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Wasif007
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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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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pipe A fills 2 units in the first hour then pipe B comes into effect in the 2nd hour. If that is the case, pipe B is supposed to drain 2 units. How can pipe B drain 3 units if there are 2 units only? but tank becomes full at the 18th hour nonetheless

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pipe A fills 2 units in the first hour then pipe B comes into effect in the 2nd hour. If that is the case, pipe B is supposed to drain 2 units. How can pipe B drain 3 units if there are 2 units only? but tank becomes full at the 18th hour nonetheless

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Your observation would be correct if not for this given statement:
"The Tank is already filled up to 1/6 of its capacity".

So if we assume that the tank is 36 units, 6 are already there at time zero,
and after 1st hour = 8 (which is 6+2)
and after 2nd hour = 5 (which is 8-3)
and so on ....
please revise my answer above for further clarification if needed,
happy to help.
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