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

Solved Examples Set 1 (Quantitative Ability)

  1. A shopkeeper bought a radio from a wholesaler for $ 250.00. In addition, he paid a sales tax of 15% on the cost price. He then sold the radio for $ 315.00. Calculate the cash profit made by the shopkeeper.

    1. $ 20.00
    2. $ 22.50
    3. $ 25.00
    4. $ 27.50
    5. $ 27.00
    cost price = $ 250
    sales tax = .15 Γ— 250 = $ 37.5
    cash profit = 315 - 250 - 37.5 = $ 27.5
  2. Which of the following expressions is equivalent to \(\frac{π‘₯^2 + 3x + 1}{π‘₯ + 1}\)?

    1. x + 2
    2. π‘₯ + 3
    3. π‘₯ + 2 - 1/(π‘₯ + 1)
    4. π‘₯ + 3 + 1/(π‘₯ + 1)
    5. π‘₯ + 4 + 5/(π‘₯ + 1)
    As \(π‘₯^2 + 3x + 1 = (π‘₯^2 + 3x + 2) -1\)
    and
    \(\frac{π‘₯^2 + 3x + 2}{x + 1} = \frac{(π‘₯ + 2)(x + 1)}{x + 1} = π‘₯ + 2\)
    Therefore,
    \(\frac{π‘₯^2 + 3x + 1}{x + 1} = \frac{π‘₯^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (π‘₯ + 2) - \frac{1}{x + 1}\)
  3. 12% of ________ = 48

    1. 250
    2. 100
    3. 400
    4. 200
    5. 300
    \(12 \text{% of } x = 48\)
    \(0.12x = 48\)
    \(x = \frac{48}{0.12} = 400\)
  4. 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.
  5. The closest approximation of \(\frac{69.28 Γ— .004}{.03}\) 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
  6. 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
    \(50 Γ— 48 \over 40\) = 60 \(km \over hr\)
  7. 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
    \( 50 Γ— 6 \over 10 \) = 30 minutes
  8. 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 Γ— 6400 = $ 2240
    If the interest rate is 4% then interest amount is
    4% of 6400 = 0.04 Γ— 6400 = $ 2560
    Now the difference of both interests = 2560 - 2240 = $ 320 per annum
    Interest for half year (6 months) = \(320 \over 2\) = $ 160
  9. 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
  10. \( {63.84 \over ?} \) = 21

    1. 3.04
    2. 3.4
    3. 30.4
    4. 300.4
    5. 0.304
    ? = \( 63.84 \over 21 \) = 3.04
  11. A man walked for 3 hours at 4.5 km/h and cycled for some time at 15 km/h. Altogether, he traveled 21 km. Find the time taken for cycling.

    1. 1/2 hour
    2. 1 hour
    3. 1 1⁄2 hours
    4. 2 hours
    5. 2 1⁄2 hours
    The man walked the distance = 3 x 4.5 = 13.5 km. The distance cycled by the man = 21 - 13.5 = 7.5 km
    As he cyled 15 km in 1 h
    he cycled 1 km in 1/15 h
    Finally, he cycled 7.5 km in 7.5/15 = 1/2 h
  12. 42.98 + ? = 107.87

    1. 64.89
    2. 65.89
    3. 64.98
    4. 65.81
    5. 63.89
    ? = 107.87 - 42.98 = 64.89
  13. If n! = n β‹… (n βˆ’ 1) β‹… (n βˆ’ 2) β‹… (n βˆ’ 3) . . . 2 β‹… 1, what is the value of \(\frac{(6!)(4!)}{(5!)(3!)}\)

    1. 5/4
    2. 8/5
    3. 10
    4. 24
    5. 1152
    \(\frac{(6!)(4!)}{(5!)(3!)}\) = \(\frac{(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2. 1)}{(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)}\) = \(\frac{6 . 4}{1}\) = 24
  14. 350 Γ— ? = 4200

    1. 12
    2. 24
    3. 15
    4. 30
    5. 16
    \( ? = {4200 \over 350} =12 \)
  15. 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 Γ— 820000 = $ 451000
  16. 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
  17. A shopkeeper sold two articles for $ 48 each. He made a 25% profit on one article and a loss of 20% on the other. What was his net gain or loss on the sale of the two articles?

    1. loss of $ 1.40
    2. gain of $ 2.40
    3. loss of $ 2.40
    4. gain of $ 1.40
    5. gain of $ 2.60
    25% profit at selling price $ 48 = 48 x .25 = $ 12
    20% loss at selling price $ 48 = 48 x 0.2 = $ 9.6
    gain = profit - loss = 12 - 9.6 = $ 2.4
  18. 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
    \(14 Γ— 10 \over 4 \) = 35 days
  19. Matthew’s age (π‘š) is three years more than twice Rita’s age (π‘Ÿ). Which equation shows the relationship between their ages?

    1. π‘š = π‘Ÿ βˆ’ 32
    2. π‘š = π‘Ÿ + 32
    3. π‘š = 2(π‘Ÿ + 3)
    4. π‘š = 2π‘Ÿ βˆ’ 3
    5. π‘š = 2π‘Ÿ + 3
    As Matthew's age (π‘š) is three more years (+3) than twice Rita's age (2π‘Ÿ). Therefore, π‘š = 2π‘Ÿ + 3.
  20. Which expression is equivalent to \(\frac{6π‘₯^2 + 4π‘₯}{2π‘₯}\)?

    1. 7x
    2. 5x2
    3. 3x + 2
    4. 6x2 + 2
    5. 3x2 + 2x
    As \(\frac{6π‘₯^2}{2π‘₯} = 3π‘₯,\) and \(\frac{4π‘₯}{2π‘₯} = 2,\) so then \(\frac{6π‘₯^2 + 4π‘₯}{2π‘₯} = 3π‘₯ + 2\)

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