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

Solved Examples Set 1 (Quantitative Ability)

1. 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
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. In the series 8, 9, 12, 17, 24 . . . the next number would be

1. 29
2. 30
3. 33
4. 35
5. 41
In the series, 8, 9, 12, 17, 24 . . .
9 − 8 = 1
12 − 9 = 3
17 − 12 = 5
24 − 17 = 7
Hence, the difference between the next term and 24 must be 9 or
x − 24 = 9, and
x = 33
Hence, the next term in the series must be 33
4. 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
5. 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.
6. 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 $$km \over hr$$
7. 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
8. $${1250 \over 25} × 0.5 = ?$$

1. 250
2. 50
3. 2.5
4. 25
5. 125
$${1250 \over 25} × 0.5 = 50 × 0.5 = 25$$
9. 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
10. A man sells two houses for $2 lac each. On one he gained 20% and on the other he lost 20%. His total profit or loss % in the transaction will be 1. 4% profit 2. 5% loss 3. no profit, no loss 4. 4% loss 5. 3% loss % loss = (% loss X % profit)/100 = (20 X 20)/100 = 4% 11. 72 + 679 + 1439 + 537+ ? = 4036 1. 1309 2. 1208 3. 2308 4. 2423 5. 1309 72 + 679 + 1439 + 537+ ? = 4036 2727 + ? = 4036 ? = 4036 - 2727 = 1309 12. If 4a + 2 = 10, then 8a + 4 = 1. 5 2. 16 3. 20 4. 24 5. 28 One may answer this question by solving 4a + 2 = 10 4a = 8 a= 2 Now, plugging in 2 for a: 8a + 4 = 8(2) + 4 = 20 A faster way of solving this is to see the relationship between the quantity 4a + 2 (which equals 10) and 8a + 4. Since 8a + 4 is twice 4a + 2, the answer must be twice 10, or 20. 13. 1015 / 0.05 / 40 = ? 1. 50.75 2. 507.5 3. 506 4. 2056 5. 5075 1015 / 0.05 / 40 = 20300 / 40 = 507.5 14. A person's net income is$ 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
15. 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
16. 350 × ? = 4200

1. 12
2. 24
3. 15
4. 30
5. 16
$$? = {4200 \over 350} =12$$
17. 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
18. 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 $$\frac{2}{2 + 2 + 2 + 3} = \frac{2}{9}$$.
19. 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%
20. $$\frac{\frac{7}{10} × 14 × 5 × \frac{1}{28}}{\frac{10}{17} × \frac{3}{5} × \frac{1}{6} × 17} =$$

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