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Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour

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Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour  [#permalink]

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New post Updated on: 26 Mar 2019, 05:29
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Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hours, Pipe C can fill Tank in 6 Hours. The Tank is already filled up to 1/6 of its capacity. Now Pipe A is opened in the First Hour alone, Pipe B is opened in the Second Hour alone and Pipe C is opened in the Third Hour alone. This cycle is repeated until the Tank gets filled. Then in How many Hours does the rest of Tank gets filled?

A. 15 Hours
B. 18 Hours
C. 20 Hours
D. 24 Hours
E. None

Originally posted by Wasif007 on 26 Mar 2019, 05:26.
Last edited by Bunuel on 26 Mar 2019, 05:29, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour  [#permalink]

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New post 26 Mar 2019, 07:25
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let the total work be 36 units (a common multiple for 18,12 & 6)
so per 1 hour, A would pump in 2 units, B would pump out 3 units & C would pump in 6 units
so 1 cycle of 3 hours would add 5 units in.

As \(\frac{1}{6}\) is already filled, so 6 units are already there.

at zero time: 6 units
after 3 hour (1 cycle): 6 + (2-3+6) = 6 + 5(1) = 11 , ... and so on .....

to fill the 36 units:
36 = 6 + 5x , where x is the number of cycles
x = 6, which represent 3 hours each, so 18 hours

so B
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Re: Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour  [#permalink]

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New post 27 Mar 2019, 19:08
Wasif007 wrote:
Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hours, Pipe C can fill Tank in 6 Hours. The Tank is already filled up to 1/6 of its capacity. Now Pipe A is opened in the First Hour alone, Pipe B is opened in the Second Hour alone and Pipe C is opened in the Third Hour alone. This cycle is repeated until the Tank gets filled. Then in How many Hours does the rest of Tank gets filled?

A. 15 Hours
B. 18 Hours
C. 20 Hours
D. 24 Hours
E. None


We can see that for 3 consecutive hours, the tank is filled 1/18 - 1/12 + 1/6 = 2/36 - 3/36 + 6/36 = 5/36 of its capacity. Since 5/6 of the tank still needs to be filled and (5/6)/(5/36) = 6, we need 3 x 6 = 18 hours to fill the rest of the tank.

Answer: B
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Re: Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour  [#permalink]

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New post 10 Aug 2019, 12:41
Wasif007 wrote:
Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hours, Pipe C can fill Tank in 6 Hours. The Tank is already filled up to 1/6 of its capacity. Now Pipe A is opened in the First Hour alone, Pipe B is opened in the Second Hour alone and Pipe C is opened in the Third Hour alone. This cycle is repeated until the Tank gets filled. Then in How many Hours does the rest of Tank gets filled?

A. 15 Hours
B. 18 Hours
C. 20 Hours
D. 24 Hours
E. None


Let's take a common multiple of 18,6 and 12. Say 36.
Capacity of tank be 36 and it is already 6 units filled.
Now A takes 12 hours which means to fill a tank of 36 units it is doing 2 units per hour. Similarly, B is doing -3 and C is doing +6.
All 3 combined are doing +2-3+6 which is +5. Our remaining capacity is 30 units. Which means we will do this for 6 times. and each time it is 3 hours. Hence 18 hours.

P.S: I approach such questions with assuming units of work. Try it. It is usually easier. Thanks!
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Re: Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour  [#permalink]

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New post 10 Aug 2019, 12:51
5/6 of the capacity of the tank is empty: 30/36
30/36- (1/18 - 1/12 + 1/6)=25/36
5/6 of the capacity of the tank will get full in 3 hours.so 6×3= 18 hours to fill the tank.
Option B

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Re: Pipe A can fill a Tank in 18 Hours, Pipe B can empty a Tank in 12 Hour   [#permalink] 10 Aug 2019, 12:51
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