BANK & INSURANCE (MIXTURE AND ALLIGATION) PART 1

Total Questions: 60

11. A vessel contains milk and water in which water is 20%. 20 liters of mixture was taken out and replaced by water and the ratio becomes 12:13. Find the initial quantity of milk in the vessel

Correct Answer: (a) 40 liters 
Solution:

Milk and water ratio = 4:1
Given,
(4x – 16)/(x + 20) = 12/13
13x – 52 = 3x + 48
⇒ 10x = 100 ⇒ x = 10
Initial quantity of milk in the vessel = 40 liters

12. A container contains pure milk. From these 4 liters of milk taken out and replaced by water. This process is repeated for one more time and the remaining milk in the container is 12.8 liters. What is the initial quantity of milk in the container?

Correct Answer: (b) 20 liters 
Solution:Let us take initial quantity of a container be x
Remaining milk = Initial (1 – Replaced/Initial)ⁿ
12.8 = x (1 – 4/x)²
12.8x = x² + 16 – 8x
5x² – 104x + 80 = 0
Simplify the above equation, we get x = 20 and 0.8 (Eliminate)

13. A trader mixes two types of rice varieties with a cost of Rs.36 and Rs.42. If he sells the mixture of Rs.44 at 10% profit, in what ratio he mixes two types of rice varieties?

Correct Answer: (a) 1:2 
Solution:

CP of Mixture = 44/110 × 100 = Rs.40

Rs.36       Rs.42
...................Rs.40
2                       4

Required ratio = 1:2

14. How many kg of rice variety 1 costing Rs.48/kg should a shopkeeper mix with 20 kg of rice variety 2 costing Rs.56 per kg so that he makes a profit of 20% on selling the mixture at Rs.62.4/kg?

Correct Answer: (d) 20 kg 
Solution:

CP of mixture = 62.4/120 × 100 = Rs.52

Rs.48      Rs.56
...................Rs.52
4                   4

Required ratio = 4:4 = 1:1
Required kg = 20/1 × 1 = 20 kg

15. How many liters of water should be added to a 60 liters mixture containing milk and water in the ratio of 2:1 such that the resultant mixture has 50% milk in it?

Correct Answer: (b) 20 liters 
Solution:Total mixture = 60 liters
Milk and water in the ratio of 2:1,
So, Milk = 40 lit, water = 20 lit
Given,
40/(20 + x) = 1/1
40 = 20 + x ⇒ x = 20 liters

16. A vessel contains 40 liters of pure wine. Find the amount of water mixing if he sells the mixture at cost price and gains 20%

Correct Answer: (c) 10 liters 
Solution:He sells the mixture at cost price and gains 20%.
Here gain will be the water. So,
80% of mixture = 40 liters
⇒ Mixture = 40/80 × 100 = 50 liters
Amount of water added = 50 – 10 = 10 liters

17. A bucket contains 60 liters of pure wine, in which x liters drawn off and replaced with water. This process is repeated for two times. Find the value of x if the final ratio of wine to water is 81:19

Correct Answer: (b) 6 liters 
Solution:Given,
Wine and water in final process (60/100 × 81 = 48.6)
and (60/100 × 19 = 11.4) respectively
48.6 = 60 (1 – x/60)²
48.6/60 = (1 – x/60)²
81/100 = (1 – x/60)²
9/10 = 1 – x/60
x/60 = 1/10 ⇒ x = 6 liters

18. Two equal vessels A and B contain 60% of sugar and 40% of sugar respectively and remaining Rava. In which 40 kg of mixture is taken out from vessel A and replaced into vessel B. Find the initial quantity of vessel if the final ratio of sugar and Rava in vessel B is 16:19

Correct Answer: (d) 100 liters 
Solution:Vessel A Sugar and Rava ratio = 3:2
Vessel B sugar and Rava ratio = 2:3
Given,
(2x + 24)/(3x + 16) = 16/19
38x + 456 = 48x + 256
10x = 200 ⇒ x = 20 liters
Initial quantity = x × 5 = 100 liters

19. A bucket contains some quantity of milk and water, in the ratio of water and milk is 3:5. 40 liters of mixture is drawn out and replaced with water and the ratio of milk and water becomes 5:11 then find the initial quantity of milk

Correct Answer: (b) 50 liters 
Solution:(5x – 25)/(3x – 15 + 40) = 5/11
55x – 275 = 15x + 125
⇒ x = 10 liters
Initial quantity of milk = 10 × 5 = 50 liters

20. There are two vessels A and B contains wine and water in the ratio of 3:2 and 4:1 respectively. In what ratio of wine and water should be mixed if the ratio of wine and water is 5:2.

Correct Answer: (e) None of these
Solution:

3/5        4/5
.....................5/7

(4/5 – 5/7)   :        (5/7 – 3/5)
⇒ 3:4

(or)

2/5        1/5

       2/7

(1/5 − 2/7) : (2/7 − 2/5)
⇒ 3:4

Required ratio = 3:4