Question

Question #f43b7

Answer 1

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

`"Volume of 80 % + Volume of 20 % = 6 L of 70 %"`

Let `"Volume of 80 %" = x color(white)(l)"L"`; then `"Volume of 20 %" = (6 - x)color(white)(l) "L"`

Also,

`"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")))`

`0.80x + 1.2 - 0.20x = 4.2`

`0.60x = 3.0`

`x = 3.0/0.60 = 5.0`

`"Volume of 80 %" = x color(white)(l)"L" = "5 L"`

`"Volume of 20 %" = "(6 - 5) L" = "1 L"`