Question 1184883: A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?
Found 4 solutions by mananth, josgarithmetic, MathTherapy, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! The first acid solution contains 25 % acid, --x liters
the second contains 40 % -- y liters
and the third contains 85 % -- z liters
x+y+z =135-------(1)
Quantities used Concentration
25%x + 40%y + 85% z = 35% (x+y+z)
25x + 40y +85z = 35x+35y+35z
50z +5y-10x =0 --------(2)
z=2y------------------(3)
substitute z=2y
in (1) we get x+3y =135
in (2) we get 105y-10x= =0
10x=105y
x=10.5 y
x+3y =135
substitute x=10.5 y
13.5y =135
y =10 liters
z=2y
z = 20 liters
x =105 liters
The first acid solution 25% , --105 liters
the second contains 40 % -- 10 liters
and the third contains 85 % -- 20 liters
Answer by josgarithmetic(39618)  (Show Source):
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?
Let amount of 40% solution to mix, be F
Then amount of 85% solution to mix = 2F
Also, amount of 25% to mix = 135 - F - 2F = 135 - 3F
We then get: .4F + .85(2F) + .25(135 - 3F) = .35(135)
.4F + 1.7F + .25(135) - .75F = .35(135)
.4F + 1.7F - .75F = .35(135) - .25(135)
1.35F = .1(135)
Amount of 40% to be mixed, or 
You should easily be able to calculate the amount of 25% and 85% solutions, to mix.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
You have received responses showing three different algebraic solutions -- one using three variables, one using two, and one using one.
Use the solution from the tutor who uses one variable. Taking the little extra time to figure out how to set up the problem using a single variable will save you a lot of time in solving the problem.
Here then is a completely different method for solving this kind of problem, without formal algebra. The words of explanation make it sound long and difficult; but the calculations are relatively simple, so if your mental math is good you can solve the problem quickly by this method.
First consider mixing the 40% and 85% acid, using twice as much of the 85% acid. That means 2/3 of this mixture will be the 85% acid; and that means the percentage of this mixture will be 2/3 of the way from 40% to 85% -- which is 70%.
Then consider mixing this 70% acid solution with the 25% acid solution to get a 35% acid solution. 35% is (35-25)/(70-25)=10/45=2/9 of the way from 25% to 70%, so 2/9 of the 135 liters total, or 30 liters, is the 70% acid solution.
Then, since the 70% acid solution contains twice as much 85% acid as 40% acid, the mixture contains 10 liters of 40% acid and 20 liters of 85% acid.
Then the amount of 25% acid solution used in the mixture is 135-30=105 liters.
ANSWERS:
25%: 105 liters
40%: 10 liters
85%: 20 liters
CHECK:
.25(105)+.40(10)+.85(20)=26.25+4+17=47.25
.35(135)=47.25
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