Solved Examples Set 3 (Quantitative Ability)

1. 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

2. 1015 / 0.05 / 40 = ?

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

   1015 / 0.05 / 40 = 20300 / 40 = 507.5

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. By selling a fan for $ 475, a person loses 5%. To get a gain of 5%, he should sell the fan for:

   1. $ 500
   2. $ 525
   3. $ 535
   4. $ 575
   5. $ 505

   cost price = 100/(100 - 5) x 475 = $ 500
   sale price = (100 + 5)/100 x 500 = $ 525

5. 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

6. 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,
   \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)

7. 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

8. 350 × ? = 4200

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

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

9. 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

10. 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%

11. 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.

12. How much would I have to pay for a book which cost $ 72 to product, if the printing company sold it to a bookseller at 20% profit and in return the bookseller sold it to me at a profit of 25%?

    1. $ 104
    2. $ 106
    3. $ 108
    4. $ 110
    5. $ 109

    Cost of the book product = $ 72
    Profit of printing company = 20% of 72 = 0.2 x 72 = 14.4
    Now the cost of the book = 72 +14.4 = $ 86.4
    Profit of the bookseller = 25% of 86.4 = 21.4
    Finally, the cost of the book = 86.4 + 21.4 = $ 108

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. 60% of 37 = ?

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

    60% of 37 = 0.6 × 37 = 22.2

15. If n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ (n − 3) . . . 2 ⋅ 1, what is the value of \(\frac{(6!)(4!)}{(5!)(3!)}\)

    1. 5/4
    2. 8/5
    3. 10
    4. 24
    5. 1152

    \(\frac{(6!)(4!)}{(5!)(3!)}\) = \(\frac{(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2. 1)}{(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)}\) = \(\frac{6 . 4}{1}\) = 24

16. 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

17. A man bought a flat for $ 820000. He borrowed 55% of this money from a bank. How much money did he borrow from the bank?

    1. $ 451000
    2. $ 452000
    3. $ 453000
    4. $ 454000
    5. $ 450000

    55% of 820000 = 0.55 × 820000 = $ 451000

18. ?% of 60 = 24

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

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

19. 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%.

20. 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

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