SOLUTION: a chemist has 10 gallons of a 30% alcohol solution. how many gallons of pure alcohol should he add to obtain a mixture of 40% alcohol

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Question 163178: a chemist has 10 gallons of a 30% alcohol solution. how many gallons of pure alcohol should he add to obtain a mixture of 40% alcohol
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.30*10+x=.40(10+x)
3+x=4+.4x
x-.4x=4-3
.6x=1
x=1/.6
x=1.667 gallons of pure alcohol is needed.
Proof:
.30*10+1.667=.40(10+1.667)
3+1.667=.40(10+1.667)
4.667=.40*11.667
4.667=4.667