SOLUTION: How many liters of 6% acid solution must be added to 12 L of a 10% acid solution to make a solution which is 9% acid? I know that the( # of liters of acid from the weak soultio

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Question 841575: How many liters of 6% acid solution must be added to 12 L of a 10% acid solution to make a solution which is 9% acid?
I know that the( # of liters of acid from the weak soultions ) + ( the # of liters of acid from the strong solution ) = ) the liters of acid I'm the mix solution )
I just don't know how to set up the equation. .
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
let x equal the number of liters of 6% solution and let y equal the number of liters of the 9% solution.

the equation should be:

.06x + .10*12 = .09*y

Since x + 12 will be equal to y, you can replace y with x + 12 to get:

.06x + .10*12 = .09*(x+12)

now you just solve for x.

start with:
.06x + .10*12 = .09*(x+12)
simplify to get:
.06x + 1.2 = .09x + .09*12
simplify further to get:
.06x + 1.2 = .09x + 1.08
subtract .06x from both sides of this equation and subtract 1.08 from both sides of this equation to get:
1.2 - 1.08 = .09x - .06x
simplify to get:
.12 = .03x
divide both sides of this equation by .03 to get:
.12/.03 = x which simplifies to x = 4

your answer should be x = 4
this means you mix 4 liters of 6% solution with 12 liters of 10% solution to get 16 liters of 9% solution.
.06*4 = .24
.10*12 = 1.2
add them together to get 1.44 liters of acid.
divide 1.44 by 16 liters total solution to get .09.
that's your solution.