SOLUTION: The radiator in a car is filled with a solution of 65 percent antifreeze and 35 percent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling
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Question 401322: The radiator in a car is filled with a solution of 65 percent antifreeze and 35 percent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the radiator is 3.2 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of coolant that must be drained and replaced with water to achieve a 50% antifreeze concentration
Now we now that the amount of pure antifreeze remaining after x amount of coolant is drained out is 0.65(3.2-x) and this has to equal the amount of pure antifreeze in the final mixture after the water is added (0.50*3.2). So our equation to solve is:
0.65(3.2-x)=0.50*3.2 simplify
2.08-0.65x=1.6 subtract 2.08 from each side
-0.65x=-0.48
x=0.738 liters ----amount of coolant that must be drained and replaced
CK
0.65*(3.2-0.738)=0.50*3.2
0.65*2.462=1.6
1.6003~~~~1.6
Hope this helps--ptaylor
