Pipes & Cisterns
Problems related to pipes & cisterns are similar to those of time and work but with a slight difference of conceptual agents like inlets & outlets.
Basic Terms:
Inlet: A pipe which is used to fill up the tank,cistern or reservoir is known as ‘Inlet’. This type of nature indicates ‘plus’ or ‘positive type’ of work done.
Outlet: A pipe which is used to empty the tank, cistern or reservoir is known as ‘Outlet’. This type of nature indicates ‘minus’ or ‘negative type’ type of work done.
Important Formulae:
| 1) If a pipe requires ‘x’ hours to fill up the tank, then part filled in 1 hr = | 1 |
| x |
| 2) If a pipe requires ‘y’ hours to empty the full tank, then part emptied in 1 hr = | 1 |
| y |
3) Net Work Done = (Sum of work done by inlets) – (Sum of work done by outlets)
4) Suppose that one pipe takes ‘x’ hours to fill up the tank and the another pipe takes ‘y’ hours to empty the full tank, then, on opening both the pipes, there are 2 possible conditions:
| For condition I: x < y, Net part filled in 1 hr = | 1 | – | 1 |
| x | y |
| For condition II: x > y, Net part emptied in 1 hr = | 1 | – | 1 |
| y | x |
Quick Tricks & Tips:
1) If two pipes take ‘x’ & ‘y’ hrs respectively to fill the tank and the third pipe takes ‘z’ hrs to empty the tank and all of them are opened together, then
| The net part filled in 1hr = | 1 | + | 1 | – | 1 |
| x | y | z |
Hence,
| The time taken to fill the tank = | 1 |
| (1/x) + (1/y) + (1/z) |
2) Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs.
| If the tank is full, the time taken by the leak to empty the tank = | hrs | ||||
| 1 | |||||
|
3) Suppose that pipe ‘A’ fills the tank as fast as the other pipe ‘B’. If pipe ‘B’ (slower) & pipe ‘A’ (faster) take ‘x’ min & ‘x/n’ min respectively to fill up an empty tank together, then
| Part of the tank filled in 1 hr = | n + 1 |
| x |
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