# Mixtures and Alligations Questions for IIFT PDF

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## Mixtures and Alligations Questions for IIFT PDF

Download important IIFT Mixtures and Alligations Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Mixtures and Alligations questions and answers for IIFT exam.

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Question 1: Consider three mixtures — 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) The same amount of water and liquid B

b) The same amount of liquids B and C

c) More water than liquid B

d) More water than liquid A

Question 2: 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) 251 : 163

b) 239 : 161

c) 220 : 149

d) 229 : 141

Question 3: 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) 2 : 3

b) 1 : 2

c) 1 : 3

d) 3 : 4

Question 4: 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 5: 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

Instructions

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

Question 6: Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?

a) 1.3

b) 1

c) 0.6

d) 2.3

Question 7: A manufacturer has 200 litres of acid solution which has 15% acid content. How many litres of acid solution with 30% acid content may be added so that acid content in the resulting mixture will be more than 20% but less than 25%?

a) More than 100 litres but less than 300 litres

b) More than 120 litres but less than 400 litres

c) More than 100 litres but less than 400 litres

d) More than 120 litres but less than 300 litres

e) None of the above

Question 8: Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?

a) 328 units

b) 368 units

c) 392 units

d) 616 units

e) None of the above

Question 9: Gopal sells fruit juice mixture using orange juice and pineapple juice. Gopal prepares this mixture by drawing out a jug of orange juice from a 10 litre container filled with orange juice, and replacing it with pineapple juice. If Gopal draws out another jug of the resultant mixture and replaces it with pineapple juice, the container will have equal volumes of orange juice and pineapple juice. The volume of the jug in litres, is

a) 2

b) < 2 and $\leq$ 2.5

c) 2.5

d) > 2.5 and $\leq$ 3

e) $\geq$ 3

Question 10: 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) 27:14

b) 27:13

c) 27:16

d) 27:18

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 $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{160}$
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 $\frac{149}{160}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C  is correct.

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 that 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, A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

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 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.

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

Let the volume of the solution with 30 % acid content lie between $v_1$ and $v_2$, where we get a 20% acid solution for $v_1$

For $v_2$, we get a 25 % acid solution as the resultant mixture.

=> $15 \% (200) + 30 \% (v_1) = 20 \% (200 + v_1)$

=> $30 + 0.3 v_1 = 40 + 0.2 v_1$

=> $0.1 v_1 = 10$ => $v_1 = 10 \times 10 = 100$ litres

Similarly, $15 \% (200) + 30 \% (v_2) = 25 \% (200 + v_2)$

=> $30 + 0.3 v_2 = 50 + 0.25 v_2$

=> $0.05 v_2 = 20$ => $v_2 = 20 \times 20 = 400$ litres

$\therefore$ For the acid content in the resultant mixture to lie between 20 % and 25 %, the volume of the 30 % concentration acid solution must lie between 100 litres and 400 litres.

Let the quantities of the chemicals X and Y, mixed to produce product M be $5c$ and $4c$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $\frac{3}{4} \times 5c = \frac{15 c}{4}$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $\frac{2}{3} \times 4c = \frac{8 c}{3}$

Quantity of B in M = $\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $9c = 864$ => $c = \frac{864}{9} = 96$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$

$\therefore$ Quantity of water added to M = $1232 – 864 = 368$ units

Let volume of jug = $v$ litre

After first replacement, volume of orange juice = $(10 – v)$ litre

Volume of pineapple juice = $v$ litre

After second replacement, volume of orange juice remaining

= $(10 – v) – (\frac{10 – v}{10} v) = \frac{(10 – v)^2}{10}$

Volume of pineapple juice remaining = $v – \frac{v^2}{10} = \frac{v (10 – v)}{10}$

Total volume of pineapple juice = $\frac{v (10 – v)}{10} + v = \frac{20v – v^2}{10}$

It is given that container has equal volumes of both juices.

=> $\frac{(10 – v)^2}{10} = \frac{20v – v^2}{10}$

=> $v^2 – 20v + 50 = 0$

=> $v = 17.07 , 2.93$

$\because$ Container is 10 litres, => $v \neq 17.07$

$\therefore v = 2.93$ litres

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, $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$
Therefore, $91X+81=78X+108$

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

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