Question 995434
This is 2 problems in 1
Let {{{ a }}} = grams of 80% alloy needed
Let {{{ b }}} = grams of 65% alloy needed
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First solve for 72% silver end result
(1) {{{ a + b = 30 }}}
(2) {{{ ( .8a + .65b ) / 30 = .72 }}}
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(2) {{{ .8a + .65b = .72*30 }}}
(2) {{{ .8a + .65b = 21.6 }}}
(2) {{{ 80a + 65b = 2160 }}}
(2) {{{ 16a + 13b = 432 }}}
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Multiply both sides of (1) by {{{ 13 }}}
and subtract (1) from (2)
(2) {{{ 16a + 13b = 432 }}}
(1) {{{ -13a - 13b = -390 }}}
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{{{ 3a = 42 }}}
{{{ a = 14 }}}
and
(1) {{{ 14 + b = 30 }}}
(1) {{{ b = 16 }}}
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Now solve for 75% alloy end result
(1) {{{ a + b = 30 }}}
(2) {{{ ( .8a + .65b ) / 30 = .72 }}}
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(2) {{{ .8a + .65b = .75*30 }}}
(2) {{{ .8a + .65b = 22.5 }}}
(2) {{{ 80a + 65b = 2250 }}}
(2) {{{ 16a + 13b = 450 }}}
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Multiply both sides of (1) by {{{ 13 }}}
and subtract (1) from (2)
(2) {{{ 16a + 13b = 450 }}}
(1) {{{ -13a - 13b = -390 }}}
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{{{ 3a = 60 }}}
{{{ a = 20 }}}
and
(1) {{{ 20 + b = 30 }}}
(1) {{{ b = 10 }}}
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From these results,
the least amount of 80% alloy needed is 14 grams ( making 72% silver )
the most amount of 80% alloy needed is 20 grams ( making 75% silver )
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You can check the answers by plugging
{{{ a }}} and {{{ b }}} back into (2) for both
72% and 75% cases