<PRE><B>Question 515491</B>
An automobile radiator contains 16L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that the mixture will be 50% antifreeze?
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Equation:
alcohol - alcohol + alcohol = alcohol
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0.30*16 - 0.30x + 1.00x = 0.50*16
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30*16 - 30x + 100x = 50*16
70x = 20*16
x = (2/7)16 = 4.57 L (amt. to be removed and replaced)
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Cheers,
Stan H.</PRE>