Solved Examples Set 1 (Quantitative Ability)

1. 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

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. 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

4. 12% of ________ = 48

   1. 250
   2. 100
   3. 400
   4. 200
   5. 300

   \(12 \text{% of } x = 48\)
   \(0.12x = 48\)
   \(x = \frac{48}{0.12} = 400\)

5. 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.

6. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

7. \( {1250 \over 25} × 0.5 = ? \)

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

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

8. A primary school had an enrollment of 850 pupils in January 1970. In January 1980 the enrollment was 1,120. What was the percentage increase for the enrollment?

   1. 31.76%
   2. 33.50%
   3. 30.65%
   4. 34.76%
   5. 30.55%

   Percentage increase for the enrollment = \(1120 - 850 \over 850\) × 100 = 31.76

9. 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

10. 5789 - 2936 + 1089 = ?

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

    5789 - 2936 + 1089 = 6878 - 2936 = 3942

11. 5.41 - 3.29 × 1.6 = ?

    1. 14.6
    2. 0.3392
    3. 0.146
    4. 3.392
    5. 1.46

    5.41 - 3.29 × 1.6 = 5.41 - 5.264 = 0.146

12. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
    What is the value of 𝑥 in the equation?

    1. 10
    2. 20
    3. 26
    4. 31
    5. 40

    By cross multiplying, 4(𝑥 – 8) =3 × 24. Thus, 4𝑥 – 32 = 72, and so 4𝑥 = 104 and 𝑥 = 26.

13. A shop owner blends three types of coffees, A, B and C, in the ratio 3:5:7. Given that type A coffee costs $ 70 per kg, type B coffee costs $ 100 per kg and type C coffee costs $ 130 per kg, calculate the cost per kg of the blended mixture.

    1. $ 106
    2. $ 108
    3. $ 109
    4. $ 110
    5. $ 105

    Cost per kg = 70 x 1/5 + 100 x 1/3 + 130 x 7/15 = $ 108 per kg

14. \( {63.84 \over ?} \) = 21

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

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

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

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

    

16. 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

17. 42.98 + ? = 107.87

    1. 64.89
    2. 65.89
    3. 64.98
    4. 65.81
    5. 63.89

    ? = 107.87 - 42.98 = 64.89

18. \( {396 \over 11} \) + 19 = ?

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

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

19. 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.

20. A and B enter into a partnership contributing $ 800 and $ 1000 respectively. At the end of 6 months they admit C, who contributes $ 600. After 3 years they get a profit of $ 966. Find the share of each partner in the profit.

    1. $ 336, $ 420, $ 210
    2. $ 360, $ 400, $ 206
    3. $ 380, $ 390, $ 196
    4. $ 345, $ 405, $ 210
    5. $ 325, $ 400, $ 200

    A shares = 800 × 3 = 2400
    B shares = 1000 × 3 = 3000
    C shares = 600 × 2 1⁄2 = 1500
    Total shares = 2400 + 3000 + 1500 = 6900
    A's profit = \(2400 \over 6900 \) × 966 = $ 336
    B's profit = \(3000 \over 6900 \) × 966 = $ 420
    C's profit = \(1500 \over 6900 \) × 966 = $ 210

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