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

Solved Examples Set 3 (Quantitative Ability)

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

    1. x + 2
    2. π‘₯ + 3
    3. π‘₯ + 2 - 1/(π‘₯ + 1)
    4. π‘₯ + 3 + 1/(π‘₯ + 1)
    5. π‘₯ + 4 + 5/(π‘₯ + 1)
    As \(π‘₯^2 + 3x + 1 = (π‘₯^2 + 3x + 2) -1\)
    and
    \(\frac{π‘₯^2 + 3x + 2}{x + 1} = \frac{(π‘₯ + 2)(x + 1)}{x + 1} = π‘₯ + 2\)
    Therefore,
    \(\frac{π‘₯^2 + 3x + 1}{x + 1} = \frac{π‘₯^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (π‘₯ + 2) - \frac{1}{x + 1}\)
  2. 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
  3. 8 : ? :: 1 : 4

    1. 24
    2. 16
    3. 0
    4. 32
    5. 20
    ? Γ— 1 = 8 Γ— 4
    ? = 32
  4. A boy scored 90 marks for his mathematics test. This was 20% more than what he had scored for the geography test. How much did he score in geography?

    1. 71 marks
    2. 73 marks
    3. 75 marks
    4. 77 marks
    5. 78 marks
    20% of x + x = 90
    0.2x + x = 90
    1.2x = 90
    x = \(90 \over 1.2\)
    x = 75
  5. Which expression is equivalent to \(\frac{6π‘₯^2 + 4π‘₯}{2π‘₯}\)?

    1. 7x
    2. 5x2
    3. 3x + 2
    4. 6x2 + 2
    5. 3x2 + 2x
    As \(\frac{6π‘₯^2}{2π‘₯} = 3π‘₯,\) and \(\frac{4π‘₯}{2π‘₯} = 2,\) so then \(\frac{6π‘₯^2 + 4π‘₯}{2π‘₯} = 3π‘₯ + 2\)
  6. 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%

  7. In the figure above, AB is one edge of a cube. If AB equals 5, what is the surface area of the cube?

    1. 25
    2. 100
    3. 125
    4. 150
    5. 300
    Since one edge of the cube is 5, all edges equal 5. Therefore, the area of one face of the cube is:
    5 Γ— 5 = 25
    Since a cube has 6 equal faces, its surface area will be:
    6 Γ— 25 = 150
  8. Matthew’s age (π‘š) is three years more than twice Rita’s age (π‘Ÿ). Which equation shows the relationship between their ages?

    1. π‘š = π‘Ÿ βˆ’ 32
    2. π‘š = π‘Ÿ + 32
    3. π‘š = 2(π‘Ÿ + 3)
    4. π‘š = 2π‘Ÿ βˆ’ 3
    5. π‘š = 2π‘Ÿ + 3
    As Matthew's age (π‘š) is three more years (+3) than twice Rita's age (2π‘Ÿ). Therefore, π‘š = 2π‘Ÿ + 3.
  9. If n! = n β‹… (n βˆ’ 1) β‹… (n βˆ’ 2) β‹… (n βˆ’ 3) . . . 2 β‹… 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
  10. 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 Γ— 250 = $ 37.5
    cash profit = 315 - 250 - 37.5 = $ 27.5
  11. After spending 88% of his income, a man had $ 2160 left. Find his income.

    1. $ 18000
    2. $ 19000
    3. $ 20000
    4. $ 22000
    5. $ 17000
    Let income = x
    x = 88% of x + 2160
    x - 0.88x = 2160
    0.12x = 2160
    x = \(216000 \over 12\) = 18000
  12. A shopkeeper sold two articles for $ 48 each. He made a 25% profit on one article and a loss of 20% on the other. What was his net gain or loss on the sale of the two articles?

    1. loss of $ 1.40
    2. gain of $ 2.40
    3. loss of $ 2.40
    4. gain of $ 1.40
    5. gain of $ 2.60
    25% profit at selling price $ 48 = 48 x .25 = $ 12
    20% loss at selling price $ 48 = 48 x 0.2 = $ 9.6
    gain = profit - loss = 12 - 9.6 = $ 2.4
  13. 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
  14. 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%.
  15. 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
    \( 15 Γ— 10 \over 8 \) = 18 3⁄4 days
  16. 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
  17. \(\frac{\frac{7}{10} Γ— 14 Γ— 5 Γ— \frac{1}{28}}{\frac{10}{17} Γ— \frac{3}{5} Γ— \frac{1}{6} Γ— 17} = \)

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

  18. 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
    \( 40 Γ— 15 Γ— 5 \over 6 Γ— 4 \) = 125 men
  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 Γ— 1600 = S$ 48
    the rest amount = 1600 - 48 = S$ 1552
    S$ 1 = \(1 \over 3.20\) = Pond 0.3125
    Now S$ 1552 = 1552 Γ— 0.3125 = Pond 485
  20. 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 Γ— width Γ— height = 6 Γ— 5 Γ— 4 = 120

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