Question 301827
Let {{{a}}} = liters of 60% solution needed
Let {{{b}}} = liters of 80% solution needed
given:
{{{.6a}}} = acid in 60% solution
{{{.8b}}} = acid in 80% solution
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(1) {{{a + b = 20}}} liters
{{{(.6a + .8b)/20 = .65}}}
{{{.6a + .8b = 13}}}
(2) {{{6a + 8b = 130}}}
Multiply both sides of (1) by {{{6}}} and
subtract from (2)
(2) {{{6a + 8b = 130}}}
(1) {{{ -6a - 6b = -120}}}
{{{2b = 10}}}
{{{b = 5}}}
{{{a = 15}}}
15 liters of 60% solution and 5 liters of 80% solution are needed