Solved Examples Set 1 (Quantitative Ability)

1. 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

2. 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.

3. 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

4. Which of the following expressions is equivalent to \(\frac{𝑥^2 + 3x + 1}{𝑥 + 1}\)?

   1. x + 2
   2. 𝑥 + 3
   3. 𝑥 + 2 - 1/(𝑥 + 1)
   4. 𝑥 + 3 + 1/(𝑥 + 1)
   5. 𝑥 + 4 + 5/(𝑥 + 1)

   As \(𝑥^2 + 3x + 1 = (𝑥^2 + 3x + 2) -1\)
   and
   \(\frac{𝑥^2 + 3x + 2}{x + 1} = \frac{(𝑥 + 2)(x + 1)}{x + 1} = 𝑥 + 2\)
   Therefore,
   \(\frac{𝑥^2 + 3x + 1}{x + 1} = \frac{𝑥^2 + 3x + 2}{x + 1} - \frac{1}{x + 1} = (𝑥 + 2) - \frac{1}{x + 1}\)

5. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

   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

6. 10 men can complete a job in 14 days. How long will it take 4 men to finish the same job if they work at the same rate?

   1. 33 days
   2. 35 days
   3. 37 days
   4. 39 days
   5. 31 days

   \(14 × 10 \over 4 \) = 35 days

7. 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

8. A group of boys were to choose between playing hockey and badminton. The number of boys choosing hockey was three times that of those choosing badminton. Asking 12 boys who chose hockey to play badminton would make the number of players for each game equal. Find the number who chose badminton originally.

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

   Let no. of boys for badminton = x
   then no. of boys for hockey = 3x
   According to the statement,
   3x - 12 = x + 12 (12 leave hockey, 12 join badminton)
   2x = 24
   x = 12
   Hence, there were 12 boys originally choosing badminton.

9. 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.

10. 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%

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 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 \(km \over hr\)

13. ?% of 60 = 24

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

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

14. 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

15. 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
    $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

16. 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

17. 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

18. 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

19. 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

20. 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

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