Solved Examples Set 2 (Quantitative Ability)

1. A man travelled 120 km to a town. He could have reached the town 4 1⁄2 hours earlier had he increased his speed by 3 km/h. Find the speed at which he travelled.

   1. 6.56 km
   2. 7.57 km
   3. 8.58 km
   4. 9.59 km
   5. 5.55 km

   Let the normal speed \(= x \text{ } \frac{km}{hr}\)
   Time taken when travelled at the normal speed \(= \frac{120}{x}\) hr
   Time taken when travelled at the increased speed \(= \frac{120}{x + 3}\) hr
   $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

2. 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 \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).

3. 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.

4. 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

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. 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%

7. 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

8. 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

9. 60% of 37 = ?

   1. 20
   2. 21
   3. 22.2
   4. 22
   5. none

   60% of 37 = 0.6 × 37 = 22.2

10. The closest approximation of \(\frac{69.28 × .004}{.03}\) 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

11. Which expression is equivalent to \(\frac{6𝑥^2 + 4𝑥}{2𝑥}\)?

    1. 7x
    2. 5x²
    3. 3x + 2
    4. 6x² + 2
    5. 3x² + 2x

    As \(\frac{6𝑥^2}{2𝑥} = 3𝑥,\) and \(\frac{4𝑥}{2𝑥} = 2,\) so then \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

12. 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%.

13. 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

14. 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

15. \( {396 \over 11} \) + 19 = ?

    1. 19.8
    2. 36
    3. 55
    4. 33
    5. 50

    \( {396 \over 11} \) + 19 = 36 + 19 = 55

16. A and B enter into a partnership contributing $ 800 and $ 1000 respectively. At the end of 6 months they admit C, who contributes $ 600. After 3 years they get a profit of $ 966. Find the share of each partner in the profit.

    1. $ 336, $ 420, $ 210
    2. $ 360, $ 400, $ 206
    3. $ 380, $ 390, $ 196
    4. $ 345, $ 405, $ 210
    5. $ 325, $ 400, $ 200

    A shares = 800 × 3 = 2400
    B shares = 1000 × 3 = 3000
    C shares = 600 × 2 1⁄2 = 1500
    Total shares = 2400 + 3000 + 1500 = 6900
    A's profit = \(2400 \over 6900 \) × 966 = $ 336
    B's profit = \(3000 \over 6900 \) × 966 = $ 420
    C's profit = \(1500 \over 6900 \) × 966 = $ 210

17. \( {63.84 \over ?} \) = 21

    1. 3.04
    2. 3.4
    3. 30.4
    4. 300.4
    5. 0.304

    ? = \( 63.84 \over 21 \) = 3.04

18. 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.

19. 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

20. 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.

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