Hi,

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?

 x + .30(15-x) = .55*15

     .70x = .25*15

        x = .25*15/.70

 

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
 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?

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:  

               

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

                = .55

Multiply both sides by 15

               .30(15-x)+x = 8.25

Solve that and get  or  quarts.

Edwin









     

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