Solved Examples Set 1 (Quantitative Ability)

1. A man takes 50 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in

   1. 1 hour
   2. 30 minutes
   3. 20 minutes
   4. 10 minutes
   5. 40 minutes

   \( 50 × 6 \over 10 \) = 30 minutes

2. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

   1. 33 days
   2. 35 days
   3. 37 days
   4. 39 days
   5. 31 days

   \(14 × 10 \over 4 \) = 35 days

3. \( {5.76 \over 1.6} - 2.4 = ? \)

   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

4. \( {63.84 \over ?} \) = 21

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

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

5. 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

6. A basket that contains 2 apples, 3 bananas, 6 oranges, and 4 pears is in the workroom. When Ms. Hutchinson went to the workroom, other workers had already taken 1 banana, 2 oranges, and 1 pear. From the remaining fruit, Ms. Hutchinson randomly took 3 pieces of fruit separately from the basket. If each fruit is equally likely to be chosen, what is the probability that the third piece was an orange if the first two she took were also oranges?

   1. 4/165
   2. 9/11
   3. 4/11
   4. 3/11
   5. 2/9

   Ms. Hutchinson randomly takes the 3 pieces of fruit from the basket, there are 2 apples, 3 -1 = 2 bananas, 6 - 2 = 4 oranges, and 4 - 1 = 3 pears. Assuming that the first 2 pieces of fruit Ms. Hutchinson takes are oranges, there will be 2 apples, 2 bananas, 4 - 2 = 2 oranges, and 3 pears left in the basket when she selects the third piece of fruit. The probability that the third piece of fruit she selects will be an orange is \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).

7. 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

8. 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

9. 
   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 × 5 = 25
   Since a cube has 6 equal faces, its surface area will be:
   6 × 25 = 150

10. 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

11. 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

12. A fruit-seller has 120 oranges. Given that he has 20% more apples than oranges and 40% less oranges than pears, find the number of apples and the number of pears the fruit seller has.

    1. 144, 200
    2. 148, 380
    3. 149, 220
    4. 140, 190
    5. 142, 190

    No. of apples = 120 + 20% of 120 = 120 + 0.2 × 120 = 144
    
    Let x = No. of pears
    x - 40% of x = 120
    x - 0.4x = 120
    0.6x = 120
    x = \(120 \over 0.6\) = 200
    Hence, no. of pears = 200

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 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

15. 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

16. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    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

17. A certain number was doubled and the result then multiplied by 3. If the product was 138, find the number.

    1. 21
    2. 23
    3. 25
    4. 27
    5. 19

    Let x be the number
    the number is doubled, 2x
    the result is multiplied by 3, 3 × 2x = 6x
    6x = 138
    x = \(138 \over 6\) = 23

18. The closest approximation of \(\frac{69.28 × .004}{.03}\) is

    1. 0.092
    2. 0.92
    3. 9.2
    4. 92
    5. 920

    This problem is most easily completed by rearranging and approximating as follows:
    (69.28 x .004)/.03 ≅ 69 x .1 = 6.9
    which is the only reasonably close answer to 9.2

19. 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,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

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