SOLUTION: I'm struggling a bit with word problems and mixtures that contain percentages. Maybe you can help. Mix a solution that is 30% alcohol with a solution that is 80% alcohol to mak

Question 1102865: I'm struggling a bit with word problems and mixtures that contain percentages. Maybe you can help.
Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 625 mL of a solution that is 60% alcohol. How much of each solution should you use?
30% Soulution: __ ML
80% Solution: __ ML
Found 2 solutions by ankor@dixie-net.com, richwmiller:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 625 mL of a solution that is 60% alcohol.
How much of each solution should you use?
:
let x = amt of 80% solution
we know the total amt will be 625, therefore
(625-x) = amt of 30% solution
:
the equation
.80x + .30(625-x) = .60(625)

Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
We want to know how many mL. at 30% to mix with 80%.
We need a total of 625 mL. at 60%
Using the method of alligation.

80 30 30/50*625=375 mL of 80% solution 60 375+250= 625 mL of 60% solution 30 20 20/50*625=250 mL of 30% solution 50

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