Question 1020061
Let {{{ x }}} = ounces of 10% alloy needed
{{{ .1x }}} = ounces of copper in 10% alloy
{{{ .9*400 = 360 }}} = ounces of copper in 90% alloy
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{{{ ( .1x + 360 ) / ( x + 400 ) = .3 }}}
{{{ .1x + 360 = .3*( x + 400 ) }}}
{{{ .1x + 360 = .3x + 120 }}}
{{{ .2x = 240 }}}
{{{ x = 1200 }}}
1200 ounces of 10% alloy is needed
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check:
{{{ ( .1x + 360 ) / ( x + 400 ) = .3 }}}
{{{ ( .1*1200 + 360 ) / ( 1200 + 400 ) = .3 }}}
{{{ ( 120 + 360 ) / 1600 = .3 }}}
{{{ 480 / 1600 = .3 }}}
{{{ 480 = .3*1600 }}}
{{{ 480 = 480 }}}
OK