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# If a portion of a half water/half alcohol mix is replaced

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If a portion of a half water/half alcohol mix is replaced  [#permalink]

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Updated on: 31 May 2012, 04:18
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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%

Originally posted by zest4mba on 02 Sep 2010, 08:24.
Last edited by Bunuel on 31 May 2012, 04:18, edited 1 time in total.
Edited the OA
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23 Dec 2010, 20:11
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1
11
rtaha2412 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%

Question on Mixtures can be easily solved using weighted averages concept discussed here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
I would also recommend that you go through the complete theory from some standard book if you are not comfortable.

This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later.

First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol.
Attachment:

Ques1.jpg [ 3.8 KiB | Viewed 26676 times ]

So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1.
Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced.

To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%
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Re: Mixture problem-Can someone explain this  [#permalink]

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19 Sep 2010, 10:51
16
6
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
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Re: Mixture problem-Can someone explain this  [#permalink]

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02 Sep 2010, 08:52
11
14
zest4mba 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%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) 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 a 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 as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.

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

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Re: Mixture problem-Can someone explain this  [#permalink]

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02 Sep 2010, 10:11
zest4mba 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%

I think the answer B can be considered only if the question was rephrased as what percentage of alcohol was replaced in the original solution with water. (20/100*100). Else the answer should be E as explained by other above.
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Re: Mixture problem-Can someone explain this  [#permalink]

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02 Sep 2010, 17:55
4
1
Yup. It is E indeed.
- If V is volume of the mixture then V/2 is alc and V/2 is water.
- Take Xml of the solution away (it takes X/2 alc with it). So the alc level now is (V-X)/2.
- Add X ml back but this solution only has X/4 alc. So new alc content = (V-X)/2 + X/4
- New alc content = 3V/10 as it is 30%.
Solving it gives X as 80%.
More or less the same approach that Bunuel took.
Thank you,
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Re: Mixture problem-Can someone explain this  [#permalink]

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19 Sep 2010, 09:52
Guys,

I'm trying to apply a shortcut provided by KillerSquirrel in this thread mixture-55090.html, but getting a wrong answer
Could you please elaborate this Mystery?
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24 Dec 2010, 12:50
1
rtaha2412 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%

If you need additional instruction on these problem types, refer to the Jeff Sackmann Total Math handout p.158-159. He gives a more detailed explanation of Bunuel's method (which I think is the quickest approach to solving these problem types).
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Re: Mixture problem-Can someone explain this  [#permalink]

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27 Sep 2011, 02:17
gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

I am not sure how can we write 50% ---------------- 25%
here ? Can someone explain this ?
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Re: Mixture problem-Can someone explain this  [#permalink]

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27 Sep 2011, 02:46
2
siddhans wrote:
gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

I am not sure how can we write 50% ---------------- 25%
here ? Can someone explain this ?

Its called Allegation... u can do google to find more about it .

Intial quantity(IQ) ....................................... replaced Quantity(RQ)

50% =========================================25%

================== 30% Result===================

subtract result -RQ ============================== subtract IQ- result
30-25 ======================================= 50-30
5% ======================================== 20%

Hope this will be clear

PS: Pm'ng u a material on Allegation. Hope that will help
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Re: Mixture problem-Can someone explain this  [#permalink]

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01 Oct 2011, 06:33
1
am I right?
let x be water, y - alcohol. so we have -
0.5x+0.5y -0.25y=0.3x+0.3y
x=1/4y
x/y=1/4
so in a new solution y 's portion is 4/5 or 80%
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30 May 2012, 21:21
1
VeritasPrepKarishma wrote:
rtaha2412 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%

Question on Mixtures can be easily solved using weighted averages concept discussed here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
I would also recommend that you go through the complete theory from some standard book if you are not comfortable.

This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later.

First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol.
Attachment:
Ques1.jpg

So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1.
Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced.

To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%

How have u used the weighted avg formula?

w1/w2 = (c2-avg) / (avg-c1)

if c1 = 25 and c2=50 and avg = 30 - here how do we calculate w1 and/ or w2?
didnot understand this part
"Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
"

Thank you
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05 Jun 2012, 05:53
2
kuttingchai wrote:
How have u used the weighted avg formula?

w1/w2 = (c2-avg) / (avg-c1)

if c1 = 25 and c2=50 and avg = 30 - here how do we calculate w1 and/ or w2?
didnot understand this part
"Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
"

Thank you

Check out this post:
http://www.veritasprep.com/blog/2012/01 ... -mixtures/

It explains how to use weighted average for such replacement questions.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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16 Dec 2013, 06:22
zest4mba 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%

Quick way

Use Smart Numbers

Give 100 for the initial amount

Then you will have 50-0.25x = 30
x = 80

So % is 80/100 is 80%

Hope it helps
Cheers!
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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29 May 2014, 11:41
Or one can use differentials to slve

Initially 50% alcohol
Then 25% alcohol

Resulting mixture 30% alcohol

Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4

Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.

Hope this helps
Cheers
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If a portion of a half water/half alcohol mix is replaced  [#permalink]

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02 Jul 2014, 17:12
Bunuel wrote:
zest4mba 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%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) 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 a 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 as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.

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

Bunuel,

We have assumed that X is removed from the total solution and then we have calculated the value of X, since solution is 50% alcohol and the question stem has asked us to calculate the amount of alcohol initially replaced, and that will be 50% of X, since we have taken X as the total amount of solution replaced/removed (50%+50%).
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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03 Jul 2014, 07:58
VeritasPrepKarishma wrote:
rtaha2412 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%

Question on Mixtures can be easily solved using weighted averages concept discussed here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
I would also recommend that you go through the complete theory from some standard book if you are not comfortable.

This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later.

First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol.
Attachment:
Ques1.jpg

So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1.
Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced.

To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%

hey karishma,
i have a doubt. here they have asked that only part of the mixture is replaced but why are you taking the original quantity of 50% alchohol while calculating ?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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03 Jul 2014, 19:46
saggii27 wrote:
VeritasPrepKarishma wrote:
rtaha2412 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%

Question on Mixtures can be easily solved using weighted averages concept discussed here:
http://gmatclub.com/forum/tough-ds-105651.html#p828579
I would also recommend that you go through the complete theory from some standard book if you are not comfortable.

This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later.

First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol.
Attachment:
Ques1.jpg

So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1.
Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced.

To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%

hey karishma,
i have a doubt. here they have asked that only part of the mixture is replaced but why are you taking the original quantity of 50% alchohol while calculating ?

They haven't given the quantity of 50% alcohol at all. All we know is that 50% alcohol is mixed with 25% alcohol to give 30% alcohol so we have found the ratio in which they must be mixed. We got this ratio as 4:1 i.e. we need to mix 4 parts of 25% alcohol with 1 part of 50% alcohol to get 30% alcohol. But since originally we had all 50% alcohol, out of every 5 parts of we must have taken out 4 parts and put 25% alcohol in its place. So we must have replaced 4/5th of the 50% alcohol solution.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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03 Jul 2014, 19:50
Karishma My point is what was the assumption, in the solution given by Bunuel what was X? X was the total amount of Mixture displaced with new mixture of different alcohol consumption. X was not the total amount of alcohol displaced.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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03 Jul 2014, 20:24
1
honchos wrote:
Karishma My point is what was the assumption, in the solution given by Bunuel what was X? X was the total amount of Mixture displaced with new mixture of different alcohol consumption. X was not the total amount of alcohol displaced.

honchos, in the previous post, I addressed saggii27's question to me.

As for your question to Bunuel, we find the value of x, which is the fraction of the 50% SOLUTION that was replaced. We found that 4/5th of the original solution was replaced. Note that we assume solutions to be homogeneous. This means that if 4/5th of the solution was replaced, 4/5th of the original alcohol was replaced and 4/5th of the original water was replaced.
Say you had 100 ml solution with 50 ml each of water and alcohol. You removed 4/5th of this solution i.e. 80 ml. So you removed 40 ml of each -water and alcohol i.e. you removed 4/5th of alcohol and 4/5th of water.
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Re: If a portion of a half water/half alcohol mix is replaced   [#permalink] 03 Jul 2014, 20:24

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