SOLUTION: In order to create 3oo milliliters of a solution that is 60% alcohol, a solution that is 50% ALCOHOL is mixed with a solution that is 80% alcohol. Ho much of each solution is requ
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Question 7769: In order to create 3oo milliliters of a solution that is 60% alcohol, a solution that is 50% ALCOHOL is mixed with a solution that is 80% alcohol. Ho much of each solution is required?
__________milliliters of the 50% solution and
__________milliliters of the 80% solution. Answer by prince_abubu(198) (Show Source):
You can put this solution on YOUR website! It looks like you want to dilute (AKA, lessen/weaken) the alcohol concentration from 80% to 60% by mixing in a 50% alcohol solution.
Let x = the mL's of the 80% alcohol.
Let 300 - x = the mL's of the 50% alcohol. (We know that the mLs of the 80% mix and the 50% mix MUST equal to 300. Let's say that x is the mL's of the 80% alcohol and y is the mL's of the 50% alcohol. Then x + y = 300 because their volumes must add up to the 300 mL. Solving for y would mean getting the expression for the amount of 50% alcohol. That would be y = 300 - x.)
So, we know that there's x mL of the 80% alchol. That's the 0.8x.
We know that the 50% alcohol is (300 - x) mL, which translates to 0.5(300 - x) Since we're adding this 50% solution to the 80%, our equation so far is 0.8x + 0.5(300 - x).
After those two are combined, we should get a mixture that is 300 mL AND 60%. So, 0.8x + 0.5(300 - x) = 0.6(300). Now we're going to solve for x, the amount of 80% solution we started with.
<---- used distributive property <--- combined like terms . So there were 100 mL of the 80% solution. If the entire mixture was 300 mL in total, then there would have been 200 mL of the 50% solution.
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