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

Solved Examples Set 3 (Quantitative Ability)

  1. Rashid's salary was reduced by 20%. In order to restore his salary at the original amount, it must be raised by

    1. 20%
    2. 22.50%
    3. 25%
    4. 26%
    5. 27%
    Let Rashid's Salary 100
    20% reduced salary is 80
    As the reduced amount is 20
    So what percentage of the present sallary is required to be equal to 20?
    ?% of 80 = 20
    ? = \(20 \over 80\) × 100 = 25%
  2. 5789 - 2936 + 1089 = ?

    1. 3942
    2. 4041
    3. 2626
    4. 3932
    5. 3940
    5789 - 2936 + 1089 = 6878 - 2936 = 3942
  3. 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
  4. \( {5.76 \over 1.6} - 2.4 = ? \)

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

    1. 19.8
    2. 36
    3. 55
    4. 33
    5. 50
    \( {396 \over 11} \) + 19 = 36 + 19 = 55
  7. A bank exchanges British currency for Singapore currency at the rate of S$ 3.20 to pond 1. Calculate, in Pond, the amount exchanged for S$ 1,600 by a customer who also had to pay an extra 3% commission for this transaction.

    1. Pond 475
    2. Pond 485
    3. Pond 495
    4. Pond 505
    5. Pond 510
    As commission is 3% of 1600 = 0.03 × 1600 = S$ 48
    the rest amount = 1600 - 48 = S$ 1552
    S$ 1 = \(1 \over 3.20\) = Pond 0.3125
    Now S$ 1552 = 1552 × 0.3125 = Pond 485
  8. 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
  9. 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.
  10. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
    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.
  11. 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
  12. 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
  13. 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%
  14. 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
  15. 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 \(km \over hr\)
  16. On a trip to visit friends, a family drives 65 miles per hour for 208 miles of the trip. If the entire trip was 348 miles and took 6 hours, what was the average speed, in miles per hour, for the rest of the trip?

    1. 44
    2. 50
    3. 51
    4. 58
    5. 60
    As the first part of the trip took \(\frac{208 \text{ miles}}{65 \text{ } \frac{miles}{hour}} = 3.2 \text{ hours},\) so the remaining 140 miles (348 - 208) took 2.8 hours (6 - 3.2). The average speed for the rest of the trip was \(\frac {140 \text{ miles}}{2.8 \text{ hours}} = 50 \) miles per hour.
  17. 350 × ? = 4200

    1. 12
    2. 24
    3. 15
    4. 30
    5. 16
    \( ? = {4200 \over 350} =12 \)
  18. After spending 88% of his income, a man had $ 2160 left. Find his income.

    1. $ 18000
    2. $ 19000
    3. $ 20000
    4. $ 22000
    5. $ 17000
    Let income = x
    x = 88% of x + 2160
    x - 0.88x = 2160
    0.12x = 2160
    x = \(216000 \over 12\) = 18000
  19. A certain number was doubled and the result then multiplied by 3. If the product was 138, find the number.

    1. 21
    2. 23
    3. 25
    4. 27
    5. 19
    Let x be the number
    the number is doubled, 2x
    the result is multiplied by 3, 3 × 2x = 6x
    6x = 138
    x = \(138 \over 6\) = 23
  20. 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

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