Question 711002
Let {{{ x }}} = gallons of 70% antifreeze to be added
{{{ 2 - x }}} = gallons of 20% antifreeze to be added
{{{ .7x }}} = pure antifreeze in 70% mixture
{{{ .2*( 2 - x ) }}} = pure antifreeze in 20% mixture
given:
{{{ ( .7x + .2*( 2 - x ) ) / 2 = .5 }}}
{{{ ( .7x + .4 - .2x ) / 2 = .5 }}}
{{{ .5x + .4 = 1 }}}
{{{ .5x = .6 }}}
{{{ x = 1.2 }}}
{{{ 2 - x = .8 }}}
1.2 gallons of 70% antifreeze needs to be added
.8 gallons of 20% antifreeze needs to be added
check:
{{{ ( .7x + .2*( 2 - x ) ) / 2 = .5 }}}
{{{ ( .7*1.2 + .2*( 2 - 1.2 ) ) / 2 = .5 }}}
{{{ ( .84 + .2*.8 ) / 2 = .5 }}}
{{{ ( .84 + .16 ) / 2 = .5 }}}
{{{ 1/2 = .5 }}}
{{{ .5 = .5 }}}
OK