Question 5494: Solution A is 8% acid and Solution B is 4% acid. If a technician wants to mix them to make 150 liters of Solution C which is 6% acid, how many liters of each should be mixed together?
thanks for the help!
Answer by Abbey(339)   (Show Source):
You can put this solution on YOUR website! Let A=volume (quantity) of 8% acid
Let B=volume of 4% solution
Amount of acid in the 8% solution is 8% of the volume, or .08A
Amount of acid in the 4% solution is 4% of the volume, or .04B
So if we want a total of 150 gallons we can say:
the volume of A + the volume of B = 150 gallons, or
A+B=150
this can be rewritten as A=150-B
and the amount of acid we want is .06(150) (6% of the 150 gallons should be acid)
so we can create an equation that states:
,08A+.04B=.06(150)
and since we know that A=150-B, we can substitute this into our equation:
,08A+.04B=.06(150)
,08(150-B)+.04B=.06(150)
Then just solve for B:
12-.08B+.04B=9
12-.04B=9
12=9+.04B
3=.04B
75=B
We need 75 gallons of solution B, so we need 75 gallons of solution A
put these answers into your equations to verify:
,08(75)+.04(75)=.06(150)
This is a true statement, so the answer is correct.
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