Solved Examples Set 3 (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. If 4a + 2 = 10, then 8a + 4 =

   1. 5
   2. 16
   3. 20
   4. 24
   5. 28

   One may answer this question by solving
   4a + 2 = 10
   4a = 8
   a= 2
   Now, plugging in 2 for a:
   8a + 4 = 8(2) + 4 = 20
   A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.

3. 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.

4. 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.

5. 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

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

7. A single discount equivalent to a discount series of 20%, 10% and 25% is

   1. 55%
   2. 54%
   3. 46%
   4. 42%
   5. 50%

   If 3 succesive discounts are a%, b% and c%
   then single discount = a + b + c – (\(ab \over 100 \) + \(bc \over 100\) + \(ca \over 100 \) – \(abc \over 10000 \))
   a = 20, b = 10, c = 25, solving we get, 46%.

8. 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

9. 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%

10. 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

11. 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

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

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

    

13. ?% of 60 = 24

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

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

14. 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
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

15. 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

16. 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

17. How much would I have to pay for a book which cost $ 72 to product, if the printing company sold it to a bookseller at 20% profit and in return the bookseller sold it to me at a profit of 25%?

    1. $ 104
    2. $ 106
    3. $ 108
    4. $ 110
    5. $ 109

    Cost of the book product = $ 72
    Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4
    Now the cost of the book = 72 +14.4 = $ 86.4
    Profit of the bookseller = 25% of 86.4 = 21.4
    Finally, the cost of the book = 86.4 + 21.4 = $ 108

18. ? × 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

19. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003

    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)

20. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

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