Question 1162778: A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 30% and the third contains 70%. He wants to use all three solutions to obtain a mixture of 81 liters containing 55% acid, using 3 times as much of the 70% solution as the 30% solution. How many liters of each solution should be used?
Found 4 solutions by ankor@dixie-net.com, MathTherapy, Alan3354, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source):
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 30%
and the third contains 70%.
He wants to use all three solutions to obtain a mixture of 81 liters containing 55% acid, using 3 times as much of the 70% solution as the 30% solution.
How many liters of each solution should be used?
:
let a = amt of 15% solution
let b = 30% solution
then, since they want 3 times as much of 70% solution:
let 3b = amt of 70% solution
and then since we know the total is 81, we know that:
a = (81-4b) the 15% solution
:
.15a + .30b + .70(3b) = .55(81)
.15a + .30b + 2.1b = 44.55
.15a + 2.5b = 44.55
replace a with (81-4b)
.15(81-4b) + 2.4b = 44.55
12.15 - .60b + 2.4b = 44.55
-.6b + 2.4b = 44.55 - 12.15
1.8b = 32.4
b = 32.4/1.8
b = 18 liters of 30% solution
then
3(18) = 54 liters of 70% solution
and
81 - 18 - 54 = 9 liters of 15% solution
:
:
:
Check see if this works:
.15(9) + .30(18) + .70(54) = .55(81)
1.35 + 5.4 + 37.8 = 44.55
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 15% acid, the second contains 30% and the third contains 70%. He wants to use all three solutions to obtain a mixture of 81 liters containing 55% acid, using 3 times as much of the 70% solution as the 30% solution. How many liters of each solution should be used?
Quite a few tutors have been PREACHING on here, the need to solve problems like these using ONE VARIABLE. I guess some who continue to use multiple
variables - the likes of the other person who responded - don't feel that they need to learn the most efficient and least complex way to solutions.
Again, I would suggest that you don't solve this problem his way!
Let the amount of 30% solution, be T
Then amount of 70% solution is: 3T
Therefore, the amount of 15% solution is: 81 - (T + 3T), or 81 - 4T
We then get the following equation: .3T + .7(3T) + .15(81 - 4T) = .55(81)
.3T + 2.1T + .15(81) - .6T = .55(81)
1.8T = .55(81) - .15(81)
1.8T = .4(81)
Amount of 30% solution to mix, or 
Do you think you can now find the 70% and 15% amounts?
Answer by Alan3354(69443)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
An algebraic solution using a single variable is certainly a good way to go with this.
If a formal algebraic solution is not required, and if you are good with mental arithmetic, then you can solve this without having to write and solve an equation.
The way the information is given makes it possible to use this alternative method on this problem.
We are told that he is using 3 times as much 70% solution as the 30% solution. That means that, among those two ingredients, 3/4 of it is the 70% solution. That in turn means that the percentage of those two together is 3/4 of the way from 30% to 70% -- that is, at 60%.
So now we can view the problem as mixing 15% acid and 60% acid to get 55% acid.
And similarly to above, since 55% is 8/9 of the way from 15% to 60%, 8/9 of the total mixture is the 60% solution.
And now we are ready to solve the problem.
8/9 of the mixture is the 60% acid, which is a combination of the 30% acid and 70% acid. That means 1/9 of the mixture is the 15% acid.
1/9 of the total 81 liters is 9 liters. So he is using 9 liters of the 15% acid solution.
That means 72 liters of the 60% acid solution. And since we know 3/4 of that mixture is the 70% acid and 1/4 of it is the 30% acid, we know he is using 54 liters of the 70% acid and 18 liters of the 30% acid.
ANSWERS:
9 liters of the 15% acid;
18 liters of the 30% acid;
54 liters of the 70% acid
CHECK:
.15(9)+.30(18)+.70(54) = 1.35+5.4+37.8 = 44.55
.55(81) = 44.55
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