November 20, 2018 November 20, 2018 09:00 AM PST 10:00 AM PST The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat. November 20, 2018 November 20, 2018 06:00 PM EST 07:00 PM EST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1826
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
04 Jul 2014, 02:16
Water .......... Alcohol ......... Total 50 .............. 50 ....................... 100 Let "x" quantity is removed 50  0.5x ....... 50  0.5x ............... 100x Same "x" quantity Alcohol is added of 25% concerntration 50  0.5x......... 50  0.5x + 0.25x .......... 100  x + x Post addition, Alcohol concentration is 30% \(50  0.25x = \frac{30}{100} * 100\) 0.25x = 20 x = 80% Answer = E
_________________
Kindly press "+1 Kudos" to appreciate



Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 486
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
04 Jul 2014, 05:33
PareshGmat wrote: Water .......... Alcohol ......... Total
50 .............. 50 ....................... 100
Let "x" quantity is removed
50  0.5x ....... 50  0.5x ............... 100x
Same "x" quantity Alcohol is added of 25% concentration
50  0.5x......... 50  0.5x + 0.25x .......... 100  x + x
Post addition, Alcohol concentration is 30%
\(50  0.25x = \frac{30}{100} * 100\)
0.25x = 20
x = 80%
Answer = E what you are trying to explain? You have calculated X. This is the question 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% Even I was saying this the question didn't asked what %age of total solution was replaced, but the question asked what %age of original alcohol was replaced. X here is the amount of total solution replaced. 50 0.5x = The amount of alcohol replaced, which is equal to 40 and is also 80% coincidentally because alcohol was 1/2 the Original mixture but neither yous nor the original solution addressed this.
_________________
Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/howtoscore750and750imovedfrom710to189016.html



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1826
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
06 Jul 2014, 17:42
honchos wrote: PareshGmat wrote: Water .......... Alcohol ......... Total
50 .............. 50 ....................... 100
Let "x" quantity is removed
50  0.5x ....... 50  0.5x ............... 100x
Same "x" quantity Alcohol is added of 25% concentration
50  0.5x......... 50  0.5x + 0.25x .......... 100  x + x
Post addition, Alcohol concentration is 30%
\(50  0.25x = \frac{30}{100} * 100\)
0.25x = 20
x = 80%
Answer = E what you are trying to explain? You have calculated X. This is the question 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% Even I was saying this the question didn't asked what %age of total solution was replaced, but the question asked what %age of original alcohol was replaced. X here is the amount of total solution replaced. 50 0.5x = The amount of alcohol replaced, which is equal to 40 and is also 80% coincidentally because alcohol was 1/2 the Original mixture but neither yous nor the original solution addressed this. Can you elaborate as to how the highlighted value was computed?
_________________
Kindly press "+1 Kudos" to appreciate



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
06 Jul 2014, 20:21
honchos 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%
Even I was saying this the question didn't asked what %age of total solution was replaced, but the question asked what %age of original alcohol was replaced. X here is the amount of total solution replaced.
50 0.5x = The amount of alcohol replaced, which is equal to 40 and is also 80% coincidentally because alcohol was 1/2 the Original mixture but neither yous nor the original solution addressed this. Honchos, it is not 80% coincidentally. It has to be 80% because 80% of the solution was replaced. Look at my explanation given in the post above: "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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 24 Nov 2014
Posts: 18

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
18 Mar 2015, 03:46
I am thinking about that for over one hour now... I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?
0.5(100−x)+0.25x=0.3*100 → x=80
First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.
Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters. In my opinion 30 liters of alcohol we being replaced by water so 3/5.
Can somebody explain that to me comprehensibly? Thanks a lot.



Math Expert
Joined: 02 Aug 2009
Posts: 7036

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
18 Mar 2015, 06:45
madmax1000 wrote: I am thinking about that for over one hour now... I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?
0.5(100−x)+0.25x=0.3*100 → x=80
First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.
Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters. In my opinion 30 liters of alcohol we being replaced by water so 3/5.
Can somebody explain that to me comprehensibly? Thanks a lot. hi madmax1000, the Q asks us how much original alcohol is replaced... as found by you and taken by you.... original alcohol =50l and after removing 80 l, 40 l of original alcohol has been replaced.. what is later added is not the replaced(original) alcohol but the replacement alcohol and what has been asked is the original alcohol, which has come down from 50 to 10... so replaced alcohol=40/50.. hope it helped..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
18 Mar 2015, 20:38
madmax1000 wrote: I am thinking about that for over one hour now... I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?
0.5(100−x)+0.25x=0.3*100 → x=80
First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.
Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters. In my opinion 30 liters of alcohol we being replaced by water so 3/5.
Can somebody explain that to me comprehensibly? Thanks a lot. Say, you have a 100 ml glass of 50% water + 50% alcohol solution. So water is 50 ml in it and alcohol is 50 ml in it homogeneously mixed. Now, of the 100 ml, say you take out, 20 ml solution (i.e. you take out 20% of the solution). How much water and how much alcohol did you take out? Since the solution has 5050 alcohol and water, the 20 ml will have 10 ml water and 10 ml alcohol, right? So of the 50 ml alcohol originally, you took out 10 ml alcohol. This means you took out 20% of the original alcohol. Of the 50 ml water originally, you took out 10 ml water. This means you took out 20% of the original water. So by taking out 20% of the original solution (which was 100 ml), you took out 20% of original water (which was 10 ml of 50 ml) and 20% of original alcohol (which was 10 ml of 50 ml). Does this make sense?
_________________
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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 24 Nov 2014
Posts: 18

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
19 Mar 2015, 02:35
yes thanks i was deducting the amount of alcohol that was taken out of the alcohol and replaced, because i thought that doesnt count..



VP
Joined: 07 Dec 2014
Posts: 1114

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
28 Jan 2016, 14:01
let x=% of alcohol replaced .5.5x+.25x=.3 x=.2/.25=4/5=80%



Intern
Joined: 05 Jan 2016
Posts: 11
Location: United States

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
08 Aug 2016, 08:02
Hi Can anyone explain the meaning of the question?? I've been trying to comprehend..say the solution is 100 ml 50 ml water and 50 ml alcohol.. now it says half water/half alcohol is replaced... as in each component is 25ml each and so 25+25=50? I dint understand this.can anyone help please?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
08 Aug 2016, 08:30
kamkaul wrote: Hi Can anyone explain the meaning of the question?? I've been trying to comprehend..say the solution is 100 ml 50 ml water and 50 ml alcohol.. now it says half water/half alcohol is replaced... as in each component is 25ml each and so 25+25=50? I dint understand this.can anyone help please? The question says: "If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution" So half water/half alcohol is the kind of original mix we have. We have a solution made of 50% alcohol and the other 50% is water. We replace some part of it with a 25% alcohol solution (so it has 25% alcohol and 75% water) How much of it do we replace, we do not know. If of the original, say 100 ml, we replace 20 ml, then we will take out 20 ml of 50% solution (so 10 ml alcohol 10 ml water) and instead put 20 ml of 25% solution (5 ml alcohol and 15 ml water). Now we would have alcohol > 40 ml + 5 ml = 45 ml and we would have water > 40 ml + 15 ml = 55 ml So the resulting solution will be 45% alcohol solution. Does this all make sense? But our resulting solution is only 30% alcohol. We solve it using the introductory concept discussed here: http://www.veritasprep.com/blog/2011/04 ... mixtures/
_________________
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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Intern
Joined: 05 Jan 2016
Posts: 11
Location: United States

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
08 Aug 2016, 08:49
Hi Karishma,
Thanks so much, I did understand the question now.. Original composition = 50% water + 50% alcohol ; after replacement 25% alcohol + 75 % water = 30% alcohol.. Now if I apply allegation: old alcohol new alcohol 50 25 30 5 20 old alcohol/new alcohol = 1/4 How can I proceed after this...? Also I am finding it difficult to construct the same through equations..could you please help?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
08 Aug 2016, 20:52
kamkaul wrote: Hi Karishma,
Thanks so much, I did understand the question now.. Original composition = 50% water + 50% alcohol ; after replacement 25% alcohol + 75 % water = 30% alcohol.. Now if I apply allegation: old alcohol new alcohol 50 25 30 5 20 old alcohol/new alcohol = 1/4 How can I proceed after this...? Also I am finding it difficult to construct the same through equations..could you please help? Did you check the post for which I provided the link above? Solution 1 > 50% alcohol Solution 2  > 25% alcohol Final Solution > 30% alcohol w1/w2 = (30  25)/(50  30) = 1/4 So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2. Hence, 4/5th of solution 1 was replaced. This is 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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 06 Oct 2015
Posts: 91
Location: Bangladesh
Concentration: Accounting, Leadership

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
09 Aug 2016, 04:38
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%... Nice explanation indeed. But I want to know what does 50% represents? Is it water or alcohol?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
09 Aug 2016, 21:28
NaeemHasan 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%... Nice explanation indeed. But I want to know what does 50% represents? Is it water or alcohol? 25% and 30% are the concentrations of alcohol. So basically we are working with alcohol. Hence 50% is also the concentration of alcohol.
_________________
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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 06 Oct 2015
Posts: 91
Location: Bangladesh
Concentration: Accounting, Leadership

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
11 Aug 2016, 09:37
VeritasPrepKarishma wrote: NaeemHasan 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%... Nice explanation indeed. But I want to know what does 50% represents? Is it water or alcohol? 25% and 30% are the concentrations of alcohol. So basically we are working with alcohol. Hence 50% is also the concentration of alcohol. Thank you. I did not understand the question at first. Thanks a ton.



Intern
Joined: 20 Aug 2016
Posts: 49
GMAT 1: 570 Q46 V23 GMAT 2: 610 Q49 V25 GMAT 3: 620 Q45 V31
WE: Information Technology (Other)

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
25 Oct 2017, 10:23
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%... Loved this solution, so easy and quick! Thanks a lot.



VP
Joined: 09 Mar 2016
Posts: 1091

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
25 Jan 2018, 04:29
VeritasPrepKarishma wrote: kamkaul wrote: Hi Karishma,
Thanks so much, I did understand the question now.. Original composition = 50% water + 50% alcohol ; after replacement 25% alcohol + 75 % water = 30% alcohol.. Now if I apply allegation: old alcohol new alcohol 50 25 30 5 20 old alcohol/new alcohol = 1/4 How can I proceed after this...? Also I am finding it difficult to construct the same through equations..could you please help? Did you check the post for which I provided the link above? Solution 1 > 50% alcohol Solution 2  > 25% alcohol Final Solution > 30% alcohol w1/w2 = (30  25)/(50  30) = 1/4 So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2. Hence, 4/5th of solution 1 was replaced. This is 80%. Hi VeritasPrepKarishmanow I got totally confused why here you write this solution w1/w2 = (30  25)/(50  30) = 1/4 here you follow weighted average formula, ok A1 = 50 A2 = 25 Aavg = 30 in this case i understand, since 25 is closer to 30 we take A2 = 25 meaning it 25 has more volume, But in this link https://gmatclub.com/forum/weightedave ... l#p2003243 you write in this way \(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)} = \frac{(50 – 30)}{(30 – 25)} = \frac{4}{1}\) here you take A1 = 25 , A2 = 50 and Aavg = 30  Why ? can you explain the difference between the first and the second solution? thank you



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8550
Location: Pune, India

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
25 Jan 2018, 06:58
dave13 wrote: VeritasPrepKarishma wrote: kamkaul wrote: Hi Karishma,
Thanks so much, I did understand the question now.. Original composition = 50% water + 50% alcohol ; after replacement 25% alcohol + 75 % water = 30% alcohol.. Now if I apply allegation: old alcohol new alcohol 50 25 30 5 20 old alcohol/new alcohol = 1/4 How can I proceed after this...? Also I am finding it difficult to construct the same through equations..could you please help? Did you check the post for which I provided the link above? Solution 1 > 50% alcohol Solution 2  > 25% alcohol Final Solution > 30% alcohol w1/w2 = (30  25)/(50  30) = 1/4 So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2. Hence, 4/5th of solution 1 was replaced. This is 80%. Hi VeritasPrepKarishmanow I got totally confused why here you write this solution w1/w2 = (30  25)/(50  30) = 1/4 here you follow weighted average formula, ok A1 = 50 A2 = 25 Aavg = 30 in this case i understand, since 25 is closer to 30 we take A2 = 25 meaning it 25 has more volume, But in this link https://gmatclub.com/forum/weightedave ... l#p2003243 you write in this way \(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)} = \frac{(50 – 30)}{(30 – 25)} = \frac{4}{1}\) here you take A1 = 25 , A2 = 50 and Aavg = 30  Why ? can you explain the difference between the first and the second solution? thank you Dave, it is just a matter of nomenclature. You can take A1, w1 as the 50% alcohol solution and A2, w2 as 25% alcohol solution or the other way around. In any case, answer will be the same. Check here: https://www.veritasprep.com/blog/2017/1 ... averages/This post deals with exactly this confusion.
_________________
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 >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12882
Location: United States (CA)

Re: If a portion of a half water/half alcohol mix is replaced
[#permalink]
Show Tags
25 Feb 2018, 15:21
Hi All, We're told that we're going to mix a certain amount of a 50% alcohol solution with a certain amount of a 25% alcohol solution and end up with a 30% alcohol solution. This is an example of a 'Weighted Average' question. Here's how we can set up that calculation using Algebra: A = number of ounces of 50% solution B = number of ounces of 25% solution A+B = total ounces of the mixed solution (.5A + .25B)/(A+B) = .3 .5A + .25B = .3A + .3B .2A = .05B 20A = 5B 4A = B A/B = 1/4 This means that for every 1 ounce of solution A, we have 4 ounces of solution B. To answer the question that's asked, imagine that you have 5 ounces of solution A. We're told to REPLACE 4 ounces of it with 4 ounces of solution B (thus, we'll end up with the 30% mixture that we're after). Since 4 ounces of the 5 ounces were replaced, 4/5 of the original alcohol and 4/5 of the original water was replaced. 4/5 = 80% Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****




Re: If a portion of a half water/half alcohol mix is replaced &nbs
[#permalink]
25 Feb 2018, 15:21



Go to page
Previous
1 2 3
Next
[ 41 posts ]



