Solved Examples Set 2 (Quantitative Ability)

1. 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
   $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

2. 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

3. 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

4. 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

5. 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

6. \( {2244 \over 0.88} = ? × 1122 \)

   1. 20.02
   2. 20.2
   3. 19.3
   4. 2.27
   5. 3.27

   \( {2244 \over 0.88} = ? × 1122 \)
   \(? = {2550 \over 1122} = 2.27 \)

7. 60% of 37 = ?

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

   60% of 37 = 0.6 × 37 = 22.2

8. On a trip to visit friends, a family drives 65 miles per hour for 208 miles of the trip. If the entire trip was 348 miles and took 6 hours, what was the average speed, in miles per hour, for the rest of the trip?

   1. 44
   2. 50
   3. 51
   4. 58
   5. 60

   As the first part of the trip took \(\frac{208 \text{ miles}}{65 \text{ } \frac{miles}{hour}} = 3.2 \text{ hours},\) so the remaining 140 miles (348 - 208) took 2.8 hours (6 - 3.2). The average speed for the rest of the trip was \(\frac {140 \text{ miles}}{2.8 \text{ hours}} = 50 \) miles per hour.

9. 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%.

10. 
    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

11. A man pays 10% of his income for his income tax. If his income tax amounts to $ 1500, what is his income?

    1. $ 13000
    2. $ 15000
    3. $ 17000
    4. $ 19000
    5. $ 11000

    Let x = income
    10% of x = $ 1500
    0.1x = $ 1500
    x = \(1500 \over 0.1\) = $ 15000

12. 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

13. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

14. 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

15. 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

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

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. A and B can reap a field in 30 days, working together. After 20 days, however, B is called away and A takes 20 days more to complete the work. B alone could do the whole work in

    1. 48 days
    2. 50 days
    3. 56 days
    4. 60 days
    5. 64 days

    (A + B)'s 20 day's work = \(1 \over 30 \) × 20 = \(2 \over 3 \)
    Remaining work = 1 - \(2 \over 3 \) = \(1 \over 3 \)
    Now, \(1 \over 3 \) work is done by A in 20 days.
    Therefore, the whole work will be done by B in 20 × 3 = 60 days.

19. 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

20. if x% of 60 = 48 then x = ?

    1. 80
    2. 60
    3. 90
    4. 40
    5. 70

    x = \( {48 × 100 \over 60} \) = 80

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