Question 662267
A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 55% antifreeze?
<pre>
When we drain out x quarts there will only be 15-x quarts left in there.

30% of those 15-x quarts is pure antifreeze. So that's .30(15-x) quarts 
of pure antifreeze in there after we have drained off x quarts.

Now we replace those x quarts with pure antifreeze.  Now the amount
of pure antifreeze is .30(15-x)+x quarts.  There are 15 quarts of liquid
in the radiator, so the fraction of the 15 quarts that is pure antifreeze
is this fraction:  

               {{{(.30(15-x)+x)/15}}}

And that fraction is to be equal to 55% or .55

               {{{(.30(15-x)+x)/15}}} = .55

Multiply both sides by 15

               .30(15-x)+x = 8.25

Solve that and get {{{75/14}}} or {{{5&5/14}}} quarts.

Edwin</pre>