Solved Examples Set 1 (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. 1015 / 0.05 / 40 = ?

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

   1015 / 0.05 / 40 = 20300 / 40 = 507.5

3. 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

4. If n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ (n − 3) . . . 2 ⋅ 1, what is the value of \(\frac{(6!)(4!)}{(5!)(3!)}\)

   1. 5/4
   2. 8/5
   3. 10
   4. 24
   5. 1152

   \(\frac{(6!)(4!)}{(5!)(3!)}\) = \(\frac{(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2. 1)}{(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)}\) = \(\frac{6 . 4}{1}\) = 24

5. 60% of 37 = ?

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

   60% of 37 = 0.6 × 37 = 22.2

6. A shopkeeper bought a radio from a wholesaler for $ 250.00. In addition, he paid a sales tax of 15% on the cost price. He then sold the radio for $ 315.00. Calculate the cash profit made by the shopkeeper.

   1. $ 20.00
   2. $ 22.50
   3. $ 25.00
   4. $ 27.50
   5. $ 27.00

   cost price = $ 250
   sales tax = .15 × 250 = $ 37.5
   cash profit = 315 - 250 - 37.5 = $ 27.5

7. 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.

8. ?% of 60 = 24

   1. 40
   2. 48
   3. 45
   4. 42
   5. 38

   ?% × 60 = 24
   \(? = {24 \over 60} × 100 \) = 40

9. 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.

10. 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 \(km \over hr\)

11. A man walked for 3 hours at 4.5 km/h and cycled for some time at 15 km/h. Altogether, he traveled 21 km. Find the time taken for cycling.

    1. 1/2 hour
    2. 1 hour
    3. 1 1⁄2 hours
    4. 2 hours
    5. 2 1⁄2 hours

    The man walked the distance = 3 x 4.5 = 13.5 km. The distance cycled by the man = 21 - 13.5 = 7.5 km
    As he cyled 15 km in 1 h
    he cycled 1 km in 1/15 h
    Finally, he cycled 7.5 km in 7.5/15 = 1/2 h

12. \( {2244 \over 0.88} = ? × 1122 \)

    1. 20.02
    2. 20.2
    3. 19.3
    4. 2.27
    5. 3.27

    \( {2244 \over 0.88} = ? × 1122 \)
    \(? = {2550 \over 1122} = 2.27 \)

13. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

14. 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

15. 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

16. 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%.

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. 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

19. 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

20. A train takes 50 minutes for a journey if it runs at 48 km/hr. The rate at which the train must run to reduce the time to 40 minutes will be

    1. 50 km/hr
    2. 55 km/hr
    3. 60 km/hr
    4. 57 km/hr
    5. 65 km/hr

    \(50 × 48 \over 40\) = 60 \(km \over hr\)

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