## How to Solve Mixture Problems Part 6

This is the 6th part and last part of the Solving Mixture Problems Series. In the previous 4 parts, we have learned how to solve mixture problems involving percent and in part 5, we have learned how to solve problems involving percents. In this post, we solve another problem involving percent.

A bakery owner wants to box assorted chocolates. Each box is to be made up of packs of black chocolates worth \$10 per pack and packs of ordinary chocolates worth \$7 each pack. How many packs of each kind should he use to make 15 packs which he can sell for \$8 per pack?

Solution

Let x = number of \$10 packs
15 – x = number of \$7 packs

Multiplying the cost per pack and the number of packs we have

(\$10)(x) = total cost of \$10 packs
(\$7)(15 – x) = total cost of the \$7 packs
(\$8)(15) = total cost of all the chocolates

Now, we know that

total cost of \$10 packs + total cost of the \$7 packs = total cost of all the chocolates.

Substituting the values, we have

(\$10)(x) + (\$7)(15 – x) = (\$8)(15).

Eliminating the dollar sign and solving for x, we have

(10)(x) + (7)(15 – x) = (8)(15)
10x + 105 – 7x = 120
3x + 105 = 120
3x = 120 – 105
3x = 15
x = 5.
This means that we need 5 packs of \$10 and 10 packs of \$7 chocolates.

Check:
(\$10)(5) + (\$7)(10) = (\$8)(15)
\$50 + \$70 = \$120
\$120 = \$120
Therefore, we are correct.