Solved Examples Set 3 (Quantitative Ability)

1. Which set of ordered pairs represents a function?

   1. {(−5,5),(4,8),(−5,−6)}
   2. {(−1,−1),(−1,6),(−1,−10)}
   3. {(−3,7),(2,5),(−7,7)}
   4. {(2,3),(−2,4),(−2,−5)}
   5. {(2,3),(3,2),(2,5)}

   For a set of ordered pairs to be a function, no single 𝑥-coordinate can be mapped to two distinct 𝑦-coordinates. This is not the case for option A, where 𝑥=−5 is mapped to both 𝑦=5 and 𝑦=−6. Similarly, in options B (𝑥=−1), D (𝑥=−2), and E (𝑥=2), an 𝑥 value is mapped to two different 𝑦 values.

2. A man earned an annual income of $ 245000 in 1990. He was allowed a deduction of $ 15000 relief for each of his three children and a personal relief of $ 30000. If he was charged a tax rate of 4% on first $ 50000 and 6% on his remaining income, calculate the total tax charged.

   1. $ 9200
   2. $ 8700
   3. $ 9500
   4. $ 9400
   5. $ 9000

   Total Income = $ 245000
   Total relief = 3 × 15000 + 30000 = $ 75000
   Rest income = 245000 - 75000 = 170000
   Tax on 1st 50000 = 0.04 × 50000 = $ 2000
   Tax on rest amount 120000 = 0.06 × 120000 = $ 7200
   Total tax = 200 + 7200 = $ 9200

3. 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

4. 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%

5. ? × 12 = 75% of 336

   1. 48
   2. 252
   3. 28
   4. 21
   5. 23

   ? × 12 = 75% of 336
   ? × 12 = 0.75 × 336
   ? × 12 = 252
   \(? = \frac{252}{12}\)
   ? = 21

6. A man is 5 times as old as his son. 2 years ago the sum of the squares of their ages was 1114. Find the present age of son.

   1. 7 years
   2. 9 years
   3. 8 years
   4. 8 1/2 years
   5. 6 years

   Let son's age = x, then
   father's age = 5x
   As before 2 years ago the sum of the squares of their ages was 1114, the equation becomes as
   \((x - 2)^2 + (5x - 2)^2 = 1114 \)
   By simplifying the equation, we have
   \(13x^2 -12x -553 = 0\)
   Now solving the equation, we have
   \(13x^2 - 12x - 553 = 0\)
   \(13x^2 - 91x + 79x -553 = 0\)
   13x(x - 7) + 79(x - 7) = 0
   (x - 7)(13x + 79) = 0
   x = 7 and x = -6.077
   As age could not be negative, hence the present age of the son is 7 years.

7. 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.

8. A fruit-seller has 120 oranges. Given that he has 20% more apples than oranges and 40% less oranges than pears, find the number of apples and the number of pears the fruit seller has.

   1. 144, 200
   2. 148, 380
   3. 149, 220
   4. 140, 190
   5. 142, 190

   No. of apples = 120 + 20% of 120 = 120 + 0.2 × 120 = 144
   
   Let x = No. of pears
   x - 40% of x = 120
   x - 0.4x = 120
   0.6x = 120
   x = \(120 \over 0.6\) = 200
   Hence, no. of pears = 200

9. 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

10. 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

11. 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

12. A bank increased the rate of interest which it paid to depositors from 3.5% to 4% per annum. Find how much more interest a man would receive if he deposited $ 64000 in the bank for 6 months at the new interest rate

    1. $ 160
    2. $ 180
    3. $ 200
    4. $ 220
    5. $ 150

    If the interest rate is 3.5% then interest amount is
    3.5% of 6400 = 0.035 × 6400 = $ 2240
    If the interest rate is 4% then interest amount is
    4% of 6400 = 0.04 × 6400 = $ 2560
    Now the difference of both interests = 2560 - 2240 = $ 320 per annum
    Interest for half year (6 months) = \(320 \over 2\) = $ 160

13. 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

14. 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.

15. 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

16. 5873 + 12034 + 1106 = ?

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

    5873 + 12034 + 1106 = 17907 + 1106 = 19013

17. Rashid's salary was reduced by 20%. In order to restore his salary at the original amount, it must be raised by

    1. 20%
    2. 22.50%
    3. 25%
    4. 26%
    5. 27%

    Let Rashid's Salary 100
    20% reduced salary is 80
    As the reduced amount is 20
    So what percentage of the present sallary is required to be equal to 20?
    ?% of 80 = 20
    ? = \(20 \over 80\) × 100 = 25%

18. 8 : ? :: 1 : 4

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

    ? × 1 = 8 × 4
    ? = 32

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. 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

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