<PRE><B>Question 181478</B>
The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water.
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The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze.
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If the capacity of the raditor is 3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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antifreeze - antifreeze + + antifreeze = antifreeze
0.65*3 - 0.65x + 0*x = 0.50*3 

Multiply thru by 100 to get:

65*3 - 65x + 0*x = 50*3

195 = 65x + 0 = 150
-65x = -45
x = 9/13 liters (amt. of liguid that must be replace)
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Cheers,
Stan H.
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