# Problem of the Week Problem D and Solution Dye Refresher

## Problem

To create an ink refresher for dye-based ink, some crafters will mix pure vegetable glycerine with water to get a mixture that is $$12\%$$ vegetable glycerine, by volume. Kathy does not have pure vegetable glycerine, but she does have

• a $$90$$ mL mixture that is $$10.5\%$$ vegetable glycerine,

• a $$120$$ mL mixture that is $$30\%$$ vegetable glycerine, and

• a $$1$$ L mixture that is $$7.5\%$$ vegetable glycerine.

Since Kathy is a math teacher, she knows she can use the contents of these three mixtures to create a mixture that is $$12\%$$ vegetable glycerine, by volume. She combines the contents of the entire $$90$$ mL mixture with the contents of the entire $$120$$ mL mixture, and then adds some of the $$1$$ L mixture. How many millilitres of the $$1$$ L mixture should she add to create a new mixture that is $$12\%$$ vegetable glycerine, by volume?

## Solution

Let $$x$$ be the number of millilitres needed from the $$1$$ L mixture.

The $$90$$ mL mixture that is $$10.5\%$$ vegetable glycerine has $$0.105\times 90 = 9.45$$ mL of vegetable glycerine.

The $$120$$ mL mixture that is $$30\%$$ vegetable glycerine has $$0.30\times 120 = 36$$ mL of vegetable glycerine.

In the $$x$$ mL from the $$1$$ L mixture, there is $$0.075\times x =~0.075x$$ mL of vegetable glycerine.

Therefore, the total amount of vegetable glycerine in the new mixture is $$9.45 + 36 + 0.075x = (45.45 + 0.075x)$$ mL.

The new mixture contains $$90 + 120 + x = (210+x)$$ mL of liquid, of which $$12\%$$ is vegetable glycerine.
Therefore, $$0.12\times (210+x) = (25.2 + 0.12x)$$ mL of the new mixture is vegetable glycerine.

Since we have shown that the amount of vegetable glycerine in the new mixture is $$(45.45 + 0.075x)$$ mL and $$(25.2 + 0.12x)$$ mL, we must have \begin{align} 45.45 + 0.075x &= 25.2 + 0.12x \\ 0.075x-0.12x &= 25.2-45.45 \\ -0.045x &= -20.25 \\ x &= 450 \end{align} Therefore, she should add $$450$$ mL of the $$1$$ L mixture that is $$7.5\%$$ vegetable glycerine.