Solved Examples Set 1 (Quantitative Ability)

1. A person's net income is $ 1373.70 and he pays an income tax of 5%. His gross income in dollars must be

   1. 1446
   2. 1118.96
   3. 1308.29
   4. 1438.25
   5. 1211.21

   Let gross income in dollars = x
   then according to the statement,
   x = 5% of x + 1373.70
   x - 0.05x = 1373.70
   0.95x = 1373.70
   x = \(137370 \over 95\) = 1446

2. ?% of 60 = 24

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

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

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. 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

5. 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

6. 60% of 37 = ?

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

   60% of 37 = 0.6 × 37 = 22.2

7. \( {396 \over 11} \) + 19 = ?

   1. 19.8
   2. 36
   3. 55
   4. 33
   5. 50

   \( {396 \over 11} \) + 19 = 36 + 19 = 55

8. 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%.

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

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

   

10. 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%

11. 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

12. Which of the following expressions is equivalent to \(\frac{𝑥^2 + 3x + 1}{𝑥 + 1}\)?

    1. x + 2
    2. 𝑥 + 3
    3. 𝑥 + 2 - 1/(𝑥 + 1)
    4. 𝑥 + 3 + 1/(𝑥 + 1)
    5. 𝑥 + 4 + 5/(𝑥 + 1)

    As \(𝑥^2 + 3x + 1 = (𝑥^2 + 3x + 2) -1\)
    and
    \(\frac{𝑥^2 + 3x + 2}{x + 1} = \frac{(𝑥 + 2)(x + 1)}{x + 1} = 𝑥 + 2\)
    Therefore,
    \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)

13. 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.

14. 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.

15. ? × 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

16. 5873 + 12034 + 1106 = ?

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

    5873 + 12034 + 1106 = 17907 + 1106 = 19013

17. Matthew’s age (𝑚) is three years more than twice Rita’s age (𝑟). Which equation shows the relationship between their ages?

    1. 𝑚 = 𝑟 − 32
    2. 𝑚 = 𝑟 + 32
    3. 𝑚 = 2(𝑟 + 3)
    4. 𝑚 = 2𝑟 − 3
    5. 𝑚 = 2𝑟 + 3

    As Matthew's age (𝑚) is three more years (+3) than twice Rita's age (2𝑟). Therefore, 𝑚 = 2𝑟 + 3.

18. A shopkeeper buys 300 identical articles at a total cost of $ 1500. He fixes the selling price of each article at 20% above the cost price and sells 260 articles at the price. As for the remaining articles, he sells them at 50% of the selling price. Calculate the shopkeeper's total profit.

    1. $ 180
    2. $ 185
    3. $ 200
    4. $ 190
    5. $ 170

    cost price of each item = \( 1500 \over 300 \) = $ 5
    selling price at 20% above the cost price = 5 + 5 × .2 = $ 6
    selling price of 260 items = 260 × 6 = $ 1560
    selling price of remaining 40 items = 40 × 6 × .5 = $ 120
    Total profit = 1560 + 120 - 1500 = $ 180

19. 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

20. 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

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