Mixture and Alligation Concepts | Tricks | Formulas for CAT Exam

Table of Content

• Concepts and Formulas
• Tricks
• Rule of Constant
• Mixing with replacement
• Two different Mixture
• Practice Problems

Questions from Mixtures and Alligation topic have increased in recent CAT exam. Usually, we apply the concepts and formulas of weighted average, percentages and ratios to solve such questions. A shortcut tricks of the criss-cross method is also very popular.

Concepts and Formulas

Say, you go to the market and buy some quantities of two different kinds of rice costing Rs. a per kg and Rs. b per kg respectively (a<b). At home, you mix both of them. Now, you want to find the cost of the resultant mixture. We call the cost per kg of the mixture to be mean or average cost (m).

It is obvious that the mean cost would be higher than a and lower than b. But, you want to calculate the actual mean cost. So let us derive the formula.

Let you buy x kg of first kind and y kg of second kind. So the total amount you spend is: Rs. ax + by. This is also the total cost of the mixture and the total amount of mixture is (x + y) kg. Therefore, mean cost (m) =\(\frac{{ax + by}}{{x + y}}\) . Observe that this is nothing but finding the weighted average. Rewriting the formula:

\(m = \frac{{ax + by}}{{x + y}}\)

Or  m(x + y)=ax + by

Or mx + my = ax + by

Or x (m-a) = y(b-m)

Or \(\frac{x}{y} = \frac{{(b – m)}}{{(m – a)}}\), i.e. we get the rule of alligation:

Let us take some problems to understand the application of the above mixture and alligation rule:

Problem 1: Two different kinds of rice costing Rs. 20 per kg and 35 per kg are mixed to get a mixture that costs Rs. 25 per kg. In what ratio the two kinds of rice have been mixed.

Applying the rule of mixture and Alligation:

\(\frac{{35 – 25}}{{25 – 20}} = \frac{{10}}{5} = \frac{2}{1}\)

Hence, the required ratio is 2:1.

Criss-cross method

Some aspirants use the above method in different format, which we call criss-cross method. Below is the format:

The working is, we take the positive difference of mean price and cheaper price and write the difference in the place of Quantity of dearer price. Similarly, take the positive difference of mean price and dearer price and write the difference in the place of Quantity of cheaper price.

For above problem, the working of criss-cross method is as follow:

So the ratio of quantity of cheaper price to quantity of dearer price = 10:5 = 2:1.
Important: The order in which the cheaper price and dearer price are written in the above method follows the order in which they are given in the question.

Rule of Constant

There are another types of questions on mixtures and alligation where the quantity of one element in the mixture does not change while adding another element to the first mixture. For such questions, I recommend an alternate method of using the rule of constant to get to the answer. The concept uses the simple understanding of percentages. Let us learn it with the help of an example.

Problem 2: How many litres of pure water should be added to 40 litres of 30% milk solution so that the resultant mixture is a 15% milk solution?

We can have three different methods to solve the above problems including the rule of alligation. Let us take all of them to strengthen our concepts.

Method I: A standard school textbook approach.

We assume that the quantity of water added to be x litres. The quantity of milk in the existing solution is 30% of 40 = 12 litres, with the addition of water, the quantity of new solution becomes (40 + x) litres. As per the problem, the percentage of milk in new solution should be 15 %. Therefore,

\(\frac{{12}}{{40 + x}} \times 100 = 15\)

Solving for x, we get x = 40 liters.

Method II: Rule of alligation.

We assume that the two solutions of milk and water are added to get the new solution and apply the approach we used in Example 3.

Taking milk as the common element in both the solutions, we have 30% milk in first and 0% milk in the second solution (i.e. pure water). On mixing them, we got 15% milk in the final solution.  Therefore,

Or the ratio of the quantity of first and second solution should be 15:15 = 1:1

Hence, 40 litres of pure water should be mixed to get the desired new solution.

Method III: Rule of constant.

In this rule, we target the element in the mixture whose amount does not change but its percentage changes because of the change in the total amount of the mixture.

As we calculated above, the quantity of milk in the first solution is 12 litres, and it will remain same in the new solution as well. That is,

12 liters = 30% of the first solution =15% of the new solution

Therefore, the quantity of new solution =  \(\frac{{100}}{{15}} \times 12\) = 80 liters. Now, the increase in the quantity of the new mixture is because of adding extra water. Hence, the quantity of water added = 80 – 40 = 40 liters

Mixing with replacements

Few varieties are created in the problems based on the replacement. For each type, we use a different formula.

Type I: When quantity withdrawn and quantity replaced are same:

From x liters of pure, a liters of milk is replaced with equal volume of water. Again b liters of mixture is replaced with b liters of water and so on. At the end of the operation, the volume of milk in the final solution is given by:

Final volume of milk = \(x\left( {1 – \frac{a}{x}} \right)\left( {1 – \frac{b}{x}} \right)\left( {1 – \frac{c}{x}} \right) \ldots .\)

And if same volume is replaced each time, i.e. a = b = c=.. , then the formula reduces to:

Final volume of milk = \(x{\left( {1 – \frac{a}{x}} \right)^n}\)where, n is the number of time the operation is repeated.

Problem 3: From the 40 litres solution of pure milk, 5 litres of milk is replaced with equal quality of water. Again 5 litres of the mixture is substituted with 5 litres of water. This operation is repeated one more time. Find the volume of milk in the final solution.

Solution:

Applying the above formula, we get:

Final volume of milk = \(40{\left( {1 – \frac{5}{{40}}} \right)^3}\)=26.8 liters (approx).

Type II: When quantity withdrawn and quantity replaced are NOT same:

Let us take an example to see the application of the method.

Problem 4: From the 40 litres solution of pure milk, 5 litres of milk is replaced with 6 litres of water. Next time, 6 litres of the mixture is replaced with 7 litres of water. Find the volume of milk in the final solution.

After the first operation, the quantity of milk and water are in the ratio 35:6. So the quantity of milk left after the first operation = 35 litres.  Now, when 6 litres of the mixture is withdrawn, the quantities of milk and water took out will be in the ratio 35:6. Therefore, the quantity of milk withdrawn = \(\frac{{35}}{{41}} \times 6\) liters.

Hence the quantity of milk left = \(40 \times \frac{{35}}{{41}}\)

If the operation is repeated n number of times, we have to repeat the same process ‘n’ number of times.

Mixing Two different Mixtures

Problem 5: In two alloys A and B, the ratio of zinc to tin is 5 : 2 and 3 : 4 respectively. How much kg of alloy A should be mixed with 21 kg of the alloy B such that in the final mixture, the quantities of zinc are tin are equal.

Solution:

The problem can be solved by using rule of alligation by making minor adjustments. Since the alloy A and B are already mixture of zinc and tin, we take only one but same element from each alloy i.e. either zinc or tin to represent the respective alloys.

Let us consider zinc only. The portion of zinc in both the alloy A and B are 5/7 and 3/7 respectively. Also, in the final mixture, the quantities of zinc are tin are equal, the portion of zinc in the final mixture = 1/2

Now, applying the rule we get:

Or the ratio in which the alloys A and B should be mixed = 1 : 3. Therefore, the required quantity of alloy A =\(\frac{1}{3} \times 21\)=7 kg.

Note: Instead of zinc, if we take tin, the result remains same.

Practice Problems

• Mixture and Alligation problems set 01
• Mixture and Alligation problems set 02
• Mixture and Alligation problems set 03

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