Solved Examples Set 1 (Quantitative Ability)

1. A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 1114. Find the present age of son.

   1. 7 years
   2. 9 years
   3. 8 years
   4. 8 1/2 years
   5. 6 years

   Let son's age = x, then
   father's age = 5x
   As before 2 years ago the sum of the squares of their ages was 1114, the equation becomes as
   \((x - 2)^2 + (5x - 2)^2 = 1114 \)
   By simplifying the equation, we have
   \(13x^2 -12x -553 = 0\)
   Now solving the equation, we have
   \(13x^2 - 12x - 553 = 0\)
   \(13x^2 - 91x + 79x -553 = 0\)
   13x(x - 7) + 79(x - 7) = 0
   (x - 7)(13x + 79) = 0
   x = 7 and x = -6.077
   As age could not be negative, hence the present age of the son is 7 years.

2. 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

3. 5789 - 2936 + 1089 = ?

   1. 3942
   2. 4041
   3. 2626
   4. 3932
   5. 3940

   5789 - 2936 + 1089 = 6878 - 2936 = 3942

4. 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

5. A man sells two houses for $ 2 lac each. On one he gained 20% and on the other he lost 20%. His total profit or loss % in the transaction will be

   1. 4% profit
   2. 5% loss
   3. no profit, no loss
   4. 4% loss
   5. 3% loss

   % loss = (% loss X % profit)/100 = (20 X 20)/100 = 4%

6. 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.

7. A man was 32 years old when his daughter was born. He is now five times as old as his daughter. How old is his daughter now?

   1. 7 years
   2. 8 years
   3. 9 years
   4. 10 years
   5. 6 years

   Let's assume the daughter is d years old now. That means that the man is now (32 + d) years old, so that
   (32 + d) = 5d
   32 = 4d
   d = 8

8. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

9. 8 : ? :: 1 : 4

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

   ? × 1 = 8 × 4
   ? = 32

10. 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

11. 350 × ? = 4200

    1. 12
    2. 24
    3. 15
    4. 30
    5. 16

    \( ? = {4200 \over 350} =12 \)

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. 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.

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

    \( 40 × 15 × 5 \over 6 × 4 \) = 125 men

15. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    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

16. 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

17. 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

18. \( {1250 \over 25} × 0.5 = ? \)

    1. 250
    2. 50
    3. 2.5
    4. 25
    5. 125

    \( {1250 \over 25} × 0.5 = 50 × 0.5 = 25 \)

19. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

20. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003

    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)

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