Solved Examples Set 3 (Quantitative Ability)

1. 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

2. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

3. 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

4. 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

5. 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

6. The amount of hot cocoa powder remaining in a can is 6 1⁄4 tablespoons. A single serving consists of 1 3⁄4 tablespoons of the powder. What is the total number of servings of the powder remaining in the can?

   1. 3 1⁄2
   2. 3 4⁄7
   3. 4 3⁄7
   4. 4 1⁄2
   5. 6

   As \(6\frac{1}{4} = \frac{25}{4}\) and \(1\frac{3}{4} = \frac{7}{4}\). Therefore,
   \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

7. 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%.

8. 5.41 - 3.29 × 1.6 = ?

   1. 14.6
   2. 0.3392
   3. 0.146
   4. 3.392
   5. 1.46

   5.41 - 3.29 × 1.6 = 5.41 - 5.264 = 0.146

9. 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

10. 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

11. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    1. 1
    2. 3
    3. 0
    4. 67
    5. 25

    \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25}\)
    \(= 25 \text{% } × {4 \over 4\text{%}} × {1 \over 25} \)
    \(= 0.25 × {4 \over 0.04} × {1 \over 25}\)
    \(= {25 \over 25}\)
    = 1

12. 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
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

13. 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

14. 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

15. 8 : ? :: 1 : 4

    1. 24
    2. 16
    3. 0
    4. 32
    5. 20

    ? × 1 = 8 × 4
    ? = 32

16. 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

17. 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.

18. 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

19. 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

20. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

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