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

Solved Examples Set 1 (Quantitative Ability)

  1. 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
  2. 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
  3. Which of the following is the largest?

    1. half of 30% of 280
    2. one-third of 70% of 160
    3. twice 50% of 30
    4. three times 40% of 40
    5. 60% of 60
    Let us calculate the value of each:
    A. 0.5 × 0.3 × 280 = 42
    B. 0.33 × 0.7 × 160 = 36.96
    C. 2 × 0.5 × 30 = 30
    D. 3 × 0.4 × 40 = 48
    E. 0.6 × 60 = 36
  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
    \(50 × 48 \over 40\) = 60 \(km \over hr\)
  5. 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
  6. if x% of 60 = 48 then x = ?

    1. 80
    2. 60
    3. 90
    4. 40
    5. 70
    x = \( {48 × 100 \over 60} \) = 80
  7. In the series 8, 9, 12, 17, 24 . . . the next number would be

    1. 29
    2. 30
    3. 33
    4. 35
    5. 41
    In the series, 8, 9, 12, 17, 24 . . .
    9 − 8 = 1
    12 − 9 = 3
    17 − 12 = 5
    24 − 17 = 7
    Hence, the difference between the next term and 24 must be 9 or
    x − 24 = 9, and
    x = 33
    Hence, the next term in the series must be 33
  8. 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.
  9. 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
  10. If 4a + 2 = 10, then 8a + 4 =

    1. 5
    2. 16
    3. 20
    4. 24
    5. 28
    One may answer this question by solving
    4a + 2 = 10
    4a = 8
    a= 2
    Now, plugging in 2 for a:
    8a + 4 = 8(2) + 4 = 20
    A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.
  11. 12% of ________ = 48

    1. 250
    2. 100
    3. 400
    4. 200
    5. 300
    \(12 \text{% of } x = 48\)
    \(0.12x = 48\)
    \(x = \frac{48}{0.12} = 400\)
  12. 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
  13. 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.
  14. 5873 + 12034 + 1106 = ?

    1. 19016
    2. 20001
    3. 19013
    4. 2018
    5. 19010
    5873 + 12034 + 1106 = 17907 + 1106 = 19013
  15. 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 \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).
  16. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075
    1015 / 0.05 / 40 = 20300 / 40 = 507.5
  17. 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
  18. Which set of ordered pairs represents a function?

    1. {(−5,5),(4,8),(−5,−6)}
    2. {(−1,−1),(−1,6),(−1,−10)}
    3. {(−3,7),(2,5),(−7,7)}
    4. {(2,3),(−2,4),(−2,−5)}
    5. {(2,3),(3,2),(2,5)}
    For a set of ordered pairs to be a function, no single 𝑥-coordinate can be mapped to two distinct 𝑦-coordinates. This is not the case for option A, where 𝑥=−5 is mapped to both 𝑦=5 and 𝑦=−6. Similarly, in options B (𝑥=−1), D (𝑥=−2), and E (𝑥=2), an 𝑥 value is mapped to two different 𝑦 values.
  19. 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
  20. 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

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