SOLUTION: How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16 liters of a mixture containing 50% alcohol?

Question 511931: How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16 liters of a mixture containing 50% alcohol?
Found 2 solutions by nerdybill, oberobic:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16 liters of a mixture containing 50% alcohol?
.
Let x = amount (liters) of 70% alcohol
then
16-x = amount (liters) of 20% alcohol
.
.70x + .20(16-x) = .50(16)
.70x + 3.2-.20x = 8
.50x + 3.2 = 8
.50x = 4.8
x = 9.6 liters (70% alcohol)
.
20% alcohol:
16-x = 16-9.6 = 6.4 liters

Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
To do mixture problems you need to keep track of how much 'pure' stuff you have or need.
In 16 liters of a 50% alcohol solution, you have .5*16 = 8 liters of 'pure' alcohol.
.
Since you have a defined amount to make, you can define your variables as follows:
x = amount of 70% alcohol to add
(16-x) = the amount of 20% alcohol to add
.
70%*x + 20%*(16-x) = 50%*16
.
.7x + .2(16-x) = 8
.
multiply by 10 to eliminate decimals
.
7x + 2(16-x) = 80
.
7x + 32 -2x = 80
.
5x = 48
.
x = 48/5 = 9 3/5
.
16-x = 80/5 - 48/5 = 32/5 = 6 2/5
.
So, you add 9 3/5 liters of of 70% and 6 2/5 liters of 20% to produce 16 liters of 50% alcohol.
.
Check the amount of 'pure' stuff to be sure this answer is correct.
.7*48/5 = 6.72
.2*32/5 = 1.28
6.72 + 1.28 = 8, which we know is the right answer from the first step.
.
Done.
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