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?
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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?
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Let x = amount (liters) of 70% alcohol
then
16-x = amount (liters) of 20% alcohol
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.70x + .20(16-x) = .50(16)
.70x + 3.2-.20x = 8
.50x + 3.2 = 8
.50x = 4.8
x = 9.6 liters (70% alcohol)
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20% alcohol:
16-x = 16-9.6 = 6.4 liters
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.
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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
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70%*x + 20%*(16-x) = 50%*16
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.7x + .2(16-x) = 8
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multiply by 10 to eliminate decimals
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7x + 2(16-x) = 80
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7x + 32 -2x = 80
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5x = 48
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x = 48/5 = 9 3/5
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16-x = 80/5 - 48/5 = 32/5 = 6 2/5
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So, you add 9 3/5 liters of of 70% and 6 2/5 liters of 20% to produce 16 liters of 50% alcohol.
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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.
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Done.
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