SOLUTION: How many liters of a 70% acid solution are needed to be mixed with 18 liters of a 10% acid solution to get a mixture which is 25% alcohol?

Question 157305: How many liters of a 70% acid solution are needed to be mixed with 18 liters of a 10% acid solution to get a mixture which is 25% alcohol?
Found 2 solutions by checkley77, josmiceli:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.7x+.1*18=.25(x+18)
.7x+1.8=.25x+4.5
.7x-.25x=4.5-1.8
.45x=2.7
x=2.7/.45
x=6 gallons of 70% acid is needed.
Proof:
.7*6+.1*18=.25(6+18)
4.2+1.8=.25*24
6=6

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
First of all, this might be a "trick" question. It talks about acid
solutions giving an alcohol solution. Maybe it's miscopied someway.
Let = liters of 70% solution needed
In words:
(alcohol in 70% solution + alcohol in 10% solution) / (total solution) = 25%

multiply both sides by

liters of 70% solution needed
check:

OK
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