# Mixed Quantitative Aptitude Questions Set 122

Directions(1-5): Find the missing term ‘?’ of the following questions.

1. 418, 208, 102, 48, ?, 5
10
30
40
20
50
Option D
*2+2
*2+4
*2+6
*2+8
*2+10
? = 20

2. 180, 191, 193, 203, 206, ?
250
215
241
200
232
Option
+11
+2
+10
+3
+9
? = 215

3. 3, 4, 9, 28, ?, 566
113
100
120
111
119
Option A
*1+1
*2+1
*3+1
*4+1
*5+1
? = 113

4. 153, 155, 160, 170, 187, ?
200
213
192
220
225
Option B
Difference of difference
? = 213

5. 3, 1.5, 1.5, 3, 12, 96, ?
1440
1123
1536
1626
1330
Option C
*0.5
*1
*2
*4
*8
*16
? = 1536

6. Two vessels A and B contains milk and water mixed in the ratio 8:5 and 5:2 respectively. Find the ratio in which these two mixture be mixed to get a new mixture containing 69(3/13) % milk.
2:7
8:9
5:6
4:5
1:3
Option A
Ratio = 2/91:1/13 = 2:7

7. A shopkeeper buys 144 items at 1.8 Rs each. But later he realized that 13(8/9)% of the total items are defected and could not be sold. He sells the remaining at Rs. 2.4 each. What is his overall gain percentage?
12(20/23)%
13(15/18)%
9(19/23)%
10(11/21)%
14(22/27)%
Option E
Total CP = 1.8 × 144 = Rs. 259.2
Total SP = ( 100– 125/900)*144*2.4 = Rs.297.6
Gain percentage = ( 297.6– 259.2)/259.6*100 = 14(22/27)%

8. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27kg of sugar costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per kg?
55 kg
42 kg
63 kg
59 kg
60 kg
Option C
Mean price = 10/ 110 ×9.24 = 10×0.84 = 8.4
9 ===== 7
==8.4
1.4====0.6
Ratio = 7:3
Required Quantity = 27/3*7 = 63 kg

9. The CP of two dozen mangoes is Rs 32 , after selling 18 mangoes at 12 Rs per dozen ,the shopkeeper reduced the rate as Rs 4 per dozen. Then find the loss percentage ?
33.3%
37.5%
41.6%
31.2%
29.3%
Option B
Total CP = 32
Total SP = 12 + 6 + 2 = 20
Loss percentage = 12/32 ×100 = 37.5%

10. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the current.
5 kmph
1 kmph
2 kmph
4 kmph
3 kmph
Option B
Let downstream speed = x
Upstream speed = y 15/ 𝑥 = 3(45/60)
=>15/𝑥 = 15 4
=>𝑥 = 4
5/ 𝑦 = 2(30/60)
=>5/ 𝑦 = 5/ 2
=> 𝑦 = 2
Speed of current = 1 kmph