Solved Examples Set 3 (Quantitative Ability)

1. 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.

2. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey.

   1. 53 km/hr
   2. 53.33 km/hr
   3. 54 1⁄4 km/hr
   4. 55 km/hr
   5. 56 km/hr

   The time, car took for the first half, \(50 \over 40 \) = 1.25 hrs
   and for the second half \(50 \over 80 \) = 0.625 hrs
   Total time = 1.25 + 0.625 = 1.875 hrs
   Average speed = \(100 \over 1.875 \) = 53.3 \(km \over hr\)

3. 8 : ? :: 1 : 4

   1. 24
   2. 16
   3. 0
   4. 32
   5. 20

   ? × 1 = 8 × 4
   ? = 32

4. \( {63.84 \over ?} \) = 21

   1. 3.04
   2. 3.4
   3. 30.4
   4. 300.4
   5. 0.304

   ? = \( 63.84 \over 21 \) = 3.04

5. A man sells two houses for $ 2 lac each. On one he gained 20% and on the other he lost 20%. His total profit or loss % in the transaction will be

   1. 4% profit
   2. 5% loss
   3. no profit, no loss
   4. 4% loss
   5. 3% loss

   % loss = (% loss X % profit)/100 = (20 X 20)/100 = 4%

6. A shopkeeper buys 300 identical articles at a total cost of $ 1500. He fixes the selling price of each article at 20% above the cost price and sells 260 articles at the price. As for the remaining articles, he sells them at 50% of the selling price. Calculate the shopkeeper's total profit.

   1. $ 180
   2. $ 185
   3. $ 200
   4. $ 190
   5. $ 170

   cost price of each item = \( 1500 \over 300 \) = $ 5
   selling price at 20% above the cost price = 5 + 5 × .2 = $ 6
   selling price of 260 items = 260 × 6 = $ 1560
   selling price of remaining 40 items = 40 × 6 × .5 = $ 120
   Total profit = 1560 + 120 - 1500 = $ 180

7. 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

8. 60% of 37 = ?

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

   60% of 37 = 0.6 × 37 = 22.2

9. 12% of ________ = 48

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

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

10. By selling 60 chairs, a man gains an amount equal to selling price of 10 chairs. The profit percentage in the transaction is

    1. 10%
    2. 15%
    3. 16.67%
    4. 20%
    5. 22%

    selling price of 60 chairs = selling price of 10 chairs
    profit of 60 chairs = profit of 10 chairs
    profit of 6 chairs = profit of 1 chair
    profit of 1 chair = profit of 1/6 chair
    profit %age = 1/6 x 100 = 16.67%

11. 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.

12. 
    In the figure above, AB is one edge of a cube. If AB equals 5, what is the surface area of the cube?

    1. 25
    2. 100
    3. 125
    4. 150
    5. 300

    Since one edge of the cube is 5, all edges equal 5. Therefore, the area of one face of the cube is:
    5 × 5 = 25
    Since a cube has 6 equal faces, its surface area will be:
    6 × 25 = 150

13. A boy of height 165 cm is replaced by another, which decreases the average height of the group of 34 students, by 1 cm. The height of the new student is

    1. 129 cm
    2. 130 cm
    3. 131 cm
    4. 132 cm
    5. 133 cm

    Total decreased height of 34 students = 1 × 34 = 34 cm
    Height of the replaced student = 165 - 34 = 131 cm

14. 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

15. At a book fair, a book was reduced in price from $ 75 to $ 60. If the first price gives a 50% profit, find the percentage profit of the book sold at the reduced price.

    1. 20%
    2. 30%
    3. 40%
    4. 50%
    5. 10%

    As $ 75 (first price) gives a profit = 50%
    $ 1 gives a profit = (50/75)%
    $ 60 (reduced price) gives profit = (50/75 x 60)% = 40%

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. 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

18. 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

19. A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 1114. Find the present age of son.

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

    Let son's age = x, then
    father's age = 5x
    As before 2 years ago the sum of the squares of their ages was 1114, the equation becomes as
    \((x - 2)^2 + (5x - 2)^2 = 1114 \)
    By simplifying the equation, we have
    \(13x^2 -12x -553 = 0\)
    Now solving the equation, we have
    \(13x^2 - 12x - 553 = 0\)
    \(13x^2 - 91x + 79x -553 = 0\)
    13x(x - 7) + 79(x - 7) = 0
    (x - 7)(13x + 79) = 0
    x = 7 and x = -6.077
    As age could not be negative, hence the present age of the son is 7 years.

20. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
    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.

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