Solved Examples Set 1 (Quantitative Ability)

1. A boy scored 90 marks for his mathematics test. This was 20% more than what he had scored for the geography test. How much did he score in geography?

   1. 71 marks
   2. 73 marks
   3. 75 marks
   4. 77 marks
   5. 78 marks

   20% of x + x = 90
   0.2x + x = 90
   1.2x = 90
   x = \(90 \over 1.2\)
   x = 75

2. 72 + 679 + 1439 + 537+ ? = 4036

   1. 1309
   2. 1208
   3. 2308
   4. 2423
   5. 1309

   72 + 679 + 1439 + 537+ ? = 4036
   2727 + ? = 4036
   ? = 4036 - 2727 = 1309

3. 60% of 37 = ?

   1. 20
   2. 21
   3. 22.2
   4. 22
   5. none

   60% of 37 = 0.6 × 37 = 22.2

4. ?% of 60 = 24

   1. 40
   2. 48
   3. 45
   4. 42
   5. 38

   ?% × 60 = 24
   \(? = {24 \over 60} × 100 \) = 40

5. A and B enter into a partnership contributing $ 800 and $ 1000 respectively. At the end of 6 months they admit C, who contributes $ 600. After 3 years they get a profit of $ 966. Find the share of each partner in the profit.

   1. $ 336, $ 420, $ 210
   2. $ 360, $ 400, $ 206
   3. $ 380, $ 390, $ 196
   4. $ 345, $ 405, $ 210
   5. $ 325, $ 400, $ 200

   A shares = 800 × 3 = 2400
   B shares = 1000 × 3 = 3000
   C shares = 600 × 2 1⁄2 = 1500
   Total shares = 2400 + 3000 + 1500 = 6900
   A's profit = \(2400 \over 6900 \) × 966 = $ 336
   B's profit = \(3000 \over 6900 \) × 966 = $ 420
   C's profit = \(1500 \over 6900 \) × 966 = $ 210

6. A fruit-seller has 120 oranges. Given that he has 20% more apples than oranges and 40% less oranges than pears, find the number of apples and the number of pears the fruit seller has.

   1. 144, 200
   2. 148, 380
   3. 149, 220
   4. 140, 190
   5. 142, 190

   No. of apples = 120 + 20% of 120 = 120 + 0.2 × 120 = 144
   
   Let x = No. of pears
   x - 40% of x = 120
   x - 0.4x = 120
   0.6x = 120
   x = \(120 \over 0.6\) = 200
   Hence, no. of pears = 200

7. A shop owner blends three types of coffees, A, B and C, in the ratio 3:5:7. Given that type A coffee costs $ 70 per kg, type B coffee costs $ 100 per kg and type C coffee costs $ 130 per kg, calculate the cost per kg of the blended mixture.

   1. $ 106
   2. $ 108
   3. $ 109
   4. $ 110
   5. $ 105

   Cost per kg = 70 x 1/5 + 100 x 1/3 + 130 x 7/15 = $ 108 per kg

8. 5873 + 12034 + 1106 = ?

   1. 19016
   2. 20001
   3. 19013
   4. 2018
   5. 19010

   5873 + 12034 + 1106 = 17907 + 1106 = 19013

9. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
   What is the value of 𝑥 in the equation?

   1. 10
   2. 20
   3. 26
   4. 31
   5. 40

   By cross multiplying, 4(𝑥 – 8) =3 × 24. Thus, 4𝑥 – 32 = 72, and so 4𝑥 = 104 and 𝑥 = 26.

10. \( {2244 \over 0.88} = ? × 1122 \)

    1. 20.02
    2. 20.2
    3. 19.3
    4. 2.27
    5. 3.27

    \( {2244 \over 0.88} = ? × 1122 \)
    \(? = {2550 \over 1122} = 2.27 \)

11. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003

    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)

12. A group of laborers accepted to do a piece of work in 20 days. 8 of them did not turn up for the work and the remaining did the work in 24 days. The original number of laborers was

    1. 47
    2. 48
    3. 49
    4. 50
    5. 51

    x laborers do work in 20 days and x-8 laborers do same work in 24 days. As the no. of laborers decrease, the no. of days increased then it becomes as
    x : x - 8 :: 24 : 20
    product of interiors = product of exteriors
    24x - 192 = 20x
    4x = 192
    x = 48

13. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

14. A man earned an annual income of $ 245000 in 1990. He was allowed a deduction of $ 15000 relief for each of his three children and a personal relief of $ 30000. If he was charged a tax rate of 4% on first $ 50000 and 6% on his remaining income, calculate the total tax charged.

    1. $ 9200
    2. $ 8700
    3. $ 9500
    4. $ 9400
    5. $ 9000

    Total Income = $ 245000
    Total relief = 3 × 15000 + 30000 = $ 75000
    Rest income = 245000 - 75000 = 170000
    Tax on 1st 50000 = 0.04 × 50000 = $ 2000
    Tax on rest amount 120000 = 0.06 × 120000 = $ 7200
    Total tax = 200 + 7200 = $ 9200

15. A single discount equivalent to a discount series of 20%, 10% and 25% is

    1. 55%
    2. 54%
    3. 46%
    4. 42%
    5. 50%

    If 3 succesive discounts are a%, b% and c%
    then single discount = a + b + c – (\(ab \over 100 \) + \(bc \over 100\) + \(ca \over 100 \) – \(abc \over 10000 \))
    a = 20, b = 10, c = 25, solving we get, 46%.

16. A and B can reap a field in 30 days, working together. After 20 days, however, B is called away and A takes 20 days more to complete the work. B alone could do the whole work in

    1. 48 days
    2. 50 days
    3. 56 days
    4. 60 days
    5. 64 days

    (A + B)'s 20 day's work = \(1 \over 30 \) × 20 = \(2 \over 3 \)
    Remaining work = 1 - \(2 \over 3 \) = \(1 \over 3 \)
    Now, \(1 \over 3 \) work is done by A in 20 days.
    Therefore, the whole work will be done by B in 20 × 3 = 60 days.

17. Which set of ordered pairs represents a function?

    1. {(−5,5),(4,8),(−5,−6)}
    2. {(−1,−1),(−1,6),(−1,−10)}
    3. {(−3,7),(2,5),(−7,7)}
    4. {(2,3),(−2,4),(−2,−5)}
    5. {(2,3),(3,2),(2,5)}

    For a set of ordered pairs to be a function, no single 𝑥-coordinate can be mapped to two distinct 𝑦-coordinates. This is not the case for option A, where 𝑥=−5 is mapped to both 𝑦=5 and 𝑦=−6. Similarly, in options B (𝑥=−1), D (𝑥=−2), and E (𝑥=2), an 𝑥 value is mapped to two different 𝑦 values.

18. 40 men can build a wall 4 metres high in 15 days. The number of men required to build a similar wall 5 metres high in 6 days is

    1. 115
    2. 125
    3. 105
    4. 135
    5. 130

    \( 40 × 15 × 5 \over 6 × 4 \) = 125 men

19. 40 arithmetic questions, each carrying equal marks, were given in a class test. A boy answered 25 questions correctly. What percentage was this? To pass a test a student must answer at least 45% of the questions correctly. Find the least number of correct answers needed to pass.

    1. 62.5%, 18
    2. 63.5%, 16
    3. 64.5%, 20
    4. 61.0%, 21
    5. 60.0%, 22

    \(x \text{% of } 40 = 25\)
    \(x \text{% } × 40 = 25\)
    \(x = {25 \over 40} × 100 \)
    x = 62.5
    
    \(x = 45 \text{% of } 40 \)
    \(x = 0.45 × 40 \)
    x = 18

20. 1.02 - 0.20 + ? = 0.842

    1. 0.222
    2. 232
    3. 2
    4. 0.022
    5. 0.012

    1.02 - 0.20 + ? = 0.842
    0.82 + ? = 0.842
    ? = 0.842 - 0.82 = 0.022

Solved Examples Set 1
Solved Examples Set 2
Solved Examples Set 3