SOLUTION: A chemist has three solutions of acid that must be mixed to obtain 20 liters of a solution that is 38% acid. Solution A is 30% acid, solution B is 20% acid and solution C is 60% ac

Question 562256: A chemist has three solutions of acid that must be mixed to obtain 20 liters of a solution that is 38% acid. Solution A is 30% acid, solution B is 20% acid and solution C is 60% acid. Because of another chemical in these solutions, the chemist must keep the ratio of solution C to solution A at 2 to 1. How many liters of each should she mix together?

Found 2 solutions by josmiceli, issacodegard:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
given:
(1)
liters of acid in mixture
(2)
(3)
------------
(3)
(2)
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)

Substitute (3) into this

and
(3)
(3)
(3)
(1)
(1)
(1)
(1)
4 liters of A
8 liters of B
8 liters of C are needed
check:
(2)
(2)
(2)
OK
(1)
(1)
(1)
OK
Answer by issacodegard(60) (Show Source): You can put this solution on YOUR website!
Hi, let A,B,C be the total volumes of the solutions, and let x,y,z be the volumes of acid in the solutions A,B, and C respectively.
Then we have the equations,
(0.3)A=x
(0.2)B=y
(0.6)C=z
A+B+C=20 (total final volume)
x+y+z=(0.38)(20) (this must be the acidic part of the final solution)
2A=C
So, we can add the first 3 equations and utilize the last two to get,
(1.5)A+(0.2)B=x+y+z=(0.38)(20)=7.6
3A+B=20
So by solving the system, you should get A=4, B=8, and thus C=8.
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