# CAT Mixture and Alligation Questions With Video Solutions [PDF]

Mixture and Alligation is one of the most important topics in the CAT Quant Section. You can check out these Mixture and Alligation questions in the **CAT Previous year’s papers. **If you want to learn the basics, you can watch these videos on **Mixture and Alligation**. This article will look into some important Mixture and Alligation Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Mixture and Alligation Most Important Questions PDF below, which is completely Free.

Download Mixture and Alligation Questions for CAT

**Question 1:Â **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?

[CAT 2004]

a)Â 2 : 3

b)Â 1 : 2

c)Â 1 : 3

d)Â 3 : 4

**1)Â AnswerÂ (A)**

**Solution:**

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.

**Question 2:Â **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)Â 2 : 3

b)Â 4 : 3

c)Â 3 : 2

d)Â 3 : 4

**2)Â AnswerÂ (D)**

**Solution:**

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2Â = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) =Â $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

**Question 3:Â **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)Â A > B

b)Â A < B

c)Â A = B

d)Â Cannot be determined

**3)Â AnswerÂ (C)**

**Solution:**

Let the volume of the cup be V.

Hence, after removing three cups of alcohol from the first container,

Volume of alcohol 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 water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V} $

Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.

**Question 4:Â **Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

**4)Â AnswerÂ (B)**

**Solution:**

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4

Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$

The total volume of water is $\frac{2X}{9}+\frac{4}{13}$

They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$

Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

**Question 5:Â **Consider three mixtures â€” the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a)Â The same amount of water and liquid B

b)Â The same amount of liquids B and C

c)Â More water than liquid B

d)Â More water than liquid A

**5)Â AnswerÂ (C)**

**Solution:**

The proportion of water in the first mixture is $\frac{1}{3}$

The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$

The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$

The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$

The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$

The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$

The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C Â is correct.

**Question 6:Â **A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

a)Â 96

b)Â 98

c)Â 86

d)Â 84

**6)Â AnswerÂ (A)**

**Solution:**

Let the price of peanuts be Rs. 100x per kg

Then, the price of walnuts = Rs. 300x per kg

Cost price of peanuts for the shopkeeper = Rs. 110x per kg

Cost price of walnuts for the shopkeeper = Rs. 360x per kg

Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x

Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts

Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656

Profit = 25%

So, cost price = Rs. 2124.80

Therefore, 6640x = 2124.80

On solving, we get x = 0.32

Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.

Hence, option A is the correct answer

**Question 7:Â **A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

a)Â 16

b)Â 26

c)Â 20

d)Â 22

**7)Â AnswerÂ (C)**

**Solution:**

Let the price of paint B be x.

Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture.

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%.

=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B.

When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.

The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal.

=> (x+x+8)/2 = 24

2x = 40

x = Rs. 20

Therefore, optionÂ C is the right answer.

**Question 8:Â **A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

a)Â 30%

b)Â 40%

c)Â 50%

d)Â 60%

**8)Â AnswerÂ (C)**

**Solution:**

Let the volume of the first and the second solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture

= (20 + 300S)

Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final solution

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = $\frac{1}{2}$ = 50%

Hence, 50 is the correct answer.

**Question 9:Â **A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

a)Â 30.3

b)Â 35.2

c)Â 25.4

d)Â 20.5

**9)Â AnswerÂ (B)**

**Solution:**

Final quantity ofÂ alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml

Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %

**Question 10:Â **There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

a)Â 251 : 163

b)Â 239 : 161

c)Â 220 : 149

d)Â 229 : 141

**10)Â AnswerÂ (B)**

**Solution:**

It is given that in drum 1, A and B are in the ratio 18 : 7.

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given thatÂ drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A

$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$

$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$

$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$

$\Rightarrow$ $x = \dfrac{239}{161}$

Therefore, we can say that in drum 2,Â A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.