Solved Examples Set 1 (Quantitative Ability)

1. \( {396 \over 11} \) + 19 = ?

   1. 19.8
   2. 36
   3. 55
   4. 33
   5. 50

   \( {396 \over 11} \) + 19 = 36 + 19 = 55

2. 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.

3. 1015 / 0.05 / 40 = ?

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

   1015 / 0.05 / 40 = 20300 / 40 = 507.5

4. 8 : ? :: 1 : 4

   1. 24
   2. 16
   3. 0
   4. 32
   5. 20

   ? × 1 = 8 × 4
   ? = 32

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. 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

7. 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%

8. 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

9. ?% of 60 = 24

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

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

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. 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

12. 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

13. A man pays 10% of his income for his income tax. If his income tax amounts to $ 1500, what is his income?

    1. $ 13000
    2. $ 15000
    3. $ 17000
    4. $ 19000
    5. $ 11000

    Let x = income
    10% of x = $ 1500
    0.1x = $ 1500
    x = \(1500 \over 0.1\) = $ 15000

14. The amount of hot cocoa powder remaining in a can is 6 1⁄4 tablespoons. A single serving consists of 1 3⁄4 tablespoons of the powder. What is the total number of servings of the powder remaining in the can?

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

    As \(6\frac{1}{4} = \frac{25}{4}\) and \(1\frac{3}{4} = \frac{7}{4}\). Therefore,
    \(\frac{6\frac{1}{4} \text{ tsp}}{1\frac{3}{4} \text{ } \frac{tsp}{ serving}} = \frac{\frac{25}{4}}{\frac{7}{4}} \text{ servings} = \frac{25}{7} \text{ servings} = 3\frac{4}{7} \text{ servings}\)

15. A car traveled 100 km with half the distance at 40 km/h and the other half at 80 km/h. Find the average speed of the car for the whole journey.

    1. 53 km/hr
    2. 53.33 km/hr
    3. 54 1⁄4 km/hr
    4. 55 km/hr
    5. 56 km/hr

    The time, car took for the first half, \(50 \over 40 \) = 1.25 hrs
    and for the second half \(50 \over 80 \) = 0.625 hrs
    Total time = 1.25 + 0.625 = 1.875 hrs
    Average speed = \(100 \over 1.875 \) = 53.3 \(km \over hr\)

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

17. 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

18. The closest approximation of \(\frac{69.28 × .004}{.03}\) is

    1. 0.092
    2. 0.92
    3. 9.2
    4. 92
    5. 920

    This problem is most easily completed by rearranging and approximating as follows:
    (69.28 x .004)/.03 ≅ 69 x .1 = 6.9
    which is the only reasonably close answer to 9.2

19. 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

20. A man takes 50 minutes to cover a certain distance at a speed of 6 km/hr. If he walks with a speed of 10 km/hr, he covers the same distance in

    1. 1 hour
    2. 30 minutes
    3. 20 minutes
    4. 10 minutes
    5. 40 minutes

    \( 50 × 6 \over 10 \) = 30 minutes

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