Mixture Problems

This chapter discusses a common type of word problem that can be solved by linear equations: mixture problems. Read the chapter, watch the videos, and work through examples. Complete the review exercise at the end of the chapter.

Mixture Problem: Chemistry

A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use?

To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution ( $x$ ) and the amount of dilute solution ( $y$ ). We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml) and the final amount of solute ((15% of 500 ml=75 ml). Our equations will look like this:

Volume equation: $x + y = 500$

Solute equation: $0.6x+0.05y=75$

To isolate a variable for substitution, we can see it's easier to start with equation 1:

$x+y=500$	subtract y from both sides:
$x=500-y$	now substitute into equation 2:
$0.6(500-y)+0.05 y=75$	distribute the 0.6:
$300-0.6 y+0.05 y=75$	collect like terms:
$300-0.55 y=75$	subtract 300 from both sides:
$-0.55 y=-225$	divide both sides by -0.55:
$\underline {y=409 m l}$	substitute back into equation for x:
$x=500-409=91 m l$

So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.