Solved Examples Set 3 (Quantitative Ability)

1. 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

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

3. 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

4. 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

5. 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 \(km \over hr\)

6. 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

7. A man saves $ 500, which is 15% of his annual income. How much does he earn in one year?

   1. $ 3542.5
   2. $ 3333.33
   3. $ 3132.3
   4. $ 3075.75
   5. $ 4444.4

   Let annual income = x
   15% of x = 500
   x = \(500 \over 15\) × 100 = \(10000 \over 3\) = 3333.33

8. 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

9. 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

10. \( {0.027 \over 90} = ? \)

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

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

11. 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

12. 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.

13. In the series 8, 9, 12, 17, 24 . . . the next number would be

    1. 29
    2. 30
    3. 33
    4. 35
    5. 41

    In the series, 8, 9, 12, 17, 24 . . .
    9 − 8 = 1
    12 − 9 = 3
    17 − 12 = 5
    24 − 17 = 7
    Hence, the difference between the next term and 24 must be 9 or
    x − 24 = 9, and
    x = 33
    Hence, the next term in the series must be 33

14. 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.

15. 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

16. A group of boys were to choose between playing hockey and badminton. The number of boys choosing hockey was three times that of those choosing badminton. Asking 12 boys who chose hockey to play badminton would make the number of players for each game equal. Find the number who chose badminton originally.

    1. 12
    2. 14
    3. 11
    4. 13
    5. 10

    Let no. of boys for badminton = x
    then no. of boys for hockey = 3x
    According to the statement,
    3x - 12 = x + 12 (12 leave hockey, 12 join badminton)
    2x = 24
    x = 12
    Hence, there were 12 boys originally choosing badminton.

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. By selling 60 chairs, a man gains an amount equal to selling price of 10 chairs. The profit percentage in the transaction is

    1. 10%
    2. 15%
    3. 16.67%
    4. 20%
    5. 22%

    selling price of 60 chairs = selling price of 10 chairs
    profit of 60 chairs = profit of 10 chairs
    profit of 6 chairs = profit of 1 chair
    profit of 1 chair = profit of 1/6 chair
    profit %age = 1/6 x 100 = 16.67%

19. 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%

20. 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.

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