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M03-04

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New post 16 Sep 2014, 00:19
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44% (02:09) correct 56% (02:10) wrong based on 279 sessions

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New post 16 Sep 2014, 00:19
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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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New post 25 Nov 2014, 05:56
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New post 25 Nov 2014, 07:07
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New post 12 Dec 2014, 12:07
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, 06:00
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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Re: M03-04  [#permalink]

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New post 06 May 2015, 08:04
great question, looks like i need more practice on mixtures.
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New post 06 May 2015, 08:52
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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, 10:38
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, 02:57
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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New post 25 Jun 2015, 15:15
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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 27 Aug 2015, 01:13
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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New post 27 Nov 2015, 04:58
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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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New post 20 Feb 2016, 21:54
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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, 07:23
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 13 Apr 2017, 17:57
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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.

Starting: 50% (S)
Added solution: 25% (A)
Resultant solution: 30% (R)

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)

A-----R--------------------S
25-----30--------------------50

Now count the percentages in between the two.

From A to R: 5
From R to S: 20

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.

Ans: E
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New post 23 Oct 2018, 21:53
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.
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New post 20 Dec 2018, 23:46
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)
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New post 03 Feb 2019, 13:25
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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. \(80\%\)


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
Image
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:
Image


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

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

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

So, 80 liters were originally removed

Answer: E

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New post 04 Feb 2019, 04:28
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Dear GMATPrepNow Brent

When I see you explaination, I always want to say: Thanks to your effort and people like you who work hard to make it easier for students.

Indeed True tutor :thumbup: :angel:
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Re: M03-04   [#permalink] 04 Feb 2019, 04:28

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