SOLUTION: A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to
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Question 1170748: A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 64 liters containing
45% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used? Found 2 solutions by Boreal, MathTherapy:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x liters of .15 or .15x pure acid
y liters of .35 or .35y pure acid
2y liters of .65 or 1.3y pure acid
total amount is 64 liters of .45 acid or 28.8 liters of pure acid
x+3y=64 liters
.15x+1.65y=28.80
15x+45y=960
15x+165y=2880 subtract second
-120y=-1920
y=16 liters of .35 (5.6 l per acid)
2y=32 l of .65 (20.8)
x=16 l of .15 (2.4)
You can put this solution on YOUR website!
A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 64 liters containing
45% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?
Let amount of 35% acid to be used be T
Then amount of 65% acid to be used is 2T, and amount of 15% acid to be used is, 64 - (T + 2T) = 64 - 3T
We then get: .35T + .65(2T) + .15(64 - 3T) = .45(64)
.35T + 1.3T + 9.6 - .45T = 28.8
1.2T = 19.2
Amount of 35% acid to be used, or
Amount of 65% acid to be used:
Amount of 15% acid to be used:
