Question 909062
Let {{{ a }}} = liters of 20% mixture needed
{{{ .2a }}} = liters of the concentrate in 20% mixture
Let {{{ b }}} = liters of 60% mixture needed
{{{ .6b }}} = liters of the concentrate in 60% mixture
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(1) {{{ a + b = 80 }}}
(2) {{{ ( .2a + .6b ) / 80 = .3 }}}
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(2) {{{ .2a + .6b = 24 }}}
(2) {{{ 2a + 6b = 240 }}}
(2) {{{ a + 3b = 120 }}}
Subtract (1) from (2)
(2) {{{ a + 3b = 120 }}}
(1) {{{ -a - b = -80 }}}
{{{ 2b = 40 }}}
{{{ b = 20 }}}
and
(1) {{{ a + b = 80 }}}
(1) {{{ a + 20 = 80 }}}
(1) {{{ a = 60 }}}
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60 liters of 20% mixture is needed
20 liters of 60% mixture is needed
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check:
(2) {{{ ( .2a + .6b ) / 80 = .3 }}}
(2) {{{ ( .2*60 + .6*20 ) / 80 = .3 }}}
(2) {{{ ( 12 + 12 ) / 80 = .3 }}}
(2) {{{ 24 = .3*80 }}}
(2) {{{ 24 = 24 }}}
OK