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I need 100 liters of 50% alcohol solution. If I only have 30% and 80% solutions in...

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mountainridge1 | eNotes Newbie

Posted June 25, 2013 at 5:45 PM via web

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I need 100 liters of 50% alcohol solution. If I only have 30% and 80% solutions in stock, how many liters of each should I mix to get the 100 liters of 50% alcohol?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 25, 2013 at 5:54 PM (Answer #1)

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30% and 80% solutions of alcohol have to be mixed to obtain a 50% solution. The volume of the 50% solution required is 100 liters.

Let the number of liters of 30% solution to be used be x, the number of solutions of 80% solution is 100 - x.

0.3*x + 0.8*(100 - x) = 0.5*100

=> 0.3*x + 80 - 0.8x = 50

=> 0.5x = 30

=> x = 60

To obtain the 100 l of 50% solution 60 liters of 30% solution and 40 liters of 80% solution are required.

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mountainridge1 | eNotes Newbie

Posted July 9, 2013 at 5:54 PM (Reply #1)

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Why is 80% represented by 100-x and why is 30% represented as x? Part of the problem is that I don't know how to convert percentages. Could you explain just a little more thouroughly? I really appriciate the help!

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