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. 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

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 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. \( {1250 \over 25} × 0.5 = ? \)

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

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

6. 15 men can complete a job in 10 days. How long will it take 8 men to finish the same job if they work at the same rate?

   1. 14 3⁄4 days
   2. 16 3⁄4 days
   3. 18 3⁄4 days
   4. 20 3⁄4 days
   5. 22 3⁄4 days

   \( 15 × 10 \over 8 \) = 18 3⁄4 days

7. By selling 60 chairs, a man gains an amount equal to selling price of 10 chairs. The profit percentage in the transaction is

   1. 10%
   2. 15%
   3. 16.67%
   4. 20%
   5. 22%

   selling price of 60 chairs = selling price of 10 chairs
   profit of 60 chairs = profit of 10 chairs
   profit of 6 chairs = profit of 1 chair
   profit of 1 chair = profit of 1/6 chair
   profit %age = 1/6 x 100 = 16.67%

8. 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

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

10. A rectangular room is 6 m long, 5 m wide and 4 m high. The total volume of the room in cubic meters is

    1. 24
    2. 30
    3. 120
    4. 240
    5. 140

    Total volume = length × width × height = 6 × 5 × 4 = 120

11. \(\frac{\frac{7}{10} × 14 × 5 × \frac{1}{28}}{\frac{10}{17} × \frac{3}{5} × \frac{1}{6} × 17} = \)

    1. 4/7
    2. 1
    3. 7/4
    4. 2
    5. 17/4

    

12. 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.

13. 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

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. 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%.

16. 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.

17. 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.

18. Rashid buys three books for $ 16 each and four books for $ 23 each, what will be the average price of books

    1. $ 18
    2. $ 20
    3. $ 22
    4. $ 24
    5. $ 16

    Price of 3 books = 3 × 16 = $ 48
    Price of 4 books = 4 × 23 = $ 92
    Total price = $ 140
    Total books = 3 + 4 = 7
    Average price of books = \(140 \over 7 \) = $ 20

19. A group of boys were to choose between playing hockey and badminton. The number of boys choosing hockey was three times that of those choosing badminton. Asking 12 boys who chose hockey to play badminton would make the number of players for each game equal. Find the number who chose badminton originally.

    1. 12
    2. 14
    3. 11
    4. 13
    5. 10

    Let no. of boys for badminton = x
    then no. of boys for hockey = 3x
    According to the statement,
    3x - 12 = x + 12 (12 leave hockey, 12 join badminton)
    2x = 24
    x = 12
    Hence, there were 12 boys originally choosing badminton.

20. 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

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