Question 123565
Let {{{x}}} = acid needed to be added
The equation in words is:
(original acid + added acid) / (total acid + water) = 60%
Originally, 
{{{.4*100 = 40}}}l of acid
{{{.6*100 = 60}}}l of water
{{{(40 + x) / (100 + x) = .6}}}
{{{40 + x = .6(100 + x)}}}
{{{40 + x = 60 + .6x}}}
{{{.4x = 20}}}
{{{x = 50}}}l of acid to be added
check:
{{{(40 + x) / (100 + x) = .6}}}
{{{(40 + 50) / (100 + 50) = .6}}}
{{{90 / 150 = .6}}}
{{{3/5 = .6}}}
{{{.6 = .6}}}
OK