Solved Examples Set 1 (Quantitative Ability)

1. How much would I have to pay for a book which cost $ 72 to product, if the printing company sold it to a bookseller at 20% profit and in return the bookseller sold it to me at a profit of 25%?

   1. $ 104
   2. $ 106
   3. $ 108
   4. $ 110
   5. $ 109

   Cost of the book product = $ 72
   Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4
   Now the cost of the book = 72 +14.4 = $ 86.4
   Profit of the bookseller = 25% of 86.4 = 21.4
   Finally, the cost of the book = 86.4 + 21.4 = $ 108

2. 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

3. 5873 + 12034 + 1106 = ?

   1. 19016
   2. 20001
   3. 19013
   4. 2018
   5. 19010

   5873 + 12034 + 1106 = 17907 + 1106 = 19013

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. 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.

6. 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.

7. ? × 12 = 75% of 336

   1. 48
   2. 252
   3. 28
   4. 21
   5. 23

   ? × 12 = 75% of 336
   ? × 12 = 0.75 × 336
   ? × 12 = 252
   \(? = \frac{252}{12}\)
   ? = 21

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. 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

10. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

11. 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.

12. 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

13. A group of laborers accepted to do a piece of work in 20 days. 8 of them did not turn up for the work and the remaining did the work in 24 days. The original number of laborers was

    1. 47
    2. 48
    3. 49
    4. 50
    5. 51

    x laborers do work in 20 days and x-8 laborers do same work in 24 days. As the no. of laborers decrease, the no. of days increased then it becomes as
    x : x - 8 :: 24 : 20
    product of interiors = product of exteriors
    24x - 192 = 20x
    4x = 192
    x = 48

14. \( {1250 \over 25} × 0.5 = ? \)

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

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

15. ?% of 60 = 24

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

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

16. 5789 - 2936 + 1089 = ?

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

    5789 - 2936 + 1089 = 6878 - 2936 = 3942

17. By selling a fan for $ 475, a person loses 5%. To get a gain of 5%, he should sell the fan for:

    1. $ 500
    2. $ 525
    3. $ 535
    4. $ 575
    5. $ 505

    cost price = 100/(100 - 5) x 475 = $ 500
    sale price = (100 + 5)/100 x 500 = $ 525

18. 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

19. 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,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

20. \( {63.84 \over ?} \) = 21

    1. 3.04
    2. 3.4
    3. 30.4
    4. 300.4
    5. 0.304

    ? = \( 63.84 \over 21 \) = 3.04

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