M17-11

It should be 0.2*10 = 0.05(10+w), so you should be equating alcohol.

Bunuel, I tried to convert this into a equation : 20% [10 Lit] = 25% [ 10 +W] and it is giving incorrect answer. After looking at clarification I can understand the solution but appreciate if you can advise what is wrong in my equation.

Hello NAvinash

IMO, you are slightly incorrect there. See the highlighted portion in RED. You correctly understood that you have to reduce the percentage of alcohol to 25% (or by 75%), but that reduction has to be in the given percentage of alcohol, that is, 25% of 20% of alcohol in the solution = 5% of alcohol in the solution.

So, your equation will be;
20% [10 L] = 5% [10+W]
Which will get you W = 30L.

I hope it helped.

I think this is a high-quality question and I agree with explanation.

HI GMATGuruNY , MentorTutoring , BrentGMATPrepNow,

Can you help me out on this problem? I tried to comprehend but missed it.

Hello again, NandishSS. There are two separate components that you have to work through to solve the question. I often teach my students to stop whenever they encounter punctuation in a Quant problem, to take a moment to process what information was given. Too many students make the mistake of trying to read everything quickly and then jumping right into the problem, perhaps feeling overwhelmed because they do not know how to put all the pieces together. Anyway, how about we put this technique to use in this question?


This information allows us to determine the number of liters of alcohol in the solution: 0.2(10) = 2.


To reduce the concentration of alcohol, we have to reduce the 20% (concentration) of the original solution by 75%, or, in other words, we need to keep 1/4 of the original concentration: 1/4 * 20 = 5. The solution after the addition of the water must have a 5% alcohol content.

Now we can put the two separate parts of the problem together. Since we are not adding any alcohol to the solution, but water only, we know that 2L of alcohol must be equivalent to the 5% concentration of alcohol in the final solution (which I will denote F).

2 = 0.05F or 2 = (5/100)F (if decimals cause you trouble, as they do for some people)

2 = (1/20)F

2 = F/20

40 = F

The final solution will be 40L in volume, of which 2L will be alcohol. Check your solution (if you were not in a time crunch):

2/40 = 1/20 = 5/100

Since 5/100 is 5%, we can now be sure about our solution. A good trap answer here would be 40 liters, since that is the volume of the final solution. But be careful and make sure you are answering the question that is being asked. Here, at a barebones level (since we have already sorted out all the details), we have,



If we started with 10L and we ended up with 40L, then we added 30L of water to the solution. Choice (C) must be the answer.

I hope that helps. Although, looking back on the official solution, I pretty much did the same thing as Bunuel, perhaps the extra explanations will allow you to better understand how to break down such a problem. If not, just let me know.

- Andrew

When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.
This way, it's very easy to keep track of each part of the mixture.
We get:


How much water should be added to reduce the concentration of alcohol in the solution by 75% ?
75% of 20 = 15
20 - 15 = 5
So, we want the RESULTING mixture to be 5% alcohol.
In other words, we want the RESULTING mixture to be 5/100 alcohol.

We can write: 2/(10 + x) = 5/100
Simplify: 2/(10 + x) = 1/20
Cross multiply: 10 + x = 40
Solve: x = 30

Answer: C

RELATED VIDEO

20% alcohol concentration reduced by 75% = 25% of 20% = 5% alcohol concentration.

Alcohol percentage in the original solution: 20%
Alcohol percentage in the added water: 0%
Alcohol percentage in the mixture: 5%

Let S = the original solution and W = the added water.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (20% and 0%) on the ends and the percentage for the mixture (5%) in the middle.
S 20%-------------5%------------0% W

Step 2: Calculate the distances between the percentages.
S 20%-----15-----5%-----5-----0% W

Step 3: Determine the ratio in the mixture.
The ratio of original solution to added water is equal to the RECIPROCAL of the distances in red.
S:W = 5:15 = 1:3

Since S:W = 1:3 = 10:30, and the volume of original solution is 10 ounces, the final mixture must be composed of 10 ounces of original solution and 30 ounces of added water.

HI MentorTutoring,

Can you help me with below? I didn't get it.



One humble Request -- Please add "@" before NandishSS

The reason I do not tag someone when I reply, NandishSS, is that I assume the person posing the question would take an interest in any responses that came through, so I simply have not felt the need. I prefer to tag someone if I need help with something (almost always technical, since my grasp of such matters is rather limited). In fact, my previous attempts to tag using the @ symbol have failed, and I ended up with nothing more than a goofy-looking post that I then had to go back and edit. So to tag someone, I actually track down a post someone else has made with a tag in it, hit the Quote button to write a response, copy the line of code, close the window, and paste the line of code into my actual response, changing the destination manually so as not to tag the wrong person. Anyway, onto the matter at hand.

If you are told to reduce something by 75%, then you can bypass the step in which you calculate 75% and then subtract that from 100% of the value. Rather, you can appreciate that reducing something by 75% means that you will be left with 25%, or 1/4, of the original amount. In the problem at hand, 1/4 of 20% is 5%. This approach can shave considerable time off a problem in which, say, multiple discounts were applied to a purchased item. To illustrate, how about the following problem?

John buys a suit that is regularly priced at $725. However, he finds the suit on a discount rack that says all items found there are 20 percent off (in addition to any other discounts). Since there is a Fourth of July sale, everything in the store is 15 percent off on top of the lowest price. Finally, John cut out a coupon from a newspaper that allows for 10 percent off any item, a discount that may also be applied in addition to any other discounts. If sales tax is 6 percent, how much did John pay for the suit after all the discounts had been applied?

Of course, you could work through the discounts one by one, but it is much faster to rationalize that if John receives discounts of 20 percent, 15 percent, and 10 percent, respectively, then he will be paying 80 percent, 85 percent, and 90 percent of the cost at each step. Finally, the tax can be added to 100 percent (of the discounted price) to bypass the step in which you would calculate 6 percent of the discounted price and then add that amount back in. Altogether, then, you could quickly calculate the answer:

\(0.8*0.85*0.9*1.06*725=470.322\)

John would end up paying $470.32 after all the discounts had been applied. I hope that helps clarify what I was aiming to get at. If not, just reply and let me know. I can assure you that, tag or no tag, if I see a response to my own with questions in it, I will write back.

- Andrew

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.