Question 891846
Let {{{ a }}} = liters of the 30% solution used
{{{ 2a }}} = liters of the 50% solution used
Let {{{ b }}} = liters of the 10% solution used
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{{{ .3a }}} = liters of acid in 30% solution
{{{ .5*2a = a }}} liters of acid in 50% solution
{{{ .1b }}} = liters of acid in 10% solution
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(1) {{{ ( .3a + a + .1b ) / 50 = .32 }}}
(2) {{{ a + 2a + b = 50 }}}
(2) {{{ 3a + b = 50 }}}
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(1) {{{ 1.3a + .1b = .32*50 }}}
(1) {{{ 1.3a + .1b = 16 }}}
(1) {{{ 13a + b = 160 }}}
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Subtract (2) from (1)
(1) {{{ 13a + b = 160 }}}
(2) {{{ -3a - b = - 50 }}}
{{{ 10a = 110 }}}
{{{ a = 11 }}}
and
{{{ 2a = 22 }}}
and, since
(2) {{{ 3a + b = 50 }}}
(2) {{{ 3*11 + b = 50 }}}
(2) {{{ b = 50 - 33 }}}
(2) {{{ b = 17 }}}
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11 liters of the 30% solution should be used
22 liters of the 50% solution should be used
17 liters of the 10% solution should be used
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check:
(1) {{{ ( .3a + a + .1b ) / 50 = .32 }}}
(1) {{{ ( .3*11 + 11 + .1*17 ) / 50 = .32 }}}
(1) {{{  3.3 + 11 + 1.7 = .32*50 }}}
(1) {{{ 16 = 16 }}}
OK