Solved Examples Set 1 (Quantitative Ability)

1. 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.

2. A shopkeeper bought a radio from a wholesaler for $ 250.00. In addition, he paid a sales tax of 15% on the cost price. He then sold the radio for $ 315.00. Calculate the cash profit made by the shopkeeper.

   1. $ 20.00
   2. $ 22.50
   3. $ 25.00
   4. $ 27.50
   5. $ 27.00

   cost price = $ 250
   sales tax = .15 × 250 = $ 37.5
   cash profit = 315 - 250 - 37.5 = $ 27.5

3. 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

4. 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

5. \( {0.027 \over 90} = ? \)

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

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

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 rectangular room is 6 m long, 5 m wide and 4 m high. The total volume of the room in cubic meters is

   1. 24
   2. 30
   3. 120
   4. 240
   5. 140

   Total volume = length × width × height = 6 × 5 × 4 = 120

8. A shop owner blends three types of coffees, A, B and C, in the ratio 3:5:7. Given that type A coffee costs $ 70 per kg, type B coffee costs $ 100 per kg and type C coffee costs $ 130 per kg, calculate the cost per kg of the blended mixture.

   1. $ 106
   2. $ 108
   3. $ 109
   4. $ 110
   5. $ 105

   Cost per kg = 70 x 1/5 + 100 x 1/3 + 130 x 7/15 = $ 108 per kg

9. Rashid buys three books for $ 16 each and four books for $ 23 each, what will be the average price of books

   1. $ 18
   2. $ 20
   3. $ 22
   4. $ 24
   5. $ 16

   Price of 3 books = 3 × 16 = $ 48
   Price of 4 books = 4 × 23 = $ 92
   Total price = $ 140
   Total books = 3 + 4 = 7
   Average price of books = \(140 \over 7 \) = $ 20

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

11. 72 + 679 + 1439 + 537+ ? = 4036

    1. 1309
    2. 1208
    3. 2308
    4. 2423
    5. 1309

    72 + 679 + 1439 + 537+ ? = 4036
    2727 + ? = 4036
    ? = 4036 - 2727 = 1309

12. 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

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

14. A man sells two houses for $ 2 lac each. On one he gained 20% and on the other he lost 20%. His total profit or loss % in the transaction will be

    1. 4% profit
    2. 5% loss
    3. no profit, no loss
    4. 4% loss
    5. 3% loss

    % loss = (% loss X % profit)/100 = (20 X 20)/100 = 4%

15. 60% of 37 = ?

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

    60% of 37 = 0.6 × 37 = 22.2

16. 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

17. 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.

18. 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

19. ?% of 60 = 24

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

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

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