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

Solved Examples Set 3 (Quantitative Ability)

  1. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)
  2. Which of the following is the largest?

    1. half of 30% of 280
    2. one-third of 70% of 160
    3. twice 50% of 30
    4. three times 40% of 40
    5. 60% of 60
    Let us calculate the value of each:
    A. 0.5 × 0.3 × 280 = 42
    B. 0.33 × 0.7 × 160 = 36.96
    C. 2 × 0.5 × 30 = 30
    D. 3 × 0.4 × 40 = 48
    E. 0.6 × 60 = 36
  3. By selling 60 chairs, a man gains an amount equal to selling price of 10 chairs. The profit percentage in the transaction is

    1. 10%
    2. 15%
    3. 16.67%
    4. 20%
    5. 22%
    selling price of 60 chairs = selling price of 10 chairs
    profit of 60 chairs = profit of 10 chairs
    profit of 6 chairs = profit of 1 chair
    profit of 1 chair = profit of 1/6 chair
    profit %age = 1/6 x 100 = 16.67%
  4. 5873 + 12034 + 1106 = ?

    1. 19016
    2. 20001
    3. 19013
    4. 2018
    5. 19010
    5873 + 12034 + 1106 = 17907 + 1106 = 19013
  5. ? × 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
  6. A man earned an annual income of $ 245000 in 1990. He was allowed a deduction of $ 15000 relief for each of his three children and a personal relief of $ 30000. If he was charged a tax rate of 4% on first $ 50000 and 6% on his remaining income, calculate the total tax charged.

    1. $ 9200
    2. $ 8700
    3. $ 9500
    4. $ 9400
    5. $ 9000
    Total Income = $ 245000
    Total relief = 3 × 15000 + 30000 = $ 75000
    Rest income = 245000 - 75000 = 170000
    Tax on 1st 50000 = 0.04 × 50000 = $ 2000
    Tax on rest amount 120000 = 0.06 × 120000 = $ 7200
    Total tax = 200 + 7200 = $ 9200
  7. 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,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)
  8. 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
  9. 350 × ? = 4200

    1. 12
    2. 24
    3. 15
    4. 30
    5. 16
    \( ? = {4200 \over 350} =12 \)
  10. 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
  11. \( {5.76 \over 1.6} - 2.4 = ? \)

    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
  12. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    1. 1
    2. 3
    3. 0
    4. 67
    5. 25
    \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25}\)
    \(= 25 \text{% } × {4 \over 4\text{%}} × {1 \over 25} \)
    \(= 0.25 × {4 \over 0.04} × {1 \over 25}\)
    \(= {25 \over 25}\)
    = 1
  13. 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
  14. In the series 8, 9, 12, 17, 24 . . . the next number would be

    1. 29
    2. 30
    3. 33
    4. 35
    5. 41
    In the series, 8, 9, 12, 17, 24 . . .
    9 − 8 = 1
    12 − 9 = 3
    17 − 12 = 5
    24 − 17 = 7
    Hence, the difference between the next term and 24 must be 9 or
    x − 24 = 9, and
    x = 33
    Hence, the next term in the series must be 33
  15. 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
  16. A man travelled 120 km to a town. He could have reached the town 4 1⁄2 hours earlier had he increased his speed by 3 km/h. Find the speed at which he travelled.

    1. 6.56 km
    2. 7.57 km
    3. 8.58 km
    4. 9.59 km
    5. 5.55 km
    Let the normal speed \(= x \text{ } \frac{km}{hr}\)
    Time taken when travelled at the normal speed \(= \frac{120}{x}\) hr
    Time taken when travelled at the increased speed \(= \frac{120}{x + 3}\) hr
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$
  17. 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 \(km \over hr\)
  18. \( {1250 \over 25} × 0.5 = ? \)

    1. 250
    2. 50
    3. 2.5
    4. 25
    5. 125
    \( {1250 \over 25} × 0.5 = 50 × 0.5 = 25 \)
  19. 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
  20. A boy of height 165 cm is replaced by another, which decreases the average height of the group of 34 students, by 1 cm. The height of the new student is

    1. 129 cm
    2. 130 cm
    3. 131 cm
    4. 132 cm
    5. 133 cm
    Total decreased height of 34 students = 1 × 34 = 34 cm
    Height of the replaced student = 165 - 34 = 131 cm

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