#
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.