In the name of ALLAH, the most beneficient, the most merciful

Solved Examples Set 3 (Quantitative Ability)

  1. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075
    1015 / 0.05 / 40 = 20300 / 40 = 507.5
  2. 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.
  3. ?% of 60 = 24

    1. 40
    2. 48
    3. 45
    4. 42
    5. 38
    ?% × 60 = 24
    \(? = {24 \over 60} × 100 \) = 40
  4. 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.
  5. \( {396 \over 11} \) + 19 = ?

    1. 19.8
    2. 36
    3. 55
    4. 33
    5. 50
    \( {396 \over 11} \) + 19 = 36 + 19 = 55
  6. 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
  7. A man bought a flat for $ 820000. He borrowed 55% of this money from a bank. How much money did he borrow from the bank?

    1. $ 451000
    2. $ 452000
    3. $ 453000
    4. $ 454000
    5. $ 450000
    55% of 820000 = 0.55 × 820000 = $ 451000
  8. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003
    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)
  9. 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
  10. Rashid buys three books for $ 16 each and four books for $ 23 each, what will be the average price of books

    1. $ 18
    2. $ 20
    3. $ 22
    4. $ 24
    5. $ 16
    Price of 3 books = 3 × 16 = $ 48
    Price of 4 books = 4 × 23 = $ 92
    Total price = $ 140
    Total books = 3 + 4 = 7
    Average price of books = \(140 \over 7 \) = $ 20
  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. 5.41 - 3.29 × 1.6 = ?

    1. 14.6
    2. 0.3392
    3. 0.146
    4. 3.392
    5. 1.46
    5.41 - 3.29 × 1.6 = 5.41 - 5.264 = 0.146
  13. If 3x = −9, then 3x3 − 2x + 4 =

    1. -83
    2. -71
    3. -47
    4. -17
    5. 61
    First solving 3x = −9, x = −3. Now plug into 3x3 − 2x + 4:
    3x3 − 2x + 4
    = 3(-3)3 − 2(-2) + 4
    = 3(−27) + 6 + 4
    = −81 + 6 + 4
    = −71
  14. A rectangular room is 6 m long, 5 m wide and 4 m high. The total volume of the room in cubic meters is

    1. 24
    2. 30
    3. 120
    4. 240
    5. 140
    Total volume = length × width × height = 6 × 5 × 4 = 120
  15. \(\frac{\frac{7}{10} × 14 × 5 × \frac{1}{28}}{\frac{10}{17} × \frac{3}{5} × \frac{1}{6} × 17} = \)

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

  16. The amount of hot cocoa powder remaining in a can is 6 1⁄4 tablespoons. A single serving consists of 1 3⁄4 tablespoons of the powder. What is the total number of servings of the powder remaining in the can?

    1. 3 1⁄2
    2. 3 4⁄7
    3. 4 3⁄7
    4. 4 1⁄2
    5. 6
    As \(6\frac{1}{4} = \frac{25}{4}\) and \(1\frac{3}{4} = \frac{7}{4}\). Therefore,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)
  17. A girl is 18 years younger than her mother. In 6 years time, the sum of their ages will be 54.How old is the girl now?

    1. 10 years
    2. 11 years
    3. 12 years
    4. 13 years
    5. 14 years
    Let girl's age = x
    then mother's age = x + 18
    After 6 years,
    x + 6 + x + 18 + 6 = 54
    2x + 30 = 54
    2x = 24
    x = 12
  18. A train takes 50 minutes for a journey if it runs at 48 km/hr. The rate at which the train must run to reduce the time to 40 minutes will be

    1. 50 km/hr
    2. 55 km/hr
    3. 60 km/hr
    4. 57 km/hr
    5. 65 km/hr
    \(50 × 48 \over 40\) = 60 \(km \over hr\)
  19. \( {5.76 \over 1.6} - 2.4 = ? \)

    1. 1.2
    2. 2.4
    3. 7.2
    4. 0.12
    5. 0.012
    \( {5.76 \over 1.6} - 2.4 = \) 3.6 - 2.4 =1.2
  20. 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 \(km \over hr\)

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