2 pipes can fill a tank in 9 minutes when opened simultaneously. If the efficiency of the first pipe is twice the efficiency of the second pipe, find the time required by the second pipe to fill the tank.

These are some important formulas for solving pipes and cistern based questions: If x hours are required to fill up a tank, then part filled in 1 hr =1/x If y hours are required to empty the tank, then part emptied in 1 hour = 1/y If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (y-x)}, provided y>x If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled... Show more

These are some important formulas for solving pipes and cistern based questions:

If x hours are required to fill up a tank, then part filled in 1 hr =1/x
If y hours are required to empty the tank, then part emptied in 1 hour = 1/y
If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (y-x)}, provided y>x
If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (x-y)}, provided x>y
Net work done = (Sum of work done by Inlets) – (Sum of work done by Outlets)
One inlet can fill the tank in x hr and the other inlet can fill the same tank in y hrs, if both the inlets are opened at the same time, the time taken to fill the whole tank = {(xy) / (y+x)}
If two pipes take x and y hours respectively to fill a tank of water and a third pipe is opened which takes z hours to empty the tank, then the time taken to fill the tank = {1 / (1/x)+(1/y)+(1/z)} and the net part of the tank filled in 1 hr = (1/x)+(1/y)-(1/z)

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