Solved Examples Set 1 (Quantitative Ability)

1. 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.

2. 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

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

4. 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

5. 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 \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).

6. 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%

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

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

   

8. 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

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. 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%

11. 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

12. If 3x = −9, then 3x³ − 2x + 4 =

    1. -83
    2. -71
    3. -47
    4. -17
    5. 61

    First solving 3x = −9, x = −3. Now plug into 3x³ − 2x + 4:
    3x³ − 2x + 4
    = 3(-3)³ − 2(-2) + 4
    = 3(−27) + 6 + 4
    = −81 + 6 + 4
    = −71

13. 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

14. 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

15. A man saves $ 500, which is 15% of his annual income. How much does he earn in one year?

    1. $ 3542.5
    2. $ 3333.33
    3. $ 3132.3
    4. $ 3075.75
    5. $ 4444.4

    Let annual income = x
    15% of x = 500
    x = \(500 \over 15\) × 100 = \(10000 \over 3\) = 3333.33

16. 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

17. 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

18. 60% of 37 = ?

    1. 20
    2. 21
    3. 22.2
    4. 22
    5. none

    60% of 37 = 0.6 × 37 = 22.2

19. A third-grade class is composed of 16 girls and 12 boys. There are 2 teacher-aides in the class. The ratio of girls to boys to teacher-aides is

    1. 16:12:1
    2. 8:6:2
    3. 8:6:1
    4. 8:3:1
    5. 4:3:1

    Girls to boys to teacher-aides are in proportion 16 to 12 to 2. Reduced to lowest terms, 16:12:2 equals 8:6:1.

20. A bank increased the rate of interest which it paid to depositors from 3.5% to 4% per annum. Find how much more interest a man would receive if he deposited $ 64000 in the bank for 6 months at the new interest rate

    1. $ 160
    2. $ 180
    3. $ 200
    4. $ 220
    5. $ 150

    If the interest rate is 3.5% then interest amount is
    3.5% of 6400 = 0.035 × 6400 = $ 2240
    If the interest rate is 4% then interest amount is
    4% of 6400 = 0.04 × 6400 = $ 2560
    Now the difference of both interests = 2560 - 2240 = $ 320 per annum
    Interest for half year (6 months) = \(320 \over 2\) = $ 160

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