Question 268025
In words:
(antifreeze started out with - antifreeze removed + antifreeze added)/(total amout of fluid finally in radiator) = 55%
Let {{{x}}} = amount of fluid to be drained
given:
Radiator started with:
{{{10}}} quarts of fluid
{{{10*.25 = 2.5}}} quarts antifreeze
{{{.25x}}} = quarts antifreeze drained
{{{x}}} = quarts of pure antifreeze added
-----------------------------------------
{{{(2.5 - .25x + x)/10 = .55}}}
(Note that radiator stated with {{{10}}} quarts and ended with {{{10}}} quarts)
{{{(1.25x + 2.5)/10 = .55}}}
{{{1.25x + 2.5 = 5.5}}}
{{{1.25x = 3}}}
{{{x = 2.4}}}
2.4 quarts of fluid must be drained and replaced with pure antifreeze
check answer:
{{{(2.5 - .25x + x)/10 = .55}}}
{{{(2.5 - .25*2.4 + 2.4)/10 = .55}}}
{{{(2.5 - .6 + 2.4)/10 = .55}}}
{{{5.5/10 = .55}}}
{{{5.5 = 5.5}}}
OK