Question 800405
Let {{{ a }}} = ml of 80% solution needed
{{{ .8a }}} = ml of acid in 80% solution
Let {{{ b }}} = ml of 20% solution needed
{{{ .2b }}} = ml of acid in 20% solution
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(1) {{{ a + b = 120 }}}
(2) {{{ ( .8a + .2b ) / 120 = .5 }}}
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(2) {{{ .8a + .2b = .5*120 }}}
(2) {{{ .8a + .2b = 60 }}}
(2) {{{ 8a + 2b = 600 }}}
(2) {{{ 4a + b = 300 }}}
Subtract (1) from (2)
(2) {{{ 4a + b = 300 }}}
(1) {{{ -a - b = -120 }}}
{{{ 3a = 180 }}}
{{{ a = 60 }}}
and, since
(1) {{{ a + b = 120 }}}
(1) {{{ b = 60 }}}
60 ml of 80% acid solution are needed
check:
(2) {{{ ( .8*60 + .2*60 ) / 120 = .5 }}}
(2) {{{ ( 48 + 12 ) / 120 = .5 }}}
(2) {{{ 60 = .5*120 }}}
(2) {{{ 60 = 60 }}}
OK