SOLUTION: You need 350 mL of a 40% alcohol solution. On hand, you have a 35% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

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Question 722539: You need 350 mL of a 40% alcohol solution. On hand, you have a 35% alcohol mixture and a 60% alcohol mixture. How much of each mixture will you need to obtain the desired solution?
Answer by fcabanski(1391) About Me  (Show Source):
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These problems involve two equations. The first equation is the raw amount. In this case call the 35% solution x and the 60% solution y, and the amount is x + y = 350.


The second equation is the amount of alcohol. The 350 mL desired solution has 40% alcohol, so it has .4 * 350 = 140mL of alcohol. The 35% solution has .35x mL of alcohol, and the 60% solution has .6y mL of alcohol. 140 = .35x + .6y


Solve the first equation for one variable in terms of the other. x = 350-y. Substitute that into the second equation, and then solve that equation for the remaining variable.


140 = .35 * (350-y) + .6y ---> 140 = 122.5 - .35y +.6y --->17.5 = .25y ---> 70 = y.


x = 350 - y = 350-70 = 280. You need 280 mL of the 35 % solution and 70 mL of the 60% solution.

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