## Mixtures & Alligations

### 1. Introduction

A mixture refers to a combination of two or more substances. Questions related to mixtures include

$1$) Two or more pure substances being mixed in a certain ratio, such as pure milk with water.
$2$) Two or more mixtures or solutions being mixed in a certain ratio, such as 40% milk solution mixed with $60 \%$ milk solution.
$3$) One solution and one pure substance. $40 \%$ milk solution mixed with pure milk.
$4$) Solutions with successive replacement.
$5$) Evaporation of substance.

These questions can be answered using the weighted average method or the alligation method, both of which are explained in the subsequent sections.

### 2. Weighted Average Method

Where $n$ substances with values of $x_{1}, x_{2}, ... , x_{n}$ are mixed in such a way that their weights in the mixture are $w_{1}, w_{2}, ...., w_{n}$ respectively, then the weighted average of the mixture is

$x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + ... + x_{n}w_{n}}{w_{1} + w_{2} + ... + w_{n}}$

Note:
1) The absolute values of weights can be simplified as long as their ratio remains the same (in Example 1).
2) Weights are quantities or volumes (like kg, litres, number of students, etc.) that influence the average.
3) When you're confused between the values ($x$ terms) and weights ($w$ terms), note that the average we are required to find is for the values ($x$ terms). The other unit in the question will constitute the weights ($w$ terms).

### Example 1

A, B and C are $20 \%, 30 \%$ and $50 \%$ milk solutions. If $20$ litres of A, $40$ litres of B and $80$ litres of C are mixed then what is the percentage of milk in the resultant solution?

### Solution

The quantities ($20, 40$ and $80$ litres) of A, B and C are the weights in this question.
The percentages ($20 \%, 30 \%$ and $50 \%$) are the x-values for which we have to find the weighted average.

Weighted average $\%$ of milk $= \dfrac{20 \times 20 + 30 \times 40 + 50 \times 80}{20 + 40 + 80} = \dfrac{5600}{140} = 40 \%$

Alternatively

As explained in Note $1$ above, weights are relative. $\therefore$ their ratio can be applied.

Ratio of quantities of A, B and C $= 20 : 40 : 80 = 1 : 2 : 4$

Weighted average $\%$ of milk $= \dfrac{20 \times 1 + 30 \times 2 + 50 \times 4}{1 + 2 + 4} = \dfrac{280}{7} = 40 \%$

Answer: $40 \%$

### Example 2

In a certain company, each employee earns exactly one of three different salaries. The monthly salaries of $28$ employees is Rs. $20,000$ each, of $98$ employees is Rs. $35,000$ each and of $x$ employees is Rs. $25,000$ each. If average monthly salary of the company is Rs. $30,000$, then find $x$.

### Solution

Note that two particular groups have $28$ and $98$ employees, wherein $14$ divides both these numbers. Let $x = 14y$ (to simplify calculations).

Ratio of weights $= 28 : 98 : 14y = 2 : 7 : y$

In averages, we have studied that if all the values are divided by the same amount, the average also gets divided by the same amount. As all salaries are multiples of 1000, we will take the salaries to be Rs. $20$, Rs. $35$ and Rs. $25$, which would make the weighted average Rs. $30$.

$x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + x_{3}w_{3}}{w_{1} + w_{2} +w_{3}}$

$\implies 30 = \dfrac{20 \times 2 + 35 \times 7 + 25 \times y}{2 + 7 + y}$

$\implies 30 (9 + y) = 285 + 25y$
$\implies 5y = 15$
$\implies y = 3$

$x = 14 \times 3 = 42$

Answer: $42$

### Example 3

The ratios of copper and tin in two alloys, A and B, are $3 : 5$ and $5 : 7$ respectively. Alloy C is formed by mixing A and B in the ratio $2 : 1$. What is the ratio of copper and tin in Alloy C?

### Solution

Two different mixtures of copper and tin are being mixed here. In such cases, we should always apply weighted average method on the portion of one of the elements only. In this example, let us take the portion of copper in each of these mixtures and apply the weighted average.

Portion of Copper in the alloys A and B are $\dfrac{3}{(3 + 5)} = \dfrac{3}{8}$ and $\dfrac{5}{(5 + 7)} = \dfrac{5}{12}$ respectively.

These alloys are mixed in the ratio of $2 : 1$

Portion of Copper in the final alloy $= \dfrac{\dfrac{3}{8} \times 2 + \dfrac{5}{12} \times 1}{2 + 1} = \dfrac{\dfrac{9 + 5}{12}}{3} = \dfrac{7}{18}$

∴ Portion of Tin in the final alloy $= 1 - \dfrac{7}{18} = \dfrac{11}{18}$

Ratio of Copper and Tin in final alloy $= 7 : 11$

Answer: $7 : 11$
The text passage contains multiple examples of mixture and alligation problems, each with a question and a corresponding solution.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## Mixtures & Alligations

### 1. Introduction

A mixture refers to a combination of two or more substances. Questions related to mixtures include

$1$) Two or more pure substances being mixed in a certain ratio, such as pure milk with water.
$2$) Two or more mixtures or solutions being mixed in a certain ratio, such as 40% milk solution mixed with $60 \%$ milk solution.
$3$) One solution and one pure substance. $40 \%$ milk solution mixed with pure milk.
$4$) Solutions with successive replacement.
$5$) Evaporation of substance.

These questions can be answered using the weighted average method or the alligation method, both of which are explained in the subsequent sections.

### 2. Weighted Average Method

Where $n$ substances with values of $x_{1}, x_{2}, ... , x_{n}$ are mixed in such a way that their weights in the mixture are $w_{1}, w_{2}, ...., w_{n}$ respectively, then the weighted average of the mixture is

$x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + ... + x_{n}w_{n}}{w_{1} + w_{2} + ... + w_{n}}$

Note:
1) The absolute values of weights can be simplified as long as their ratio remains the same (in Example 1).
2) Weights are quantities or volumes (like kg, litres, number of students, etc.) that influence the average.
3) When you're confused between the values ($x$ terms) and weights ($w$ terms), note that the average we are required to find is for the values ($x$ terms). The other unit in the question will constitute the weights ($w$ terms).

### Example 1

A, B and C are $20 \%, 30 \%$ and $50 \%$ milk solutions. If $20$ litres of A, $40$ litres of B and $80$ litres of C are mixed then what is the percentage of milk in the resultant solution?

### Solution

The quantities ($20, 40$ and $80$ litres) of A, B and C are the weights in this question.
The percentages ($20 \%, 30 \%$ and $50 \%$) are the x-values for which we have to find the weighted average.

Weighted average $\%$ of milk $= \dfrac{20 \times 20 + 30 \times 40 + 50 \times 80}{20 + 40 + 80} = \dfrac{5600}{140} = 40 \%$

Alternatively

As explained in Note $1$ above, weights are relative. $\therefore$ their ratio can be applied.

Ratio of quantities of A, B and C $= 20 : 40 : 80 = 1 : 2 : 4$

Weighted average $\%$ of milk $= \dfrac{20 \times 1 + 30 \times 2 + 50 \times 4}{1 + 2 + 4} = \dfrac{280}{7} = 40 \%$

Answer: $40 \%$

### Example 2

In a certain company, each employee earns exactly one of three different salaries. The monthly salaries of $28$ employees is Rs. $20,000$ each, of $98$ employees is Rs. $35,000$ each and of $x$ employees is Rs. $25,000$ each. If average monthly salary of the company is Rs. $30,000$, then find $x$.

### Solution

Note that two particular groups have $28$ and $98$ employees, wherein $14$ divides both these numbers. Let $x = 14y$ (to simplify calculations).

Ratio of weights $= 28 : 98 : 14y = 2 : 7 : y$

In averages, we have studied that if all the values are divided by the same amount, the average also gets divided by the same amount. As all salaries are multiples of 1000, we will take the salaries to be Rs. $20$, Rs. $35$ and Rs. $25$, which would make the weighted average Rs. $30$.

$x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + x_{3}w_{3}}{w_{1} + w_{2} +w_{3}}$

$\implies 30 = \dfrac{20 \times 2 + 35 \times 7 + 25 \times y}{2 + 7 + y}$

$\implies 30 (9 + y) = 285 + 25y$
$\implies 5y = 15$
$\implies y = 3$

$x = 14 \times 3 = 42$

Answer: $42$

### Example 3

The ratios of copper and tin in two alloys, A and B, are $3 : 5$ and $5 : 7$ respectively. Alloy C is formed by mixing A and B in the ratio $2 : 1$. What is the ratio of copper and tin in Alloy C?

### Solution

Two different mixtures of copper and tin are being mixed here. In such cases, we should always apply weighted average method on the portion of one of the elements only. In this example, let us take the portion of copper in each of these mixtures and apply the weighted average.

Portion of Copper in the alloys A and B are $\dfrac{3}{(3 + 5)} = \dfrac{3}{8}$ and $\dfrac{5}{(5 + 7)} = \dfrac{5}{12}$ respectively.

These alloys are mixed in the ratio of $2 : 1$

Portion of Copper in the final alloy $= \dfrac{\dfrac{3}{8} \times 2 + \dfrac{5}{12} \times 1}{2 + 1} = \dfrac{\dfrac{9 + 5}{12}}{3} = \dfrac{7}{18}$

∴ Portion of Tin in the final alloy $= 1 - \dfrac{7}{18} = \dfrac{11}{18}$

Ratio of Copper and Tin in final alloy $= 7 : 11$

Answer: $7 : 11$
user
The text passage contains multiple examples of mixture and alligation problems, each with a question and a corresponding solution.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS