BANK & INSURANCE (MIXTURE AND ALLIGATION) PART 2

Total Questions: 60

21. In a college lab, two vessels contain mixture of acid and water, the ratio of acid and water in the first vessel and second vessel is 3 : 5 and 3 : 2 respectively. Find the ratio in which the quantity of these vessels be mixed such that the new mixture has acid and water in the ratio 5 : 7.

Correct Answer: (b) 22 : 5
Solution:

Applying allegation on the proportion of acid, we get:

A      B
1/5     3/5

   5/12

11/60   1/24

Hence, the required ratio = 11/60 : 1/24 = 22 : 5

22. Mixture ‘A’ contains only acid and water in the ratio 11:4, respectively. 100 litres of mixture ‘B’ (acid + water) contains 30 litres more acid than water in it. If mixtures ‘A’ and ‘B’ are mixed together, then acid forms 70% of the resultant mixture. Find the quantity of water in mixture ‘A’.

Correct Answer: (c) 40 litres
Solution:

Let the quantity of acid in mixture ‘A’ = ‘11y’ litres
Then, quantity of water in mixture ‘A’ = (4/11) × 11y = ‘4y’ litres

Let the quantity of water and acid in mixture ‘B’ be ‘x’ litres and (x + 30) litres, respectively.

So, x + x + 30 = 100
2x + 30 = 100
x = (100 − 30) ÷ 2 = 35

So, quantity of water and acid in mixture ‘B’ is 35 litres and 65 litres, respectively.

According to the question,
(11y + 35 + 30) ÷ (11y + 35 + 30 + 4y + 35) = 0.7

(11y + 65) = (15y + 100) × 0.7
11y + 65 = 10.5y + 75
0.5y = 5 − 10
So, y = 5 ÷ 0.5 = 10

So, quantity of water in mixture ‘A’ = 4 × 10 = 40 litres
Hence, option c.

23. Amit has a 70-litre mixture of milk and wine containing 60% wine. He wants to add some milk to equally balance the quantities of milk and wine however, instead of adding more milk he adds same quantity of wine. Find the ratio of the quantities of wine to that of milk in the mixture.

Correct Answer: (c) 2:1
Solution:

Present quantity of wine in the mixture = 70 × 0.6 = 42 litres
Present quantity of milk in the mixture = 70 − 42 = 28 litres

Quantity of milk to be added = 42 − 28 = 14 litres

Required ratio = (42 + 14) : 28 = 2 : 1
Hence, option c.

24. A shopkeeper bought ‘x - 44’ kg of rice at Rs. 60 per kg and 44 kg of rice at Rs. ‘x’ per kg and mixed them. He sold the mixture at Rs. 92.3 per kg and thus making a profit of 30%. Find the value of ‘x’.

Correct Answer: (a) 80 
Solution:

Cost price of one kg of mixture = (92.3 / 1.30) = Rs. 71

ATQ
[(60 × (x − 44) + (44 × x)) / (x − 44 + 44)] = 71

60x − 2640 + 44x = 71x
33x = 2640
So, x = 80

25. A mixture of milk, water and honey contains milk and water in the ratio 7:11 respectively while quantity of honey in it is 4 litres more than that of water. If 14 litres of honey is added, then the quantity of honey becomes twice that of milk. Find the original quantity of the mixture.

Correct Answer: (d) 178 litres 
Solution:

Let the quantity of milk in the mixture = 7y litres
Then, quantity of water = 7y × (11/7) = 11y litres

Quantity of honey in the mixture = (11y + 4) litres

According to the question,
11y + 4 + 14 = 7y × 2
11y + 18 = 14y
3y = 18
So, y = 18 ÷ 3 = 6

And so, original quantity of the mixture = 7y + 11y + 11y + 4 = 29y + 4 = 29 × 6 + 4 = 178 litres

Hence option d.

26. A 200 litre mixture of water and acid contains 24 litres more water than acid. If 25% of the mixture is replaced with ‘k’ litres of acid, then quantity of water in the resultant mixture would be 20% more than that of acid in it. Find the value of ‘k’.

Correct Answer: (a) 4 
Solution:

Let the original quantity of acid in the mixture

= ‘y’ litres
Then, original quantity of water in the mixture
= (y + 24) litres

We have, y + y + 24 = 200
2y + 24 = 200
So, y = (200 − 24) ÷ 2 = 88 litres

And so, original quantities of acid and water in the mixture are 88 litres and 112 litres, respectively

After removing 25% of the mixture,
Remaining quantity of water = 112 × 0.75 = 84 litres
Remaining quantity of acid = 88 × 0.75 = 66 litres

Final quantity of acid = 84 ÷ 1.2 = 70 litres
So, k = 70 − 66 = 4

27. Can ‘A’ and Can ‘B’ contain pure milk and pure water, respectively. 60% of content from Can ‘A’ is poured into Can ‘B’. Now 40% of mixture formed in Can ‘B’ is poured back into Can ‘A’. If the ratio of quantity of milk to that of water in Can ‘A’ is now 4:5, then find the ratio of quantity of contents in Can ‘A’ and Can ‘B’, respectively.

Correct Answer: (c) 1:2
Solution:

Let the quantity of milk in can ‘A’ be ‘100x’ litres and quantity of water in can ‘B’ be ‘100y’ litres.

So, quantity of mixture in can ‘B’ when 60% of content from can ‘A’ is poured into can ‘B’ = (100y + 60x) litres

And, quantity left in mixture ‘A’ = 100x − 60x = ‘40x’ litres

So, quantity of milk in can ‘A’ when 40% of mixture in can ‘B’ is poured back into can ‘A’ = (0.4 × 100x) + (0.4 × 60x) = 40x + 24x = ‘64x’ litres

Quantity of water in can ‘A’ = 0.4 × 100y = ‘40y’ litres

ATQ;
(64x/40y) = (4/5)
Or, (x/y) = (1/2)

So, required ratio = 1 : 2
Hence, option c.

28. A jar contains a mixture of two liquids ‘P’ and ‘Q’ in the ratio 8:7, respectively. 30 litres of this mixture is replaced with 12 litres of liquid ‘B’ such that ratio of quantities of liquid ‘P’ and ‘Q’ in the final mixture becomes 8:9, respectively. Find the quantity of liquid ‘B’ in the final mixture.

Correct Answer: (b) 54 litres 
Solution:

Let initial quantity of liquid ‘P’ and ‘Q’ in the jar be ‘8x’ litres & ‘7x’ litres, respectively.

Quantity of liquid ‘P’ taken out = (8/15) × 30 = 16 litres
Quantity of liquid ‘Q’ taken out = (7/15) × 30 = 14 litres

Quantity of liquid ‘P’ in the final mixture = (8x − 16) litres
Quantity of liquid ‘Q’ in the final mixture = 7x − 14 + 12 = (7x − 2) litres

Given, (8x − 16)/(7x − 2) = (8/9)
72x − 144 = 56x − 16
16x = 128
x = 8

So, the quantity of liquid ‘B’ in the final mixture = 7x − 2 = 7 × 8 − 2 = 54 litres

29. A 208 litre mixture of petrol and kerosene contains 32 litres more kerosene than petrol. If half of the mixture was replaced with ‘x’ litres of petrol, then ratio of quantity of petrol to kerosene would become 5:6 respectively. Find the value of ‘x’.

Correct Answer: (b) 6 
Solution:

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

Then, original quantity of kerosene = (y + 32) litres

So, y + y + 32 = 2y + 32 = 208
So, y = (208 − 32) ÷ 2 = 88

Original quantities of petrol and kerosene in the mixture were 88 litres and 120 litres

After removing half of the mixture,
Quantity of petrol left = 88 − 88 × 0.5 = 44 litres

Quantity of kerosene left = 120 − 120 × 0.5 = 60 litres

According to the question,
(44 + x) : (60) = 5 : 6

So, (44 + x) = 60 × (5/6) = 50
Therefore, x = 50 − 44 = 6

Hence, option b.

30. A mixture (milk + water) of 108 litres contains milk and water in the ratio 7:5, respectively. 48 litres of mixture is replaced with 15 litres of water. Find the percentage of quantity of water in the final mixture.

Correct Answer: (b) (160/3)% 
Solution:

Initial quantity of Milk = (7/12) × 108 = 63 litres
Initial quantity of Water = (5/12) × 108 = 45 litres

Final quantity of Milk = 63 − (7/12) × 48 = 63 − 28 = 35 litres

Final quantity of Water = 45 − (5/12) × 48 = 45 − 20 + 15 = 40 litres

The percentage of quantity of water in the final mixture = [40/(40 + 35)] × 100 = (160/3)%

Hence, option b.