# How to Solve Mixture Problems Part 2

In the **previous post**, we have learned the basics of mixture problems. We have learned that if solutions are added, then the pure content of the combined solution is equal to the sum of the all the amount of pure content in the added solutions.

In this post, we are going to discuss two more mixture problems. We have already finished two examples in the previous part, so we start with Example 3.

**Example 3**

How many liters of 80% alcohol solution must be added to 60 liters of 40% alcohol solution to produce a 50% alcohol solution.

*Solution and Explanation*

The first thing that you will notice is that we donâ€™t know the amount of liquid with 80% alcohol solution. So, if we let x = volume of the solution of liquid with 80% alcohol content. So, the amount of pure alcohol content is 80% times x.

We also know that the amount of alcohol in the second solution is 60% times 40.

Now, if we let solution 1 be equal to the solution with 80% alcohol, solution 2 with 40% alcohol, and solution 3 be the combined solutions, we have

amt of alchohol in solution 1 = 0.8x

amt of alcohol in solution 2 = (0.4)(60)

amt of alchohol in solution 3 = 0.50(0.8x + 60)

Note that we have already **converted the percentages to decimals** in the calculation above: 80% = 0.8, 40% = 0.4, and 50% = 0.5.

amt. of alcohol in solution 1 + amt. of alcohol in solution 2 = amt. of alcohol in solution 3

Solving, we have

0.8x + (0.4)(60) = 0.50(x + 60)

0.8x + 24 = 0.5x + 30

0.8x – 0.5x = 30 – 24

0.3x = 6

To get rid of the decimal, we multiply both sides by 10.

3x = 60

x = 60/3

x = 20.

This means that we need 20 liters of solution 1, the solution containing 80% alcohol. We need to combine this to solution 2, to get solution 3 which has a 50% alcohol content.

In problems like this, you can check your answers by substituting the value of x to the original equation.

0.8x + (0.4)(60) = 0.50(x + 60)

0.8(20)+ (0.4)(60) = 0.50(20 + 60)

16 + 24 = 0.50(80)

40 = 40

Indeed, the amount of alcohol in the left hand side is the same as the amount of alcohol in the right hand side.