SOLUTION: how many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution?

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Question 353272: how many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution?
Found 3 solutions by Fombitz, stanbon, ewatrrr:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let A+be the number of liters of 14% alcohol.
14A%2B50%2820%29=30%28A%2B20%29
14A%2B50%2820%29=30A%2B30%2820%29
16A=20%2820%29=400
highlight%28A=25%29 liters


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution?
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Equation:
alcohol + alcohol = alcohol
0.50*20 + 0.14x = 0.30(20+x)
Multiply thru by 100 to get:
50*20 + 14x = 30*20 + 30x
16x = 20*20
x = 25 L (amt. of 14% solution needed for the mixture)
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Cheers,
Stan H.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
Let x represent the "total" amount of the 14% solution to be added.
Then (20L + x) would be the total amount of the final 30% solution.
.
question states
.14x + .50*20L = .30( 20L + x)
10L - 6L = .30x - .14x
.
4L = .16x
x = 25L of the 14% solution needs to be added
.
Check our answer
.14*25 + .50*20 = .30*45
3.5 + 10 = 13.5