Mixed Quantitative Aptitude Questions Set 152

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

1. 36% of 245 − 40% of 210 = 10 − ?
4.4
3.2
6.5
5.8
2.9
Option D
36% of 245 − 40% of 210 = 10 − ?
=>? = 10 – 88.2 + 84 = 5.8

2. 3( 1/7 ) + 2( 3/5) + 𝟕( 1/5) – 5(3/7)− 18/35 = 35/?
3
7
2
5
4
Option D
3( 1/7 ) + 2( 3/5) + 𝟕( 1/5) – 5(3/7)− 18/35 = 35/?
35/? = 263/35 – 18/35
=> ? = 35×35/245 = 5

3. √1225 ÷ (343)^1/3 × 45% of 760 = ?
1727
1820
1893
1989
1710
Option E
√1225 ÷ (343)^1/3 × 45% of 760 = ?
? = 35 ÷ 7 × 342 = 1710

4. 175% of 460 + 110% of 170 + 2^? = 1000
7
2
1
5
3
Option E
175% of 460 + 110% of 170 + 2^? = 1000
=>2^? = 1000 − 805 − 187 = 8
=> ? = 3

5. 18^7.9 × 3^0.1 × 6^0.1 ÷ (3^4 × 6^4) = 18^?
4
2
3
1
5
Option A
18^7.9 × 3^0.1 × 6^0.1 ÷ (3^4 × 6^4) = 18^?
=>18^? = 18^7.9 × 18^0.1 ÷ 18^4
=> ? = 4

6. Cost price of article A is double then that of article B and shopkeeper mark up both the article 20% more than the cost price. If at the time of sale shopkeeper gave Rs. 9 discount and earn 17% profit on total. Find the cost price of article A?
263
368
289
345
200
Option E
Let C.P. of A = 2x
C.P. of B = x
Total cost price = 3x
Mark up price = 3x × 12/10 = 3.6x
ATQ, 3.6x – 9 = 3x × 1.17
⇒ 3.6x – 3.51x = 9
⇒ 0.09x = 9
⇒ x = 100
C.P. of article A = 200

7. Ravi invested Rs 18000 in scheme ‘A’ which offers 15% p.a. at simple interest and Rs 15000 in scheme ‘B’ which offers 18% p.a at compound interest. Find the difference between the interest earn from these two schemes after two years?
400
440
450
452
486
Option E
Interest earn from scheme ‘A’ = 18000 × 15 × 2/100 = 5400
Interest earn from Scheme ‘B’ = 15000 [(1 + 18/100)^2 − 1] = 15000 [ 3924/10000] = 5886
Required difference = 5886 – 5400 = 486

8. A tank can be filled with water by two pipes A and B together in 36 minutes. If the pipe B was closed after 30 minutes, the tank is filled in 40 minutes. The pipe B can alone fill the tank in?
50
70
60
80
90
Option E
In 30 min the part of the tank will be filled by both tap = 30/36 = 5/6
Required tap = 1 – 5/6 = 1/6
1/ 6 part of the tank will be filled by tank A in 10 min.
∴ tap A will take 60 min.
∴ tap B will take time to fill the tank = 1/36 – 1/60 = 1/90
Required time = 90 min.

9. 40 men, working 8 hours a day can do a piece of work in 15 days. Find the number of days in which second group of 60 men working 4 hrs a day can do twice the work. Assume that 3 men of the first group do as much work in 2 hour as 4 men of the second group do in 3 hrs.
70
60
80
50
100
Option C
Let efficiency of men of first group is M1 and second group is M2
3M1 × 2 = 4M2 × 3
M1 = 2M2
work = 40M1 × 8 × 15
According to question,
40M1 × 8 × 15 × 2 = 60M2 × 4 × d
d = 80 days

10. An alloy of copper and aluminum has 40% copper. Another alloy of Copper and Zinc has Copper and Zinc in the ratio 2: 7. These two alloys are mixed in ratio 5:3. Quantity of aluminum is what percent more/less than the quantity of copper in final alloy.
14.4%
13.8%
12.5%
11.6%
10.9%
Option C
Let quantity → 5x and 3x
Quantity of copper in final alloy = 2/5 × 5x + 2/9 × 3x = 8x/3
Quantity of Aluminium = 3/5 × 5x = 3x
Required percentage = [3x – 8x/3]/ 8x/3 × 100 = 12.5%