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Mixtures & Alligations

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CAT 2025 Lesson : Mixtures & Alligations - Weighted Average & 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}}

⇒

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

⇒

30 (9 + y) = 285 + 25y

⇒

5y = 15

⇒

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

3. Alligation Method

Where only two substances are mixed, we almost always use the alligation method to compute the weighted average due to its ease. The property used by the alligation method is outlined below.

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

⇒

x_{A}w_{1} + a_{A}w_{2} = x_{1}w_{1} + x_{2}w_{2}

⇒

x_{A}w_{1} - x_{1}w_{1} = x_{2}w_{2} - x_{A}w_{2}

⇒

w(x_{A} - x_{1}) = w_{2}(x_{2} - x_{A})

⇒

\dfrac{w_{1}}{w_{2}} = \dfrac{(x_{2} - x_{A})}{(x_{A} - x_{1})}

Note that the above is just the proof for the figure that follows, which is the alligation method.

Average of two terms will always lie between them.

\therefore x_{A}

will always be between

x_{1}

and

x_{2}

.

Irrespective of

x_{1}

being greater than or less than

x_{2}

, we simply need to find their difference (the absolute value) in the calculations shown below.

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