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# A cask initially contains pure alcohol up to the brim. The cask can be

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Intern
Joined: 03 Jul 2015
Posts: 35

Kudos [?]: 21 [0], given: 27

A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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26 Dec 2015, 10:29
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55% (hard)

Question Stats:

61% (02:04) correct 39% (01:37) wrong based on 85 sessions

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A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask

a 4/5
b. 3/5
c. 4/6
d. 3/6
e. 2/6
plz explain someone
[Reveal] Spoiler: OA

Kudos [?]: 21 [0], given: 27

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1723 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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26 Dec 2015, 11:16
anik19890 wrote:
A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask

a 4/5
b. 3/5
c. 4/6
d. 3/6
e. 2/6
plz explain someone

Take these questions step by step.

You are given that the initial quantity of the cask is 15 l. So for the changing of the liquids,

The first time you remove 5 l of pure alcohol---> 10 l of alcohol remaining and you add 5 l of water. Ratio of alcohol in the remaining mixture = 10/(10+5) = 10/15 =2/3, similarly water = 1/3.

The second time you remove 5 l of the mixture having alcohol = 5*2/3 = 10/3 --> remaining alcohol = 10-10/3 =20/3

Similarly, the amount of water removed = 5*1/3=5/3 ---> remaining water = 5-5/3=10/3 but you also add 5 l more of water --> total water now = 5+10/3 = 25/3

Finally, ratio asked = 20/3 / 25/3 = 20/25 = 4/5

A is the correct answer.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1723 [0], given: 792

Intern
Joined: 03 Jul 2015
Posts: 35

Kudos [?]: 21 [0], given: 27

Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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26 Dec 2015, 11:22
Engr2012 wrote:
anik19890 wrote:
A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask

a 4/5
b. 3/5
c. 4/6
d. 3/6
e. 2/6
plz explain someone

Take these questions step by step.

You are given that the initial quantity of the cask is 15 l. So for the changing of the liquids,

The first time you remove 5 l of pure alcohol---> 10 l of alcohol remaining and you add 5 l of water. Ratio of alcohol in the remaining mixture = 10/(10+5) = 10/15 =2/3, similarly water = 1/3.

The second time you remove 5 l of the mixture having alcohol = 5*2/3 = 10/3 --> remaining alcohol = 10-10/3 =20/3

Similarly, the amount of water removed = 5*1/3=5/3 ---> remaining water = 5-5/3=10/3 but you also add 5 l more of water --> total water now = 5+10/3 = 25/3

Finally, ratio asked = 20/3 / 25/3 = 20/25 = 4/5

A is the correct answer.

the second time when 5 l is removed why 5 is multiplied?

Kudos [?]: 21 [0], given: 27

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1723 [0], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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26 Dec 2015, 11:29
anik19890 wrote:
Engr2012 wrote:
anik19890 wrote:
A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask

a 4/5
b. 3/5
c. 4/6
d. 3/6
e. 2/6
plz explain someone

Take these questions step by step.

You are given that the initial quantity of the cask is 15 l. So for the changing of the liquids,

The first time you remove 5 l of pure alcohol---> 10 l of alcohol remaining and you add 5 l of water. Ratio of alcohol in the remaining mixture = 10/(10+5) = 10/15 =2/3, similarly water = 1/3.

The second time you remove 5 l of the mixture having alcohol = 5*2/3 = 10/3 --> remaining alcohol = 10-10/3 =20/3

Similarly, the amount of water removed = 5*1/3=5/3 ---> remaining water = 5-5/3=10/3 but you also add 5 l more of water --> total water now = 5+10/3 = 25/3

Finally, ratio asked = 20/3 / 25/3 = 20/25 = 4/5

A is the correct answer.

the second time when 5 l is removed why 5 is multiplied?

It's because you are told that 5 l of whatever is remaining in the cask is removed. Before the first one, the entire liquid was alcohol but now once we started to mix water with alcohol, whatever you remove now will be a mixture of water and alcohol. This mixture will have water and alcohol in the ratio determined by their ratio before the second one.

This is the reason why you have 5 l of mixture removed out of which 1/3 is water and 2/3 is alcohol.

Hope this helps
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html
Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Kudos [?]: 1723 [0], given: 792

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Joined: 09 Sep 2013
Posts: 16709

Kudos [?]: 273 [0], given: 0

Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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21 Jan 2017, 16:13
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Concentration: Strategy, International Business
Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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17 Aug 2017, 00:46
Engr2012 wrote:
anik19890 wrote:
A cask initially contains pure alcohol up to the brim. The cask can be emptied by removing exactly 5 liters at a time . Each time this is done, the cask must be filled back to the brim with water. The capacity of the cask is 15 liters. When the cask is completely emptied and filled back to the brim two times, what is the ratio of alcohol to water in the cask

a 4/5
b. 3/5
c. 4/6
d. 3/6
e. 2/6
plz explain someone

Take these questions step by step.

You are given that the initial quantity of the cask is 15 l. So for the changing of the liquids,

The first time you remove 5 l of pure alcohol---> 10 l of alcohol remaining and you add 5 l of water. Ratio of alcohol in the remaining mixture = 10/(10+5) = 10/15 =2/3, similarly water = 1/3.

The second time you remove 5 l of the mixture having alcohol = 5*2/3 = 10/3 --> remaining alcohol = 10-10/3 =20/3

Similarly, the amount of water removed = 5*1/3=5/3 ---> remaining water = 5-5/3=10/3 but you also add 5 l more of water --> total water now = 5+10/3 = 25/3

Finally, ratio asked = 20/3 / 25/3 = 20/25 = 4/5

A is the correct answer.

Thanks for giving the solution.

However , I am confused how have you covered the condition "when the cask is fully emptied and filled to brim two times"

Doesn't the solution provided indicates that from cask 5lts solution is replaced with 5lts water twice.

Kudos [?]: 9 [0], given: 74

Senior Manager
Joined: 29 Jun 2017
Posts: 345

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WE: Engineering (Transportation)
Re: A cask initially contains pure alcohol up to the brim. The cask can be [#permalink]

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17 Aug 2017, 05:37
5
KUDOS
Simplest solution: A is answer

Volume can = 15 Ltr initially full alcohol

1st mix = 5 ltr alcohol removed , 10 ltr remained

Volume Alcohol to Can is 10 /15

2nd Mix - Now 5 ltr mix is removed in that alcohol will be out by (10/15) * 5 = 10/3 ltr and remaining will be 10-10/3 = 20/3

Volume of Alcohol to can will be (20/3) / 15 = 4/9
means 4 is alcohol and 9 is mixture ( 4 ltr alcohol + 5 ltr water)

4/5 is the Answer A
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Give Kudos for correct answer and/or if you like the solution.

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Re: A cask initially contains pure alcohol up to the brim. The cask can be   [#permalink] 17 Aug 2017, 05:37
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# A cask initially contains pure alcohol up to the brim. The cask can be

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