Solved Examples Set 3 (Quantitative Ability)

1. 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
   $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

2. A man bought a flat for $ 820000. He borrowed 55% of this money from a bank. How much money did he borrow from the bank?

   1. $ 451000
   2. $ 452000
   3. $ 453000
   4. $ 454000
   5. $ 450000

   55% of 820000 = 0.55 × 820000 = $ 451000

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. 40 men can build a wall 4 metres high in 15 days. The number of men required to build a similar wall 5 metres high in 6 days is

   1. 115
   2. 125
   3. 105
   4. 135
   5. 130

   \( 40 × 15 × 5 \over 6 × 4 \) = 125 men

5. 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

6. Which expression is equivalent to \(\frac{6𝑥^2 + 4𝑥}{2𝑥}\)?

   1. 7x
   2. 5x²
   3. 3x + 2
   4. 6x² + 2
   5. 3x² + 2x

   As \(\frac{6𝑥^2}{2𝑥} = 3𝑥,\) and \(\frac{4𝑥}{2𝑥} = 2,\) so then \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

7. 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%

8. 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

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 shop owner blends three types of coffees, A, B and C, in the ratio 3:5:7. Given that type A coffee costs $ 70 per kg, type B coffee costs $ 100 per kg and type C coffee costs $ 130 per kg, calculate the cost per kg of the blended mixture.

    1. $ 106
    2. $ 108
    3. $ 109
    4. $ 110
    5. $ 105

    Cost per kg = 70 x 1/5 + 100 x 1/3 + 130 x 7/15 = $ 108 per kg

11. A man was 32 years old when his daughter was born. He is now five times as old as his daughter. How old is his daughter now?

    1. 7 years
    2. 8 years
    3. 9 years
    4. 10 years
    5. 6 years

    Let's assume the daughter is d years old now. That means that the man is now (32 + d) years old, so that
    (32 + d) = 5d
    32 = 4d
    d = 8

12. 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

13. 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.

14. \(\frac{\frac{7}{10} × 14 × 5 × \frac{1}{28}}{\frac{10}{17} × \frac{3}{5} × \frac{1}{6} × 17} = \)

    1. 4/7
    2. 1
    3. 7/4
    4. 2
    5. 17/4

    

15. After spending 88% of his income, a man had $ 2160 left. Find his income.

    1. $ 18000
    2. $ 19000
    3. $ 20000
    4. $ 22000
    5. $ 17000

    Let income = x
    x = 88% of x + 2160
    x - 0.88x = 2160
    0.12x = 2160
    x = \(216000 \over 12\) = 18000

16. 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

17. 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

18. 60% of 37 = ?

    1. 20
    2. 21
    3. 22.2
    4. 22
    5. none

    60% of 37 = 0.6 × 37 = 22.2

19. 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 \(km \over hr\)

20. ? × 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

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