<PRE><B>Question 132499</B>
How much pure water must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% antifreeze solution?
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40% solution DATA:
Amt. = 2 liters ; amt. of active ingredient = 0.4*2 = 0.8 liters
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Pure water DATA:
Amt. = x liters ; amt of active ingredient = 0*x = 0 liters
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Mixture DATA:
Amt. = (2+x) liters ; amt of active = 0.25(2+x) = 0.5+0.25x liters
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EQUATION:
active + active = active
0.8 + 0 = 0.5+0.25x
0.3 = 0.25x 
x = 1.2 liters ( amt. of water needed for the mixture )
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Cheers,
Stan H.
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