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

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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%60 \% milk solution.
3) One solution and one pure substance. 40%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 nn substances with values of x1,x2,...,xnx_{1}, x_{2}, ... , x_{n} are mixed in such a way that their weights in the mixture are w1,w2,....,wnw_{1}, w_{2}, ...., w_{n} respectively, then the weighted average of the mixture is

xA=x1w1+x2w2+...+xnwnw1+w2+...+wnx_{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 (xx terms) and weights (ww terms), note that the average we are required to find is for the values (xx terms). The other unit in the question will constitute the weights (ww terms).

Example 1

A, B and C are 20%,30%20 \%, 30 \% and 50%50 \% milk solutions. If 2020 litres of A, 4040 litres of B and 8080 litres of C are mixed then what is the percentage of milk in the resultant solution?

Solution

The quantities (20,4020, 40 and 8080 litres) of A, B and C are the weights in this question.
The percentages (20%,30%20 \%, 30 \% and 50%50 \%) are the x-values for which we have to find the weighted average.

Weighted average %\% of milk =20×20+30×40+50×8020+40+80=5600140=40%= \dfrac{20 \times 20 + 30 \times 40 + 50 \times 80}{20 + 40 + 80} = \dfrac{5600}{140} = 40 \%

Alternatively

As explained in Note 11 above, weights are relative. ∴\therefore their ratio can be applied.

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

Weighted average %\% of milk =20×1+30×2+50×41+2+4=2807=40%= \dfrac{20 \times 1 + 30 \times 2 + 50 \times 4}{1 + 2 + 4} = \dfrac{280}{7} = 40 \%

Answer: 40%40 \%

Example 2

In a certain company, each employee earns exactly one of three different salaries. The monthly salaries of 2828 employees is Rs. 20,00020,000 each, of 9898 employees is Rs. 35,00035,000 each and of xx employees is Rs. 25,00025,000 each. If average monthly salary of the company is Rs. 30,00030,000, then find xx.

Solution

Note that two particular groups have 2828 and 9898 employees, wherein 1414 divides both these numbers. Let x=14yx = 14y (to simplify calculations).

Ratio of weights =28:98:14y=2:7:y= 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. 2020, Rs. 3535 and Rs. 2525, which would make the weighted average Rs. 3030.

xA=x1w1+x2w2+x3w3w1+w2+w3x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + x_{3}w_{3}}{w_{1} + w_{2} +w_{3}}

⇒ 30=20×2+35×7+25×y2+7+y30 = \dfrac{20 \times 2 + 35 \times 7 + 25 \times y}{2 + 7 + y}

⇒ 30(9+y)=285+25y30 (9 + y) = 285 + 25y
⇒ 5y=155y = 15
⇒ y=3 y = 3

x=14×3=42x = 14 \times 3 = 42

Answer: 4242

Example 3

The ratios of copper and tin in two alloys, A and B, are 3:53 : 5 and 5:75 : 7 respectively. Alloy C is formed by mixing A and B in the ratio 2:12 : 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 3(3+5)=38\dfrac{3}{(3 + 5)} = \dfrac{3}{8} and 5(5+7)=512\dfrac{5}{(5 + 7)} = \dfrac{5}{12} respectively.

These alloys are mixed in the ratio of 2:12 : 1

Portion of Copper in the final alloy =38×2+512×12+1=9+5123=718= \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 =1−718=1118= 1 - \dfrac{7}{18} = \dfrac{11}{18}

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

Answer: 7:117 : 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.

xA=x1w1+x2w2w1+w2x_{A} = \dfrac{x_{1}w_{1} + x_{2}w_{2}}{w_{1} + w_{2}}

⇒ xAw1+aAw2=x1w1+x2w2x_{A}w_{1} + a_{A}w_{2} = x_{1}w_{1} + x_{2}w_{2}

⇒ xAw1−x1w1=x2w2−xAw2x_{A}w_{1} - x_{1}w_{1} = x_{2}w_{2} - x_{A}w_{2}

⇒ w(xA−x1)=w2(x2−xA)w(x_{A} - x_{1}) = w_{2}(x_{2} - x_{A})

⇒ w1w2=(x2−xA)(xA−x1)\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. ∴xA\therefore x_{A} will always be between x1x_{1} and x2x_{2}.

Irrespective of x1x_{1} being greater than or less than x2x_{2}, we simply need to find their difference (the absolute value) in the calculations shown below.



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