# Mixed Quantitative Aptitude Questions Set 154

1. Two partners invest Rs 12500 and Rs 8500 respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be divided in the ratio of their capital. If one partner gets Rs 240 more than the other, find the total profit made in the business.

3113
3000
3500
3300
3150
Option E
(Profit of A) : (Profit of B) = 12,500 : 8,500 = 125 : 85 = 25 : 17
40% of total profit = 240 Ã— (25 + 17)/(25 âˆ’ 17) = 1260
100% profit = 1260/40 Ã— 100 = 3150

2. Aman invests a certain sum in scheme A at compound interest (compounded annually) of 10% per annum for 2 years. In scheme B he invests at simple interest of 8% per annum for 2 years. He invests in schemes A and B in the ratio of 1 : 2.The difference between the interest earned from both the schemes is Rs 990. Find the amount invested in scheme A.
8900
7890
9000
7007
9008
Option C
Let in both schemes he invested Rs. P and 2P respectively.
ATQ, |P [(1 + 10/100)^2 âˆ’ 1] â€“ (2P Ã— 8 Ã— 2)/100 | = 990
â‡’ | 21P/100 âˆ’ 32P/100| = 990
â‡’ P = 99000/11
â‡’ P = 9000

3. Xâ€™s age 3 years ago was three time the present age of Y. At present, Zâ€™s age is twice the age of Y. Also, Z is 12 years younger than X. What is the present age of Z?
10
15
14
18
12
Option D
Let present ages of all the three are X, Y and Z respectively.
X = 3Y + 3 â€¦(i)
Z = 2Y â€¦(ii)
X = Z + 12 â€¦(iii)
From equations (i), (ii) and (iii)
X â€“ 3Y = 3 and X â€“ 2Y = 12
After solving these two resultant equations, we get
Y = 9 years
Zâ€™s present age = 18 years.

4. A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour?
10/97
13/92
11/90
15/91
17/93
Option D
Required probability = 6C2 Ã— 5C1/ 15C3 = 15/91

5. A bag contains 5 red balls, 6 yellow and 3 green balls. If two balls are picked at random, what is the probability that both are red or both are green in colour?
1/3
1/4
1/7
1/9
1/2
Option C
Required probability = 5C2/ 14C2 + 3C2/14C2 = 10/91 + 3/91 = 13/91 = 1/7

6. Directions(6-10): Find the missing term ‘?’ of the following series.

7. 282, 286, 302, ?, 402, 502
300
317
338
320
313
Option C
+(2)^2, +(4)^2, +(6)^2, +(8)^2, +(10)^2
? = 338

8. 2187, 729, 243, 81, 27, 9, ?
4
6
5
3
7
Option D
/3,/3,/3,/3,/3,/3
? = 3

9. 384, 381, 372, 345, 264, ?
21
29
23
27
25
Option A
-3,-9,-27,-81,-243
? = 21

10. 5, 9, 18, 34, 59, 95, ?
120
144
163
135
170
Option B
+2^2,+3^2,+4^2,+5^2,+6^2,+7^2
? = 144

11. 8, 15, 36, 99, 288, ?
800
891
828
855
885
Option D
+7,+21,+63,+189,+567
? = 855

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