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A container holds 4 quarts of alcohol and 4 quarts of water. How many
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A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume? A. 4/3 B. 5/3 C. 7/3 D. 8/3 E. 10/3
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Originally posted by shrive555 on 10 Jun 2011, 00:08.
Last edited by Bunuel on 06 Apr 2016, 21:52, edited 1 time in total.
Renamed the topic and edited the question.




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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 12:20
rohit8865 wrote: shrive555 wrote: A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?
A. 4/3 B. 5/3 C. 7/3 D. 8/3 E. 10/3 Can anyone will give equation for alligation method??? An alternate to the alligation method is the more direct/algebraic method: Let x be the amount of water to be added. New total amount of water = 4+x Total amount of alcohol = 4 New total = 4+4+x=8+x Final ratio required (for water) = 5/(5+3)=5/8 Thus, (4+x)/(8+x) = 5/8 > solving for x you get x = 8/3. D is thus the correct answer. Hope this helps.




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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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10 Jun 2011, 00:46
Currently its is 4 units of Alcohol and 4 units of water to make the ratio 3/5 ; how much water has to be added?
4/4+x =3/5 20 = 12+3x x= 8/3 Its D



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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10 Jun 2011, 04:40
Let the number of quarts that should be added to get the required ratio = x So total quarts of water = (x + 4) But the original number of quarts of alcohol remains the same, so we have : 4/(x+4) = 3/5 Crossmultiply: Now we have 20 = 3x + 12 => 3x = 20 12 => x = 8/3 Answer  D
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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10 Jun 2011, 20:38
D.... 4/(4+x) = 3/5 Solve for x
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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06 Apr 2016, 20:08
shrive555 wrote: A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?
4/3 5/3 7/3 8/3 10/3 Isn't 4 quarts to 4 quarts equivalent to a ratio of 1:1? Which is the same unit of 3:5. Wouldn't the equation then be 1/1+X=3/5? Any help in this matter is greatly appreciated.



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 00:47
shrive555 wrote: A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?
A. 4/3 B. 5/3 C. 7/3 D. 8/3 E. 10/3 Can anyone will give equation for alligation method???



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 02:08
rohit8865 wrote: shrive555 wrote: A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?
A. 4/3 B. 5/3 C. 7/3 D. 8/3 E. 10/3 Can anyone will give equation for alligation method??? For this particularproblem , algebric method is the best... However for your convenience I am posting Alligation method  Attachment:
1.PNG [ 1.87 KiB  Viewed 7328 times ]
Attachment:
2.PNG [ 4.14 KiB  Viewed 7320 times ]
Rest I leave for you to solve.... PS : USe alligation rule when percentage is given ( My personal opinion ) it is less time consuming and convenient....Abhishek
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A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 11:24
Abhishek009 wrote: rohit8865 wrote: shrive555 wrote: A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?
A. 4/3 B. 5/3 C. 7/3 D. 8/3 E. 10/3 Can anyone will give equation for alligation method??? For this particularproblem , algebric method is the best... However for your convenience I am posting Alligation method  Attachment: 1.PNG Attachment: 2.PNG Rest I leave for you to solve.... PS : USe alligation rule when percentage is given ( My personal opinion ) it is less time consuming and convenient....AbhishekAbhi thanks 4 reply... But i could not get (50>12.5>62.5)part.Per me (5012.5=37.5 as 37.5 comes from 3/8 ,similarly 5/8=67.5...fair....but not understood 50>12.5>62.5 also water is added so it must be 100 instead 50%) plz explain 4 me...may be i m missing something... thanks again..



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 11:32
rohit8865 wrote: Abhi thanks 4 reply... But i could not get (50>12.5>62.5)part. plz explain 4 me...may be i m missing something... thanks again.. My pleasure , Quote: 4 quarts of alcohol and 4 quarts of water So, Total quantity is 8 units (say) Out of that 50% or 4 units is water and 50% or 4 units is alcohol You want to come to  Quote: 3 parts alcohol to 5 parts water by volume (3/8*100) 37.5 units or 3 out of 8 units of water and (5/8*100) 62.5 units or 5 out of 8 units of alcohol Hope I am clear this time...
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 11:39
Abhishek009 wrote: rohit8865 wrote: Abhi thanks 4 reply... But i could not get (50>12.5>62.5)part. plz explain 4 me...may be i m missing something... thanks again.. My pleasure , Quote: 4 quarts of alcohol and 4 quarts of water So, Total quantity is 8 units (say) Out of that 50% or 4 units is water and 50% or 4 units is alcohol You want to come to  Quote: 3 parts alcohol to 5 parts water by volume (3/8*100) 37.5 units or 3 out of 8 units of water and (5/8*100) 62.5 units or 5 out of 8 units of alcohol Hope I am clear this time...%age part is ok...but confusion is for Avg. part.... 5012.5=37.5 its ok but 12.550=62.5(here i m not clear) hope u got this time.. thanks



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A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 12:13
Quote: Hope I am clear this time... %age part is ok...but confusion is for Avg. part.... 5012.5=37.5 its ok but 12.550=62.5(here i m not clear)
hope u got this time..
thanks
Ok rohit8865Let me try once again , I hope you are familiar with this form Attachment:
Capture.PNG [ 12.14 KiB  Viewed 7243 times ]
Here you have c , d , d  m & m  c Does it get a bit easier now ?I suggest you go through this excellent video (from PerfectScores.com)  https://www.youtube.com/watch?v=x5ycYnbnXy8Do search Youtube with the search item " alligation and mixture aptitude " you will get many inputs... Also do find time to go through these pdf  http://awabsir.com/socialactivity/wpc ... redoc.pdfhttp://fi.ge.pgstatic.net/attachments/3 ... 2eed81.pdfHappy preparations Abhishek
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 19:54
Abhishek009 wrote: Quote: Hope I am clear this time... %age part is ok...but confusion is for Avg. part.... 5012.5=37.5 its ok but 12.550=62.5(here i m not clear)
hope u got this time..
thanks
Ok rohit8865Let me try once again , I hope you are familiar with this form Attachment: Capture.PNG Here you have c , d , d  m & m  c Does it get a bit easier now ?I suggest you go through this excellent video (from PerfectScores.com)  https://www.youtube.com/watch?v=x5ycYnbnXy8Do search Youtube with the search item " alligation and mixture aptitude " you will get many inputs... Also do find time to go through these pdf  http://awabsir.com/socialactivity/wpc ... redoc.pdfhttp://fi.ge.pgstatic.net/attachments/3 ... 2eed81.pdfHappy preparations AbhishekAbhi thats what i m asking too.. as per diagram in your first post. m=12.5 & c=50 but from wherm where you calculated 62.5 because mc inthat case is 12.550=... thanks



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 20:48
[quote="subhashghosh"]Let the number of quarts that should be added to get the required ratio = x
So total quarts of water = (x + 4)
But the original number of quarts of alcohol remains the same, so we have :
4/(x+4) = 3/5
Crossmultiply:
Now we have 20 = 3x + 12
=> 3x = 20 12
=> x = 8/3
What is wrong when I do as below :
Let the number of quarts that should be added to get the required ratio = x
So total quarts of water = (x + 4)
But the original number of quarts of alcohol remains the same, so we have :
(x+4)/8 = 5/8 { vol of water/total vol}
Crossmultiply:
=> x = 1
I am not getting the answer.
I have got many correct answer using this method. Please clear my doubt , what is the point I am missing. Help appreciated.



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 20:50
What is wrong when I do as below :
Let the number of quarts that should be added to get the required ratio = x
So total quarts of water = (x + 4)
But the original number of quarts of alcohol remains the same, so we have :
(x+4)/8 = 5/8 { vol of water/total vol}
Crossmultiply:
=> x = 1
I am not getting the answer.
I have got many correct answer using this method. Please clear my doubt , what is the point I am missing. Help appreciated.



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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25 Apr 2016, 21:06
robu wrote: What is wrong when I do as below :
Let the number of quarts that should be added to get the required ratio = x
So total quarts of water = (x + 4)
But the original number of quarts of alcohol remains the same, so we have :
(x+4)/8 = 5/8 { vol of water/total vol}
Crossmultiply:
=> x = 1
I am not getting the answer.
I have got many correct answer using this method. Please clear my doubt , what is the point I am missing. Help appreciated. you are just missing that now the toal volume is also increased i.e 8+x after addition..



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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26 Apr 2016, 06:54
robu wrote: What is wrong when I do as below :
Let the number of quarts that should be added to get the required ratio = x
So total quarts of water = (x + 4)
But the original number of quarts of alcohol remains the same, so we have :
(x+4)/8 = 5/8 { vol of water/total vol}
Crossmultiply:
=> x = 1
I am not getting the answer.
I have got many correct answer using this method. Please clear my doubt , what is the point I am missing. Help appreciated. Total new volume of the mixture will now become 8+x from 8 after the addition of x units of water. Refer to the solution acontainerholds4quartsofalcoholand4quartsofwaterhowmany114969.html#p1677059



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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30 Sep 2017, 23:03
VeritasPrepKarishma Hello Karishma. I recently read your blog posts on weighted averages and found the w1/w2 method and the scale method to be very useful, fast and accurate. So i've just started solving mixture problems using your techniques as much as possible. Will take sometime for me to master it. Could you please explain the how you can solve this sum using your method? Thanks



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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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01 Oct 2017, 00:12
sameerkamath21 wrote: VeritasPrepKarishma Hello Karishma. I recently read your blog posts on weighted averages and found the w1/w2 method and the scale method to be very useful, fast and accurate. So i've just started solving mixture problems using your techniques as much as possible. Will take sometime for me to master it. Could you please explain the how you can solve this sum using your method? Thanks Right now the concentration of water in the mix is 4/8. We want to add pure water (8/8) to it and make the concentration of water in the final solution (5/8). w1/w2 = (8/8  5/8)/(5/8  4/8) = 3/1 So for every 3 parts of the original mix, we should add 1 part of pure water. The original mix is 8 quarts. So we should add 8/3 quarts of pure water. Answer (D)
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many
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04 Oct 2017, 10:04
VeritasPrepKarishma Thank you so much Karishma. Just one doubt. The A2 and A1 that you used in your blog post about the same topic keep changing as per the question right? And what will w1/w2 be in this case?As in acc to this question?




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