Solved Examples Set 3 (Quantitative Ability)

1. 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%.

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. 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

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

   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

5. 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

6. 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%

7. 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

8. \( {5.76 \over 1.6} - 2.4 = ? \)

   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

9. 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.

10. 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.

11. 5789 - 2936 + 1089 = ?

    1. 3942
    2. 4041
    3. 2626
    4. 3932
    5. 3940

    5789 - 2936 + 1089 = 6878 - 2936 = 3942

12. 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

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. 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

15. 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%

16. 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

17. 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.

18. A certain solution is to be prepared by combining chemicals X, Y and Z in the ratio 18:3:2. How many liters of the solution can be prepared by using 36 liters of X?

    1. 46 liters
    2. 47 liters
    3. 45 liters
    4. 49 liters
    5. 44 liters

    As total ratio is 18 +3 + 2 = 23
    Let total solution is x liters
    Then \(18 \over 23\) x = 36
    x = \(36 × 23 \over 18\) = 46 liters

19. 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

20. 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

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