Solved Examples Set 1 (Quantitative Ability)

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

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

   

2. 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

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. 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

5. 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.

6. A man pays 10% of his income for his income tax. If his income tax amounts to $ 1500, what is his income?

   1. $ 13000
   2. $ 15000
   3. $ 17000
   4. $ 19000
   5. $ 11000

   Let x = income
   10% of x = $ 1500
   0.1x = $ 1500
   x = \(1500 \over 0.1\) = $ 15000

7. 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 \(\frac{6𝑥^2 + 4𝑥}{2𝑥} = 3𝑥 + 2\)

8. A shopkeeper buys 300 identical articles at a total cost of $ 1500. He fixes the selling price of each article at 20% above the cost price and sells 260 articles at the price. As for the remaining articles, he sells them at 50% of the selling price. Calculate the shopkeeper's total profit.

   1. $ 180
   2. $ 185
   3. $ 200
   4. $ 190
   5. $ 170

   cost price of each item = \( 1500 \over 300 \) = $ 5
   selling price at 20% above the cost price = 5 + 5 × .2 = $ 6
   selling price of 260 items = 260 × 6 = $ 1560
   selling price of remaining 40 items = 40 × 6 × .5 = $ 120
   Total profit = 1560 + 120 - 1500 = $ 180

9. 60% of 37 = ?

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

   60% of 37 = 0.6 × 37 = 22.2

10. 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%.

11. 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%

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

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

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

14. 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.

15. 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

16. A shopkeeper bought a radio from a wholesaler for $ 250.00. In addition, he paid a sales tax of 15% on the cost price. He then sold the radio for $ 315.00. Calculate the cash profit made by the shopkeeper.

    1. $ 20.00
    2. $ 22.50
    3. $ 25.00
    4. $ 27.50
    5. $ 27.00

    cost price = $ 250
    sales tax = .15 × 250 = $ 37.5
    cash profit = 315 - 250 - 37.5 = $ 27.5

17. 1015 / 0.05 / 40 = ?

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

    1015 / 0.05 / 40 = 20300 / 40 = 507.5

18. A rectangular room is 6 m long, 5 m wide and 4 m high. The total volume of the room in cubic meters is

    1. 24
    2. 30
    3. 120
    4. 240
    5. 140

    Total volume = length × width × height = 6 × 5 × 4 = 120

19. \( {1250 \over 25} × 0.5 = ? \)

    1. 250
    2. 50
    3. 2.5
    4. 25
    5. 125

    \( {1250 \over 25} × 0.5 = 50 × 0.5 = 25 \)

20. 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

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