Question 544050
Let {{{a}}} = ounces of 70% solution needed
Let {{{b}}} = ounces of 20% solution needed
given:
{{{ .7a }}} = alcohol in 70% solution
{{{ .2b }}} = alcohol in 20% solution
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In words:
( ounces of alcohol in 70% and 20% solutions) / ( ounces in final mixture ) = 50%
given:
(1) {{{ a + b = 20 }}}
(2) {{{ ( .7a + .2b ) / 20 = .5 }}}
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(2) {{{ .7a + .2b = 10 }}}
(2) {{{ 7a + 2b = 100 }}}
Multiply both sides of (1) by {{{2}}}
and subtract (1) from (2)
(2) {{{ 7a + 2b = 100 }}}
(1) {{{ -2a - 2b = -40 }}}
{{{ 5a = 60 }}}
{{{ a = 12 }}}
and, since
(1) {{{ a + b = 20 }}}
(1) {{{ 12 + b = 20 }}}
(1) {{{ b = 8 }}}
12 ounces of 70% solution and 8 ounces of 20% solution are needed
check answer:
(2) {{{ ( .7a + .2b ) / 20 = .5 }}}
(2) {{{ ( .7*12 + .2*8 ) / 20 = .5 }}}
(2) {{{ ( 8.4 + 1.6 ) / 20 = .5 }}}
(2) {{{ 10/20 = .5 }}}
OK