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A cask initially contains pure alcohol up to the brim. The c
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14 Apr 2019, 10:29
14 Apr 2019, 10:29
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48% (08:35) correct
51% (02:11) wrong based on 31 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) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) 4 : 5
(E) 5 : 4
(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) 4 : 5
(E) 5 : 4
Re: A cask initially contains pure alcohol up to the brim. The c
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13 Oct 2019, 02:26
13 Oct 2019, 02:26
Any explanation? Actually I do not understand the question that much, if every time I empty 5 liters of the cask I must fill it back with water how can the whole cask be completely emptied?
Re: A cask initially contains pure alcohol up to the brim. The c
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18 Oct 2019, 10:04
18 Oct 2019, 10:04
2
I got confused with the 'completely' too. If its completely emptied in the first round then there is no alcohol at all due to which we would end up with all water. The options do not correspond to this understanding.
So, I think completely here means 5 litres removed.
In that case lets look at alcohol and water content at each time.
Beginning,
Vol (alcohol) = 15
Vol (water) = 0
After 1st removal
5L alcohol removed so, Vol (alcohol) = 15-5 = 10L
5L water added, Vol (water) = 5L
Current ratio (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 since the ratio was 2:1, during the time of removal. Thus, 5L mixture would contain 10/3 l aclohol
water removed = 5/3
5l water is then added;
Final volumes:
Alocohol = 15-5-10/3 = 20/3
Water = 0 + 5 - 5/3 + 5 = 25/3
ratio = (20/3)/(25/3) = 20/25 = 4/5
So, I think completely here means 5 litres removed.
In that case lets look at alcohol and water content at each time.
Beginning,
Vol (alcohol) = 15
Vol (water) = 0
After 1st removal
5L alcohol removed so, Vol (alcohol) = 15-5 = 10L
5L water added, Vol (water) = 5L
Current ratio (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 since the ratio was 2:1, during the time of removal. Thus, 5L mixture would contain 10/3 l aclohol
water removed = 5/3
5l water is then added;
Final volumes:
Alocohol = 15-5-10/3 = 20/3
Water = 0 + 5 - 5/3 + 5 = 25/3
ratio = (20/3)/(25/3) = 20/25 = 4/5
