How many liters of pure water should be mixed with a 8-L solution of 90% acid to produce a mixture that is 90% water?

Question

The answer has to be as a integer or as a decimal.

Answer 1

90(8) + 0(W) = 10(8+W) = 80+ 10W = 720

90% acid + 0% acid = 10% acid

10W = 720-80 = 640

W = 64 Liters of water

Add 64 liters of water to 8 liters of 90% and you get 72 liters of 10% acid = 90% water

alternatively,

10(8) + 100W = 90(8+W) = 720+90W

10% water + 100%water = 90% water

10W = 720-80 = 640