Solved Examples Set 2 (Quantitative Ability)

1. 12% of ________ = 48

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

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

2. A shopkeeper sold two articles for $ 48 each. He made a 25% profit on one article and a loss of 20% on the other. What was his net gain or loss on the sale of the two articles?

   1. loss of $ 1.40
   2. gain of $ 2.40
   3. loss of $ 2.40
   4. gain of $ 1.40
   5. gain of $ 2.60

   25% profit at selling price $ 48 = 48 x .25 = $ 12
   20% loss at selling price $ 48 = 48 x 0.2 = $ 9.6
   gain = profit - loss = 12 - 9.6 = $ 2.4

3. 5789 - 2936 + 1089 = ?

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

   5789 - 2936 + 1089 = 6878 - 2936 = 3942

4. 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

5. 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,
   \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

6. 8 : ? :: 1 : 4

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

   ? × 1 = 8 × 4
   ? = 32

7. 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

8. 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

9. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

10. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

11. 40 arithmetic questions, each carrying equal marks, were given in a class test. A boy answered 25 questions correctly. What percentage was this? To pass a test a student must answer at least 45% of the questions correctly. Find the least number of correct answers needed to pass.

    1. 62.5%, 18
    2. 63.5%, 16
    3. 64.5%, 20
    4. 61.0%, 21
    5. 60.0%, 22

    \(x \text{% of } 40 = 25\)
    \(x \text{% } × 40 = 25\)
    \(x = {25 \over 40} × 100 \)
    x = 62.5
    
    \(x = 45 \text{% of } 40 \)
    \(x = 0.45 × 40 \)
    x = 18

12. 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

13. 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

14. \( {0.027 \over 90} = ? \)

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

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

15. 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%

16. Which set of ordered pairs represents a function?

    1. {(−5,5),(4,8),(−5,−6)}
    2. {(−1,−1),(−1,6),(−1,−10)}
    3. {(−3,7),(2,5),(−7,7)}
    4. {(2,3),(−2,4),(−2,−5)}
    5. {(2,3),(3,2),(2,5)}

    For a set of ordered pairs to be a function, no single 𝑥-coordinate can be mapped to two distinct 𝑦-coordinates. This is not the case for option A, where 𝑥=−5 is mapped to both 𝑦=5 and 𝑦=−6. Similarly, in options B (𝑥=−1), D (𝑥=−2), and E (𝑥=2), an 𝑥 value is mapped to two different 𝑦 values.

17. 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
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

18. 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

19. 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

20. 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

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