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

Solved Examples Set 3 (Quantitative Ability)

  1. 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
  2. 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
  3. 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.
  4. 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 \(km \over hr\)
  5. A man pays 10% of his income for his income tax. If his income tax amounts to $ 1500, what is his income?

    1. $ 13000
    2. $ 15000
    3. $ 17000
    4. $ 19000
    5. $ 11000
    Let x = income
    10% of x = $ 1500
    0.1x = $ 1500
    x = \(1500 \over 0.1\) = $ 15000
  6. 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
  7. A person's net income is $ 1373.70 and he pays an income tax of 5%. His gross income in dollars must be

    1. 1446
    2. 1118.96
    3. 1308.29
    4. 1438.25
    5. 1211.21
    Let gross income in dollars = x
    then according to the statement,
    x = 5% of x + 1373.70
    x - 0.05x = 1373.70
    0.95x = 1373.70
    x = \(137370 \over 95\) = 1446
  8. 12% of ________ = 48

    1. 250
    2. 100
    3. 400
    4. 200
    5. 300
    \(12 \text{% of } x = 48\)
    \(0.12x = 48\)
    \(x = \frac{48}{0.12} = 400\)
  9. 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
  10. 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
  11. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    1. 1
    2. 3
    3. 0
    4. 67
    5. 25
    \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25}\)
    \(= 25 \text{% } Γ— {4 \over 4\text{%}} Γ— {1 \over 25} \)
    \(= 0.25 Γ— {4 \over 0.04} Γ— {1 \over 25}\)
    \(= {25 \over 25}\)
    = 1
  12. \( {396 \over 11} \) + 19 = ?

    1. 19.8
    2. 36
    3. 55
    4. 33
    5. 50
    \( {396 \over 11} \) + 19 = 36 + 19 = 55
  13. 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
  14. 40 men can build a wall 4 metres high in 15 days. The number of men required to build a similar wall 5 metres high in 6 days is

    1. 115
    2. 125
    3. 105
    4. 135
    5. 130
    \( 40 Γ— 15 Γ— 5 \over 6 Γ— 4 \) = 125 men
  15. A can do a piece of work in 10 days and B can do it in 15 days. The number of days required by them to finish it, working together is

    1. 8
    2. 7
    3. 6
    4. 4
    5. 3
    A's 1 day work = \(1 \over 10\)
    B's 1 day work = \(1 \over 15\)
    Now both A and B's 1 day work = \({1 \over 10} + {1 \over 15}\) = \(3 + 2 \over 30\) = \(1 \over 6\)
    Hence the work by both A and B will be completed in 6 days.
  16. 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
  17. 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 Γ— width Γ— height = 6 Γ— 5 Γ— 4 = 120
  18. A basket that contains 2 apples, 3 bananas, 6 oranges, and 4 pears is in the workroom. When Ms. Hutchinson went to the workroom, other workers had already taken 1 banana, 2 oranges, and 1 pear. From the remaining fruit, Ms. Hutchinson randomly took 3 pieces of fruit separately from the basket. If each fruit is equally likely to be chosen, what is the probability that the third piece was an orange if the first two she took were also oranges?

    1. 4/165
    2. 9/11
    3. 4/11
    4. 3/11
    5. 2/9
    Ms. Hutchinson randomly takes the 3 pieces of fruit from the basket, there are 2 apples, 3 -1 = 2 bananas, 6 - 2 = 4 oranges, and 4 - 1 = 3 pears. Assuming that the first 2 pieces of fruit Ms. Hutchinson takes are oranges, there will be 2 apples, 2 bananas, 4 - 2 = 2 oranges, and 3 pears left in the basket when she selects the third piece of fruit. The probability that the third piece of fruit she selects will be an orange is \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).
  19. 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 = \(1120 - 850 \over 850\) Γ— 100 = 31.76
  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