Solved Examples Set 3 (Quantitative Ability)

1. ?% of 60 = 24

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

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

2. 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

3. 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

4. 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

5. \( {63.84 \over ?} \) = 21

   1. 3.04
   2. 3.4
   3. 30.4
   4. 300.4
   5. 0.304

   ? = \( 63.84 \over 21 \) = 3.04

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. 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

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

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

   

9. 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

10. A group of laborers accepted to do a piece of work in 20 days. 8 of them did not turn up for the work and the remaining did the work in 24 days. The original number of laborers was

    1. 47
    2. 48
    3. 49
    4. 50
    5. 51

    x laborers do work in 20 days and x-8 laborers do same work in 24 days. As the no. of laborers decrease, the no. of days increased then it becomes as
    x : x - 8 :: 24 : 20
    product of interiors = product of exteriors
    24x - 192 = 20x
    4x = 192
    x = 48

11. If 3x = −9, then 3x³ − 2x + 4 =

    1. -83
    2. -71
    3. -47
    4. -17
    5. 61

    First solving 3x = −9, x = −3. Now plug into 3x³ − 2x + 4:
    3x³ − 2x + 4
    = 3(-3)³ − 2(-2) + 4
    = 3(−27) + 6 + 4
    = −81 + 6 + 4
    = −71

12. Which of the following expressions is equivalent to \(\frac{𝑥^2 + 3x + 1}{𝑥 + 1}\)?

    1. x + 2
    2. 𝑥 + 3
    3. 𝑥 + 2 - 1/(𝑥 + 1)
    4. 𝑥 + 3 + 1/(𝑥 + 1)
    5. 𝑥 + 4 + 5/(𝑥 + 1)

    As \(𝑥^2 + 3x + 1 = (𝑥^2 + 3x + 2) -1\)
    and
    \(\frac{𝑥^2 + 3x + 2}{x + 1} = \frac{(𝑥 + 2)(x + 1)}{x + 1} = 𝑥 + 2\)
    Therefore,
    \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)

13. A can do a piece of work in 10 days and B can do it in 15 days. The number of days required by them to finish it, working together is

    1. 8
    2. 7
    3. 6
    4. 4
    5. 3

    A's 1 day work = \(1 \over 10\)
    B's 1 day work = \(1 \over 15\)
    Now both A and B's 1 day work = \({1 \over 10} + {1 \over 15}\) = \(3 + 2 \over 30\) = \(1 \over 6\)
    Hence the work by both A and B will be completed in 6 days.

14. 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.

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. 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

17. A person's net income is $ 1373.70 and he pays an income tax of 5%. His gross income in dollars must be

    1. 1446
    2. 1118.96
    3. 1308.29
    4. 1438.25
    5. 1211.21

    Let gross income in dollars = x
    then according to the statement,
    x = 5% of x + 1373.70
    x - 0.05x = 1373.70
    0.95x = 1373.70
    x = \(137370 \over 95\) = 1446

18. 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%.

19. 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 \(km \over hr\)

20. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

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