Solved Examples Set 3 (Quantitative Ability)

1. \( {0.027 \over 90} = ? \)

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

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

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

3. if x% of 60 = 48 then x = ?

   1. 80
   2. 60
   3. 90
   4. 40
   5. 70

   x = \( {48 × 100 \over 60} \) = 80

4. 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

5. 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

6. At a book fair, a book was reduced in price from $ 75 to $ 60. If the first price gives a 50% profit, find the percentage profit of the book sold at the reduced price.

   1. 20%
   2. 30%
   3. 40%
   4. 50%
   5. 10%

   As $ 75 (first price) gives a profit = 50%
   $ 1 gives a profit = (50/75)%
   $ 60 (reduced price) gives profit = (50/75 x 60)% = 40%

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

8. If 4a + 2 = 10, then 8a + 4 =

   1. 5
   2. 16
   3. 20
   4. 24
   5. 28

   One may answer this question by solving
   4a + 2 = 10
   4a = 8
   a= 2
   Now, plugging in 2 for a:
   8a + 4 = 8(2) + 4 = 20
   A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.

9. 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

10. 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

11. 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

12. A and B enter into a partnership contributing $ 800 and $ 1000 respectively. At the end of 6 months they admit C, who contributes $ 600. After 3 years they get a profit of $ 966. Find the share of each partner in the profit.

    1. $ 336, $ 420, $ 210
    2. $ 360, $ 400, $ 206
    3. $ 380, $ 390, $ 196
    4. $ 345, $ 405, $ 210
    5. $ 325, $ 400, $ 200

    A shares = 800 × 3 = 2400
    B shares = 1000 × 3 = 3000
    C shares = 600 × 2 1⁄2 = 1500
    Total shares = 2400 + 3000 + 1500 = 6900
    A's profit = \(2400 \over 6900 \) × 966 = $ 336
    B's profit = \(3000 \over 6900 \) × 966 = $ 420
    C's profit = \(1500 \over 6900 \) × 966 = $ 210

13. Matthew’s age (𝑚) is three years more than twice Rita’s age (𝑟). Which equation shows the relationship between their ages?

    1. 𝑚 = 𝑟 − 32
    2. 𝑚 = 𝑟 + 32
    3. 𝑚 = 2(𝑟 + 3)
    4. 𝑚 = 2𝑟 − 3
    5. 𝑚 = 2𝑟 + 3

    As Matthew's age (𝑚) is three more years (+3) than twice Rita's age (2𝑟). Therefore, 𝑚 = 2𝑟 + 3.

14. A certain number was doubled and the result then multiplied by 3. If the product was 138, find the number.

    1. 21
    2. 23
    3. 25
    4. 27
    5. 19

    Let x be the number
    the number is doubled, 2x
    the result is multiplied by 3, 3 × 2x = 6x
    6x = 138
    x = \(138 \over 6\) = 23

15. After spending 88% of his income, a man had $ 2160 left. Find his income.

    1. $ 18000
    2. $ 19000
    3. $ 20000
    4. $ 22000
    5. $ 17000

    Let income = x
    x = 88% of x + 2160
    x - 0.88x = 2160
    0.12x = 2160
    x = \(216000 \over 12\) = 18000

16. 1.02 - 0.20 + ? = 0.842

    1. 0.222
    2. 232
    3. 2
    4. 0.022
    5. 0.012

    1.02 - 0.20 + ? = 0.842
    0.82 + ? = 0.842
    ? = 0.842 - 0.82 = 0.022

17. 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

18. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

    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

19. ?% of 60 = 24

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

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

20. 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.

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