Solved Examples Set 2 (Quantitative Ability)

1. 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

2. 12% of ________ = 48

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

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

3. 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

4. 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

5. 1015 / 0.05 / 40 = ?

   1. 50.75
   2. 507.5
   3. 506
   4. 2056
   5. 5075

   1015 / 0.05 / 40 = 20300 / 40 = 507.5

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

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

   

8. \( {63.84 \over ?} \) = 21

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

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

9. 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

10. \( {396 \over 11} \) + 19 = ?

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

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

11. 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,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

12. ?% of 60 = 24

    1. 40
    2. 48
    3. 45
    4. 42
    5. 38

    ?% × 60 = 24
    \(? = {24 \over 60} × 100 \) = 40

13. 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%

14. 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

15. 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

16. 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

17. 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

18. A group of laborers accepted to do a piece of work in 20 days. 8 of them did not turn up for the work and the remaining did the work in 24 days. The original number of laborers was

    1. 47
    2. 48
    3. 49
    4. 50
    5. 51

    x laborers do work in 20 days and x-8 laborers do same work in 24 days. As the no. of laborers decrease, the no. of days increased then it becomes as
    x : x - 8 :: 24 : 20
    product of interiors = product of exteriors
    24x - 192 = 20x
    4x = 192
    x = 48

19. 8 : ? :: 1 : 4

    1. 24
    2. 16
    3. 0
    4. 32
    5. 20

    ? × 1 = 8 × 4
    ? = 32

20. 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

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