SOLUTION: How much pure water must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% anitfreeze solution?
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Question 132499This question is from textbook Holt Algebra 1
: How much pure water must be mixed with 2 liters of a 40% solution of antifreeze to get a 25% anitfreeze solution?
This question is from textbook Holt Algebra 1
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
2*.4=.25(x+2)
.8=.25x+.5
.25x=.8-.5
.25x=.3
x=.3/.25
x=1.2 liters of pure water is needed.
proof
.2*.4=.25(1.2+2)
.8=.25*3.2
.8=.8
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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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