Solved Examples Set 1 (Quantitative Ability)

1. 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

2. A man takes 50 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in

   1. 1 hour
   2. 30 minutes
   3. 20 minutes
   4. 10 minutes
   5. 40 minutes

   \( 50 × 6 \over 10 \) = 30 minutes

3. 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.

4. 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

5. 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

6. \( {396 \over 11} \) + 19 = ?

   1. 19.8
   2. 36
   3. 55
   4. 33
   5. 50

   \( {396 \over 11} \) + 19 = 36 + 19 = 55

7. Matthew’s age (𝑚) is three years more than twice Rita’s age (𝑟). Which equation shows the relationship between their ages?

   1. 𝑚 = 𝑟 − 32
   2. 𝑚 = 𝑟 + 32
   3. 𝑚 = 2(𝑟 + 3)
   4. 𝑚 = 2𝑟 − 3
   5. 𝑚 = 2𝑟 + 3

   As Matthew's age (𝑚) is three more years (+3) than twice Rita's age (2𝑟). Therefore, 𝑚 = 2𝑟 + 3.

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

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

   x = \( {48 × 100 \over 60} \) = 80

9. A man saves $ 500, which is 15% of his annual income. How much does he earn in one year?

   1. $ 3542.5
   2. $ 3333.33
   3. $ 3132.3
   4. $ 3075.75
   5. $ 4444.4

   Let annual income = x
   15% of x = 500
   x = \(500 \over 15\) × 100 = \(10000 \over 3\) = 3333.33

10. A bank increased the rate of interest which it paid to depositors from 3.5% to 4% per annum. Find how much more interest a man would receive if he deposited $ 64000 in the bank for 6 months at the new interest rate

    1. $ 160
    2. $ 180
    3. $ 200
    4. $ 220
    5. $ 150

    If the interest rate is 3.5% then interest amount is
    3.5% of 6400 = 0.035 × 6400 = $ 2240
    If the interest rate is 4% then interest amount is
    4% of 6400 = 0.04 × 6400 = $ 2560
    Now the difference of both interests = 2560 - 2240 = $ 320 per annum
    Interest for half year (6 months) = \(320 \over 2\) = $ 160

11. 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

12. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

13. 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

14. 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%.

15. How much would I have to pay for a book which cost $ 72 to product, if the printing company sold it to a bookseller at 20% profit and in return the bookseller sold it to me at a profit of 25%?

    1. $ 104
    2. $ 106
    3. $ 108
    4. $ 110
    5. $ 109

    Cost of the book product = $ 72
    Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4
    Now the cost of the book = 72 +14.4 = $ 86.4
    Profit of the bookseller = 25% of 86.4 = 21.4
    Finally, the cost of the book = 86.4 + 21.4 = $ 108

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. 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%

18. Which expression is equivalent to \(\frac{6𝑥^2 + 4𝑥}{2𝑥}\)?

    1. 7x
    2. 5x²
    3. 3x + 2
    4. 6x² + 2
    5. 3x² + 2x

    As \(\frac{6𝑥^2}{2𝑥} = 3𝑥,\) and \(\frac{4𝑥}{2𝑥} = 2,\) so then \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

19. 60% of 37 = ?

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

    60% of 37 = 0.6 × 37 = 22.2

20. The closest approximation of \(\frac{69.28 × .004}{.03}\) is

    1. 0.092
    2. 0.92
    3. 9.2
    4. 92
    5. 920

    This problem is most easily completed by rearranging and approximating as follows:
    (69.28 x .004)/.03 ≅ 69 x .1 = 6.9
    which is the only reasonably close answer to 9.2

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