Question 418175
Let {{{x}}} = quarts that must be drained off
In words:
(quarts of antifreeze in final solution)/(quarts of final solution) = 70%
given:
{{{ .55*20 = 11}}} quarts of antifreeze in radiator originally
{{{.55x}}} = quarts of antifreeze in solution  drained off
{{{11 - .55x}}} = quarts of antifreeze in solution left after
draining off {{{x}}} quarts
{{{.8x}}} = quarts of antifreeze in solution added
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{{{ (11 - .55x + .8x)/(20 + x - x) = .7 }}}
{{{ 11 + .25x = .7*20 }}}
{{{ .25x = 14 - 11 }}}
{{{x = 3/.25 }}}
{{{x = 12}}}
12 quarts must be drained off and 12 quarts of 80% solution
must be added
check answer:
{{{ (11 - .55x + .8x)/ 20 = .7 }}}
{{{ (11 - .55*12 + .8*12)/ 20 = .7 }}}
{{{ (11 - 6.6 + 9.6) / 20 = .7}}}
{{{ 14 = .7*20}}}
{{{14 = 14}}}
OK