Question 774957
Let {{{ x }}} = ounces of 30% alloy needed
{{{ .3x }}} = ounces of copper in 30% alloy
{{{ .6*600 = 360 }}} ounces of copper in 60% alloy
------------------
{{{ ( .3x + 360 ) / ( x + 600 ) = .4 }}}
{{{ .3x + 360 = .4*( x + 600 ) }}}
{{{ .3x + 360 = .4x + 240 }}}
{{{ .1x = 360 - 240 }}}
{{{ .1x = 120 }}}
{{{ x = 1200 }}}
1200 ounces of 30% alloy are needed
check:
{{{ ( .3x + 360 ) / ( x + 600 ) = .4 }}}
{{{ ( .3*1200 + 360 ) / ( 1200 + 600 ) = .4 }}}
{{{ ( 360 + 360 ) / 1800 = .4 }}}
{{{ 720 = .4*1800 }}}
{{{ 720 = 720 }}}
OK