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

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

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!

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

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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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

Posted from my mobile device

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.

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

\(Vinitial=Total-Removed\)

\(Vfinal=Total-Removed+Replaced\)

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

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

Ans (A)

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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Same Thoughts, couldnt even begin with the problem.
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Each time 5 liters is removed and filled back with water.
5 liters is 5/15 or 1/3 part of 15 liters

So, Qty of Alcohol after 2 replacement => 15*(2/3)*(2/3) = 20/3 L
Then, Qty of water after 2 replacement => 15 - 20/3 = 25/3 L

Therefore, A:W = 20/3 : 25/3 = 20:25 = 4:5

Given: 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.
Asked: 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

1st removal and filling: -
Alcohol = 15 - 5 = 10 liters
Water = 5 liters

2nd removal and filling: -
Alcohol removed = 10/3
Water removed = 5/3
Water filled = 5
Alcohol = 10 - 10/3 = 20/3
Water = 25/3
Alcohol: Water = 20/25 = 4/5

IMO A

Can anyone please explain the question

Posted from my mobile device

“Completely emptied twice” doesn’t mean empty out 15 (5 at a time) twice

We are only doing 2 replacements



Posted from my mobile device

Hi Bunuel / ScottTargetTestPrep / others
I have followed the same procedure listed in the forum discussion. However, I have arrived at E as the answer.

I felt confused between
A cask initially contains pure alcohol up to the brim (ie 15L of Alcohol)
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 (ie 10 of Alcohol)

I worked with the latter to arrive at 1/3. Do you think that the question is poorly worded? If no, could you please let me know where I'm going wrong?

Thanks

elPatron434, 1/3 is probably the biggest trap answer here. After the first round of emptying and replacing you'd have 10L of alcohol and 5L of water. I'm pretty sure you got this right.
Now, when you empty 5L again, these 5L won't be just of alcohol. In fact these 5 L would be divided amongst Alc and water in their current ratio (2:1).
So you need to subtract 10/3 from Alc and 5/3 from Water. Now add back 5 to water and you'll have your answer!

Although, I got the answer right because I understood what the question meant, I think it is poorly worded, especially the following part -
"When the cask is completely emptied and filled back to the brim two times". Completely emptied distorts the meaning.

Thanks Brian123
The reason I got 1/3 was because I chose the initial volume to be 10L as opposed to 15L(As it says its empty and 2 replacements are done)

Initial alcohol = 10L
After first replacement, alcohol = 5L and water =5L
In the second replacement, 2.5 L of water and alcohol are removed. 5L of water added.
Composition now is 2.5L alcohol and 7.5L water.

I guess I got confused by completely emptied

I am not able to get one particular condition in the question. We can never empty the cask as we are taking out only 5 litres at a time and also refilling it by same amount. Also, if it was to be completely emptied somehow, we will be left with no alcohol. If we simply take two iterations of the process, we can reach the OA.

My approach:

Volume of removal of liquids from the cask: 5l

Total volume: 15l

Volume which remains: 10l

10l/15l = 2/3. All contents will be reduced to their 2/3rd amounts. Since we are refilling with water, the only reference we have is alcohol. Thus:

Volume of alcohol remaining:
1st iteration: 15×(2/3)
2nd iteration: 15×(2/3)^2
Alcohol left: 15×(4/9)
Ratio of alcohol to mixture: 15×(4/9) ÷ 15 = 4/9
Alcohol to water ratio = 4/5 or 4:5

Posted from my mobile device

Verbiage is terrible. Please remove "completely emptied"...couldn't even begin to figure out how that would happen. Otherwise an easy 600 level problem

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