Question 301358
Let {{{a}}} = liters of 60% solution needed
Let {{{b}}} = liters of 80% solution needed
given:
(1) {{{a + b = 20}}} liters
{{{.6a + .8b = .65*20}}}
{{{.6a + .8b = 13}}}
(2) {{{6a + 8b = 130}}}
Multiply both sides of (1) by {{{6}}} and subtract (1) from (2)
{{{6a + 8b = 130}}}
{{{-6a - 6b = -120}}}
{{{2b = 10}}}
{{{b = 5}}}
and, since
{{{a + b = 20}}}
{{{a + 5 = 20}}}
{{{a = 15}}}
15 liters of 60% solution and 5 liters of 80% solution are needed
check:
In words:
(liters of acid you end up with) / (final total liters of solution) = 65%
{{{(.6*15 + .8*5)/20 = .65}}}
{{{9 + 4 = .65*20}}}
{{{13 = 13}}}
OK