Question 1108141
Let {{{ a }}} = liters of 30% sloution needed
Let {{{ b }}} = liters of 60% solution needed
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(1) {{{ a + b = 30 }}}
(2) {{{ ( .3a + .6b ) / 30 = .4 }}}
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(2) {{{ .3a + .6b = 12 }}}
(2) {{{ 3a + 6b = 120 }}}
(2) {{{ a + 2b = 40 }}}
Subtract (1) from (2)
(2) {{{ a + 2b = 40 }}}
(1) {{{ -a - b = -30 }}}
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{{{ b = 10 }}}
and
(1) {{{ a + 10 = 30 }}}
(1) {{{ a = 20 }}}
He needs 20 liters of 30% solution
and 10 liters of 60% solution
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check:
(2) {{{ ( .3a + .6b ) / 30 = .4 }}}
(2) {{{ ( .3*20 + .6*10 ) / 30 = .4 }}}
(2) {{{ ( 6 + 6 ) / 30 = .4 }}}
(2) {{{ 12 = .4*30 }}}
(2) {{{ 12 = 12 }}}
OK