# Mixed Quantitative Aptitude Questions Set 124

Directions(1-5): What approximate value will come in place of question mark ‘?’ in the following questions.

1. 25.6% of 250 + √? = 119
2110
2255
3025
2995
3030
Option C
250 × 25.6 100 + √? = 119
⇒ 64 + √? = 119
⇒ √? = 119 − 64 = 55
⇒ ? = 55 × 55 = 3025

2. [30% 𝑜𝑓 {(80% 𝑜𝑓 850) ÷ 34}] = ?
8
6
9
5
4
Option B
? = [ 30/100 × {( 80/100 × 850) ÷ 34}]
= [ 30/100 × {680 ÷ 34}] = [ 30/100 × 20] = 6

3. 65% 𝑜𝑓 240+ ?% 𝑜𝑓 150 = 210
52
61
44
58
36
Option E
65/100 𝑜𝑓 240 + ?/ 100 𝑜𝑓 150 = 210
⇒ 156 + 1.5 ×? = 210
=> ? = [210−156]/1.5 = 36

4. 1425 + 8560 + 1680 ÷ 200 = ?
7788
8595
9110
8800
9993
Option E
? = 1425 + 8560 + 1680 ÷ 200
= 1425 + 8560 + 1680/200
= 9985 + 8.4 = 9993.4 == 9993

5. ? % of 800 = 293 – 22% of 750
12
18
16
14
10
Option C
[800 ×?]/ 100 = 293 − 750 × 22/100
=> 8 × ? = 293 − 165 = 128
=> ? = 128/8 = 16

6. The retail price of a water geyser is Rs. 1265. If the manufacturer gains 10%, the wholesale dealer gains 15% and the retailer gains 25%, then the cost of the product is:
Rs. 800
Rs. 900
Rs. 600
Rs. 400
Rs. 500
Option A
Cost price = 100/110 × 100/115 × 100/125 × 1265
= Rs. 800

7. A pipe can fill a cistern in 6 hrs. Due to a leak in its bottom, it is filled in 7 hrs. When the cistern is full, in how much time will it be emptied by the leak?
47 hrs.
40 hrs.
50 hrs.
42 hrs.
38 hrs.
Option D
In one hour 1/ 6 – 1/ 7
= 1/ 42 of the cistern is empty.
The whole cistern will be emptied in 42 hrs.

8. Ram travels a certain distance at 3 km/h and reaches 15 minutes late. If he travels at 4 km/h, he reaches 15 minutes earlier. The distance he has to travel is:
6 km
7 km
8 km
9 km
4 km
Option A
Let D be the required distance
So, D/3 − D/4 = 15+15/60
D = 6 km

9. A sum is invested for 3 years at compound interest at 5%, 10% and 20% respectively. In three years, if the sum amounts to Rs. 16,632, then find the sum.
Rs.14,000
Rs.16,000
Rs.10,000
Rs.12,000
Rs.18,000
Option D
Let, P be the sum.
16632 = P (1 + 5/ 100) (1 + 10 /100) (1 + 20/ 100) 16632
16632 = P × 21/20 × 11/10 × 6/5
P = Rs.12,000

10. A father’s age is three times the sum of the ages of his two children, but 20 years hence his age will be equal to the sum of their ages. Then, the father’s age is:
10 years
30 years
50 years
40 years
20 years
Option B
Let the father’s age be x years and age of his children be a and b years (𝑎 + 𝑏) = 𝑥/3
And (𝑎 + 𝑏) + 20 + 20 = 𝑥 + 20
=> 𝑥/3 + 20 = 𝑥
=> 𝑥 = 30 years