# Question f43b7

Nov 28, 2016

You must mix 5 L of 80 % alcohol and 1 L of 20 % alcohol.

#### Explanation:

I assume that there is no change of volume on mixing.

Then

$\text{Volume of 80 % + Volume of 20 % = 6 L of 70 %}$

Let $\text{Volume of 80 %" = x color(white)(l)"L}$; then $\text{Volume of 20 %" = (6 - x)color(white)(l) "L}$

Also,

$\text{Volume of alcohol in 80 %"color(white)(l) + "Volume of alcohol in 20 %" = "Volume of alcohol in 70 %}$

0.80 × x color(red)(cancel(color(black)("L"))) + 0.20(6 - x) color(red)(cancel(color(black)("L"))) = 0.70 × 6 color(red)(cancel(color(black)("L")))#

$0.80 x + 1.2 - 0.20 x = 4.2$

$0.60 x = 3.0$

$x = \frac{3.0}{0.60} = 5.0$

$\text{Volume of 80 %" = x color(white)(l)"L" = "5 L}$

$\text{Volume of 20 %" = "(6 - 5) L" = "1 L}$