Solved Examples Set 3 (Quantitative Ability)

1. 5789 - 2936 + 1089 = ?

   1. 3942
   2. 4041
   3. 2626
   4. 3932
   5. 3940

   5789 - 2936 + 1089 = 6878 - 2936 = 3942

2. 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,
   \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

3. 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

4. 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

5. \( {396 \over 11} \) + 19 = ?

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

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

6. 5.41 - 3.29 × 1.6 = ?

   1. 14.6
   2. 0.3392
   3. 0.146
   4. 3.392
   5. 1.46

   5.41 - 3.29 × 1.6 = 5.41 - 5.264 = 0.146

7. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

   1. 33 days
   2. 35 days
   3. 37 days
   4. 39 days
   5. 31 days

   \(14 × 10 \over 4 \) = 35 days

8. 
   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

9. 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

10. ? × 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

11. A retailer bought a compact disc from a manufacturer for $ 200. In addition to that, he paid a 15% sales tax. If he sold the disc to a customer for $ 260, calculate the cash profit he made.

    1. $ 30.00
    2. $ 35.00
    3. $ 32.50
    4. $ 28.00
    5. $ 30.50

    price of a compact disc with sales tax = 200 + 0.15 × 200
    = 200 + 30 = $ 230
    As the selling price of the disc = $ 260
    Hence, cash profit = 260 - 230 = $ 30

12. 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.

13. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    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

14. 350 × ? = 4200

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

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

15. ?% of 60 = 24

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

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

16. 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 \(km \over hr\)

17. 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

18. \( {1250 \over 25} × 0.5 = ? \)

    1. 250
    2. 50
    3. 2.5
    4. 25
    5. 125

    \( {1250 \over 25} × 0.5 = 50 × 0.5 = 25 \)

19. 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
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

20. 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.

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