Mock Question Bank for Quantitative Aptitude - Testbook.com

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Insurance Exam Papers - Mock Question

Bank for Quantitative Aptitude

Answer Key:

Q1. A Q21. B Q41. B Q61. B Q81. A Q2. C Q22. C Q42. E Q62. C Q82. D Q3. C Q23. D Q43. D Q63. D Q83. D Q4. B Q24. A Q44. A Q64. B Q84. C Q5. C Q25. C Q45. B Q65. D Q85. C Q6. A Q26. B Q46. B Q66. C Q86. A Q7. D Q27. B Q47. A Q67. C Q87. C Q8. C Q28. B Q48. C Q68. C Q88. D Q9. B Q29. C Q49. D Q69. B Q89. D Q10. D Q30. A Q50. A Q70. B Q90. B Q11. B Q31. C Q51. C Q71. C Q91. C Q12. D Q32. D Q52. A Q72. C Q92. A Q13. C Q33. D Q53. C Q73. B Q93. B Q14. B Q34. A Q54. B Q74. D Q94. D Q15. C Q35. B Q55. B Q75. B Q95. D Q16. C Q36. B Q56. C Q76. A Q96. D Q17. C Q37. A Q57. D Q77. D Q97. A Q18. D Q38. E Q58. B Q78. B Q98. B Q19. B Q39. E Q59. A Q79. C Q99. C Q20. D Q40. C Q60. B Q80. A Q100. B

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Insurance Exam Papers - Mock Question

Bank for Quantitative Aptitude

Solution 1.

Total number of alphabets, n(S) = 26

Total number of characters which are not vowels, n(E) = 21

𝑃(𝐸) =𝑛(𝐸)

𝑛(𝑆)=

21

26

Solution 2.

Let the marks scored in mathematics be x

Average marks in six subjects is 36.5

We know that Average = Sum of marks/number of subject

⇒ 36.5 = Sum of marks/6

⇒ Sum of marks = 36.5 × 6

⇒ Sum of marks = 219

Now, the average marks in 5 subjects is 39

∴ Sum of marks in 5 subjects = (Average in 5 subjects) × number of subjects

⇒ Sum of marks in 5 subjects = 39 × 5 = 195

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∴ Marks scored in mathematics = Sum of marks in 6 subjects – Sum of marks in 5

subjects

⇒ x = 219 – 195

⇒ x = 24

Solution 3.

Let the amount earned per day by a man and a woman be ‘x’ and ‘y’ respectively.

Amount earned by 5 men and 7 women in 7 days = (5x + 7y) × 7 = Rs.6440

⇒ 5x + 7y = 6440/7 = 920 ….(1)

Amount earned by 15 men and 17 women in 8 days = (15x + 17y) × 8 = Rs.20160

⇒ 15x + 17y = 20160/8 = 2520 ….(2)

Solving (1) and (2)

We get,

x = Rs. 100 and y = Rs. 60

Amount earned by 21 men and 29 women in one day = 21x + 29y = 21 × 100 + 29 × 60 =

Rs.3840

Time taken by 21 men and 29 women to earn Rs.65280 = 65280/3840 = 17 days

Solution 4.

25% of P is 14 more than 10% of Q.

⇒ 0.25P = 14 + 0.1Q



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⇒ 5P = 280 + 2Q

10% of P is 20 less than 20% of Q.

⇒ 0.1P = 0.2Q - 20

⇒ P = 2Q – 200

⇒ 5P = 10Q – 1000

Equating 5P from both equations,

⇒ 280 + 2Q = 10Q – 1000

⇒ Q = 1280/8 = 160

⇒ P = 2Q – 200 = 320 – 200 = 120

∴ PQ = 120 × 160 = 19200

Solution 5.

According to the given information,

P = Rs. 18750 R = 20% T = 4 years

We know that, for Compound interest ,

A = P (1 +𝑅

100)

𝑇

⇒ 18750 ( 1 + 20

100)

4

⇒ 18750 (120

100 )

4

⇒ 18750 × (6

5 )

4



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⇒ 18750 × 1296

625

⇒ 30 × 1296

⇒ 38880

Hence, the amount will become Rs. 38880 in four years .

Solution 6.

Let original amounts of milk and water be M and N.

⇒ (M+5)/N = 3/7

⇒ 7M + 35 = 3N

And, M/(N+5) = 1/4

⇒ 4M = N + 5

Solving the two equations, we get M = 10 and N = 35

∴ Ratio is 2:7

Solution 7.

Let the time taken by C1 be ‘t’ hrs.

The distance travelled by both the cars is the same. (∵ Distance = Speed × Time)

⇒Speed of car C1 × time taken by car C1 = Speed of car C2× time taken by car C2

⇒31 × t = 47 × ( t – 2.5 )

⇒ 31t = 47t – 117.5



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⇒ t = 7.34 hrs

Distance to the place = 31 × 7.34 = 227.65 km

∴ The distance to the place = 227.65 km

Solution 8.

Total number of balls = 12

∴ Total number of possibilities = 12C2

Possibility of the two balls to be black = 5C2

∴ Probability = 5C2/12C2 = 5/33

Solution 9.

Let the number to be subtracted be ‘a’.

Given, ratio is 15:17.

When ‘a’ is subtracted from both the numerator and denominator the ratio becomes 5:6

∴15−𝑎

17−𝑎=

5

6

⇒ 90 – 6a = 85 – 5a

⇒ a = 5

Solution 10.

As per the given data,



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4 , 2 , -2 , -10 , -26 , ?

The series is followed by the below pattern

(1 × 4) – 2 = 2

(1 × 2) – 4 = - 2

(1 × -2) – 8 = -10

(1 × -10) – 16 = -26

(1 × -26) – 32 = -58

∴The answer is -58

Solution 11.


499, 721 , 485 , 279 , 91 , ?


103 – 13 = 999

93 – 23 = 721

83 – 33 = 485

73 – 43 = 279

63 – 53 = 91

53 – 63 = -91

∴The answer is -91



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Solution 12.


1 , 2 , 10 , 19 , 83 , ?


1 + 12 = 2

2 + 22 = 10

10 + 32 = 19

19 + 43 = 83

83 + 53 = 108

∴The answer is 108

Solution 13.


100 ,101 ,98 ,103 , 96 , 105 , ?


100 + 1 = 101

101 – 3 = 98

98 + 5 = 103

103 – 7 = 96

96 + 9 = 105

105 – 11 = 94



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∴The answer is 94

Solution 14.


1 ,26 , 151, ?

The series is followed by the below pattern ,

1 + 52 = 26

26 + 53 = 151

151 + 54 = 776


Solution 15.


9 , 17 , 32 , 61 , ?


(9 × 2) – 1 = 17

(17 × 2) – 2 = 32

(32 × 3) – 3 = 61

(61 × 2) – 4 = 118




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Solution 16.

I. x2 – 5x – 104 = 0

⇒ x2 – 13x + 8x – 104 = 0

⇒ x(x – 13) + 8(x – 13) = 0

⇒ (x – 13)(x + 8) = 0

Then, x = + 13 or x = - 8

II. y2 – 29y + 204 = 0

⇒ y2 – 17y – 12y + 204 = 0

⇒ y(y – 17) – 12(y – 17) = 0

⇒ (y – 17)(y – 12) = 0

Then, y = + 17 or y = + 12

So, when x = + 13, x < y for y = + 17 and x > y for y = + 12

And when x = - 8, x < y for y = + 17 and x < y for y = + 12

∴ So, we can observe that no clear relationship cannot be determined between x and y.

Solution 17.

I. a2 – 22a – 48 = 0

⇒ a2 – 24a + 2a – 48 = 0

⇒ a(a – 24) + 2(a – 24) = 0

⇒ (a – 24)(a + 2) = 0



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Then, a = + 24 or a = - 2

II. b2 – 37b + 312 = 0

⇒ b2 – 13y – 24y + 312 = 0

⇒ b(b – 13) – 24(b – 13) = 0

⇒ (b – 13)(b – 24) = 0

Then, b = + 13 or b = + 24

So, when a = + 24, a > b for b = + 13 and a = b for b = + 24

And when a = - 2, a < b for b = + 13 and a < b for b = + 24

∴ So, we can observe that no clear relationship cannot be determined between a and b.

Solution 18.

I. a2 + 7a – 294 = 0

⇒ a2 + 21a – 14a – 294 = 0

⇒ a(a + 21) – 14(a + 21) = 0

⇒ (a + 21)(a – 14) = 0

Then, a = - 21 or a = + 14

II. b2 + 45b + 504 = 0

⇒ b2 + 24b + 21b + 504 = 0

⇒ b(b + 24) + 21(b + 24) = 0

⇒ (b + 24)(b + 21) = 0



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Then, b = - 24 or b = - 21

So, when a = - 21, a > b for b = - 24 and a = b for b = - 21

And when a = + 14, a > b for b = - 24 and a > b for b = - 21

∴ So, we can observe that a ≥ b.

Solution 19.

I. 5x2 – 63x – 98 = 0

⇒ 5x2 – 70x + 7x – 98 = 0

⇒ 5x(x – 14) + 7(x – 14) = 0

⇒ (x – 14)(5x + 7) = 0

Then, x = + 14 or x = - 7/5

II. y2 + 23y + 76 = 0

⇒ y2 + 19y + 4y + 76 = 0

⇒ y(y + 19) + 4(y + 19) = 0

⇒ (y + 19)(y + 4) = 0

Then, y = - 19 or y = - 4

So, when x = + 14, x > y for y = - 19 and x > y for y = - 4

And when x = - 7/5, x > y for y = - 19 and x > y for y = - 4

∴ So, we can observe that x > y.



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Solution 20.

I. x2 – 31x – 140 = 0

⇒ x2 – 35x + 4x – 140 = 0

⇒ x(x – 35) + 4(x – 35) = 0

⇒ (x – 35)(x + 4) = 0

Then, x = + 35 or x = - 4

II. y2 + 36y + 128 = 0

⇒ y2 + 32y + 4y + 128 = 0

⇒ y(y + 32) + 4(y + 32) = 0

⇒ (y + 32)(y + 4) = 0

Then, y = - 32 or y = - 4

So, when x = + 35, x > y for y = - 32 and x > y for y = - 4

And when x = - 4, x > y for y = - 32 and x = y for y = - 4

∴ So, we can observe that x ≥ y.

Solution 21.

Let the number of boys be x

And that of girl be y

Then, total score of boys = 73x

Total score of girls = 70y



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According to the given information:

⇒ (73x+70y)/(x+y) = 71.8

⇒ 73x + 70y = 71.8x + 71.8y

⇒ 1.2x = 1.8y

⇒ x/y = 1.2/1.8 = 2/3

∴ Percentage of girls in the class = 3

5× 100 = 60%

∴ The percentage of the girls in the class is 60

Solution 22.

Let the sum of heights of the other five friends be N. We have

N + 162 = 6(167)

⇒ N = 840

The new average after Penshu leaves = N/5 = 840/5 = 168 cm

Solution 23.

Let S.P. = 100

Then C.P is 50% of 100 = 50

Let S.P. is x% of C.P.

Then,

S.P. = (C.P. × x)/100



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⇒ 100 = (50 × x)/100

⇒ x = 200

So S.P. is 200% of C.P.

Solution 24.

Since, the first 8 questions are compulsory, the student has to choose only remaining 7

questions out of 12 in the examination.

Hence, required number of ways = 12C7

Solution 25.

Total marks obtained by P, Q, R and S in Science is 480

⇒ P + Q + R + S = 480

P secured one-third marks of the total of Q, R and S

P = (Q + R + S)/3 = (360 – P)/3

⇒ 4P = 480

⇒ P = 120

Average marks obtained by Q and R is 12 more than that secured by S

(Q + R)/2 = S + 12

⇒ Q + R = 2S + 12 = 480 – P – S

⇒ 3S = 468 – 120

⇒ S = 336/3 = 112



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∴ both statements I and II together are necessary to answer the question

Solution 26.

We know Speed × Time = Distance

If S is his normal speed then new speed is 0.75 S.

So, T must be changed to T/0.75 to keep the distance constant.

∴ 0.75 S × (1/0.75) × T = D

⇒ (1/0.75) T = T + 20

⇒ T/3 = 20

⇒ T = 60 minutes

Hence he would take 1 hour to reach if walked with normal speed S.

Solution 27.

Part of work done by Ajit in one day = 1

4× (

1

3)= 1/12

Part of work done by Sujit in one day = 1

6× (

1

4)= 1/24

Part of work done by both Ajit and Sujit in one day = 1/12 + 1/24 = 1/8

So, they both will complete wok in 8 days.

Let x be total paid amount

∵ Ajit is paid Rs. 120 for 8 days = 8×𝑥×1

12= 120

⇒ x = 180



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So total paid amount = Rs. 180

Study the following table carefully and answer the following questions:

Classification of 100 candidates based on the marks obtained by them in technical part

and the verbal part in a placement written test of 70 marks is:

Marks /

Part

60 and

above

50 and

above

40 and

above

30 and

above

20 and

above

Technical

part 8 13 26 45 67

Aptitude part 10 16 27 36 78

Aggregate

(average) 9 15 27 41 73

Solution 28.

30 % of 100 = (30/100 × 70) = 21

Hence required number, X = number of candidates scoring less than 21marks in

aggregate

X = 100 – 20 and above marks in aggregate = 100 – 73 = 27

Solution 29.

57% of 70 = 39.9 = 40

Required number = number of candidates scoring 40 and above marks in technical part

= 26



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Solution 30.

Difference = number of candidates scoring 30 and above in technical part – number of

students scoring 30 and above in aggregate = 45 – 41 = 4

Solution 31.

57 % of 70 marks = 40

And 30 % of 70 = 21

Number of candidates having 40 marks or above in aptitude = 27

Number of candidates having 20 or above marks in aggregate = 73

Therefore required percentage = (27/73 × 100) = 36.98 = 37 %

Solution 32.

Since 67 candidates get 20 marks or above in technical and out of these, 45 candidates

get 30 marks or above , therefore to select top candidates in technical , the qualifying

marks should be in the range 20 - 30

Directions: Study the following Pie-chart carefully and answer the questions given

below.

Survey conducted in an engineering college of 1000 students to find out %age of number

of students in different engineering courses.



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Percentage of girls in respective courses.

Computer Sciences – 20%

Mechanical – 5%

Electronics – 30%

Information Technology – 25%

Industrial Production – 10%

Civil – 8%

Solution 33.

Girls in Mechanical + Girls in Civil = 5% of 18% of 1000 + 8% of 15% of 1000

15%

14%

18%

16%

18%

19%

Civil Industrial Production Information Technology

Electronics Mechanical Computer Sciences



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= 9 + 12

= 21

Girls in Computer Sciences + Girls in Information Technology

= 20% of 19% of 1000 + 25% of 18% of 1000

= 38 + 45

= 83

Ratio = 21 : 83

= 21 : 83

Solution 34.

Mechanical engineering students = 18%

Computer Science Students = 19%

Required % = 18/19 × 100

= 95%

Solution 35.

Total number of boys,

= [80% of 19% + 95% of 18% + 70% of 16% + 75% of 18% + 90% of 14% + 92% of 15%] ×

1000

= 834



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Total number of girls = 1000 – 834 = 166

Required difference = 834 – 166 = 668

Solution 36.

Girls in electronics = 30% of 16% of 1000

Boys in mechanical engineering = 95% of 18% of 1000

Required % = (30% of 16%)/(95% of 18%) × 100

= (0.3 × 0.16)/(0.95 × 0.18) × 100

= 28%

Solution 37.

Number of girls in information technology = 25% of 18% of 1000

Number of girls in Industrial production = 10% of 14% of 1000

Required Ratio = (25 × 18) : (10 × 14)

= 45 : 14

Solution 38.

Given that a town has 700 women, 1700 men and 1100 children

∴The total population of the town = 700 + 1700 + 1100

= 3500

It is mentioned that there are no eunuchs in the town



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∴The percentage of men’s population = (men’s population/total population) × 100

= (1700/3500) × 100

= 48.57

Solution 39.

Let Adrian travels by car for ‘x’ hours.

⇒ Time for which he travels by bus = (17 – x) hrs

We know that,

Distance = Speed × Time

⇒ Distance travelled by car = 64 × x = 64x km

Distance travelled by bus = 48 × (17 – x) = (816 – 48x) km

According to the question,

64x + 816 – 48x = 952

⇒ 16x = 136

⇒ x = 8.5 hrs

Distance travelled by car = 64 × 8.5 = 544 km

∴ Distance travelled by car = 544 km

Solution 40.

P is 23% of Q. Q is 67% of R.



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⇒ P = 0.23Q and Q = 0.67R

⇒ P = 0.1541R

⇒ 3P = 0.4623R

R does not have any factor of 2 or 5.

For 3P to be natural number, R should be at least 10000.

Directions: In the following pie-chart percent of students enrolled in different cultural

activities of a school has been shown. You are required to study the pie-chart carefully

and answer the questions given below.

Number of Students = 1800

Karate10%

Cricket7%

Painting32%

Music8%

Sports5%

Dance38%



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Solution 41.

∵ Total number of Students = 1800

∵ Percentage of students who opted for dancing = 38%

∴ Number of students who opted for dancing = 38% of 1800 = 684

Solution 42.


∵ Percentage of students who opted for painting = 32%

∵ Percentage of students who opted for Music= 8%

∴ Number of students who opted for painting = 32% of 1800 = 576

∴ Number of students who opted for Music = 8% of 1800 = 144

∴ Required ratio = 576:144 = 4:1

Solution 43.


∴ Number of students who opted for Karate = 10% of 1800 = 180

∴ Number of students who opted for Sports = 5% of 1800 = 90

∴ Required percentage =180−90

90× 100 =

90

90× 100 = 100%



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Solution 44.


∵ Percentage of students who opted for cricket and painting together = 7 + 32 = 39%

∴ Number of students who opted for cricket and painting together = 39% of 1800 = 702

Solution 45.


∵ Percentage of students who opted for cricket and dancing together = 7 + 38 = 45%

∵ Percentage of students who opted for painting and music together = 32 + 8 = 40%

∴ Number of students who opted for cricket and dancing together = 45% of 1800 = 810

∴ Number of students who opted for painting and music together = 40% of 1800 = 720

∴ Required percentage =810−720

720× 100 = 12.5%

Directions: Study the given line graph and pie chart carefully and answer the questions

given below it.



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Solution 46.

Total expenditure of D = 30000

Expense of D on food = 20% of 30000 = 6000

Given,

24000

28000

25000

30000

15000

32000

0

5000

10000

15000

20000

25000

30000

35000

A B C D E F

Persons

Monthly expenditure of six persons A, B, C, D, E and F

Food20%

Clothes12%

Rent15%Education

16%Fuel5%

Telephone and electricity bills

10%

Others22%

Breakyp of expenses of D



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Let the average expense of all six people on food be X.

Given,

6000 = 𝑋 (1 +20

100) ⇒ 𝑋 = 5000

Solution 47.

Total expenditure of D = Rs. 30000

% expenditure on rent, education and telephone and electricity bills = 41% of 30000 =

12300

Solution 48.

Total expenditure of B = 28000

Expenditure on clothes = 10% of 28000 = 2800

Total expenditure of F = 32000

Expenditure on clothes = 15% of 32000 = 4800

% by which expenditure of F on clothes is greater than that of B = 4800−2800

2800× 100 =

71.42%

Solution 49.


Expense on fuel = 5% of 30000 = 1500



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Since the expense on fuel is same for all of them, we count anybody with expenditure

less than 30000 towards the answer. Hence, except D and F, all the other four persons

have fuel expenses greater than 5%.

Solution 50.


Expense on others = 22% of 30000 = 6600

Given, the expenses of A on others = 6600 (1 +10

100) = 7260

Total expenses of A = 24000

% of expenses of A on others = 7260

24000× 100 = 30.25%

Solution 51.

Let the average after 18th innings be x runs.

We know,

Average of quantities =sum of all quantities/no. of quantities

So, the total score after 18th innings = 18x

A batsman makes a score of 120 runs in the 18th innings and thus increases his average

by 5.

So, the average after 17th innings = (x – 5) runs.

We can write now,

17(x – 5) +120 = 18x



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⇒ 17x – 85 + 120 = 18x

⇒ x = 120 – 85

⇒ x = 35

∴ His average after 18th inning = 35 runs.

Solution 52.

When a die is thrown, we have S = {1, 2, 3, 4, 5, 6}

n(S) = 6

Let, E = event of getting an even number = {2, 4, 6}

n(E) = 3

∴ Probability of occurrence of event;

P(E) = 𝑛(𝐸)

𝑛(𝑆) = 3/6 = ½

Solution 53.

We have,

P + 0.39P = 182 + 5(P/4)

⇒ 182 = 1.39P – 1.25P

⇒ P = 182/0.14 = 1300

When 1300 is divided by 7, remainder will be 5.



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Solution 54.

Let the speed of boat in still water be U km/hr, and speed of stream be V km/hr.

⇒ Upstream speed = (U – V) km/hr

⇒ Downstream speed = (U + V) km/hr

The speed of stream is one third of the upstream speed of the boat.

⇒ U – V = 3V

⇒ U = 4V

⇒ Downstream speed = (U + V) km/hr = (U + U/4) km/hr = 5U/4 km/hr

When travelling downstream, the boat covers 46.5 km in three hours.

We know, distance = speed × time

⇒ 46.5 = 5U/4 × 3

⇒ U = 15.5/1.25 = 12.4

∴ Speed of boat in still water is 12.4 km/hr.

Directions: Read the following table carefully and answer the questions given below.

Highest marks and average marks obtained by students in subjects over the years

The maximum marks in each subject is 100.

Subjects

Year English Bengali Maths Science Geography

High Avg High Avg High Avg High Avg High Avg

2004 90 65 78 55 95 70 90 70 75 50



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2005 85 72 85 50 92 65 89 72 60 58

2006 78 60 75 58 80 68 92 68 68 56

2007 75 58 80 65 90 72 94 65 74 61

2008 75 55 85 60 94 74 90 70 76 60

2009 80 68 82 55 88 66 88 75 78 64

Solution 55.

From the table,

Total highest marks of the five subjects in 2008 = 75 + 85 + 94 + 90 +76 = 420

Hence, the average highest marks = 420/5 = 84

∴ The average highest mark of the five subjects in 2008 was 84.

Solution 56.

From the table,

The difference in the highest marks in Science between 2006 and 2009 = 78 – 68 = 10

Again, the difference in the average marks in English between 2007 and 2009 = 68 – 58

= 10

∴ The difference in the highest marks in Geography between 2006 and 2009 was exactly

equal to the difference in the average marks in English between 2007 and 2009.

Solution 57.



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From the table,

The highest marks of Bengali in 2004 = 78

The highest marks of Bengali in 2005 = 85

Hence, the increase in highest marks of Bengali from 2004 to 2005 = 85 – 78 = 7

The required percentage = [(7/78) × 100]% = 8.97% ≈ 9%

∴ The approximate percentage increase in highest marks in Bengali from 2004 to 2005

= 9.

Solution 58.

From the table,

The total highest marks in Math in 2004, 2005 and 2006 = 95 + 92 + 80 = 267

Hence, the average highest marks in Math in 2004, 2005 and 2006 = 267/3 = 89

The highest marks in Science in 2005 = 89

∴ The average highest marks in Math in 2004, 2005 and 2006 was exactly equal to the

highest marks in Science in 2005.

Solution 59.

Let us assume the initial amount of whisky in the jar to be ‘1’

∴ It has initially 0.4 litres of alcohol in it

Let x amount of whisky be removed from the jar

∴ the amount of alcohol remaining will be 0.4 – 0.4x



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After that x amount of liquid with 19% alcohol is added

Thus final amout of alcohol = 0.4 – 0.4x + 0.19x

But it is given as 37%

∴ 0.4 – 0.4x + 0.19x = 0.37

∴ -0.21x = -0.03

∴ x = 1

7

∴ The quantity of whisky replaced is 1/7.

Solution 60.

We know, Selling price = Cost Price × (1 + Profit Percentage/100)

And, Selling price = Cost Price × (1 - Loss Percentage/100)

Total selling price in sum of individual cases and overall case should be same.

⇒ 1300 × (1 + 20/100) + 1300 × (1 – N/100) = 2600 × (1 + 7/100)

⇒ 1.2 + 1 – 0.01N = 2.14

⇒ 0.06 = 0.01N

⇒ N = 6

Directions: Study the following Pie-chart carefully and answer the questions given

below.

Budget allocation to different sectors is given below. The overall budget was 1900 billion

rupees.



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Solution 61.

% of budget allocated to defense = 34%

% of budget allocated to sports = 6%

% of budget allocated to health = 24%

% of budget allocated to food = 8%

Ratio of budget allocated to defense and sports together to the budget allocated to health

and food together = (34 + 6)/(24 + 8) = 5 : 4

Solution 62.

Current budget amount = 1900 billion rupees

Let the % increase be ‘a’ %

Defense34%

Social security

12%

sports6%

food and agriculture

8%

Health24%

Education16%



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% of budget allocated to education = 16%

% of budget allocated to social security = 12%

Given, current budget allocated to education is equal in amount to the new budget

allocated to social security.

⇒ 16% of 1900 = 12% of (1900 + a% of 1900)

⇒ 16 × 19 = 12 × 19(1 + a/100)

⇒ 4/3 = 1 + a/100

⇒ a = 100/3 = 33.33%

Solution 63.

(a) % of budget allocated to defense = 34%


% budget allocated to social security = 12%

Difference between defense budget and sum of budgets allocated to social security and

sports taken together = 34% – (6 + 12)% = 16%

(b) % of budget allocated to health = 24%

% of budget allocated to food and agriculture = 8%

Difference between budget allocated to health and budget allocated to food and

agriculture

= 24% - 8% = 16%

As we can see a = b.



36 | P a g e

Solution 64.

Total budget = 1900 billion rupees

% increase in total budget = 20%

New total budget = 1900 + 20% of 1900

New total budget = 1900 × 1.2 = 2280 billion rupees

Budget allocated to research is 114 billion rupees.

% of budget allocated to research =114/2280 × 100

% of budget allocated to research = 5%

% of budget remaining = 95%

Initial % of budget allocated to education = 16%

When the budget was 100%, % of budget allocation to education is 16%

Thus when the budget is 95%, % of budget allocation to education =16/100 × 95 = 15.2%

Increase in budget for education = 15.2% of 2280 – 16% of 1900

Increase in budget for education = 42.56 billion rupees

Solution 65.

% of budget allocated to defense = 34%


% of budget allocated to health = 24%

% of budget allocated to social security = 12%



37 | P a g e

Difference of % of budget allocated = (34 + 6) – (24 + 12) = 4%

Total budget = Rs. 1900 billion

Thus difference between sum of budgets allocated to defense and sports and sum of

budgets allocated to health and social security = 4% of 1900 billion = 76 billion rupees

Solution 67.

Nancy works for 1 hour on Monday, 2 hours on Tuesday, 3 hours on Wednesday and so

on.

In this manner, she works for 5 hours on Friday.

Her wages for Friday are Rs. 4000.

⇒ Wage per hour = Rs. 4000/5 = Rs. 800

Total hours worked in week = (1 + 2 + 3 + 4 + 5 + 6) = 21

∴ Total wages earned = Rs. 800 × 21 = Rs. 16800

Solution 68.

Let share of Danny be Rs. T.

The total wage they get is Rs. 900 only as work done is same.

So, share of Gary will be Rs. (900-T).

1 hour and 20 minutes = 4/3 hours

We know, time taken to finish work is inversely proportional of wages(for a fixed

amount of time in which work is done).



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⇒ 3 × T = (4/3) × 900

⇒ T = 400

∴ Share of Danny will be Rs. 400.

Solution 69.

Let the number to be subtracted be ‘a’.

Given, ratio is 15:17.

When ‘a’ is subtracted from both the numerator and denominator the ratio becomes 5:6

∴15−𝑎

17−𝑎=

5

6

⇒ 90 – 6a = 85 – 5a

⇒ a = 5

Solution 70.

⇒ let the initial price of T-shirt be Rs 100.

⇒ after first discount of 20%, new price of T-shirt will be Rs 80.

⇒ after successive discount of 15% on new price of T-shirt i.e. Rs 80, the final price of T-

shirt will be Rs 68.

∴ Overall discount = (100 – 68) = Rs 32 i.e. 32% discount.

Solution 71.



39 | P a g e

Let the average after 15th innings be x runs.

We know,


So, the total score after 15th innings = 15x

A batsman makes a score of 75 runs in the 15th innings and thus increases his average by

3.

So, the average after 14th innings = (x – 3) runs.

We can write now,

14(x – 3) +75 = 15x

⇒ 14x – 42 + 75 = 15x

⇒ x = 75 – 42

⇒ x = 33

∴ His average after 15th inning = 33 runs.

Solution 72.

The car owner buys petrol at Rs. 8.00, Rs. 10.00 and Rs. 12.50 per litre for three

successive years.

In 1st year total petrol bought = 5000/8.00 = 625 litres

In 2nd year total petrol bought = 5000/10 = 500 litres

In 3rd year total petrol bought = 5000/12.5 = 400 litres

So, total petrol bought in successive three years = 625 + 500 + 400 = 1525 litres



40 | P a g e

Total amount spend for petrol in three years = Rs. 5000 × 3 = Rs. 15000

We know,


∴ The average cost per litre of petrol = Rs. 15000/1525 = Rs. 9.83

Solution 73.

The word contains 13 words from which we have ‘MSCLLNS’ consonants and ‘IEAEOU’

vowels.

There are 7 consonants, in which L and S occur 2 times hence,

Number of ways of arranging these letters = 7! / (2! × 2! ) = 1260

6 vowels in which E repeat twice can be arranged in 6! / 2! = 360

Required ways = 1260 × 360 = 453600

Solution 74.

Mean proportional between(x + 1) and (2x + 3) is (x + 3)

∴√(x + 1)(2x + 3) = x + 3

2x2 + 5x + 3 = x2 + 6x + 9

(x - 3)(x + 2) = 0

x =3, -2

∴Taking positive value of x, x= 3



41 | P a g e

Solution 75.

Let the income of Shivani and Anita be x, 3x respectively

Let the Expenditure of Shivani and Anita be 3y, 4y respectively

∴Savings of Shivani = x – 3y

∴Savings of Anita = 3x – 4y

As the savings of both are equal

x – 3y = 3x – 4y

y = 2x

Ratio between income and expenditure of shivani = x/3y = x/(3 × 2x) = 1/6 = 1:6

Directions: Study the following graph carefully to answer these questions.

A report on the no. of people (In thousands) of two states affected by flood over the

years:



42 | P a g e

Solution 76.

From the graph,

The total no. of people of both states affected by flood in 2006 = (15 + 38) thousands

= 53 thousands.

The total no. of people of both states affected by flood in 2010 = (36 + 34) thousands

= 70 thousands.

∴ The required percentage = [(53/70) × 100]% = 75.71% ≈ 76%.

Solution 77.

From the graph,

0

5

10

15

20

25

30

35

40

45

2005 2006 2007 2008 2009 2010

State A

State B



43 | P a g e

The no. of people affected by flood in state A in 2008 = 18000

The percentage of women affected was 40 in state A.

So, the no. of women affected by flood in state A in 2008 = 18000 × (40/100) = 7200

And,

The no. of people affected by flood in state B in 2008 = 32000

The percentage of women affected was 27 in state B.

So, the no. of women affected by flood in state B in 2008 = 32000 × (27/100) = 8640

∴ The required ratio = 7200 : 8640 = 5 : 6.

Solution 78.

From the graph,


Percentage of people rescued in State A in 2009 = 65

So, the no. of people rescued in State A in 2009 = 30000 × (65/100) = 19500

And,

The no. of people affected by flood in state B in 2009 = 28000

Percentage of people rescued in State B in 2009 = 76

So, the no. of people rescued in State B in 2009 = 28000 × (76/100) = 21280

∴ The total no. of people rescued in that year = 19500 + 21280 = 40780.

Solution 79.



44 | P a g e

From the graph,


The total no. of people affected by flood in state A over given the years

= (24 + 15 + 12 + 18 + 30 + 34) thousands = 135000

So, the average no. of people affected by flood in state A = 135000/6 = 22500

∴ The required percentage = [(18000/22500) × 100]% = 80%.

Solution 80.

From the graph,

The total no. of people affected by flood in State B over given the years

= (35 + 38 + 42 + 32 + 28 + 34) thousands = 209000

Percentage of women affected in State B in all years together = 42

Sp, the percentage of men affected in State B in all years together = 100 – 42 = 58

∴ The required difference = 209000 × (58 – 42)% = 209000 × (16/100) = 33440.

Directions: Following 7 questions are based on the following pie chart which gives the

expenditure incurred in printing a magazine.



45 | P a g e

Solution 81.

∵ Percentage of expenditure incurred for content developer = 30%

Since 100% represents 360°

∴ 30% represents = 360/100 × 30 = 108°

Solution 82. ∵ Percentage of expenditure incurred for transportation charges = 4%

Since 100% represents 360°

∴ 4% represents = 360/100 × 4 = 14.4°

Solution 83.

∵ Percentage of expenditure incurred for content developer = 30%

24%

10%

2%4%12%

18%

30%Printing Costs

Paper Costs

Miscellaneous

Transportation

Binding

Promotion Costs

Editorial Contgent Development



46 | P a g e

∵ Percentage of expenditure incurred for cost on miscellaneous = 2%

Since editorial content development cost is Rs. 30,000

∴ 30% is equivalent to Rs. 30000

∴ 2% is equivalent to = 30000/30 × 2 = R.2000

Solution 84.

Since the Binding cost is Rs. 12000

∵ Percentage of expenditure incurred for Binding cost = 12%

∴ 12% is equivalent to Rs.12000

Or, 100% is equivalent to Rs. 100000

∴ Total cost incurred for manufacturing magazines = Rs 100000

Since the print-run is 12500 copies

∴ Sale price per copy = 100000/12500 = Rs. 8

Solution 85.

As per previous question Since the Binding cost is Rs. 12000

∵ Percentage of expenditure incurred for Binding cost = 12%

∴ 12% is equivalent to Rs.12000

Or, 100% is equivalent to Rs. 100000

∴ Total cost incurred for manufacturing magazines = Rs 100000



47 | P a g e

Since total cost incurred for manufacturing magazines is Rs 100000 and the print-run is

50000 copies

∴ Sale price per copy = 100000/50000 = Rs. 2

Solution 86.

Since promotional costs for a given issue of the magazine is Rs. 9,000

∵ Percentage of expenditure incurred for promotional costs = 18%

Since 18% is equivalent to Rs. 9000

∴ 100% is equivalent to = 9000/18 × 100 = Rs.50000

∴ Total expenditure in bringing out that issue of the magazine = Rs. 50000

Solution 87.

From previous question since promotional costs for a given issue of the magazine is Rs.

9,000

∵ Percentage of expenditure incurred for promotional costs = 18%

Since 18% is equivalent to Rs. 9000

∴ 100% is equivalent to = 9000/18 × 100 = Rs.50000

∴ Total expenditure in bringing out that issue of the magazine = Rs. 50000

∵ Percentage of expenditure incurred for editorial content development = 30%

∴ Cost of editorial content development = 30% of 50000 = Rs. 15000



48 | P a g e

Solution 88.

Loan amount = price of the cell phone – down payment

= Rs (7500 – 1750) = Rs 5750

Now according to the question,

5750 + 5750 ×3

12×

𝑟

100= {2000 + 2000 ×

2

12×

𝑟

100} + {2000 + 2000 ×

1

12×

𝑟

100} + 2000

⇒ 5750 +575

40𝑟 = 6000 +

2000

1200𝑟(1 + 2)

⇒ (575/40)r = 250 + 5r

⇒ (375/40)r = 250

⇒ r = (250 × 40)/375 = 26.67%

Hence, the required rate of interest imposed by the shopkeeper is 26.67%.

Solution 89.

The pattern of given series is:

→ 498 - (2 x 2) = 494

→ 494 - (3 x 3) = 485

→ 485 + (5 x 5) = 510

→ 510 + (7 x 7) = 559*

→ 559 - (11 x 11) = 438

→ 438 - (13 x 13) = 269

So the missing number is 559.



49 | P a g e

(Note: Here, 2,3,5,7,11,13,17 are Prime Numbers)

Solution 90.

The pattern of given series is:

→ 82 × 9 = 576 ∗

→ 92 × 10 = 810

→ 102 × 11 = 1100

→ 112 × 12 = 1452

→ 122 × 13 = 1872

So the missing number is 576.

The following pie-chart shows the percentage distribution of the expenditure incurred in

producing a phone. Study the pie-chart and the answer the questions based on it.

Various Expenditures (in percentage) Incurred in producing a phone.



50 | P a g e

Solution 91.

Let the amount of Royalty to be paid for these books be Rs. r.

Then, 10 : 15 = 10000 : r

⇒ r = Rs. (10000 x 15)

= Rs. 150000.

Solution 92.

Promotion30%

Transport10%

Royalty15%

Industrial Production

10%

Packaging20%

Material Cost15%

Various Expenditures (in percentage) Incurred in producing a phone



51 | P a g e

Clearly, marked price of the book = 140% of C.P.

Also, cost of paper = 20% of C.P

Let the cost of paper for a single phone be Rs. n.

Then, 140 : 20 = 28000 : n

⇒ n = Rs.(20 x 200)

= Rs. 4000

Solution 93.

For the publisher to earn a profit of 30%. So, S.P. = 130% of C.P.

Also Transportation Cost = 10% of C.P.

Let the S.P. of 1000 books be Rs. x.

Then, 10 : 130 = 100000 : x

⇒ x = Rs.(13 ×100000)

= Rs. 1300000.

Therefore S.P. of one phone

= Rs.(1300000/1000)

= Rs. 1300

Solution 94.

Royalty on phone= 15% of total expenditure.

Transport cost on phone = 10% of total expenditure.

Difference =5% of total expenditure.

=9000 ×5%

=450



52 | P a g e

Solution 95.

Material Cost of phone= 15% of C.P.

Industrial production cost on phone = 10% of C.P.

Difference = (15% of C.P.) - (10% of C.P) = 5% of C.P.

Therefore Percentage difference

= (Difference/Material cost) x 100 %

=(5/15 × 100)%

=33 1/3%

Solution 96.

Work done by john = 1/12 × no. of days he worked

Work done by jame = 1/16 × no. of days he worked

Ratio of work done by jame and john both = 1/12 × no. of days he worked: 1/16 × no. of

days he worked

Given that, they worked together for same days

∴ Ratio of work = 1/12: 1/16 = 4/3

∴ Jame’s share = 3/ (4+3) ×9051 = 3/7 × 9051 = 3879

Solution 97.

Let ‘x’ be the amount at 6% p.a and (7000 – x) at 5% p.a

We know that

Simple interest = P× R× T/100

Where P = principal amount , R= rate of interest and T = total number of years

1900 = (x×5×6)/100+ ((7000-x)×5×5)/100

∴ 1900 = 0.3x + 1750 – 0.25x

∴ x = 150/0.05

∴ x = 3000



53 | P a g e

∴ The amount paid at 5% p.a is (7000-3000)=Rs. 4,000

Solution 98.

Let C.P. of each energy drink be ‘x’.

C.P. of 10 energy drinks = 10x

S.P. of 8 energy drinks = 10x

⇒ S.P. of each energy drink = 10𝑥

8

∴ Profit % = 𝑆.𝑃.−𝐶.𝑃.

𝐶.𝑃.×100 =

10𝑥

8−𝑥

𝑥×100 =

2

8×100 = 25%

Solution 99.

Let the S.P. of each article be ‘x’.

S.P. of 600 articles = 600x

C.P. of 480 articles = 600x

⇒ C.P. of each article = 600𝑥

480 = 5x/4 = 1.25x

∴ Loss % = 𝐶.𝑃.−𝑆.𝑃.

𝐶.𝑃.×100 =

1.25𝑥−𝑥

1.25𝑥 ×100=

0.25

1.25×100 =20%

Solution 100.

Let C.P. of 1m rope be ‘x’.

C.P. of 78m rope = 78x

S.P. of 78m rope = C.P. of 78m rope +Profit/Gain

⇒ S.P. of 78m rope = C.P. of 78m rope + C.P. of 22m rope =78x+22x = 100x

S.P. of 1m rope = 100𝑥

78

∴ Gain % = 𝑆.𝑃.−𝐶.𝑃.

𝐶.𝑃.×100 =

100𝑥

78−𝑥

𝑥×100 =

22

78×100 = 28

8

39%



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