Aptitude Pipes And Cisterns


Rule 1:

If a pipe can fill a tank in x hours, then the part filled in 1 hour =1/x.

Example : A pipe can fill a tank in 10 hours. Find the part of tank filled in one hour.

Solution : Applying rule 1 , we have the part filled in 1 hour = 1/10.

Rule 2:

If a pipe can empty a tank in y hours, then the part of the full tank emptied in 1 hour = 1/y.

ExampleA pipe can empty a tank in 15 hours. Find the part of the tank emptied in one hour.

Solution :Applying rule 2, we have the part emptied in 1 hour = 1/15.

Rule 3:

If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then the net part filled in 1 hour = (1/x-1/y)

Rule 4:

If a pipe can fill a tank in x hours and a nother pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then the net part emptied in 1 hour =(1/y-1/x)







Subscribe with us to get latest topic update






Choose a Language



Connection failed: SQLSTATE[28000] [1045] Access denied for user 'studentdb'@'localhost' (using password: YES)

Subscribe