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

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

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New post 19 Nov 2019, 01:11
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Competition Mode Question



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


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

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New post 19 Nov 2019, 01:52
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1
Initially:
Alcohol=15l
water=0l

after 1st removal:
alcohol removed=5l
alcohol remaining=10l
water added=5l
alcohol:water=10:5=2:1

after 2nd removal:
solution removed= 5l
amount of alcohol in soln removed=5*2/3=10/3
amount of water in soln removed=5*1/3=5/3
alcohol remaining=10-10/3=20/3
water remaining=5-5/3+5=25/3
alcohol:water=20:25=4:5

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

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New post 19 Nov 2019, 02:16
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Initial
Vol (alcohol) = 15
Vol (water) = 0

1st removal we get
5L alcohol removed so, Vol (alcohol) = 15-5 = 10L
5L water added, Vol (water) = 5L (alcohol:water) = 10/5 = 2:1

After 2nd removal
5 Litre of total liquid (mixture) removed from the cask
alcohol removed = 5/3 + 5/3 as 2:1 is the ratio .
5L mixture would contain 10/3 l alcohol
water removed = 5/3
5l water is then added;

Final volumes: Alcohol = 15-5-10/3 = 20/3
Water =0+ 5 - 5/3 + 5 = 25/3
ratio = (20/3)/(25/3) = 20/25 = 4/5

IMO A

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

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New post 19 Nov 2019, 03:11
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

Initial Amount of Alcohol = 15 Liters
After removing 5 liters of alcohol, amount left = 10 liters
Now, 5 liters of water is added hence \(\frac{1}{3}rd\) volume of 15 liters is water.
Again cask is emptied by 5 liters(mix of alcohol and water) and 5 liters of water is added.
Hence total water portion now = \(\frac{1}{3}*10 + 5 = \frac{25}{3}\)
Alcohol portion = \(15 - \frac{25}{3} = \frac{20}{3}\)

Ratio of alcohol to water = \(\frac{20}{3}:\frac{25}{3} = \frac{4}{5}\)

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

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New post 19 Nov 2019, 06:03
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The language of the question seems ambiguous. If cask is emptied completely and 5 litres of water is added, there will be only water present.

If cask is not emptied completely and if the iteration only consists of removing 5 litres of initial liquid (alcohol) and then adding 5 litres of water up to the brim and subsequent removal of 5 litres of mix of water and alcohol.

————— Alcohol ——— Water
Initial ——— 15 ————— 0 —
Step1 ——15-5=10 ————5 —
Ratio of alcohol: water = 10 : 5 = 2 : 1

After second iteration,
When 5 litres of mixture is removed.
Alcohol removed = 2/3*5 = 10/3
Water removed = 1/3*5 = 5/3

Alcohol remaining = 10 - 10/3 = 20/3
Water remaining = 5 - 5/3 + 5 (since 5 litres of water is added) = 25/3

Ratio of alcohol : water = 20/3 : 25/3 = 4 : 5 or 4/5

IMO Option A

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

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New post 19 Nov 2019, 08:43
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I would be hoping to see alternative approaches to solve this question.

We have a cask that contains 15liters of wine. 5 liters of wine is taken out and replaced with 5 liters of water. So we have a mixture of Wine and Water made up of 10liters of Wine and 5liters of water.
5 liters of the resulting mixture is taken out and replaced with 5 liters of water. Effectively (2*5)/3 liters of Wine are removed and 5/3 of water is removed.
The volume of wine left in the cask=10-10/3 = 20/3
The volume of water now in the cask = 5+(5-5/3) = 5 + 10/3 = 25/3
Ratio of Wine to water = 20/3:25/3 = 20:25 = 4:5

The answer is option A in my view.
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A cask initially contains pure alcohol up to the brim. The cask can be  [#permalink]

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New post 20 Nov 2019, 04:59
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Bunuel wrote:

Competition Mode Question



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


\(Cfinal=Cinitial(Vinitial/Vfinal)^n\)

\(Vinitial=Total-Removed\)

\(Vfinal=Total-Removed+Replaced\)

\(Cf_a=1(\frac{15-5}{15-5+5})^2…Cf_a=(\frac{2}{3})^2…Cf_a=\frac{4}{9}…Cf_w=\frac{5}{9}\)

\(\frac{Alcohol}{Water}=\frac{4}{5}\)

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

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New post 22 Nov 2019, 12:47
Bunuel wrote:

Competition Mode Question



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


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After 5 liters of alcohol are removed, we have 5 water and 10 alcohol, so the ratio is W : A = 1 : 2

After another 5 liters are removed, we have:

x + 2x = 5

3x = 5

x = 5/3

So, x = 5/3 liters of water is removed and 2x = 10/3 liters of alcohol is removed.

So the amount of water and alcohol after 5 more liters of water are put back is:

Water: 5 - 5/3 + 5 = 10 - 5/3 = 30/3 - 5/3 = 25/3

Alcohol: 10 - 10/3 = 30/3 - 10/3 = 20/3

Thus, the ratio of alcohol to water is 20/3:25/3 = 20:25 = 4:5

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

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New post 22 Nov 2019, 14:29
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The usage of the word "completely" is confusing in this question. It seems to imply the cask is being entirely emptied from 15L to 0L.

Should it read something like: "When 5 liters are removed from the cask and it is filled back to the brim two times..."?

Let me know if I am simply mistaken! :cool:
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Re: A cask initially contains pure alcohol up to the brim. The cask can be  [#permalink]

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New post 11 Dec 2019, 21:16
jnootenboom wrote:
The usage of the word "completely" is confusing in this question. It seems to imply the cask is being entirely emptied from 15L to 0L.

Should it read something like: "When 5 liters are removed from the cask and it is filled back to the brim two times..."?

Let me know if I am simply mistaken! :cool:



Same Thoughts, couldnt even begin with the problem.
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Re: A cask initially contains pure alcohol up to the brim. The cask can be   [#permalink] 11 Dec 2019, 21:16
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