Solved Examples Set 1 (Quantitative Ability)

1. A person's net income is $ 1373.70 and he pays an income tax of 5%. His gross income in dollars must be

   1. 1446
   2. 1118.96
   3. 1308.29
   4. 1438.25
   5. 1211.21

   Let gross income in dollars = x
   then according to the statement,
   x = 5% of x + 1373.70
   x - 0.05x = 1373.70
   0.95x = 1373.70
   x = \(137370 \over 95\) = 1446

2. 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

3. 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

4. A primary school had an enrollment of 850 pupils in January 1970. In January 1980 the enrollment was 1,120. What was the percentage increase for the enrollment?

   1. 31.76%
   2. 33.50%
   3. 30.65%
   4. 34.76%
   5. 30.55%

   Percentage increase for the enrollment = \(1120 - 850 \over 850\) × 100 = 31.76

5. 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%.

6. 5789 - 2936 + 1089 = ?

   1. 3942
   2. 4041
   3. 2626
   4. 3932
   5. 3940

   5789 - 2936 + 1089 = 6878 - 2936 = 3942

7. A man was 32 years old when his daughter was born. He is now five times as old as his daughter. How old is his daughter now?

   1. 7 years
   2. 8 years
   3. 9 years
   4. 10 years
   5. 6 years

   Let's assume the daughter is d years old now. That means that the man is now (32 + d) years old, so that
   (32 + d) = 5d
   32 = 4d
   d = 8

8. 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

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. If 4a + 2 = 10, then 8a + 4 =

    1. 5
    2. 16
    3. 20
    4. 24
    5. 28

    One may answer this question by solving
    4a + 2 = 10
    4a = 8
    a= 2
    Now, plugging in 2 for a:
    8a + 4 = 8(2) + 4 = 20
    A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.

11. 12% of ________ = 48

    1. 250
    2. 100
    3. 400
    4. 200
    5. 300

    \(12 \text{% of } x = 48\)
    \(0.12x = 48\)
    \(x = \frac{48}{0.12} = 400\)

12. 60% of 37 = ?

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

    60% of 37 = 0.6 × 37 = 22.2

13. if x% of 60 = 48 then x = ?

    1. 80
    2. 60
    3. 90
    4. 40
    5. 70

    x = \( {48 × 100 \over 60} \) = 80

14. 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

15. if a > b and b > c then:

    1. a = c
    2. a > c
    3. c > a
    4. a < c
    5. none

    As a > b > c so a > c

16. A certain solution is to be prepared by combining chemicals X, Y and Z in the ratio 18:3:2. How many liters of the solution can be prepared by using 36 liters of X?

    1. 46 liters
    2. 47 liters
    3. 45 liters
    4. 49 liters
    5. 44 liters

    As total ratio is 18 +3 + 2 = 23
    Let total solution is x liters
    Then \(18 \over 23\) x = 36
    x = \(36 × 23 \over 18\) = 46 liters

17. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

    1. 33 days
    2. 35 days
    3. 37 days
    4. 39 days
    5. 31 days

    \(14 × 10 \over 4 \) = 35 days

18. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    1. 1
    2. 3
    3. 0
    4. 67
    5. 25

    \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25}\)
    \(= 25 \text{% } × {4 \over 4\text{%}} × {1 \over 25} \)
    \(= 0.25 × {4 \over 0.04} × {1 \over 25}\)
    \(= {25 \over 25}\)
    = 1

19. A third-grade class is composed of 16 girls and 12 boys. There are 2 teacher-aides in the class. The ratio of girls to boys to teacher-aides is

    1. 16:12:1
    2. 8:6:2
    3. 8:6:1
    4. 8:3:1
    5. 4:3:1

    Girls to boys to teacher-aides are in proportion 16 to 12 to 2. Reduced to lowest terms, 16:12:2 equals 8:6:1.

20. \(\frac{\frac{7}{10} × 14 × 5 × \frac{1}{28}}{\frac{10}{17} × \frac{3}{5} × \frac{1}{6} × 17} = \)

    1. 4/7
    2. 1
    3. 7/4
    4. 2
    5. 17/4

    

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