Question 742877
Let {{{ a }}} = mL of 10% solution needed
Let {{{ b }}} = mL of 30% solution needed
{{{ .1a }}} = mL of alcohol in 10% solution
{{{ .3a }}} = mL of alcohol in 30% solution
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(1) {{{ ( .1a + .3b ) / 200 = .25 }}} mL
(2) {{{ a + b = 200 }}} mL
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(1) {{{ .1a + .3b = 50 }}}
(1) {{{ a + 3b = 500 }}}
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Subtract (2) from (1)
(1) {{{ a + 3b = 500 }}}
(2) {{{ -a - b = -200 }}}
{{{ 2b = 300 }}}
{{{ b = 150 }}}
and, since
(2) {{{ a + 150 = 200 }}}
(2) {{{ a = 50 }}}
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50 mL of 10% solution is needed
150 mL of 30% solution is needed
check:
(1) {{{ ( .1*50 + .3*150 ) / 200 = .25 }}}
(1) {{{ ( 5 + 45 ) / 200 = .25 }}}
(1) {{{ 50 = .25*200 }}}
(1) {{{ 50 = 50 }}}
OK