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# Aptitude - Pipes and Cistern - Discussion

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"When ambition ends, happiness begins."
- (Proverb)
15.

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

 [A]. 10 [B]. 12 [C]. 14 [D]. 16

Answer: Option B

Explanation:

 Part filled in 2 hours = 2 = 1 6 3

 Remaining part = 1 - 1 = 2 . 3 3

 (A + B)'s 7 hour's work = 2 3

 (A + B)'s 1 hour's work = 2 21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

 = 1 - 2 = 1 6 21 14

C alone can fill the tank in 14 hours.

 M.V.Krishna said: (Wed, Apr 6, 2011 04:21:19 AM) Part filled in 2 hours= 2/6. This step is not clear. Please explain.

 Dinesh said: (Wed, Feb 15, 2012 11:49:15 AM) A&B&C can fill a tank together in 6 hrs but they worked together for 2 hrs only so u hv to calculate the time finished in 2 hrs so A&B&C=1/6 For two hrs =2/6=1/3

 Ramya said: (Mon, Jul 23, 2012 10:27:26 PM) (A+B+C) - (A+B) can give you the answer. (A+B+C) =1/6 and (A+B+C) in 2hrs=2/6 and remaining part 1-2/6=2/3. So (A+B) in 7 hrs is 2/ (3*7) =2/21. 1/6-2/21=1/14 so answer is 14.

 Sharvari said: (Sat, Aug 25, 2012 12:12:40 AM) How to calculate (A+B)in 7 hours ??

 Nfl said: (Fri, Jan 11, 2013 08:56:54 PM) @Sharvari. A+B+C = 2/6= 1/3 ..remaining part of cistern filled by A+B by 7 hours. Remaining part 2/3 in 7 hours.. for exact 1 hour by A+B= 2/3*1/7=1/21.

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