Alligation or Mixture Questions and Answers

Q: In what ratio should water be mixed with a liquid costing ₹12 per litre so that by selling the mixture at ₹13.75 per litre the seller gains 25%?

Correct answer: 1 : 11First convert the selling price and profit into a cost price. Selling at 13.75 with a gain of 25% means \[ \text{CP} = \frac{13.75 \times 100}{125} = 11 \text{ per litre}. \] Now apply alligation between water, which costs nothing, and the liquid at 12, with a mean of 11: \[ \frac{\text{water}}{\text{liquid}} = \frac{12 - 11}{11 - 0} = \frac{1}{11}. \] The ratio is 1 : 11. Checking: 11 litres of liquid cost 132, and the 12-litre mixture therefore costs 11 per litre; selling it at 13.75 brings in 165, a gain of 33 on 132, which is 25%. The step most often skipped is the first one - alligating on the selling price rather than converting to cost price gives the wrong ratio, because the profit has not been removed.

Q: In a mixture of 90 L the ratio of acid and water is 2 : 1. If the ratio of acid and water is to be 1 : 2, then the amount of water (in liters) to be added to the mixture is?

So amount of acid = 2/ 2+1 × 90 = 60L Amount of water = 90 - 60 = 30L To make acid to water ratio = 1:2, we simply need To make water double of acid. By direct observation, we can say that we should have 60L × 2 = 120L make the required ratio We have 30L, we need (120-30) = 90L more water

Q: Two types of rice costing ₹25 and ₹40 per kg are mixed in the ratio 2 : 3. What is the cost of the mixture per kg?

Correct answer: ₹34The cost of a mixture is the weighted average of the component prices: \[ \text{mean} = \frac{2 \times 25 + 3 \times 40}{2 + 3} = \frac{50 + 120}{5} = \frac{170}{5} = 34. \] The mixture costs 34 per kg. The distractor 32.50 is the simple average of 25 and 40, correct only for equal quantities. Since more of the dearer rice is used, the answer must lie above the midpoint - which rules out 32.50 and 33 immediately, and is worth checking before any calculation. This is the direct form of the problem; alligation solves the inverse, finding the ratio when the mean is given. The relation is the same read either way, and being able to move in both directions is what the topic amounts to: \[ \frac{q_{1}}{q_{2}} = \frac{p_{2} - m}{m - p_{1}}. \] Substituting the answer back confirms it: \( \frac{40-34}{34-25} = \frac{6}{9} = \frac{2}{3} \).

Q: A 40-litre mixture of milk and water contains 10% water. How much water must be added to make the water 20% of the new mixture?

Correct answer: 5 litresThe mixture contains \( 40 \times 0.10 = 4 \) litres of water and 36 litres of milk. The milk is unchanged by adding water, so work from it. If water is to be 20% of the new mixture, milk must be 80%, and \[ 36 = 0.80 \times \text{new total} \;\Rightarrow\; \text{new total} = \frac{36}{0.8} = 45 \text{ litres}. \] The water to add is therefore \[ 45 - 40 = 5 \text{ litres}. \] Checking: 45 litres containing 9 of water is 20% water, and \( 9 - 4 = 5 \) litres were added. The method to carry away is to anchor on the quantity that does not change and express the target percentage in terms of it rather than of the component being added. Working forwards from the water - asking what makes 4 litres into 20% - gives 20 litres total and the wrong answer, because it ignores that the added water also enlarges the total.

Q: A container has 50 litres of a mixture of milk and water in the ratio 3 : 2. If 10 litres of the mixture is removed and replaced with water, what is the new ratio?

Correct answer: 12 : 13The original 50 litres contains \( 50 \times \frac{3}{5} = 30 \) litres of milk and 20 of water. Removing 10 litres of the mixture removes milk and water in their current 3 : 2 proportion - 6 litres of milk and 4 of water - leaving 24 of milk and 16 of water. Adding 10 litres of water brings the water to \( 16 + 10 = 26 \), so the new ratio is \[ 24 : 26 = 12 : 13. \] The essential point is that what is drawn off is a sample of the mixture, carrying both components in proportion. Removing 10 litres of pure milk would be a different problem and would leave 20 : 30, or 2 : 3. As a check, the total stays at 50 litres throughout, and \( 24 + 26 = 50 \).

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91.

A pot contains 40 liters of juice. How much juice will be there in the pot, if 4 liters of juice was removed and was replaced by water and this process was further repeated twice?

A.32 liter
B.32.29 liter
C.28 liter
D.29.16 liter

Answer: D

When a fraction of a solution is repeatedly removed and replaced, the original liquid left after \( n \) operations is

\[ A\left(1 - \frac{x}{A}\right)^{n} \]

Here \( A = 40 \), \( x = 4 \) and \( n = 3 \) (once, then twice more):

\[ 40\left(1 - \frac{4}{40}\right)^{3} = 40(0.9)^{3} \]

\[ = 40 \times 0.729 = 29.16 \text{ litres} \]

Simply subtracting \( 3 \times 4 = 12 \) litres would give 28, but each later removal takes out some water as well as juice.

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92.

In what ratio should water be mixed with a liquid costing ₹12 per litre so that by selling the mixture at ₹13.75 per litre the seller gains 25%?

A.1 : 11
B.1 : 4
C.2 : 9
D.1 : 6

Answer: B

Correct answer: 1 : 11

First convert the selling price and profit into a cost price. Selling at 13.75 with a gain of 25% means \[ \text{CP} = \frac{13.75 \times 100}{125} = 11 \text{ per litre}. \] Now apply alligation between water, which costs nothing, and the liquid at 12, with a mean of 11: \[ \frac{\text{water}}{\text{liquid}} = \frac{12 - 11}{11 - 0} = \frac{1}{11}. \] The ratio is 1 : 11. Checking: 11 litres of liquid cost 132, and the 12-litre mixture therefore costs 11 per litre; selling it at 13.75 brings in 165, a gain of 33 on 132, which is 25%. The step most often skipped is the first one - alligating on the selling price rather than converting to cost price gives the wrong ratio, because the profit has not been removed.

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93.

How much milk should be added in a milk solution to make milk quantity in it 75%, if 80 L of milk solution has 45% milk in it?

A.96 liters
B.75 liters
C.30 liters
D.110 liters

Answer: A

The 80 L contains \( 45\% \times 80 = 36 \) L of milk. Adding \( x \) L of pure milk raises both the milk and the total.

\[ \frac{36 + x}{80 + x} = 0.75 \]

\[ 36 + x = 60 + 0.75x \quad\Longrightarrow\quad 0.25x = 24 \]

\[ x = 96 \text{ litres} \]

The non-milk part stays fixed at 44 L, and it must end up as 25% of the total - so the total becomes 176 L, confirming the 96 L added.

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94.

3 types of Sugar got mixed by accident in a shop. First was of the rate Rs. 145 per kg and Second of Rs. 165 per kg. Only thing the shop owner knew that quantities of 3 sugar types were in the ratio 2:1:3 respectively. He finally sold the mix at rate of Rs. 180 per kg. What was price of 3rd type?

A.Rs. 208.33
B.Rs. 210
C.Rs. 215.67
D.Rs. 196.58

Answer: A

Let 3rd type sugar rate be Rs. S

Quantity ratio = 2:1:3

This means if we take 2kg of Type 1 and 1 kg of Type 2, then

we must take 3kg of Type 3.

Also, the mix will have 2kg+1kg+3kg 6 kg quantity.

(145 × 2) + (165 × 1) + (3 × S)= 6 × 180

S = Rs. 208.33 per kg = 3rd type price per kg

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95.

A mixture consists of some amount of sandalwood oil and 240 liters of water. It is priced at Rs. 275 per liter. Sandalwood oil is priced Rs. 325 per liter. How much oil is there in the mixture?

A.720 liters
B.1000 liters
C.1320 liters
D.1200 liters

Answer: C

The water contributes nothing to the value, so all the worth comes from the oil.

\[ 325x = 275(x + 240) \]

\[ 325x = 275x + 66{,}000 \]

\[ 50x = 66{,}000 \quad\Longrightarrow\quad x = 1320 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{oil}}{\text{water}} = \tfrac{275-0}{325-275} = \tfrac{275}{50} = 5.5 \), and \( 5.5 \times 240 = 1320 \).

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96.

In a mixture of 90 L the ratio of acid and water is 2 : 1. If the ratio of acid and water is to be 1 : 2, then the amount of water (in liters) to be added to the mixture is?

A.20L
B.40L
C.90L
D.100L

Answer: C

So amount of acid = 2/ 2+1 × 90 = 60L

Amount of water = 90 - 60 = 30L

To make acid to water ratio = 1:2, we simply need

To make water double of acid.

By direct observation, we can say that we should

have 60L × 2 = 120L make the required ratio

We have 30L, we need (120-30) = 90L more water

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97.

Two types of rice costing ₹25 and ₹40 per kg are mixed in the ratio 2 : 3. What is the cost of the mixture per kg?

A.₹34
B.₹32.50
C.₹35
D.₹33

Answer: B

Correct answer: ₹34

The cost of a mixture is the weighted average of the component prices: \[ \text{mean} = \frac{2 \times 25 + 3 \times 40}{2 + 3} = \frac{50 + 120}{5} = \frac{170}{5} = 34. \] The mixture costs 34 per kg. The distractor 32.50 is the simple average of 25 and 40, correct only for equal quantities. Since more of the dearer rice is used, the answer must lie above the midpoint - which rules out 32.50 and 33 immediately, and is worth checking before any calculation. This is the direct form of the problem; alligation solves the inverse, finding the ratio when the mean is given. The relation is the same read either way, and being able to move in both directions is what the topic amounts to: \[ \frac{q_{1}}{q_{2}} = \frac{p_{2} - m}{m - p_{1}}. \] Substituting the answer back confirms it: \( \frac{40-34}{34-25} = \frac{6}{9} = \frac{2}{3} \).

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98.

A 40-litre mixture of milk and water contains 10% water. How much water must be added to make the water 20% of the new mixture?

A.5 litres
B.4 litres
C.8 litres
D.10 litres

Answer: B

Correct answer: 5 litres

The mixture contains \( 40 \times 0.10 = 4 \) litres of water and 36 litres of milk. The milk is unchanged by adding water, so work from it. If water is to be 20% of the new mixture, milk must be 80%, and \[ 36 = 0.80 \times \text{new total} \;\Rightarrow\; \text{new total} = \frac{36}{0.8} = 45 \text{ litres}. \] The water to add is therefore \[ 45 - 40 = 5 \text{ litres}. \] Checking: 45 litres containing 9 of water is 20% water, and \( 9 - 4 = 5 \) litres were added. The method to carry away is to anchor on the quantity that does not change and express the target percentage in terms of it rather than of the component being added. Working forwards from the water - asking what makes 4 litres into 20% - gives 20 litres total and the wrong answer, because it ignores that the added water also enlarges the total.

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99.

A container has 50 litres of a mixture of milk and water in the ratio 3 : 2. If 10 litres of the mixture is removed and replaced with water, what is the new ratio?

A.12 : 13
B.3 : 4
C.5 : 6
D.2 : 3

Answer: A

Correct answer: 12 : 13

The original 50 litres contains \( 50 \times \frac{3}{5} = 30 \) litres of milk and 20 of water. Removing 10 litres of the mixture removes milk and water in their current 3 : 2 proportion - 6 litres of milk and 4 of water - leaving 24 of milk and 16 of water. Adding 10 litres of water brings the water to \( 16 + 10 = 26 \), so the new ratio is \[ 24 : 26 = 12 : 13. \] The essential point is that what is drawn off is a sample of the mixture, carrying both components in proportion. Removing 10 litres of pure milk would be a different problem and would leave 20 : 30, or 2 : 3. As a check, the total stays at 50 litres throughout, and \( 24 + 26 = 50 \).

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100.

To gain profit, a milkman adds 1.5 liter of water to a certain amount of milk. He sells the adulterated milk at Rs.28 per liter and pure milk costs Rs.36 per liter. Find the amount of milk.

A.5.25
B.6
C.5.5
D.5

Answer: A

The water is free, so "no loss" means the revenue from the whole adulterated quantity just covers the cost of the pure milk.

\[ 28(m + 1.5) = 36m \]

\[ 28m + 42 = 36m \quad\Longrightarrow\quad 8m = 42 \]

\[ m = 5.25 \text{ litres} \]

Selling at a price below cost is profitable only because the 1.5 L of water is sold as though it were milk.

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