 x 12 = Rs.4.2 crores. Answer is 4.

9. In problem 8 above, if 40% of the investment is borrowed from Financial Institutions at an interest rate of 18%, what will the profit after interest earned by the company be?

1. 0.3 cr. 2. 2.2 cr. 3. 0.85 cr. 4. 1.15 cr.

Ans: As the sale is Rs.15 crores, the investment would be 2.5 x 15 = Rs.37.5 crs. The amount borrowed from financial institutions is 0.40 x 37.5 = Rs.15 crores. The annual interest on this borrowing is 18% of 15 crs. = Rs.2.7 crs. The profit earned is Rs.3 crs. and the profit after interest will be Rs.0.3 crs. Answer is 1.

10. If an annual net profit (profit after interest) of Rs.2 crores is to be achieved and PNV wants to borrow 40% of the investment at an interest rate of 18%, what is the total investment required?

1. Rs.200 cr. 2. Rs.20 cr. 3. Rs.250 cr. 4. Indeterminate.

Ans: If the investment is I, the sales possible annually is I/2.5.

The gross profit will be I/2.5 x 0.25/1.25.

The amount to be borrowed from the financial institutions is Rs.0.41.

The annual interest on this amount is 0.18 x 0.41.

The profit after interest is I/2.5 x 0.25/1.25 - 0.18 x 0.41 = 2 crores.

On solving I = Rs.250 crores. Answer is 3.

Example 3: DIRECTIONS for questions 11 to 16.

XYZ Co. Ltd earned a profit of 20% in 1989. In 1990 the sales of the company improved by Rs.6,000/- and the profit increased to 25%. Capitalising on the high demand for its product, the company earned the highest profit of 50% in 1991 but maintained the sales at the same level as


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1990. In spite of the severe recession during 1992, the companys sales improved impressively by 20% but the profit slumped to the lowest ever figure of 12.5%. The amount of profit earned in 1992 was the same as the amount of profit earned in 1989.

11. What was the sales of XYZ in 1990?

1. Rs.42,000 2. Rs.36,000 3. Rs.27,500 4. Rs.30,000

12. What was the profit earned in 1990?

1. Rs.6,000 2. Rs.8,000 3. Rs.10,000 4. Rs.11,000

13. The average percentage profit for four years is

1. 25% 2. 20% 3. 28% 4. 30%

14. The total profit earned for four years is

1. Rs.20,000 2. Rs.24,000 3. Rs.41,000 4. Rs.18,000

15. If the cost of each item produced in 1991 was the same as that in 1990, the percentage decrease in quantity sold in 1991 as compared to 1990 was

1. 18% 2. 16.66% 3. 22.5% 4. 25%

16. In problem 15 above the percentage increase in the price in 1991 as compared to 1990 was

1. 20% 2. 25% 3. 30% 4. 35%

Solution: The given information can be compiled as follows (assuming sales in 1989 as S)

Sales % profit Profit

1989 S 20% 1.20.20S

1990 S + 6,000 25% 1.256,000)+0.25(S

1991 S + 6,000 50% 1.56,000)0.50(S +

1992 1.20 (S + 6,000) 12.5% 1.1256,000)1.2(Sx0.125 +

It is given that the profit earned in 1992 is equal to that in 1989.

Therefore, 1.2S 0.20

= 1.125

6000)+ (S 1.2 x 0.125

,

On solving S = 24,000


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The details can now be written as follows:

Sales % profit Profit

1989 24,000 20% 4,000

1990 30,000 25% 6,000

1991 30,000 50% 10,000

1992 36,000 12.5% 4,000

11. Sales in 1990 was Rs. 30,000. Answer is 4.

12. Profit earned in 1990 was Rs. 6,000. Answer is 1.

13. The total sales for four years is Rs. 1,20,000 and the total profit is Rs. 24,000. Therefore the total cost is Rs. 96,000. The % profit is 25. Answer is 1.

14. Total profit earned for four years is Rs. 24,000. Answer is 2.

15. The cost of goods in 1990 was 30,000 - 6,000 = 24,000.

If cost of each item is Rs. 200 (assume), the quantity sold in 1990 was 24,000/200 = 120.

Similarly the cost of goods in 1991 was 30,000 - 10,000 = 20,000. As the cost of each item remains same (at Rs. 200 in this case), the quantity sold in 1991 was 20,000/200 = 100.

Hence the percentage decrease in quantity sold in 1991 as compared to 1990 was (120 - 100)/120 = 16.66%. Answer is 2.

16. From the above answer, the price of each item in 1990 was 30,000/120 = 250.

The price of each item in 1991 was 30,000/100 = 300.

The increase in price from 1990 to 1991 is (300 - 250)/250 = 20%. Answer is 1.


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Practice Exercise 1

DIRECTIONS for questions 1 to 5.

Both Amar and Prem individually invested certain amounts in a bank at a rate of interest of 5%. Interest at the end of each year will be accumulated into the principal amount. At the end of the first year, Amar withdrew 50% of the accumulated amount, while Prem withdrew Rs.11,500. At the end of the third year, Prem had an accumulated amount equal to that which Amar had at the end of the second year. Amar withdrew Rs.4,050 at the end of the second year. The total interest earned by Amar is Rs.3,950 for three years.

1. What is the amount initially invested by Amar?

1. 35,000 2. 40,000 3. 31,500 4. 30,000

2. The amount initially deposited by Prem is

1. 35,000 2. 40,000 3. 30,000 4. 48,000

3. The total interest earned by Prem for three years is

1. 3,550 2. 3,950 3. 1,975 4. 4,350

4. Interest earned by Amar in the third year is

1. 1235 2. 900 3. 1050 4. 1800

5. If Prem had not withdrawn any amount in three years he would have had an accumulated amount of

1. 34,725 2. 43,475 3. 45,447 4. 39,912

DIRECTIONS for questions 6 to 10.

Ramkumar invests 60% of his retirement money in fixed deposits that earn an interest of 15% at the end of every year. He decides to take out this interest amount at the end of every year to meet his personal expenses. He invests the balance 40% of the retirement money in shares. At the end of the first year he sells off all the shares and realises a profit of 25%. Inspired by the return on investment on shares, Ramkumar decides to take out 25% of the money in fixed deposits and invest it in shares along with the balance money. At the end of the second year he sells all the shares and incurs a loss of 10%. Taken aback by the loss, Ramkumar puts the amount withdrawn by him earlier from fixed deposits back into it. He reinvests the balance amount again in shares. After selling all the shares at the end of the third year he realises a profit of 10%. The profit earned by Ramkumar in the three years of his transactions in shares is Rs.18,900.

6. What is the retirement benefit that Ramkumar obtained?

1. 2.5 lakhs 2. 2 lakhs 3. 3.2 lakhs 4. 1.6 lakhs

7. What is the total amount of interest earned by Ramkumar on fixed deposits?

1. 49,500 2. 27,000 3. 52,000 4. 38,900


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8. What is the loss incurred by Ramkumar in transacting in shares in the second year?

1. 16,500 2. 15,000 3. 13,000 4. 18,700

9. What was Ramkumars investment in shares in the third year?

1. 87,000 2. 1,05,000 3. 98,500 4. 72,250

10. The average annual return on investment earned by Ramkumar in three years is

1. 10.8% 2. 15.4% 3. 19.2% 4. 9.2%

DIRECTIONS for questions 11 to 15:

A company manufactures two products X and Y. Both these products are to be processed on machines A, B, C in that order. Product X requires 10 hours on machine A, 20 hours on machine B and 15 hours on machine C. Product Y requires 12 hours on machine A, 15 hours on machine B and 24 hours on machine C. There are two machines of type A, 3 machines of type B and 4 machines of type C. The company works for 8 hours a day.

11. If the company makes only product X, how many numbers of X can be made in a month consisting of 25 working days?

1. 55 2. 30 3. 72 4. 35

12. In problem 11 above, how many numbers of product Y alone can be produced in the month?

1. 33 2. 41 3. 55 4. 72

13. The company has to produce 20 numbers of product X in a month. If there are 25 working days in the month, how many numbers of product Y can be made with the remaining capacity?

1. 24 2. 33 3. 13 4. 16

14. If products X and Y are to be dispatched in pairs, how many pairs of X and Y can be produced in a month of 25 days?

1. 17 2. 18 3. 12 4. 27

15. In problem 14 above, what is the percentage unutilized capacity?

1. 9.33% 2. 21% 3. 10.5% 4. 8%

DIRECTIONS for questions 16 - 20:

Ghosh Babu deposited a certain amount of money in a bank in 1986. The bank calculated interest on the balance in the account at 10 percent simple interest, and credited it to the account once a year. At the end of the first year, Ghosh Babu withdrew the entire interest and 20 percent of the initial amount. Again, at the end of the second year, he withdrew the interest and 50 percent of the remaining amount. At the end of the third year, he withdrew the interest and 50 percent of the remaining amount. Finally, at the end of the fourth year, Ghosh Babu closed the account and collected the entire balance of Rs.11,000.


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16. The initial amount in rupees, deposited by Ghosh Babu was

1. 25,000 2. 75,000 3. 50,000 4. None of these.

17. The total interest, in rupees, collected by Ghosh Babu was

1. 12,000 2. 20,000 3. 4,000 4. 11,000

18. The year, at the end of which, Ghosh Babu withdrew the maximum amount was

1. First 2. Second 3. Third 4. Fourth

19. The year, at the end of which, Ghosh Babu collected the maximum interest was


20. The year, at the end of which, Ghosh Babu withdrew the smallest amount was



Prakash has to decide whether or not to test a batch of 1,000 widgets before sending them to the buyer. In case he decides to test, he has two options : (a) Use test I (b) Use test II. Test I costs Rs.2 per widget. However, the test is not perfect. It sends 20% of the bad ones to the buyer as good. Test II costs Rs.3 per widget. It brings out all the bad ones. A defective widget identified before sending can be corrected at a cost of Rs.25 per widget. All defective widgets are identified at the buyers end and a penalty of Rs.50 per defective widget has to be paid by Prakash.

21. Prakash should not test if the number of bad widgets in the lot is

1. less than 100 2. more than 200

3. between 120 and 190 4. Cannot be answered.

22. If the number of defective widgets in the lot is between 200 and 400, Prakash

1. Should use either Test I or Test II 2. Should use Test I only

3. Should use Test II only 4. Cannot decide.

23. If there are 200 defective widgets in the lot, Prakash

1. Should use either Test I or Test II 2. Should use Test I or not use any test

3. Should use Test II or not use any test 4. Cannot decide.

24. If Prakash is told that the lot has 160 defective widgets he should

1. Use Test I 2. Use Test II only 3. No test 4. Use Test I or No test.

25. If there are 120 defective widgets in the lot, Prakash


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1. Should use either Test I or not test 2. Should use Test II or not test

3. Should use Test I or Test II 4. Should use Test I only.


On 27-11-92, Sanjay bought 500 shares of company A at a price of Rs. 80/-, 100 shares of company B at a price of Rs. 45 and 200 shares of company C at a rate of Rs. 50/-. The sensitive index on 27-11-92 was 2,000. Sanjay knows that the price of share A is directly proportional and share price of B is inversely proportional to the sensitive index. The share price of C is proportional to the square root of the share price of A.

26. If Sanjay again buys 500 shares of A at an index of 2400, what will be his average price of share A?

1. Rs.92/- 2. Rs.98/- 3. Rs.96/- 4. Rs.12/-

27. If Sanjay sells 200 shares of C when the index is 2,500, what is his gain in rupees?

1. 2,500 2. 3,825 3. 1,110 4. 2,000

28. What is the total value of shares possessed by Sanjay at an index of 3,000?

1. Rs.75,250 2. Rs.82,000 3. Rs.48,250 4. Rs.72,000

29. Sanjay predicts that the index would reach a peak of 2400 and would fall down to 1800 in the coming month. What will be the profit earned by Sanjay if he sells all his shares at the peak index and buys back the same shares at the lowest index?

1. Rs.14,800 2. Rs.12,700 3. Rs.13,850 4. Rs.18,750

30. At what index will the price of C double?

1. 4,000 2. 8,000 3. 1,000 4. 6,400


x y = x2 + x2 y2

x y = 2y

x

+ y

31. What is the value of (2 3) 4?

1. 47/13 2. 13/2 3. 57/9 4. 65/7

32. Given x = 5, for what value of y does (y 3) x become equal to 5?

1. 27/7 2. 0 3. 17/3 4. 19/4

33. If x y = m , x y = n, what is m n for x = 2, y = 1?


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1. 412 2. 278 3. 197 4. -4

34. In problem 33 above if y = 1, then for what value of x will m be equal to n?

1. 2 2. -2 3. 1 4. -4

35. When will x y be equal to x + y?

1. Only when y is zero. 2. Only when y is equal to x2.

3. When y is equal to either zero or x . 4. None of these.


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CHAPTER-3

Blood Relationships In the study of Blood relations we come across three major types of problems.

The relation will be described with you as a centre and is given in a round about manner i.e. one needs to under go a series of relationships before arriving at the conclusion.

The relation will be given between the two people in a round about manner.

Here certain codes are used to indicate certain relations. One needs to decode it and come to the conclusion.

In order to solve the problems related to blood relation we need to know the following relations

Mothers mother : Grand mother

Fathers mother : Grand mother

Mothers father : Grand father

Fathers father : Grand father

Grandmothers brother : Grand uncle

Grandmothers sister : Grand aunt

Grandfathers brother : Grand uncle

Grandfathers sister : Grand aunt

Fathers son : Brother

Mothers son : Brother

Mothers daughter : Sister

Fathers daughter : Sister

Mothers brother : Uncle

Fathers brother : Uncle

Mothers sister : Aunt

Sons wife : Daughter-in-law

Daughters husband : Son-in-law

Husbands sister : Sister-in-law

Wifes sister : Sister-in-law

Husbands brother : Brother-in-law

Wifes brother : Brother-in-law


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Sisters son : Nephew

Brothers son : Nephew

Sisters daughter : Niece

Brothers daughter : Niece

Uncle or Aunts son : Cousin

Uncle or Aunts daughter : Cousin

Sisters husband : Brother-in-law

Brothers wife : Sister-in-law

Children of same parents : Siblings (brother and brother, sister and sister, brother and sister)

Spouse means either husbands wife or wife husban4.

The graph given below will help us to get the concepts properly.

Draw family tree in which stem represents mother and father, roots, grant parents, branches, children, and leaves, the children of the children. Aunt can be mothers or fathers sisters. Uncles are fathers or mothers brothers. These are common relations and one can easily spot out the relationship.


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Practice Exercise 2

1. If a+b means a is the husband of b, a b means a is the sister of b and a b means a is the son of b which of the following shows that A is the daughter of B?

1. A D B 2. D B C A 3. B x C A 4. C + B A 2. The man who is receiving the hat is the brother of uncles daughter. What is the relation

of the speaker and the man getting the hat.

1. Son in law 2. Cousin 3. Brother 4. Nephew

3. Pointing to a girl in the photograph Umesh said Her mothers brother is the only son of my mothers father. How is girls mother related to Umesh?

1. Mother 2. Sister 3. Aunt 4. Grandmother

4. Dennis said to Raman that the boy who has won the game is the younger of the two brothers of the daughter of my fathers wife. How is the boy related to Dennis?

1. Son in law 2. Brother 3. Cousin 4. Niece

5. Pointing to a lady in a photograph John said, She is the daughter of my grandfathers only son. How is John related to the lady?

1. Father in law 2. Son in law 3. Brother 4. Husband

6. Introducing a man a woman said, he is the only son of my mothers mother. How is the woman related to the man?

1. Mother 2. Aunt 3. Sister in law 4. Niece

7. Pointing to his sons photograph Ralph said to a woman His mother is the only daughter of your mother. How is the woman related to Ralph?

1. Wife 2. Brother 3. Uncle 4. Nephew

8. Introducing a man to her husband a wife said that his brothers father is the only son of my grand father. How is the wife related to the man?

1. Mother in law 2. Uncle 3. Sister 4. Daughter in law

9. Pointing to a lady Mohan said She is the daughter of a woman who is the mother of the husband of my mother. How is the lady related to Mohan?

1. Aunt 2. Mother 3. Daughter in law 4. Sister in law

10. Pointing to a man in group Sheela said He is the brother of the daughter of the wife of my husban4. How is the man in group related to Sheela?

1. Son 2. Uncle 3. Sister 4. Niece


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CHAPTER-4

Calendars General Information about Calendars

1. A period of one year comprises of 12 months.

2. Total number of weeks in a year is 52.

3. A normal year (Non Leap Year) consist of 365 days, where as a Leap Year consist of 366 days.

4. Number of days in a month is either 31 days or 30 days depending upon what month we are taking into consideration but only in the month of February we find 28 days and if it is a leap year then the month of February contains 29 days.

Concept of Odd Days

Consider that 2006 February 22nd was a Thursday. If it is required to be find the day on which the next 5th would fall, first calculate the

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