<PRE><B>Question 219499</B>
This assumes that the radiator will hold whatever added
antifreeze you try to put in. 
In words:
(final liters of antifreeze)/(final liters of total mixture) = 50%
Let {{{a}}} = liters of antifreeze to be added
{{{(.2*10 + a)/(10 + a) = .5}}}
Multiply both sides by {{{10 +a}}}
{{{2 + a = .5*(10 + a)}}}
{{{2 + a = 5 + .5a}}}
{{{.5a = 5 - 2}}}
{{{.5a = 3}}}
{{{a = 6}}}
6 liters of antifreeze needs to be added to make mixture 50%
check answer:
{{{(.2*10 + 6)/(10 + 6) = .5}}}
{{{(2 + 6)/16 = .5}}}
{{{8/16 = .5}}}
{{{.5 = .5}}}
OK</PRE>