If a portion of a half water/half alcohol mix is replaced : GMAT Problem Solving (PS)
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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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02 Sep 2010, 07:24
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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

Last edited by Bunuel on 31 May 2012, 03:18, edited 1 time in total.
Edited the OA
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Re: Mixture problem-Can someone explain this [#permalink]

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02 Sep 2010, 07:52
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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, 09:11
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, 16:55
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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,
Hemanth
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Re: Mixture problem-Can someone explain this [#permalink]

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19 Sep 2010, 08:52
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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Re: Mixture problem-Can someone explain this [#permalink]

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19 Sep 2010, 09:51
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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%...
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Re: Mixture problem-Can someone explain this [#permalink]

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19 Sep 2010, 09:57
Thanks!
Kudos to you!

I misprinted digits in my calculations and it spoiled my result
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Re: Mixture problem-Can someone explain this [#permalink]

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19 Sep 2010, 14:51
Nice one. I got E using algebraic method. I have learn now allegation method and will try that in next problem
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23 Dec 2010, 19:11
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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 16053 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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: Current MBA Student Joined: 19 Nov 2009 Posts: 127 Concentration: Finance, General Management GMAT 1: 720 Q49 V40 Followers: 13 Kudos [?]: 350 [0], given: 210 Re: mixture problem [#permalink] ### Show Tags 24 Dec 2010, 11:50 rtaha2412 wrote: I do not know how to solve mixture problems. Please advise 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). Senior Manager Joined: 08 Nov 2010 Posts: 417 WE 1: Business Development Followers: 7 Kudos [?]: 105 [0], given: 161 Re: Mixture problem-Can someone explain this [#permalink] ### Show Tags 27 Feb 2011, 03:55 Karishma once u get this system - its amazing. so much easier and safer. thanks! Bunuel - can you please fix the OA in the original post? i wasted 15 min, trying to understand how the answer is B and not E. thanks guys. _________________ Senior Manager Joined: 29 Jan 2011 Posts: 367 Followers: 0 Kudos [?]: 201 [0], given: 87 Re: Mixture problem-Can someone explain this [#permalink] ### Show Tags 27 Sep 2011, 01:17 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 ? Current Student Joined: 26 May 2005 Posts: 565 Followers: 18 Kudos [?]: 203 [0], given: 13 Re: Mixture problem-Can someone explain this [#permalink] ### Show Tags 27 Sep 2011, 01:46 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 Senior Manager Joined: 23 Oct 2010 Posts: 386 Location: Azerbaijan Concentration: Finance Schools: HEC '15 (A) GMAT 1: 690 Q47 V38 Followers: 21 Kudos [?]: 322 [0], given: 73 Re: Mixture problem-Can someone explain this [#permalink] ### Show Tags 01 Oct 2011, 05:33 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% _________________ Happy are those who dream dreams and are ready to pay the price to make them come true I am still on all gmat forums. msg me if you want to ask me smth Manager Joined: 28 Jul 2011 Posts: 240 Followers: 3 Kudos [?]: 119 [1] , given: 16 Re: mixture problem [#permalink] ### Show Tags 30 May 2012, 20:21 1 This post received KUDOS VeritasPrepKarishma wrote: rtaha2412 wrote: I do not know how to solve mixture problems. Please advise 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13667 [1] , given: 222 Re: mixture problem [#permalink] ### Show Tags 05 Jun 2012, 04:53 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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05 Jun 2012, 20:25
VeritasPrepKarishma wrote:
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.

Thank you , Yes i got the answer. I went thr' similar post that you answered previosuly and got the answer
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Re: If a portion of a half water/half alcohol mix is replaced [#permalink]

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12 Oct 2013, 01:38
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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, 05:22
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!
J
Re: If a portion of a half water/half alcohol mix is replaced   [#permalink] 16 Dec 2013, 05:22

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