# How to Solve Mixture Problems Part 5

In the previous posts, we have learned how to solve mixture problems involving percentages and liquid mixture problems. In this post, we are going to solve mixture problems involving prices. Although these two types of problems are different, they are very similar when you set up the equation. Below is our first problem.

Problem

A seller mixes 20 kilograms of candy worth 80 pesos per kilogram to candies worth 50 pesos per kilogram. He sold at 60 pesos per kilogram.

After selling all the candies he discovered that he had no gain or loss. How much of the 50-pesos per kilogram candies did he use?

Solution and Explanation

Let x = number of kilograms of candy worth 50-pesos per kilogram.

The total price of 20 kilograms of candy at 80 pesos per kilogram is (20 kg)(80 pesos/kg) = 1600 pesos.

The total price of x kilograms of candy at 50 pesos per kilogram is (x kg)(50 pesos/kg) pesos.

When we add these 2, the number of kilograms of candy is x + 20 (can you see why?) and it is sold at 60 pesos. So, its total price is (x + 20)(60 pesos/kg).

Using these facts, we have the following equation:

total price of 80pesos/kg candy + total price of 50pesos/kg candies = total price of 60pesos/kg candies.

Substituting the expressions above, we have

(20 kg)(80 pesos/kg) + (x kg)(50 pesos/kg) = (x + 20 kg)(60 pesos/kg)

1600 + 50x = 60(x + 20)
1600 + 50x = 60x + 1200
1600 – 1200 = 60x – 50x
400 = 10x
40 = x.

Therefore, he used 40 kilograms of candy worth 50 pesos per kilogram.

Check:

1600 + 50(40) = 60(40 + 20)
1600 + 2000 = 60(60)
3600 = 3600.

This means that we are correct.