Mixture Problems Using C1*V1=C2*V2

[DanielK799]

For whatever reason, certain mixture problems have been giving me a bunch of anxiety and throwing me off.

I've read a few different techniques- and I *thought* I understood the method using the following formula:

C1*V1=C2*V2

But I've had a couple of problems using it and I was hoping someone could maybe point out where I'm going wrong.

Here's an example (following problem is from GMAT Club); https://gmatclub.com/forum/d01-183492.html#p2505776

How many liters of pure alcohol must be added to a 40-liter solution that is 10% alcohol by volume in order to double the alcohol proportion?

A. 4
B. 5
C. 10
D. 20
E. 40



I previously used that formula for removal/replacement mixture problems. Does this approach not work well for simply adding to it?

I used the following variables, which gave me a negative number

C1= 1/10
V1= 40
C2= 1/5
V2= 40+x

[Sage Pearce-Higgins]

Good question. It looks like there are useful changes that you can make to your approach both to this problem, and to others. First, let's look at why your formula approach didn't work out.

You stated the formula C1*V1=C2*V2 but you didn't tell me what these variables stand for! And you didn't mention under which circumstances they apply! That means that this formula is worse than useless - it's positively dangerous. It's like having a tool that you don't know how to use - you're likely to injure yourself.

From your later workings, I infer:
C1= 1/10 concentration of alcohol at the start
V1= 40 volume of solution at the start
C2= 1/5 concentration of alcohol after the change
V2= 40+x volume of liquid after the change
So, following the earlier formula, you multiplied the concentration of alcohol by the volume of solution. Think for a moment what meaningful (actually pretty useful) result you're going to get. Go on, cover the screen and think about it before reading on.







The result of this product is the amount of alcohol in the solution to start with. Consequently, your formula states that the amount of alcohol at the start is the same as the amount of alcohol in the solution after you've added alcohol to it. That's nonsense!!

In conclusion, I have no idea what formula you're referring to; I can encourage you that I've scored very well on quant sections by not using such formulas (or perhaps because I don't); and I'm guessing that you're no massive fan of algebra in general so it's good news that you can discard such an approach.

Here's an alternative. This is what I literally thought as I solved the problem in 20 seconds:
40 litres that's 10% alcohol, so there must be 4 litres of alcohol in the solution. So if I add 4 more litres that should double the amount of alcohol. Answer A. But wait! There's a trap...if I add 4 litres then I've increased the volume of the solution, so to make a 20% solution I'll need to add a bit more than 4 litres, so it must be answer choice B

Please pause and compare this approach to the approach you took. I can spell out a few principles that I followed, but it will be more useful if you can identify them yourself.

1) Understand the problem. That might sound obvious, but I'm pretty sure that you (and most GMAT test-takers) don't understand the problem before they jump in with a formula or some method that they think they "should" use before even understanding what's going on.
2) Read the whole problem, including the answers. I saw straight away that the answers are spread out, so we probably don't need to be very precise with our calculations.
3) Use logic, not Math. Literally, the only Math I needed to do was 10% of 40, which I can do in my head.
4) Don't use formulas that you don't understand. This test is designed to catch out people who use formulas without thinking. Whatever approaches that you take on this test, if you understand how they work and what they mean then you're much more likely to apply them appropriately.

[DanielK799]

I think my reply was deleted, but I added the source to the problem/info I quoted (GMAT Club)... which I think is an approved source?

Thanks so much for the thorough response. I apologize for not explaining the formula better from the get go.

As you recommended, I do try not to be reliant on formulas or be set in particular ways when solving problems- it's something I'm trying to get better at. I also have a tendency to jump into the problem before looking at the answers to see if I can approximate and/or plug back it.

I guess I was trying to understand the formula/concept better by using it on a simple problem, like this one, so I could then apply it to a more complex mixture problem in the future.


I'll copy and paste the explanation of the formula/concept below along with another sample problem. I'm not disregarding what you said at all- so please don't interpret it that way. I'm just trying to figure out how to approach more complex mixture problems (I'm kind of scarred from my previous GMAT where I got a tough mixture problem and it totally threw me off and messed up my time management).


From GMAT Club

Important Points to Remember:

1. When a fraction of a homogenous solution is removed, the percentage of either part does not change. If milk:water = 1:1 in initial solution, it remains 1:1 in the leftover solution.

2. When you add one component to a solution, the amount of the other component does not change. In milk and water solution, if you add water, amount of milk is still the same (not percentage but amount). If milk:water = 1:1 in 10 liters of solution, it means, milk = 5 liters and water = 5 liters. Now, if you add 2 liters of water, amount of water = 7 liters but amount of milk is still 5 liters. The percentage of milk has changed but the amount of milk is still the same.

3. Amount of A = Concentration of A * Volume of the mixture

Amount= C∗V

In a 10 liter mixture of milk and water, if milk is 50%, amount of milk = 50%*10 = 5 liter

When you add water to this solution, the amount of milk does not change (as discussed in point 2 above). The concentration of milk changes of course since the solution is diluted.

Amount of milk before addition = Amount of milk after addition

So Initial Concentration of milk * Initial Volume of solution = Final Concentration of milk * Final Volume of solution

Ci * Vi = Cf * Vf

Or

Cf = Ci * (Vi/Vf)

Remember, this is the relation between the initial and final concentration of milk since the amount of milk remains the same. The amount of water does not remain the same since more water is added. Hence, this relation does not hold for water.

Go through these points repeatedly till you are very comfortable with them!

Example 2: 10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

(A) 3%
(B) 20%
(C) 25%
(D) 36%
(E) 40%

Solution: In each step, we are replacing the solution with water. Every time we remove p% of the solution, the amount of alcohol goes down but the concentration of alcohol in the mixture remains the same (point 1 above). When we add water, the amount of alcohol remains the same.

Let’s try and perform the steps to see what happens:

Step 1: 10% of a 50% alcohol solution is removed – In the leftover solution, concentration of alcohol remains the same i.e. 50%. If initial volume of the solution was 10 liters, new volume is 9 liters.

Step 2: Water is added to the solution to replace the 10% shortfall – the concentration of alcohol changes now (but the amount of alcohol is still the same). Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this step Cf1 = (50%)*(9/10) (using point 3)

Step 3: 10% of the solution with concentration of alcohol = Cf1 is removed – In the leftover solution, concentration of alcohol is still Cf1. The volume of the solution reduces to 9 liters again.

Step 4: Water is added to the solution to replace the 10% shortfall – The concentration of alcohol changes now. Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this step Cf2 = Cf1*(9/10) = (50%)*(9/10) *(9/10)

Step 5: 10% of the solution with concentration of alcohol = Cf2 is removed – In the leftover solution, concentration of alcohol is still Cf2. The volume of the solution reduces to 9 liters again.

Step 6: Water is added to the solution to replace the 10% shortfall – The concentration of alcohol changes now. Also, the volume of the solution is 10 liters again. In this new solution,

The concentration of alcohol after this final step Cf3 = Cf2*(9/10) = 0.5*(9/10) *(9/10) *(9/10)

The concentration changes only when water is added. Each time water is added, the concentration becomes (9/10)th of the previous concentration.

Final concentration of alcohol = (50%)∗(910)∗(910)∗(910)=36.45%

After going through that again, I'm pretty sure that's only for replacement/removal.

The particular problem I cited, was an addition. And looking back, I could have gone with the "see-saw method" with weighted averages.

10% on the left, 100% on the right, and 20% in between for the final solution. Which would have ultimately led to a ratio of 8:1:9... and then 5L of the 100% solution.

Obviously not as quick as doing it intuitively like you mentioned, but I was just trying to understand the actual concept so I can apply it to more complex problems.

[Sage Pearce-Higgins]

Thanks for posting the source - it looks like our spam detector flagged it up (your post wasn't deleted, just queued).

I would encourage you to choose your resources carefully. The problem you cite is, in my opinion, not a particularly good problem, both in the way it's phrased, and in the fact that it doesn't state that the answers are approximate (something that GMAT observes pretty carefully).

It's also telling that, after introducing some formulas, the explanation takes a route of approaching the problem logically (this is a good thing) rather than relying on formulas. The other glaring omission in my view is that the explanation doesn't mention the answers. GMAT is a multi-choice test and if you don't look at the answer choices on quant problems you are taking a different test, one I sometimes call the "GMAT-plus", which is considerably harder than the usual GMAT.

I appreciate your desire for a deeper understanding of the concepts, but you also need to train yourself to think in practical, faster ways. For example, I solved the above problem by thinking: there are 3 replacements of mixture by water, so I might think that I simply take out 5l of the alcohol each time, meaning that the resulting solution would be 35% alcohol. However, I know that's a trap, because after the first replacement the mixture will have a lower alcohol concentration, meaning that I will take out slightly less than 5l of alcohol, so the answer must be a bit higher than 35.

As for the 'teeter-totter' method, that works for problems in which we combine mixtures together, not replacement problems such as the above. It's a good method to apply to certain types of problem, such as PS 107, PS 227, and DS 429 from OG2020. Check out the Statistics chapters in our books on this. However, in my experience this strategy is only useful if students actually understand what's going on, i.e. they could explain it to someone.

[DanielK799]

You're absolutely right.

There's multiple ways to get to the answer. And I'm slowly learning that the GMAT tests our ability to choose the most efficient route to the right answer moreso than our ability to produce the right answer.

I think I'll go back to sticking to mostly OG2020 problem sets and using the ManhattanPrep Navigator for the explanations... the GmatClub problem explanations mostly consist of crazy algebra and just shatter my confidence :/ I was just using them because I had a free trial to their tests/quizzes and wanted to mix things up.

Thanks again for your help

[Sage Pearce-Higgins]

You're welcome. All the best with your studies.