In the name of ALLAH, the most beneficient, the most merciful

Solved Examples Set 3 (Quantitative Ability)

  1. 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
  2. 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
  3. 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
  4. 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%.
  5. 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%
  6. if x% of 60 = 48 then x = ?

    1. 80
    2. 60
    3. 90
    4. 40
    5. 70
    x = \( {48 × 100 \over 60} \) = 80
  7. ?% of 60 = 24

    1. 40
    2. 48
    3. 45
    4. 42
    5. 38
    ?% × 60 = 24
    \(? = {24 \over 60} × 100 \) = 40
  8. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003
    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)
  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. \( {2244 \over 0.88} = ? × 1122 \)

    1. 20.02
    2. 20.2
    3. 19.3
    4. 2.27
    5. 3.27
    \( {2244 \over 0.88} = ? × 1122 \)
    \(? = {2550 \over 1122} = 2.27 \)
  11. 60% of 37 = ?

    1. 20
    2. 21
    3. 22.2
    4. 22
    5. none
    60% of 37 = 0.6 × 37 = 22.2
  12. 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
  13. A man pays 10% of his income for his income tax. If his income tax amounts to $ 1500, what is his income?

    1. $ 13000
    2. $ 15000
    3. $ 17000
    4. $ 19000
    5. $ 11000
    Let x = income
    10% of x = $ 1500
    0.1x = $ 1500
    x = \(1500 \over 0.1\) = $ 15000
  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. 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
  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 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
  18. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075
    1015 / 0.05 / 40 = 20300 / 40 = 507.5
  19. 8 : ? :: 1 : 4

    1. 24
    2. 16
    3. 0
    4. 32
    5. 20
    ? × 1 = 8 × 4
    ? = 32
  20. At a book fair, a book was reduced in price from $ 75 to $ 60. If the first price gives a 50% profit, find the percentage profit of the book sold at the reduced price.

    1. 20%
    2. 30%
    3. 40%
    4. 50%
    5. 10%
    As $ 75 (first price) gives a profit = 50%
    $ 1 gives a profit = (50/75)%
    $ 60 (reduced price) gives profit = (50/75 x 60)% = 40%

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