# How many liters of a solution containing 45% acid must be mixed with a solution containing 60% acid to obtain 40 liters of a 48% acid solution?

Nov 29, 2015

"32 L" -> 45%
"8 L" -> 60%

#### Explanation:

In order to solve this problem, you need to set up two equations with two unknowns, the volume of the solutions and the volume of the solutes.

Use the percent concentration by volume and the volume of the target solution to figure out how much acid it must contain

40 color(red)(cancel(color(black)("L solution"))) * "48 L acid"/(100color(red)(cancel(color(black)("L")))) = "19.2 L acid"

Now, if $x$ represents the volume of the 45% acid solution and $y$ the volume of the 60% solution, then you can say that

$x + y = 40$

${\overbrace{\frac{45}{100} \cdot x}}^{\textcolor{red}{\text{acid from 1"^"st" "solution")) + overbrace(60/100 * y)^(color(green)("acid from 2"^"nd" "solution")) = overbrace(19.2)^(color(blue)("acid in target solution}}}$

This means that you have

$x = 40 - y$

Which gets you

$\frac{45}{100} \cdot \left(40 - y\right) + \frac{60}{100} \cdot y = 19.2$

$1800 - 45 y + 60 y = 1920$

$15 y = 1920 - 1800 \implies y = \frac{120}{15} = 8$

Therefore, $x$ will be equal to

$x = 40 - 8 = 32$

So, to get $\text{40 L}$ of a 48% acid solution, you need to mix $\text{32 L}$ of 45% acid solution and $\text{8 L}$ of 60% acid solution.