Reasoning and Quantitative aptitude Pipes and Cisterns

PREREQUISITES:

1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as aninlet.

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as anoutlet.

2. If a pipe can fill a tank in x hours, then:

part filled in 1 hour =1.

x

3. If a pipe can empty a tank in y hours, then:

part emptied in 1 hour =1.

y

4. 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 =1-1

.x y

5. 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 emptied in 1 hour = 1-1 .

6.Time for filling , (Filling pipe is bigger in size.)

F = (e * f)/(e - f)

7.Time for emptying , (emptying pipe is bigger in size.)

E = (f * e)/(f - e)

8.Pipes 'A' & 'B' can fill a tank in f1hrs & f2hrs respectively.Another pipe 'C' can empty thefull tank in 'e'hrs.If the three pipes are opened simultaneously then the tank is filled in

F = L/[(L/f1) + (L/f2) - (L/e)]

9.Two taps 'A' & 'B' can fill a tank in 't1' & 't2' hrs respectively.Another pipe 'C' can empty thefull tank in 'e'hrs.If the tank is full & all the three pipes are opened simultaneously . Then thetank will be emptied in,

E = L/[(L/e) - (L/f1) - (L/f2)]

10.A filling tap can fill a tank in 'f'hrs.But it takes 'e'hrs longer due to a leak at the bottom.Theleak will empty the full tank in ,

E = [t(f * e) * tf]/[t(f + e) - tf]

11.Capacity of the tank is ,

F = (f * e)/(e - f)