SOLUTION: 19. A chemist has a can full of paint thinner that is 70% alcohol. After replacing 7 L of the solution with 7 L of pure alcohol, the resulting solution is 87.5% alcohol. How many

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: 19. A chemist has a can full of paint thinner that is 70% alcohol. After replacing 7 L of the solution with 7 L of pure alcohol, the resulting solution is 87.5% alcohol. How many      Log On

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Question 1191542: 19. A chemist has a can full of paint thinner that is 70% alcohol. After
replacing 7 L of the solution with 7 L of pure alcohol, the resulting
solution is 87.5% alcohol. How many liters does the can hold?

Found 3 solutions by josgarithmetic, math_tutor2020, greenestamps:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
C, the capacity filled for can, volume.

Amount alcohol before the removal and replacement, 0.7C

Remove 7 L of the original volume, 0.7%28C-7%29 liters of just alcohol

Replace with 7 L of pure alcohol, 0.7%28C-7%29%2B7 liters alcohol.
This is contained within C liters of solution.

highlight_green%28%280.7%28C-7%29%2B7%29%2FC=0.875%29-------simplify, and solve for C.

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

x = total capacity of the paint can in liters
The goal of course is to find the numeric value of x.

If x < 7, then there isn't enough room to remove 7 L or add 7 L. We cross this possibility off the list.

If x = 7, then we're done and there's no need to do any algebra.
However, it would mean replacing all 7 L with pure alcohol would lead to 100% alcohol solution (not 87.5%).
So we ignore this case as well.

As you can see, the only possibility here is that x > 7.

Dumping 7 L of solution means we dump 7/x of the solution and keep the remaining 1 - 7/x = (x-7)/x of it.
For example, if say x = 21, then we dump 7/x = 7/21 = 1/3 of the container and keep the remaining (x-7)/x = (21-7)/21 = 14/21 = 2/3 of it.

Initially, the chemist has 0.7x liters of alcohol in the 70% solution. After dumping 7/x of the solution, s/he keeps the remaining 0.7x*(x-7)/x = 0.7(x-7) = 0.7x-4.9 liters of pure alcohol.

Also, the initial amount x drops to x*(x-7)/x = x-7
Or you can just remark that x naturally drops to x-7 after draining 7 L away.

Afterward, the chemist adds 7 L of pure alcohol
The 0.7x-4.9 bumps up to 0.7x-4.9+7 = 0.7x+2.1
The x-7 goes back to x-7+7 = x again as expected.

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Let's recap:

0.7x = starting amount of pure alcohol
x = capacity of the paint can = starting amount of solution

After dumping 7 L,
The chemist has 0.7x-4.9 L of pure alcohol
Out of x-7 liters total

After adding 7 L of pure alcohol,
The chemist has 0.7x+2.1 L of pure alcohol
Out of x liters total

Review these steps carefully if you get stuck anywhere.
Also, feel free to ask me about a certain portion or step if something still isn't making sense.

Once you see how I'm getting those items, we can move onto the next part.

-------------------------------------------

The conclusion reached so far is that we have 0.7x+2.1 L of pure alcohol out of x liters of solution.

The ratio of these two items (pureAlcohol/totalSolution) is desired to be 0.875 which is the decimal form of 87.5%

So,
(pureAlcohol)/(totalSolution) = percent of alcohol
(pureAlcohol)/(totalSolution) = 0.875
(0.7x+2.1)/(x) = 0.875
0.7x+2.1 = 0.875x
2.1 = 0.875x-0.7x
2.1 = 0.175x
0.175x = 2.1
x = 2.1/0.175
x = 12
The paint can has a full capacity of 12 liters

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Checking the answer:

The chemist starts with 12 liters of solution.
70% of which is pure alcohol, so he or she has 0.7*12 = 8.4 L of pure alcohol

The chemist dumps 7L which is 7/12 of the solution
Therefore, (7/12)*8.4 = 4.9 liters of pure alcohol are dumped and 8.4 - 4.9 = 3.5 liters of pure alcohol are kept.

This is out of 12-7 = 5 liters of solution left.
Notice that (3.5)/5 = 0.7 to indicate that we still have a 70% solution. No amount of dumping/draining will mean the alcohol concentration changes.
The only way to lower the alcohol content is to dilute it (with perhaps water), and the only way to raise the alcohol content is of course to add more pure alcohol.

Now to add the 7L of pure alcohol
The 3.5 L of pure alcohol bumps up to 3.5+7 = 10.5 L
The 5 L goes back to 5+7 = 12.

Lastly, notice that
(10.5 L of pure alcohol)/(12 L total solution) = 0.875 = 87.5% solution

We've confirmed the answer.


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Answer: 12 liters

Edit: The tutor @ josgarithmetic has a much more simpler approach so it's best to go for that. Solving that equation leads to C = 12.

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


You can solve the problem even more easily without algebra.

Although the wording disguises the problem well, this is essentially a mixture problem: you are mixing 70% alcohol with 100% alcohol to get a mixture that is 87.5% alcohol.

To solve the problem, you only need to find where the 87.5% of the final mixture lies between the 70% and 100% of the two ingredients.

Simple arithmetic shows that 87.5% is 17.5/30 = 7/12 of the way from 70% to 100%; that means 7/12 of the mixture is the 100% alcohol.

Then, since that 7/12 of the mixture is 7 liters, the capacity of the can is 12 liters.

ANSWER: 12 liters

The words of explanation make this seem like a lengthy solution; but the calculations alone are quick and easy:

87.5-70=17.5
100-70=30
17.5/30=35/60=7/12
7/12 = 7/x
x=12