Question 416360
Let {{{a}}} = ounces of $4.30/oz alloy needed
Let {{{b}}} = ounces of $1.80/ oz alloy needed
In words:
(cost of 1st alloy + cost of 2nd alloy) / (total ounces of both) = cost/ounce
(1) {{{ (4.3a + 1.8b)/200 = 2.5 }}}
(2) {{{a + b = 200}}}
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(1) {{{ (4.3a + 1.8b)/200 = 2.5 }}}
(1) {{{ 4.3a + 1.8b = 500 }}}
(1) {{{43a + 18b = 5000}}}
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Multiply both sides of (2) by {{{18}}} and
subtract from (1)
(1) {{{43a + 18b = 5000}}}
(2) {{{-18a - 18b = -3600}}}
{{{ 25a = 1400}}}
{{{ a = 56}}}
and,
{{{b = 200 - 56}}}
{{{b = 144}}}
56 ounces of $4.30/oz alloy are needed
144 ounces of $1.80/ oz alloy are needed
check answer:
(1) {{{ (4.3a + 1.8b)/200 = 2.5 }}}
(1) {{{ (4.3*56 + 1.8*144)/200 = 2.5 }}}
(1) {{{ (4.3*56 + 1.8*144)/200 = 2.5 }}}
{{{ 240.8 + 259.2 = 500 }}}
{{{ 500 = 500 }}}
OK