Solved Examples Set 3 (Quantitative Ability)

1. \( {1250 \over 25} × 0.5 = ? \)

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

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

2. 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

3. 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

4. 12% of ________ = 48

   1. 250
   2. 100
   3. 400
   4. 200
   5. 300

   \(12 \text{% of } x = 48\)
   \(0.12x = 48\)
   \(x = \frac{48}{0.12} = 400\)

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. A girl is 18 years younger than her mother. In 6 years time, the sum of their ages will be 54.How old is the girl now?

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

   Let girl's age = x
   then mother's age = x + 18
   After 6 years,
   x + 6 + x + 18 + 6 = 54
   2x + 30 = 54
   2x = 24
   x = 12

7. 40 arithmetic questions, each carrying equal marks, were given in a class test. A boy answered 25 questions correctly. What percentage was this? To pass a test a student must answer at least 45% of the questions correctly. Find the least number of correct answers needed to pass.

   1. 62.5%, 18
   2. 63.5%, 16
   3. 64.5%, 20
   4. 61.0%, 21
   5. 60.0%, 22

   \(x \text{% of } 40 = 25\)
   \(x \text{% } × 40 = 25\)
   \(x = {25 \over 40} × 100 \)
   x = 62.5
   
   \(x = 45 \text{% of } 40 \)
   \(x = 0.45 × 40 \)
   x = 18

8. 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.

9. \( {5.76 \over 1.6} - 2.4 = ? \)

   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

10. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

    1. 33 days
    2. 35 days
    3. 37 days
    4. 39 days
    5. 31 days

    \(14 × 10 \over 4 \) = 35 days

11. 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.

12. 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

13. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey.

    1. 53 km/hr
    2. 53.33 km/hr
    3. 54 1⁄4 km/hr
    4. 55 km/hr
    5. 56 km/hr

    The time, car took for the first half, \(50 \over 40 \) = 1.25 hrs
    and for the second half \(50 \over 80 \) = 0.625 hrs
    Total time = 1.25 + 0.625 = 1.875 hrs
    Average speed = \(100 \over 1.875 \) = 53.3 \(km \over hr\)

14. 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

15. 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

16. 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

17. 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.

18. if x% of 60 = 48 then x = ?

    1. 80
    2. 60
    3. 90
    4. 40
    5. 70

    x = \( {48 × 100 \over 60} \) = 80

19. 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

20. 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%.

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