SOLUTION: How many liters of a 40% antifreeze solution must be mixed with 5 liters of a 70% antifreeze solution to make a mixture that is 60% antifreeze?
I can't figure out how to set thi
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I can't figure out how to set thi
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Question 271287: How many liters of a 40% antifreeze solution must be mixed with 5 liters of a 70% antifreeze solution to make a mixture that is 60% antifreeze?
I can't figure out how to set this up. Found 2 solutions by stanbon, CharlesG2:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How many liters of a 40% antifreeze solution must be mixed with 5 liters of a 70% antifreeze solution to make a mixture that is 60% antifreeze?
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Equation:
alcohol + alcohol = alcohol
0.40x + 0.70*5 = 0.60(x+5)
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Multiply thru by 100
40x + 70*5 = 60x + 60*5
-20x = -10*5
x = 2.5 liters (amt. of 40% solution needed in the mixture)
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Cheers,
Stan H.
You can put this solution on YOUR website! How many liters of a 40% antifreeze solution must be mixed with 5 liters of a 70% antifreeze solution to make a mixture that is 60% antifreeze?
I can't figure out how to set this up.
mixture A + mixture B = mixture C
OR
% in A * A + % in B * B = % in C * C
40% * A + 70% * B = 60% * C
A = x, B = 5 liters, C = 5+x liters
0.40x + 0.70 * 5 = 0.60(5+x)
0.40x + 3.5 = 3 + 0.60x
0.5 = 0.20x
2.5 liters = x
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