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"
