In the name of ALLAH, the most beneficient, the most merciful

# Solved Examples Set 1 (Quantitative Ability)

1. A man sells two houses for $2 lac each. On one he gained 20% and on the other he lost 20%. His total profit or loss % in the transaction will be 1. 4% profit 2. 5% loss 3. no profit, no loss 4. 4% loss 5. 3% loss % loss = (% loss X % profit)/100 = (20 X 20)/100 = 4% 2. Which of the following is the largest? 1. half of 30% of 280 2. one-third of 70% of 160 3. twice 50% of 30 4. three times 40% of 40 5. 60% of 60 Let us calculate the value of each: A. 0.5 × 0.3 × 280 = 42 B. 0.33 × 0.7 × 160 = 36.96 C. 2 × 0.5 × 30 = 30 D. 3 × 0.4 × 40 = 48 E. 0.6 × 60 = 36 3. 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 $$km \over hr$$ 4. 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 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. 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 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. $${2244 \over 0.88} = ? × 1122$$ 1. 20.02 2. 20.2 3. 19.3 4. 2.27 5. 3.27 $${2244 \over 0.88} = ? × 1122$$ $$? = {2550 \over 1122} = 2.27$$ 9. 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 10. 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
11. 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.
12. If 4a + 2 = 10, then 8a + 4 =

1. 5
2. 16
3. 20
4. 24
5. 28
One may answer this question by solving
4a + 2 = 10
4a = 8
a= 2
Now, plugging in 2 for a:
8a + 4 = 8(2) + 4 = 20
A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20.
13. 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 = $108 per kg 14. By selling a fan for$ 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
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 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
17. 350 × ? = 4200

1. 12
2. 24
3. 15
4. 30
5. 16
$$? = {4200 \over 350} =12$$
18. A girl is 18 years younger than her mother. In 6 years time, the sum of their ages will be 54.How old is the girl now?

1. 10 years
2. 11 years
3. 12 years
4. 13 years
5. 14 years
Let girl's age = x
then mother's age = x + 18
After 6 years,
x + 6 + x + 18 + 6 = 54
2x + 30 = 54
2x = 24
x = 12
19. 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.
20. 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