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M03-04

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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. \(80\%\)
[Reveal] Spoiler: OA

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Official Solution:

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%.


Answer: E
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DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope it helps.
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New post 12 Dec 2014, 11:07
Bunuel wrote:
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. \(80\%\)



Bunuel,

Kindly correct me if i am wrong, isn't the question supposed to ask what percentage of the ORIGINAL SOLUTION (not alcohol) was replaced?

As per the solution, "x" is a portion of the solution replaced = 0.8 which eventually after multiplying by 100 gives 80%.

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New post 13 Dec 2014, 05:00
earnit wrote:
Bunuel wrote:
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. \(80\%\)



Bunuel,

Kindly correct me if i am wrong, isn't the question supposed to ask what percentage of the ORIGINAL SOLUTION (not alcohol) was replaced?

As per the solution, "x" is a portion of the solution replaced = 0.8 which eventually after multiplying by 100 gives 80%.


I think the following discussion should help: if-a-portion-of-a-half-water-half-alcohol-mix-is-replaced-100271.html
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New post 26 Feb 2015, 04:49
Bunuel wrote:
Official Solution:

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

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New post 26 Feb 2015, 04:53
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

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.
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New post 06 May 2015, 07:04
great question, looks like i need more practice on mixtures.

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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!

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New post 06 May 2015, 09:38
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.

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New post 07 May 2015, 01:57
meshackb wrote:
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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This helped me solve the question in a more intuitive way (according to me of course)
Assume Original Solution= 100 liters
Alcohol:Water = 50:50

Now we remove a certain amount from this solution which I define as 'x'
Alcohol:Water= 50 - (x/2) :50 - (x/2)

Next we add a solution that is 25% alcohol to it
Alcohol:Water= 50 - (x/2) + 25x/100 :50 - (x/2) + 75x/100

Simplify to get
Alcohol:Water= 50 - (x/4):50 + (x/4)

Now our target solution has Alcohol:Water = 30:70

So we can set either of our equations to get the answer
50 - (x/4) = 30 or
50 + (x/4) = 70

Solve for x to get 80

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New post 03 Aug 2015, 11:05
I got this question wrong in my gmatclub test.
Also i am not able to understand the official solution for this

Bunuel can you please help here ?

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 ??
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New post 17 Aug 2015, 01:55
adityadon wrote:
I got this question wrong in my gmatclub test.
Also i am not able to understand the official solution for this

Bunuel can you please help here ?

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 ??


Hope alternative solution HERE help to understand better.
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New post 27 Aug 2015, 00:13
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.

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adityadon wrote:
I got this question wrong in my gmatclub test.
Also i am not able to understand the official solution for this

Bunuel can you please help here ?

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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Re: M03-04 [#permalink]

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New post 20 Feb 2016, 20:54
Allegation is probably the fastest way to do this for non-math geniuses.

The distance from the average of the original solution is 20 units and the distance from the new solution is 5 units.
Reducing to 4:1

So out of 5 total parts only 1/5th remains so 80% was replaced.

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New post 25 Jun 2016, 06:23
I think this is a high-quality question and I agree with explanation. Can we solve like this :

Let V - original volume
Let L - replacement volume

50% V = 25% L + 30% V ----- eqn 1

50% V is the total volume of alcohol
25% L is the amount of volume of alcohol in replacement
30% V is the final volume of alcohol

Solving eqn 1, we get

L = 4/5 V = 80% of V

Is this correct ?

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New post 16 Sep 2016, 04:19
as per the screenshot
it comes to be 3:1
but answer is wrong :( :(
please assist
>> !!!

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Re: M03-04   [#permalink] 16 Sep 2016, 04:19

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