1. ## Mixtures Problem

Hey,
I could really use some help solving this mixtures problem. Normally I can just work it out but this one is a little bit different than most.

A solution of 90% alcohol is mixed with a solution of 30% alcohol that is smaller by 20 liters. After 15 liters of water evaporates from the solution, the alcohol content is 72%. How many liters of solution is in the final mixture.

Could really use help working out the method used to solve this type of problem

EDIT:
Solution 1: 90% Alcohol, Volume x
Solution 2: 30% Alcohol, Volume x-20
Total Solution: 72% Alcohol, Volume x+x-20-15 OR 2x-35
This is all the information given. I need to determine that volume in liters of the total solution.

2. ## Re: Mixtures Problem

Originally Posted by asdewq
Hey,
I could really use some help solving this mixtures problem. Normally I can just work it out but this one is a little bit different than most.

A solution of 90% alcohol is mixed with a solution of 30% alcohol that is smaller by 20 liters. After 15 liters of water evaporates from the solution, the alcohol content is 72%. How many liters of solution is in the final mixture.

Could really use help working out the method used to solve this type of problem
Hi asdewq!

Suppose we call the volume of the first solution x (in liters).
Then this solution contains 90% x = 0.9x liters of alcohol.

Can you say what the volume of the 2nd solution is?
And how much alcohol it contains?

3. ## Re: Mixtures Problem

That is exactly the problem, the only given is that it is 20 liters smaller than the first solution. It contains 30% alcohol

Edit: Basically my givens are
Solution 1: 90% alcohol, volume x
Solution 2: 30% alcohol, volume x-20
Final Solution: 72% alcohol, volume x+x-20-15 (2x-35)

4. ## Re: Mixtures Problem

Well, if the first solution contains x liters, then the second solution contains (x-20) liters.
That is 20 liters less than the unknown amount of the first solution.

How much liters of alcohol would that make?

5. ## Re: Mixtures Problem

Originally Posted by asdewq
Edit: Basically my givens are
Solution 1: 90% alcohol, volume x
Solution 2: 30% alcohol, volume x-20
Final Solution: 72% alcohol, volume x+x-20-15 (2x-35)
Exactly!

Now how much liters of alcohol is in each solution?

6. ## Re: Mixtures Problem

x + x - 20, - 15 or 2x-35
However, that does not give me a number of liters in the final solution.

Working it out through trial and error is simple if tedious. I just need to find a way to do it algebraically so when faced with similar questions I don't have to waste all that time

Edit: If you are asking however, how many liters are given for each solution, then that is exactly the question I am supposed to solve. None are given

7. ## Re: Mixtures Problem

Hello, asdewq!

This is a tricky one . . .

A solution of 90% alcohol is mixed with a solution of 30% alcohol that is smaller by 20 liters.
After 15 liters of water evaporates from the solution, the alcohol content is 72%.
How many liters of solution is in the final mixture?
We will consider the amount of water at each stage.

Let $\displaystyle x$ = liters of the 90% alcohol solution.
This contains $\displaystyle 0.1x$ liters of water.

Let $\displaystyle x-20$ = liters of the 30% alcohol solution.
This contains $\displaystyle 0.7(x-20)$ liters of water.

Then 15 liters of water is removed.
The mixture contains: $\displaystyle 0.1x + 0.7(x-20) - 15 \:=\:0.8x - 29$ liters of water. .[1]

The final mixture contains: $\displaystyle x + (x-20) - 15 \:=\:2x-35$ liters .(a)
. . which is 28% water.
The mixture contains: $\displaystyle 0.28(2x-35) \:=\:0.56x-9.8$ liters of water. .[2]

We just described the final amount of water in two ways.

There is our equation! . . . .$\displaystyle 0.8x - 29 \:=\:0.56x - 9,8$

Solve for $\displaystyle x\!:\;x \,=\,80$

Substitute into (a): .$\displaystyle 2(80) - 35 \:=\:125$ liters.