<PRE><B>Question 67890</B>
A radiator contains 6L of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 50% antifreeze solution?
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Amt of 25% solution = 6 liters; amt of active ingredient= 0.25*6= 1.5 liters
Amt of 25% removed is x liters; amt of active ingre removed is 0.25x liters
Amt replaced is x liters; amt of active ingredient added  = x liters
Resulting amount is 6 liters ; amt of active ingredient = 0.50(6)= 3 liters
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EQUATION:
active -active + active = active
1.5 - 0.25x + x = 3
0.75x=1.5
x=2 liters
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2 liters of the 25% solution should be drained out and replaced 
with 2 liters of 100% antifreeze.
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Cheers
Stan H. </PRE>