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If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\) B. \(20\%\) C. \(66\%\) D. \(75\%\) E. \(80\%\)

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\) B. \(20\%\) C. \(66\%\) D. \(75\%\) E. \(80\%\)

Initial solution is "half water/half alcohol mix" means that it's 50% alcohol solution.

Let the portion replaced be \(x\) and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of that portion will be \(0.5(1-x)\) and the amount of alcohol added will be \(0.25x\), so total amount of alcohol will be \(0.5(1-x)+0.25x\). On the other hand, since finally 30% alcohol solution was obtained then the amount of alcohol in the end was \(0.3*1\).

So \(0.5(1-x)+0.25x=0.3\) \(\rightarrow\) \(x=0.8\), or 80%.

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\) B. \(20\%\) C. \(66\%\) D. \(75\%\) E. \(80\%\)

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\) B. \(20\%\) C. \(66\%\) D. \(75\%\) E. \(80\%\)

As I've been doing more and more practice quant problems, it seems that I've been getting faster and faster by moving away from pure theory and trying to get an intuitive notion for the problem.

So the way I approached this was by looking at the percentages 50%, 25%, and 30%. I saw that 30% is 4/5ths of the way from 50% to 25%, so I looked at the answers and saw that one of the potential answers was 80% = 4/5ths. So I plugged in what I chose for smart numbers:

Let's say the jug holds 10 gallons of liquid, so 5 gallons is alcohol. If we remove 80%, we now have 1 gallon of alcohol and 1 gallon of water. If we add back 8 gallons of the new solution, there should be 6 gallons of water and 2 of alcohol. Therefore, the total would be 3 gallons of alcohol for 10 gallons of liquid = 30%. Done!

The term "replaced" is new to me and I wasn't sure how to approach this problem. Is this a term I might see on the GMAT? It seems like a science term that probably has a specific meaning, can anyone fill me in on what that is? Thanks! My assumption was that it means combining equal volumes.

The term "replaced" is new to me and I wasn't sure how to approach this problem. Is this a term I might see on the GMAT? It seems like a science term that probably has a specific meaning, can anyone fill me in on what that is? Thanks! My assumption was that it means combining equal volumes.

Here replace means substituting some amount of mixture with equal amount of another mixture.
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I think this is a high-quality question and I agree with explanation. The question should read " what portion of the original alcohol solution" was replaced? the word "solution" is needed for total clarity as it could otherwise imply what portion of the original alcohol qty was replaced.

I am assuming initial alochol was 50 and water was 50 and Total was 100 Now we replace X quantity . How to frame equation from now onwards for this ??

Hi aditya, I too just did this Q in club tests. i would say whenever we have Qs involving mixtures, average method is best.. for example..

we know initial % of alcohol is 50 %.. 25% alcohol is added to get a new mixture of 30%.. so we have to check how apart the two solutions are from final.. 50-30:30-25 = 20:5=4:1 so 4/5 of the second alcohol solution will be added to get 30% new sol.. ans 80%
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I always get excited with problems likes these! They're weighted heavily and categorized as difficult, but it seems to just be a weighted average problem in disguise. So if you know how to work with weighted averages, these types of problems are like free chicken!

We're given 3 solution concentrations to work with.

Draw them out on a number line in increasing order, with the resultant solution in the middle (we put the resultant in the middle since it's an average of the other two solutions)

Now, set them into a ratio. 20:5. Now simplify to 4:1.

**This means that 4 parts of the 25% solution were replaced for every 1 part of the existing solution. In other words, \(\frac{4}{5}\) replaced. (part over whole of 4:1 ratio) and \(\frac{4}{5}\) expands to \(\frac{80}{100}\) or 80%

I used to get tripped up on setting the ratio up backwards, at first glance it looks like it would be 5:20, since the space between A-R is 5, and R-S is 20, but think about it logically. The resultant (or average) solution is much closer to the "added" solution, meaning the resultant solution contains more of the added solution than the original solution, which is what brought "R" closer to "A" on our number line.

Once you get the hang of setting these weighted averages problems up like this, it becomes super easy.

For all the people asking why are we taking 80% as the alcohol percent when we get 80% for the solution replaced?

The answer is, if you replace x% of a solution with something else, then any smaller portion of the whole (lets say 50ml of the total 100ml solution) will also be replaced with the same %. For example: Assume that one has Solution A: 100ml. Solution A is composed of 60ml Rum and 40ml Water. One removes a portion of the Rum mixture, replaces it with something more potent. Regardless of how much replacement has been done, since the solution is always homogeneous (evenly mixed) the removal will affect all individual portions as the same.

So if 50% Rum mixture is replaced, then 50% of the water inside the mixture is also replaced.

Not an exact answer, but should get the wheels rolling for people trying to search for an answer as to why getting 80% for whole means the same to alcohol as well.

Let mixture volume be = 100 ml Therefore : 50 ml alcohol + 50 ml water X= total volume of alcohol Yr = Volume of alcohol replaced Ya = Volume of alcohol added => X - Yr + Ya = 30 ( resultant will be 30ml alcohol out of 100 ml) 50 - (50/100)Yr + (25/100)Ya = 30 = 20 = (25/100) Y (amount replaced and added are same, so denoting as Y) Y = 80 ( This means you are replacing 80ml of the solution and adding 80 ml of 25% alcohol ) Hence: 80% (E)

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\) B. \(20\%\) C. \(66\%\) D. \(75\%\) E. \(80\%\)

When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

Since we're asked to find a PERCENTAGE, we can assign a "nice" value to the original volume. So, let's say we start with 100 liters

As you can see, 50 liters is alcohol and 50 liters is water

Now let's remove x liters of the mixture from the container. Half of removed x liters will be alcohol, half of the removed x liters will be water In other words, we're removing 0.5x liters of alcohol, and 0.5x liters of water. So, the resulting mixture looks like this:

We're going to replace the x liters of missing solution with x liters of 25% alcohol solution

When we add the volumes of alcohol and water, we get a final mixture that looks like this:

Finally, we want the final mixture to be 30% alcohol. In other words, we want: (volume of alcohol)/(total volume of mixture) = 30/100 (aka 30%) We get: (50-0.25x)/100 = 30/100 Cross multiple: (50-0.25x)(100) = (100)(30) Simplify: 5000 - 25x = 3000 Solve to get: x = 80