# Mixtures and Allegations CAT

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Mixtures and Allegations in CAT comes under the topic of Ratios and Proportions. Though in CAT mixtures and Allegations is sub-topic, the number of questions asked from this topic is significantly good and we can’t ignore mixtures and Allegations problems in CAT.

Please go through CAT previous solved papers and also take an free mock for CAT to learn more on CAT Quantitative Aptitude.

You can download the Mixtures and Allegations CAT questions PDF or you can go through the details below.

Mixtures and Allegations CAT Questions with answers:

Question 1:

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

A. 2 : 3
B. 1 : 2
C. 1 : 3
D. 3 : 4

Question 2:

Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

A. 2 : 3
B. 4 : 3
C. 3 : 2
D. 3 : 4

Question 3:

I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

A. 90
B. 85
C. 100
D. 105

Question 4:

There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

A. A > B
B. A < B
C. A = B
D. Cannot be determined

Question 5:

Fresh grapes contain 90% water while dry grapes contain 20% water. What is the weight of dry grapes obtained from 20 kg fresh grapes?
A. 2 kg
B. 2.5 kg
C. 2.4 kg
D. None of these

Solutions for Mixtures and Allegations CAT:

Solutions:

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Fraction of A in contained 1 = $$\frac{5}{6}$$
Fraction of A in contained 2 = $$\frac{1}{4}$$
Let the ratio of liquid required from containers 1 and 2 be x:1-x
x($$\frac{5}{6}$$) + (1-x)($$\frac{1}{4}$$) = $$\frac{1}{2}$$
$$\frac{7x}{12}$$ = $$\frac{1}{4}$$
=> x = $$\frac{3}{7}$$
=> Ratio = 3:4

Let’s say number of coins are 2.5x , 3x and 4x
So total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210
So x = 42
And number of 1 rs. coins = 2.5x = 105

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,
Volume of alochol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of alcohol contained is $$\frac{3V}{500+3V}*V$$
Hence, after adding back 3 cups of the mixture, amount of alcohol in the first container is $$500-3V+\frac{9V^2}{500+3V} = \frac{500*500}{500+3V}$$
Amount of water contained in the second container is $$500 – \frac{3*500*V}{500+3V} = \frac{500*500}{500+3V}$$
So, the required proportion of alcohol in the first container and water in the second container are equal.

=> Weight of dry grapes = $$\frac{10}{0.8}$$ = 12.5 gm