# Mixtures Problem

• February 4th 2013, 03:39 PM
asdewq
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.
• February 4th 2013, 03:53 PM
ILikeSerena
Re: Mixtures Problem
Quote:

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?
• February 4th 2013, 04:01 PM
asdewq
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)
• February 4th 2013, 04:03 PM
ILikeSerena
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?
• February 4th 2013, 04:05 PM
ILikeSerena
Re: Mixtures Problem
Quote:

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?
• February 4th 2013, 04:29 PM
asdewq
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
• February 4th 2013, 07:56 PM
Soroban
Re: Mixtures Problem
Hello, asdewq!

This is a tricky one . . .

Quote:

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

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

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

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

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

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

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

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