Question 214190
This problem assumes that his radiator is big
enough to hold the extra antifreeze he's adding,
just to be aware
Let {{{a}}} = quarts of antifreeze he needs to add
In words:
(original amount of antifreeze + antifreeze to be added)/(final quarts of mixture in radiator) = 20%
--------------------------
Given:
{{{.1*8 = .8}}} quarts of original atifreeze in radiator
{{{(.8 + a)/(8 + a) = .2}}}
{{{.8 + a = .2*(8 + a)}}}
{{{.8 + a = 1.6 + .2a}}}
{{{.8a = .8}}}
{{{a = 1}}}
He must add 1 quart of antifreeze (assuming the radiator can hold
{{{8 + 1 = 9}}} quarts)
check answer:
Now he has {{{.8 + 1 = 1.8}}} quarts of antifreeze in radiator
and 9 quarts total mixture
{{{1.8/9 = .2}}} He has 20% antifreeze
OK