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

# Solved Examples Set 1 (Quantitative Ability)

1. 8 : ? :: 1 : 4

1. 24
2. 16
3. 0
4. 32
5. 20
? × 1 = 8 × 4
? = 32
2. $$25 \text{% of }{4 \over 4\text{%}} \text{ of }{1 \over 25} = ?$$

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
3. 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
4. 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
5. 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 6. 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% 7. 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 8. 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 9. 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 10. 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
11. 42.98 + ? = 107.87

1. 64.89
2. 65.89
3. 64.98
4. 65.81
5. 63.89
? = 107.87 - 42.98 = 64.89
12. 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
13. 1015 / 0.05 / 40 = ?

1. 50.75
2. 507.5
3. 506
4. 2056
5. 5075
1015 / 0.05 / 40 = 20300 / 40 = 507.5
14. A boy scored 90 marks for his mathematics test. This was 20% more than what he had scored for the geography test. How much did he score in geography?

1. 71 marks
2. 73 marks
3. 75 marks
4. 77 marks
5. 78 marks
20% of x + x = 90
0.2x + x = 90
1.2x = 90
x = $$90 \over 1.2$$
x = 75
15. $${63.84 \over ?}$$ = 21

1. 3.04
2. 3.4
3. 30.4
4. 300.4
5. 0.304
? = $$63.84 \over 21$$ = 3.04
16. 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
$$\frac{120} {x} - \frac{120}{x + 3} = 4.5$$ $$120(x + 3) − 120x = 4.5x(x + 3)$$ $$360 = 4.5x(x + 3)$$ $$720 = 9x(x + 3)$$ $$80 = x(x + 3)$$ $$x^2 + 3x - 80 = 0$$ $$x = \frac{-3 \pm \sqrt{3^2-4 × (-80)}}{2} = \frac{-3 \pm \sqrt{329}}{2}$$ $$= \frac{-3 \pm 18.14}{2} = 7.57 \text{ (ignoring the negative value)}$$
17. $${1250 \over 25} × 0.5 = ?$$

1. 250
2. 50
3. 2.5
4. 25
5. 125
$${1250 \over 25} × 0.5 = 50 × 0.5 = 25$$
18. 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
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. A man earned an annual income of $245000 in 1990. He was allowed a deduction of$ 15000 relief for each of his three children and a personal relief of $30000. If he was charged a tax rate of 4% on first$ 50000 and 6% on his remaining income, calculate the total tax charged.

1. $9200 2.$ 8700
3. $9500 4.$ 9400
5. $9000 Total Income =$ 245000
Total relief = 3 × 15000 + 30000 = $75000 Rest income = 245000 - 75000 = 170000 Tax on 1st 50000 = 0.04 × 50000 =$ 2000
Tax on rest amount 120000 = 0.06 × 120000 = $7200 Total tax = 200 + 7200 =$ 9200