Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 09 Feb 2010
Posts: 39

If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
Updated on: 31 May 2012, 03:18
Question Stats:
58% (01:55) correct 42% (02:02) wrong based on 743 sessions
HideShow timer Statistics
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%
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by zest4mba on 02 Sep 2010, 07:24.
Last edited by Bunuel on 31 May 2012, 03:18, edited 1 time in total.
Edited the OA




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10680
Location: Pune, India

Re: mixture problem
[#permalink]
Show Tags
23 Dec 2010, 19:11
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/toughds105651.html#p828579I 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 32736 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%
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




Manager
Joined: 11 Jul 2010
Posts: 150

Re: Mixture problemCan someone explain this
[#permalink]
Show Tags
19 Sep 2010, 09:51
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%...




Math Expert
Joined: 02 Sep 2009
Posts: 65242

Re: Mixture problemCan someone explain this
[#permalink]
Show Tags
02 Sep 2010, 07:52
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 \(0.5(1x)\) and the amount of alcohol added will be \(0.25x\), so total amount of alcohol will be \(0.5(1x)+0.25x\). 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 \(0.5(1x)+0.25x=0.3\) > \(x=0.8\), or 80%. Answer: E.
_________________



Intern
Joined: 06 Jul 2010
Posts: 6

Re: Mixture problemCan someone explain this
[#permalink]
Show Tags
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.



Manager
Status: Keep fighting!
Affiliations: IIT Madras
Joined: 31 Jul 2010
Posts: 164
WE 1: 2+ years  Programming
WE 2: 3+ years  Product developement,
WE 3: 2+ years  Program management

Re: Mixture problemCan someone explain this
[#permalink]
Show Tags
02 Sep 2010, 16:55
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 (VX)/2.  Add X ml back but this solution only has X/4 alc. So new alc content = (VX)/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



Manager
Joined: 18 Jun 2010
Posts: 233
Schools: Chicago Booth Class of 2013

Re: Mixture problemCan someone explain this
[#permalink]
Show Tags
19 Sep 2010, 08:52
Guys, I'm trying to apply a shortcut provided by KillerSquirrel in this thread mixture55090.html, but getting a wrong answer Could you please elaborate this Mystery?



Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 89
Concentration: Finance, General Management

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.158159. He gives a more detailed explanation of Bunuel's method (which I think is the quickest approach to solving these problem types).



Manager
Joined: 29 Jan 2011
Posts: 226

Re: Mixture problemCan 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: 428

Re: Mixture problemCan 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 3025 ======================================= 5030 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: 312
Location: Azerbaijan
Concentration: Finance

Re: Mixture problemCan 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: 152

Re: mixture problem
[#permalink]
Show Tags
30 May 2012, 20:21
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/toughds105651.html#p828579I 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 = (c2avg) / (avgc1) 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: 10680
Location: Pune, India

Re: mixture problem
[#permalink]
Show Tags
05 Jun 2012, 04:53
kuttingchai wrote: How have u used the weighted avg formula?
w1/w2 = (c2avg) / (avgc1)
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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



VP
Joined: 06 Sep 2013
Posts: 1491
Concentration: Finance

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
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 500.25x = 30 x = 80 So % is 80/100 is 80% Hence E is the answer Hope it helps Cheers! J



VP
Joined: 06 Sep 2013
Posts: 1491
Concentration: Finance

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
29 May 2014, 10:41
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 J



Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 448
Location: India
GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49
GPA: 3.3

If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
02 Jul 2014, 16:12
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 \(0.5(1x)\) and the amount of alcohol added will be \(0.25x\), so total amount of alcohol will be \(0.5(1x)+0.25x\). 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 \(0.5(1x)+0.25x=0.3\) > \(x=0.8\), or 80%. Answer: E. 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%).



Intern
Joined: 15 Jul 2012
Posts: 31

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
03 Jul 2014, 06:58
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/toughds105651.html#p828579I 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 ?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10680
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
03 Jul 2014, 18:46
saggii27 wrote: 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/toughds105651.html#p828579I 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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 448
Location: India
GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49
GPA: 3.3

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
03 Jul 2014, 18:50
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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10680
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
03 Jul 2014, 19:24
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
03 Jul 2014, 19:24



Go to page
1 2 3
Next
[ 45 posts ]

