<PRE><B>Question 412260</B>
How much pure antifreeze must be added to 12 gallons of 20% antifreeze to make a 50% antifreeze solution?
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Equation:
active + active = active
0.20*12 + 1.00x = 0.50(12+x)
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Multiply thru by 100 to get:
20*12 + 100x = 50*12 + 50x
50x = 30*12
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x = (3/5)12
x = 36/5
x = 7 1/5 gallons (amt. of pure antifreeze needed)
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Cheers,
Stan H.
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