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

Solved Examples Set 3 (Quantitative Ability)

  1. 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.
  2. If n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ (n − 3) . . . 2 ⋅ 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
  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. 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
    \( 40 × 15 × 5 \over 6 × 4 \) = 125 men
  5. 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 × 6400 = $ 2240
    If the interest rate is 4% then interest amount is
    4% of 6400 = 0.04 × 6400 = $ 2560
    Now the difference of both interests = 2560 - 2240 = $ 320 per annum
    Interest for half year (6 months) = \(320 \over 2\) = $ 160
  6. A single discount equivalent to a discount series of 20%, 10% and 25% is

    1. 55%
    2. 54%
    3. 46%
    4. 42%
    5. 50%
    If 3 succesive discounts are a%, b% and c%
    then single discount = a + b + c – (\(ab \over 100 \) + \(bc \over 100\) + \(ca \over 100 \) – \(abc \over 10000 \))
    a = 20, b = 10, c = 25, solving we get, 46%.
  7. 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
  8. A certain solution is to be prepared by combining chemicals X, Y and Z in the ratio 18:3:2. How many liters of the solution can be prepared by using 36 liters of X?

    1. 46 liters
    2. 47 liters
    3. 45 liters
    4. 49 liters
    5. 44 liters
    As total ratio is 18 +3 + 2 = 23
    Let total solution is x liters
    Then \(18 \over 23\) x = 36
    x = \(36 × 23 \over 18\) = 46 liters
  9. \( {63.84 \over ?} \) = 21

    1. 3.04
    2. 3.4
    3. 30.4
    4. 300.4
    5. 0.304
    ? = \( 63.84 \over 21 \) = 3.04
  10. 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.
  11. \( {5.76 \over 1.6} - 2.4 = ? \)

    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
  12. Which of the following expressions is equivalent to \(\frac{𝑥^2 + 3x + 1}{𝑥 + 1}\)?

    1. x + 2
    2. 𝑥 + 3
    3. 𝑥 + 2 - 1/(𝑥 + 1)
    4. 𝑥 + 3 + 1/(𝑥 + 1)
    5. 𝑥 + 4 + 5/(𝑥 + 1)
    As \(𝑥^2 + 3x + 1 = (𝑥^2 + 3x + 2) -1\)
    and
    \(\frac{𝑥^2 + 3x + 2}{x + 1} = \frac{(𝑥 + 2)(x + 1)}{x + 1} = 𝑥 + 2\)
    Therefore,
    \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)
  13. 5789 - 2936 + 1089 = ?

    1. 3942
    2. 4041
    3. 2626
    4. 3932
    5. 3940
    5789 - 2936 + 1089 = 6878 - 2936 = 3942
  14. 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
  15. 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
  16. 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 \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).
  17. 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
  18. 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
  19. 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
  20. 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

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