Question 988394
Let {{{ a }}} = ml of 60% solution needed
Let {{{ b }}} = ml of 80% solution needed
{{{ .6a }}} = ml of acid in 60% solution
{{{ .8b }}} = ml of acid in 80% solution
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(1) {{{ a + b = 200 }}}
(2) {{{ ( .6a + .8b ) / 200 = .75 }}}
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(2) {{{ .6a + .8b = .75*200 }}}
(2) {{{ .6a + .8b = 150 }}}
(2) {{{ 6a + 8b = 1500 }}}
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Multiply both sides of (1) by {{{ 6 }}}
and subtract (1) from (2)
(2) {{{ 6a + 8b = 1500 }}}
(1) {{{ -6a - 6b = -1200 }}}
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{{{ 2b = 300 }}}
{{{ b = 150 }}}
and
(1) {{{ a + b = 200 }}}
(1) {{{ a + 150 = 200 }}}
(1) {{{ a = 50 }}}
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50 ml of 60% solution and
150 ml of 80% solution are needed
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check:
(2) {{{ ( .6a + .8b ) / 200 = .75 }}}
(2) {{{ ( .6*50 + .8*150 ) / 200 = .75 }}}
(2) {{{ ( 30 + 120 ) / 200 = .75 }}}
(2) {{{ 150 = .75*200 }}}
(2) {{{ 150 = 150 }}}
OK