SOLUTION: can someone help me? this is my question: how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of 25% solution to give 30% solution?

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Question 253008: can someone help me? this is my question:
how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of 25% solution to give 30% solution?
Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of 25% solution to give 30% solution?
:
Let x = amt of 80% alcohol required to accomplish this
:
A simple "amt of alcohol equation"
.80x + .25(6) = .30(x+6)
:
.8x + 1.5 = .3x + 1.8
:
.8x - .3x = 1.8 - 1.5
:
.5x = .3
x =
x = .6 gal of 80% solution required
:
:
Check solution in original equation
.8(.6) + .25(6) = .3(.6 + 6)
.48 + 1.5 = .3(6.6)

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
how many gallons of a mixture containing 80% alcohol should be added to 6 gallons of 25% solution to give 30% solution?
--------------------------
6*0.25 + x*0.8 = (6+x)*0.3
1.5 + 0.8x = 1.8 + 0.3x
0.5x = 0.3
x = 0.6 gallons added