Solved Examples Set 2 (Quantitative Ability)

1. 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.

2. 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.

3. 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

4. A person's net income is $ 1373.70 and he pays an income tax of 5%. His gross income in dollars must be

   1. 1446
   2. 1118.96
   3. 1308.29
   4. 1438.25
   5. 1211.21

   Let gross income in dollars = x
   then according to the statement,
   x = 5% of x + 1373.70
   x - 0.05x = 1373.70
   0.95x = 1373.70
   x = \(137370 \over 95\) = 1446

5. \( {2244 \over 0.88} = ? × 1122 \)

   1. 20.02
   2. 20.2
   3. 19.3
   4. 2.27
   5. 3.27

   \( {2244 \over 0.88} = ? × 1122 \)
   \(? = {2550 \over 1122} = 2.27 \)

6. \(25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?\)

   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

7. \( {𝑥 - 8 \over 24} = {3 \over 4} \)
   What is the value of 𝑥 in the equation?

   1. 10
   2. 20
   3. 26
   4. 31
   5. 40

   By cross multiplying, 4(𝑥 – 8) =3 × 24. Thus, 4𝑥 – 32 = 72, and so 4𝑥 = 104 and 𝑥 = 26.

8. 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 \(km \over hr\)

9. 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.

10. 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

11. 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

12. 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

13. On a trip to visit friends, a family drives 65 miles per hour for 208 miles of the trip. If the entire trip was 348 miles and took 6 hours, what was the average speed, in miles per hour, for the rest of the trip?

    1. 44
    2. 50
    3. 51
    4. 58
    5. 60

    As the first part of the trip took \(\frac{208 \text{ miles}}{65 \text{ } \frac{miles}{hour}} = 3.2 \text{ hours},\) so the remaining 140 miles (348 - 208) took 2.8 hours (6 - 3.2). The average speed for the rest of the trip was \(\frac {140 \text{ miles}}{2.8 \text{ hours}} = 50 \) miles per hour.

14. \( {0.027 \over 90} = ? \)

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

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

15. 1015 / 0.05 / 40 = ?

    1. 50.75
    2. 507.5
    3. 506
    4. 2056
    5. 5075

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

16. 8 : ? :: 1 : 4

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

    ? × 1 = 8 × 4
    ? = 32

17. 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%

18. 12% of ________ = 48

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

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

19. 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.

20. If 3x = −9, then 3x³ − 2x + 4 =

    1. -83
    2. -71
    3. -47
    4. -17
    5. 61

    First solving 3x = −9, x = −3. Now plug into 3x³ − 2x + 4:
    3x³ − 2x + 4
    = 3(-3)³ − 2(-2) + 4
    = 3(−27) + 6 + 4
    = −81 + 6 + 4
    = −71

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