Mixture and Allegation

Concepts

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Mixture and Alligation Concepts for GRE

Mixture and alligation problems appear frequently in GRE Quantitative Reasoning, challenging candidates to solve for unknown quantities in mixtures with different properties. This guide provides the fundamental concepts, formulas, examples, and practice questions to help you master these types of problems.

1. Mixture Problems

In mixture problems, two or more substances with varying concentrations, prices, or other characteristics are combined. The goal is to find the resulting mixture’s concentration or the required amounts of each component to achieve a desired result.

Key Formula for Mixture Problems

When two substances with different concentrations are mixed, the resultant concentration $C_{\text{final}}$ is calculated as:

$C_{\text{final}} = \frac{C_1 \times Q_1 + C_2 \times Q_2}{Q_1 + Q_2}$

where:

$C_1$ and $C_2$ are the concentrations of the two components.
$Q_1$ and $Q_2$ are the quantities of each component.

Example 1: Mixture Problem

Problem: A chemist has a 20% saline solution and a 50% saline solution. How much of each solution must be mixed to obtain 10 liters of a 30% saline solution?

Solution:

Let $x$ be the quantity (in liters) of the 20% solution, and $10 - x$ be the quantity of the 50% solution.
Using the formula for resultant concentration:
$0.20x + 0.50(10 - x) = 0.30 \times 10$
Simplify:
$0.20x + 5 - 0.50x = 3$

$-0.30x = -2$

$x = \frac{2}{0.3} = 6.67$
Therefore, $x = 6.67$ liters of the 20% solution and $10 - 6.67 = 3.33$ liters of the 50% solution are needed.

2. Alligation Method

The alligation method is a simplified approach to solving mixture problems, particularly useful when two ingredients with known concentrations or prices are mixed to reach a target concentration or price.

Alligation Rule

The formula for the alligation rule is as follows:

$\text{Ratio of Quantities} = \frac{|C_2 - C_{\text{final}}|}{|C_1 - C_{\text{final}}|}$

where:

$C_1$ and $C_2$ are the concentrations or prices of the two components.
$C_{\text{final}}$ is the target concentration or price.

3. Additional Examples

Example 3: Mixture Problem

Problem: A milkman has milk with 10% fat content and another with 30% fat content. How much of each type should be mixed to create 50 liters of milk with 20% fat content?

Solution:

Let $x$ be the quantity of 10% fat milk, and $50 - x$ be the quantity of 30% fat milk.
Using the concentration formula:
$0.10x + 0.30(50 - x) = 0.20 \times 50$
Expanding and solving:
$0.10x + 15 - 0.30x = 10$

$-0.20x = -5$

$x = 25$
Thus, mix 25 liters of 10% fat milk with 25 liters of 30% fat milk.

Example 4: Alligation Problem

Problem: A perfumer has essential oils priced at $8 per ounce and$ 15 per ounce. He wants to create a blend priced at $12 per ounce. In what ratio should he mix these oils?

Solution:

Set up the ratio:
$\text{Ratio of Quantities} = \frac{|15 - 12|}{|8 - 12|} = \frac{3}{4}$
Therefore, he should mix the oils in a $3:4$ ratio to achieve the desired price.

4. Practice Questions

Practice Question 1:

A solution contains 30% alcohol. How many liters of pure alcohol should be added to 20 liters of this solution to make it a 40% alcohol solution?

Practice Question 2:

A jeweler has two gold alloys: one is 30% pure gold, and the other is 70% pure gold. In what ratio should they be mixed to get an alloy that is 50% pure gold?

Practice Question 3:

A beverage company has two fruit juice concentrates, one with 25% fruit and the other with 55% fruit. How many liters of each should be mixed to make 60 liters of a concentrate that is 40% fruit?

Conclusion

Mastering mixture and alligation techniques enables you to efficiently tackle GRE questions on combined solutions or costs. Practice with various scenarios to strengthen your skills and improve your problem-solving speed.

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Test – 1

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