Solved Examples Set 3 (Quantitative Ability)

1. A bank exchanges British currency for Singapore currency at the rate of S$ 3.20 to pond 1. Calculate, in Pond, the amount exchanged for S$ 1,600 by a customer who also had to pay an extra 3% commission for this transaction.

   1. Pond 475
   2. Pond 485
   3. Pond 495
   4. Pond 505
   5. Pond 510

   As commission is 3% of 1600 = 0.03 × 1600 = S$ 48
   the rest amount = 1600 - 48 = S$ 1552
   S$ 1 = \(1 \over 3.20\) = Pond 0.3125
   Now S$ 1552 = 1552 × 0.3125 = Pond 485

2. 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

3. 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.

4. 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

5. 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

6. 5873 + 12034 + 1106 = ?

   1. 19016
   2. 20001
   3. 19013
   4. 2018
   5. 19010

   5873 + 12034 + 1106 = 17907 + 1106 = 19013

7. 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

8. 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

9. 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 \(km \over hr\)

10. A girl is 18 years younger than her mother. In 6 years time, the sum of their ages will be 54.How old is the girl now?

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

    Let girl's age = x
    then mother's age = x + 18
    After 6 years,
    x + 6 + x + 18 + 6 = 54
    2x + 30 = 54
    2x = 24
    x = 12

11. \( {5.76 \over 1.6} - 2.4 = ? \)

    1. 1.2
    2. 2.4
    3. 7.2
    4. 0.12
    5. 0.012

    \( {5.76 \over 1.6} - 2.4 = \) 3.6 - 2.4 =1.2

12. 350 × ? = 4200

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

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

13. 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

14. ?% of 60 = 24

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

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

15. 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

16. A third-grade class is composed of 16 girls and 12 boys. There are 2 teacher-aides in the class. The ratio of girls to boys to teacher-aides is

    1. 16:12:1
    2. 8:6:2
    3. 8:6:1
    4. 8:3:1
    5. 4:3:1

    Girls to boys to teacher-aides are in proportion 16 to 12 to 2. Reduced to lowest terms, 16:12:2 equals 8:6:1.

17. 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.

18. On a trip to visit friends, a family drives 65 miles per hour for 208 miles of the trip. If the entire trip was 348 miles and took 6 hours, what was the average speed, in miles per hour, for the rest of the trip?

    1. 44
    2. 50
    3. 51
    4. 58
    5. 60

    As the first part of the trip took \(\frac{208 \text{ miles}}{65 \text{ } \frac{miles}{hour}} = 3.2 \text{ hours},\) so the remaining 140 miles (348 - 208) took 2.8 hours (6 - 3.2). The average speed for the rest of the trip was \(\frac {140 \text{ miles}}{2.8 \text{ hours}} = 50 \) miles per hour.

19. 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.

20. 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

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