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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. \(80\%\)

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 \(x\) andthe 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%.

Answer: E

Highlighted above - Why did we consider 1 Unit ??

Any reason? I tried with other numbers, not getting the required answer

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 \(x\) andthe 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%.

Answer: E

Highlighted above - Why did we consider 1 Unit ??

Any reason? I tried with other numbers, not getting the required answer

Thanks

You should get the correct answer with any number. I chose 1 unit to make it easier.
_________________

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.
_________________

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 ??
_________________

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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