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Alligation or Mixture Questions and Answers

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41.

Find the ratio in which wheat of inferior quality (Rs.14/kg) be mixed with wheat of superior quality (Rs.28/kg) so that the shopkeeper gains Rs.2 by selling the resulting mixture at Rs.20/kg.

Answer: B

Let the resulting mixture be 1 kg, and x kg be the amount of wheat of inferior quality.
Therefore, (1-x)kg is the amount of wheat of superior quality.
As the shopkeeper gains Rs.2, the cost of the mixture is Rs.18
14*x + 28*(1-x) = 18
14x - 28x + 28 = 18
14x = 10
x = 5/7
(1 - x) = 2/7
x : (1-x) = 5/7 : 2/7
= 5 : 2

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42.

Vessel ‘A’ contains 150 litres of mixture of milk and water in the ratio of 3:1. Vessel ‘B’ contains 200 litres of the mixture in the ratio of 5:2. If the mixture from both the vessels are poured into vessel ‘C’, find the ratio of the resulting mixture.

Answer: D

As vessel A contains 150 litres of mixture of milk and water in the ratio 3:1, the quantity of milk is 3*150/4 litres, and that of water is 150/4 litres.
Similarly, as vessel B contains 200 litres of mixture of milk and water in the ratio 5:2, the quantity of milk is 5*200/7, and that of water is 2*200/7 litres.
After mixing the total quantity is 350 litres.
The quantity of milk is {(3*150/4) + (5*200/7)} = 7150/28
The quantity of water is {(150/4) + (2*200/7)} = 2650/28
The ratio of the resulting mixture = 7150/2650 = 143 : 53

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43.

A jar is filled of liquid which is 3 parts water and 5 parts alcohol. How much of this mixture should be drawn out and replaced such that this mixture may contain half
water and half alcohol?

Answer: B

Let the jar initially contain 8 litres of mixture:3 litres water and 5 litres alcohol

Let x litres of this mixture is drawn out and is replaced by x litres of water.

Amount of water now in the solution: 3-(3/8)x+x.

Amount of alcohol now in the solution: 5-(5/8)x

Desired ratio: 1/1

=>3-(3/8)x+x=5-(5/8)x

x=8/5 litres

%x=((8/5)/8)*100=20%

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44.

Presently a mixture in a tub contains substances A and B in the ratio of 7:5. If 9 litres of this solution is drawn out and B is added to the mixture in certain quantity the resulting mixture becomes 7:9. How much litres of liquid A was kept in the tub initially?

Answer: D

Currently the mixture is in ratio of A/B: 7:5

=> This means fraction of A in the solution: 7/12

=> Fraction of B in the solvent: 5/12

=>New concentration of A= 7-(7/12)9

=>New concentration of B= 5-(5/12)9+x

=> (7-(7/12)9)/(5-(5/12)9+x) = 7/9

=>x=21

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45.

A container has 40L milk. 4L milk is taken out and replaced with water, the process done three times. What is the amount of milk now left?

Answer: C

In regard to the problem let after n times liquid in the container is x and liquid removed is y

=>Liquid in container= x

=>Liquid removed= y

After one trial, quantity of pure liquid= x- (1-y/x)

After n trials quantity of pure liquid = x - (1 -y/x)n

Therefore, Putting the values ( x = 40L y = 4L N=3) and solving

=> 40- (1-4/40)3

= 29.16L

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46.

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

Answer: C

Take the quantities as 1, 1 and 2 kg, four kilograms in all.

\[ 126 + 135 + 2p = 4 \times 153 = 612 \]

\[ 2p = 612 - 261 = 351 \quad\Longrightarrow\quad p = \text{Rs. } 175.50 \]

The third variety carries half the weight of the mixture, which is why its price does not have to be far above 153 to lift the average.

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47.

A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

Answer: C

Let the can hold \( 12k \) litres, with A \( = 7k \).

Drawing off 9 litres removes A in proportion, and adding B keeps the total at \( 12k \):

\[ \text{A left} = 7k - 9 \times \frac{7}{12} = 7k - 5.25 \]

The new ratio 7 : 9 makes A \( \tfrac{7}{16} \) of the total:

\[ \frac{7k - 5.25}{12k} = \frac{7}{16} \quad\Longrightarrow\quad 112k - 84 = 84k \]

\[ 28k = 84 \quad\Longrightarrow\quad k = 3 \]

\[ \text{A initially} = 7 \times 3 = 21 \text{ litres} \]

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48.

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

Answer: B

The target is 12 L with water : milk \( = 3 : 5 \), so water is \( \tfrac{3}{8} \times 12 = 4.5 \) L.

Let \( x \) litres come from the first can (25% water) and \( 12 - x \) from the second (50% water):

\[ 0.25x + 0.5(12 - x) = 4.5 \]

\[ 0.25x + 6 - 0.5x = 4.5 \quad\Longrightarrow\quad -0.25x = -1.5 \]

\[ x = 6 \quad\Longrightarrow\quad 6 \text{ litres from each can} \]

The target of 37.5% water is exactly midway between 25% and 50%, which is why the two cans contribute equally.

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49.

In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

Answer: C

The alligation rule gives the ratio in which two ingredients must be mixed:

\[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}} \]

\[ \frac{\text{Rs. 15 pulses}}{\text{Rs. 20 pulses}} = \frac{20 - 16.50}{16.50 - 15} = \frac{3.50}{1.50} = \frac{7}{3} \]

The mean of 16.50 lies much nearer 15, so the cheaper variety dominates - consistent with the 7 : 3 split.

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50.

How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?

Answer: D

Work back from the selling price to the cost of the mixture.

\[ \text{CP} = \frac{9.24}{1.10} = \text{Rs. } 8.40 \text{ per kg} \]

Let \( x \) kg of the Rs. 9 sugar be mixed with 27 kg of the Rs. 7 sugar:

\[ \frac{9x + 7(27)}{x + 27} = 8.40 \]

\[ 9x + 189 = 8.4x + 226.8 \quad\Longrightarrow\quad 0.6x = 37.8 \]

\[ x = 63 \text{ kg} \]

The gain is stated on the mixture, so the selling price must be converted to a cost price before any mixing calculation.

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