SOLUTION: A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out so that the radiator

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out so that the radiator      Log On


   

Question 765298: A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out so that the radiator will then contain the desire 50% antifreeze solution.
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The question really should be changed. How about this way:

A 5 gallon radiator containing a mixture of water and antifreeze solution. When tested , it was found to have only 40% antifreeze. How much must be drained out AND REPLACED WITH PURE ANTIFREEZE so that the radiator will then contain 5 GALLONS OF THE desired 50% antifreeze solution?

Let v = the amount of solution to remove and then replaced with pure antifreeze

%285%2A40-v%2A40%2Bv%2A100%29%2F5=50
Note that the desired final volume must be the same as the starting volume in the five-gallon tank.

Arithmetic steps to solving:
(5*40+(100-40)v)/5=50
5%2A40%2B60v=5%2A50
60v=5%2A50-5%2A40
60v=50
highlight%28v=5%2F6%29, meaning, take out five-sixths of a gallon and replace with 100% antifreeze.