Question 632119
Let {{{ x }}} = ounces of 20% alloy needed
{{{ .2x }}} = ounces of copper in {{{ x }}} ounces
{{{ .8*500 = 400 }}} ounces of copper in 80% alloy
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{{{ ( .2x + 400 ) / ( x + 500 ) = .3 }}}
{{{ .2x + 400 = .3*( x + 500 ) }}}
{{{ .2x + 400 = .3x + 150 }}}
{{{ .1x = 400 - 150 }}}
{{{ .1x = 250 }}}
{{{ x = 2500 }}}
2500 ounces of 20% alloy are needed
check:
{{{ ( .2*2500 + 400 ) / ( 2500 + 500 ) = .3 }}}
{{{ ( 500 + 400 ) / 3000 = .3 }}}
{{{ 900 = .3*3000 }}}
{{{ 900 = 900 }}}
OK