<PRE><B>Question 96487</B>
How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
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Antifreeze DATA:
Amount = 1 gallon :Amt of active ingredient = 1 gallon
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Water DATA:
Amount = x gallons ; Amt of active ingredient = 0 gallons
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Mixture DATA:
Amount = x+1 gallons ; Amt of active ingredient = 0.60(x+1) = 0.6x + 0.6
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EQUATION:
active + active = active
1 + 0 = 0.6x+0.6
0.4 = 0.6x
x = (2/3) gallon of water must be added
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Cheers,
Stan H.</PRE>