Solved Examples Set 3 (Quantitative Ability)

1. 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

2. 
   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

3. \( {0.027 \over 90} = ? \)

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

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

4. 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.

5. 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

6. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

   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

7. A girl is 18 years younger than her mother. In 6 years time, the sum of their ages will be 54.How old is the girl now?

   1. 10 years
   2. 11 years
   3. 12 years
   4. 13 years
   5. 14 years

   Let girl's age = x
   then mother's age = x + 18
   After 6 years,
   x + 6 + x + 18 + 6 = 54
   2x + 30 = 54
   2x = 24
   x = 12

8. 350 × ? = 4200

   1. 12
   2. 24
   3. 15
   4. 30
   5. 16

   \( ? = {4200 \over 350} =12 \)

9. Rashid's salary was reduced by 20%. In order to restore his salary at the original amount, it must be raised by

   1. 20%
   2. 22.50%
   3. 25%
   4. 26%
   5. 27%

   Let Rashid's Salary 100
   20% reduced salary is 80
   As the reduced amount is 20
   So what percentage of the present sallary is required to be equal to 20?
   ?% of 80 = 20
   ? = \(20 \over 80\) × 100 = 25%

10. 8 : ? :: 1 : 4

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

    ? × 1 = 8 × 4
    ? = 32

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. 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

13. 60% of 37 = ?

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

    60% of 37 = 0.6 × 37 = 22.2

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. A train takes 50 minutes for a journey if it runs at 48 km/hr. The rate at which the train must run to reduce the time to 40 minutes will be

    1. 50 km/hr
    2. 55 km/hr
    3. 60 km/hr
    4. 57 km/hr
    5. 65 km/hr

    \(50 × 48 \over 40\) = 60 \(km \over hr\)

16. 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,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

17. Which of the following is the largest?

    1. half of 30% of 280
    2. one-third of 70% of 160
    3. twice 50% of 30
    4. three times 40% of 40
    5. 60% of 60

    Let us calculate the value of each:
    A. 0.5 × 0.3 × 280 = 42
    B. 0.33 × 0.7 × 160 = 36.96
    C. 2 × 0.5 × 30 = 30
    D. 3 × 0.4 × 40 = 48
    E. 0.6 × 60 = 36

18. 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

19. 15 men can complete a job in 10 days. How long will it take 8 men to finish the same job if they work at the same rate?

    1. 14 3⁄4 days
    2. 16 3⁄4 days
    3. 18 3⁄4 days
    4. 20 3⁄4 days
    5. 22 3⁄4 days

    \( 15 × 10 \over 8 \) = 18 3⁄4 days

20. 12% of ________ = 48

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

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

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