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Mixture and alligation Advance Concepts

Rule of Constant

Example 4: 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 replacement

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Type I: When quantity withdrawn and quantity replaced are same: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

Example 5: 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.

Example 6: 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.



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