BANK & INSURANCE (MIXTURE AND ALLIGATION) PART 2

Total Questions: 60

31. A 200 litre mixture contains 60% milk and rest water in it. At first, 50% of the mixture is replaced with 8 litres of water and then 36 litres of the resultant mixture is replaced with ‘x’ litres of milk such that ratio of quantity of milk to that of water in the final mixture becomes 11:8, respectively. Find the value of ‘x’.

Correct Answer: (c) 4
Solution:

Ratio of quantity of milk to that of water, in the initial mixture = 60 : 40 = 3 : 2

After replacing 50% of the mixture with 8 litres of water, quantity of milk in the mixture = (200 − 100) × (3/5) = 60 litres

After replacing 50% of the mixture with 8 litres of water, quantity of water in the mixture = (200 − 100) × (2/5) + 8 = 48 litres

So, ratio of quantity of milk to that of water, now = 60 : 48 = 5 : 4

After replacing 36 litres of the mixture with ‘x’ litres of milk, quantity of milk = [(60 + 48) − 36] × (5/9) + x = (40 + x) litres

After replacing 36 litres of the mixture with ‘x’ litres of milk, quantity of water = [(60 + 48) − 36] × (4/9) = 32 litres

So, (40 + x) : (32) = 11 : 8
320 + 8x = 352
x = (352 − 320) ÷ 8
So, x = 4

Hence, option c.

32. Container has certain quantity of milk in it. 20% of the milk gets replaced with same quantity of water and the same process gets repeated for two more times. Find the initial quantity of milk in the container if the quantity of milk in the final mixture becomes 102.4 litres

Correct Answer: (a) 200 litres
Solution:

Let the initial quantity of milk in the container be ‘5x’ litres

So, quantity of milk that has been taken out = 5x × 0.2 = ‘x’ litres

According to question:
102.4 = 5x × {1 − (x/5x)}³

102.4 = 5x × (64/125)

x = (102.4/64) × 25
x = 40

So, initial quantity of milk in the container = 5x
= 40 × 5 = 200 litres

Hence, option a.

33. If the average weight of 30% of the students of a class is increased by 20% and the average weight of remaining students is increased by 30%, then find the overall percentage increase in the average weight of the class.

Correct Answer: (b) 27% 
Solution:

Let the overall percentage increase in the average weight of the class be ‘x’
By alligation

20%           30%
..............................x
30%           70%
3            7

So, {(30 − x)/(x − 20)} = (3/7)
210 − 7x = 3x − 60
10x = 270
x = 27

So, overall percent increase in average weight of the class = 27%
Hence, option b.

34. A mixture contains milk and water in the ratio of 7:5 respectively. If 144 litres of this mixture is taken out and replaced with 72 litres of milk and 36 litres of water, the ratio of quantity of milk and water in the resultant mixture will be 11:7, respectively. Find the quantity of milk in the original mixture

Correct Answer: (b) 210 litres 
Solution:

Quantity of milk in 144 litres of mixture
= (7/12) × 144 = 84 litres

Quantity of water in 144 litres of mixture
= 144 − 84 = 60 litres

Let quantity of milk and water in the original mixture be ‘7x’ litres and ‘5x’ litres, respectively

So, {(7x − 84 + 72)/(5x − 60 + 36)} = (11/7)
49x − 84 = 55x − 264
6x = 180
x = 30

So, quantity of milk in the original mixture = 30 × 7 = 210 litres

35. Jar ‘A’ contains 75% milk and rest water whereas jar ‘B’ contains 60% milk and rest water. Both jars are of equal capacity and are full to its brim. The contents of both jars are emptied in a container that contains 50 litres water such that the quantity of milk in the container become 60% of total mixture in it. Find the initial quantity of milk in jar ‘A’.

Correct Answer: (b) 150 litres 
Solution:

Let the initial quantity of mixture in jar ‘A’ and jar ‘B’ be 100x litres each

Quantity of milk in jar ‘A’ = 100x × 0.75 = 75x litres
Quantity of water in jar ‘A’ = 100x × 0.25 = 25x litres

Quantity of milk in jar ‘B’ = 100x × 0.6 = 60x litres
Quantity of water in jar ‘B’ = 100x × 0.4 = 40x litres

Total quantity of milk in the two jars = 75x + 60x = 135x litres

Total quantity of mixture in the container = 135x ÷ 0.6 = 225x litres

ATQ:
225x − 135x = 40x + 25x + 50
90x = 65x + 50
25x = 50
So, x = 2

So, quantity of milk in jar ‘A’ = 75 × 2 = 150 litres
Hence, option b.

36. In what ratio a seller should mix two varieties of sugar costing Rs. 20 per kg and Rs. 36 per kg so that he earns a 50% profit on selling the mixture at Rs. 45 per kg?

Correct Answer: (c) 3:5
Solution:

Cost price of the mixture = 45 ÷ 1.5 = Rs. 30 per kg

Let the mixture costing Rs. 20 per kg be mixture ‘A’ and the mixture costing Rs. 36 per kg be mixture ‘B’.

Using alligation, we have:

A           B
Rs. 20/kg     Rs. 36/kg

      Rs. 30/kg

36 − 30 = 6               30 − 20 = 10

So, required ratio = 6 : 10 = 3 : 5
Hence, option c.

37. A mixture contains milk and water; in the ratio 8:11, respectively. If 40% of this mixture is replaced with same quantity of milk, then the quantity of milk in the resultant mixture will be 87 litres more than that of water. Find the quantity of the original mixture.

Correct Answer: (b) 285 litres 
Solution:

Let the original quantity of milk in the mixture = ‘8y’ litres

Then, original quantity of water in the mixture
= (11/8) × 8y = ‘11y’ litres

After replacing 40% of the mixture with equal quantity of milk,
Quantity of milk in the resultant mixture = 8y × 0.6 + (8y + 11y) × 0.4 = 4.8y + 7.6y = ‘12.4y’ litres

Quantity of water in the resultant mixture = 11y × 0.6 = ‘6.6y’ litres

According to the question,
12.4y − 6.6y = 87
Or, 5.8y = 87
So, y = 87 ÷ 5.8 = 15

So, original quantity of the mixture = 8y + 11y = 19y = 19 × 15 = 285 litres
Hence, option b.

38. Two vessels contain a mixture of milk and water. In the first vessel the ratio of milk to water is 7:3 and in the second vessel the ratio is 4:1. A 30 litre cask is filled from these vessels, taking x litres from vessel 1, so as to contain a mixture of milk and water in the ratio of 11:4. If x/2 litres of mixture from vessel 1 are mixed again with the contents of the cask, find the resulting ratio of milk to water in the cask.

Correct Answer: (e) None of these
Solution:

The capacity of the cask is 30 litres.
The ratio of milk to water in the cask is 11 : 4.

Thus, the cask contains 22 litres milk and 8 litres water.

According to the question:
7p + 4q = 22
3p + q = 8

p = 2
q = 2

Hence 2 × 10 = 20 litre are taken from the first vessel and 2 × 5 = 10 litres from the second vessel.

The value of x = 20

If 10 litres from vessel 1 are mixed again to the contents of the cask, the contents of cask become 29 litres milk and 11 litres water.

Hence, the ratio is 29 : 11
Hence, option e.

39. A packet of mother diary milk was found to be adulterated. It contains milk and water in ratio 7 : 5. If we take 9 liters of the mixture from the packet then the mixture is mixed with 9 liters of water then the ratio becomes 7 : 9. Calculate how many liters of milk was in the packet initially?

Correct Answer: (d) 21 liters 
Solution:

The packet initially contains 7x and 5x liters of milk and water respectively.

Milk is considered as A

Quantity of A in mixture left = (7x − (7/12) × 9) lit
= 7x − (21/4) liters

Quantity of B in mixture left = 5x − (5/12 × 9)
= 5x − (15/4)

Therefore, (7x − 21/4)/(5x − 15/4 + 9) = 7/9

(28x − 21)/(20x + 21) = 7/9

(252x − 189) = 140x + 147

112x = 336
x = 3

So, the packet contains 21 liters of A i.e the packet contains 21 liters of milk.
Hence, option d.

40. A street vendor has 2 containers of mixtures of milk and water. The first one contains 40% water and the rest milk. The second contains 60% water and the rest milk. How much mixture should be mixed from each of the cans so as to get 12 litres of mixture such that ratio of water to milk is 4:5?

Correct Answer: (a) 9.33 litres; 2.67 litres
Solution:Let x and 12-x liters of mixture be mixed from 1st and 2nd container respectively:
Amount of milk in 1st container = 0.6 x liters
Amount of water in 1st container = 0.4 x liters
Amount of milk in 2nd container = 0.4(12-x)
Amount of water in 2nd container = 0.6(12-x)
(0.4x + 0.6(12 - x))/(0.6x + 0.4(12 - x)) = 4/5
Solving it for x, we get x = 9.33 liters
And 12-x = 2.67 liters. Hence, option a.