Solved Examples Set 3 (Quantitative Ability)

1. 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 \(km \over hr\)

2. 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.

3. A primary school had an enrollment of 850 pupils in January 1970. In January 1980 the enrollment was 1,120. What was the percentage increase for the enrollment?

   1. 31.76%
   2. 33.50%
   3. 30.65%
   4. 34.76%
   5. 30.55%

   Percentage increase for the enrollment = \(1120 - 850 \over 850\) × 100 = 31.76

4. 5873 + 12034 + 1106 = ?

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

   5873 + 12034 + 1106 = 17907 + 1106 = 19013

5. 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.

6. ? × 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

7. 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.

8. 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
   $$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$

9. 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

10. 12% of ________ = 48

    1. 250
    2. 100
    3. 400
    4. 200
    5. 300

    \(12 \text{% of } x = 48\)
    \(0.12x = 48\)
    \(x = \frac{48}{0.12} = 400\)

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 basket that contains 2 apples, 3 bananas, 6 oranges, and 4 pears is in the workroom. When Ms. Hutchinson went to the workroom, other workers had already taken 1 banana, 2 oranges, and 1 pear. From the remaining fruit, Ms. Hutchinson randomly took 3 pieces of fruit separately from the basket. If each fruit is equally likely to be chosen, what is the probability that the third piece was an orange if the first two she took were also oranges?

    1. 4/165
    2. 9/11
    3. 4/11
    4. 3/11
    5. 2/9

    Ms. Hutchinson randomly takes the 3 pieces of fruit from the basket, there are 2 apples, 3 -1 = 2 bananas, 6 - 2 = 4 oranges, and 4 - 1 = 3 pears. Assuming that the first 2 pieces of fruit Ms. Hutchinson takes are oranges, there will be 2 apples, 2 bananas, 4 - 2 = 2 oranges, and 3 pears left in the basket when she selects the third piece of fruit. The probability that the third piece of fruit she selects will be an orange is \(\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}\).

14. 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

15. 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

16. \( {0.027 \over 90} = ? \)

    1. 0.0003
    2. 0.03
    3. 3
    4. 0.00003
    5. 0.003

    \( {0.027 \over 90} = {27 \over 1000 × 90} = {3 \over 10000} = 0.0003 \)

17. 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

18. 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

19. 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

20. 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 \(km \over hr\)

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