Mixture and Alligation - Solved Questions and Answers

Mixture is a combining two or more ingredients to create a new blend.

Alligation is used to determine the proportion in which two or more ingredients with different values (such as prices, concentrations, or qualities) must be mixed to achieve a desired average value.

Mixture and Alligation questions and answers are provided below for you to learn and practice.

Question 1: From a vessel of 20 liters of pure milk, 1 liter is taken out and replaced with water, so as to keep the volume constant at 20 liters. This process is repeated 5 times. Find the percentage of pure milk left in the vessel after 5 replacements. 

Solution: 

Here, we need to apply the formula [1 - (R / P)] ⁿ
P = Initial quantity of pure element = 20 liters
R = Quantity replaced every time = 1 liter (volume removed and replaced each time)
n = Number of replacements = 5

So, the quantity of pure milk after 5 replacements = [1 - (1 / 20)]⁵
Quantity of pure milk after 5 replacements (0.95)⁵ = 0.7737809375
Therefore, percentage of pure milk left in the vessel after 5 replacements = (0.7737809375) x 100 = 77.378 %   

Question 2: A dishonest shopkeeper mixed cheaper quality rice, priced at Rs. 10 / KG with good quality rice, priced at Rs. 25 / KG, and sold the mixture at Rs. 15 / KG. Find the ratio in which he mixes the two qualities of rice. 

Solution:  

Thus, the ratio of quantities of cheaper and good quality rice = 10: 5 = 2: 1

Question 3: A grocer mixes two varieties of rice costing $15 per kg and $20 per kg in the ratio 2:3. What is the price per kg of the resulting mixture?

Solution:

Using the alligation formula:

Mean Price = \frac{(15 \times 2) + (20 \times 3)}{2 + 3} = \frac{30 + 60}{5} = \frac{90}{5} = 18

Therefore, the price per kg of the mixture is $18.

Question 4: A vessel contains 60 liters of milk. 12 liters of milk are taken out and replaced with water. If this process is repeated once more, how much milk is now in the vessel?

Solution:

First Removal: Milk left after first replacement = 60 \times \left(1 - \frac{12}{60}\right) = 60 \times \frac{48}{60} = 48 Litres

Second Removal: Milk left after second replacement = 48 \times \left(1 - \frac{12}{60}\right) = 48 \times \frac{48}{60} = 38.4 liters.

Therefore, the milk remaining in the vessel is 38.4 liters.

Question 5: How much water must be added to 40 liters of a 20% alcohol solution to make it a 10% alcohol solution?

Solution:

20% of 40 liters = \frac{20}{100} \times 40 = 8 liters.

Now, the total solution = 40 + 𝑥 liters.

New concentration of alcohol:

\frac{8}{40 + x} \times 100 = 10

Solve for 𝑥: 8 = 0.1 × (40 + 𝑥) ⇒ 8 = 4 + 0.1𝑥 ⇒ 𝑥 = 40 liters.

So, 40 liters of water must be added.

Question 6: A 20-liter mixture contains milk and water in the ratio 3:1. How much water must be added to make the ratio 1:1?

Solution:

Amount of milk in the mixture: \frac{3}{4} \times 20 = 15 liters.

Amount of water in the mixture: 20 - 15 = 5 liters.

Let 𝒙 liters of water be added to make the ratio 1:1

Set up the equation: \frac{15}{5 + x} = 1 \Rightarrow 15 = 5 + x \Rightarrow x = 10

Therefore, 10 liters of water must be added.

Question 7: A grocer mixes two varieties of sugar costing $12 per kg and $18 per kg in the ratio 3:2. What is the price per kg of the resulting mixture?

Solution:

Using the alligation formula:
Mean Price = (12×3)+(18×2)3+2=\frac{36+36}{5}=\frac{72}{5}=14.4
Therefore, the price per kg of the mixture is $14.4.

Question 8: A vessel contains 100 liters of juice. 20 liters of juice are taken out and replaced with water. If this process is repeated once more, how much juice is now in the vessel?

Solution:

First Removal:
Juice left after first replacement = 100 \times \left(1 - \frac{20}{100}\right) = 100 \times \frac{80}{100} = 80

Second Removal:
Juice left after second replacement = 80 \times \left(1 - \frac{20}{100}\right) = 80 \times \frac{80}{100} = 64
Therefore, the juice remaining in the vessel is 64 liters.