Aptitude · Pipes and Cisterns · Model 2
Split the work by pipe, not by time
When one pipe is shut before the tank is full, the instinct is to hunt for how full it was at that moment. You never need to know. The pipe that stays open works the whole time, and what is left over belongs to the pipe that left.
Set the target time and find the closing moment →01 The idea
Two pipes, two totals, one subtraction
A cistern is fed by A, which alone takes 25 hours, and B, which alone takes 20. Both are opened together, but A is closed at some point and the tank is still full after exactly 12 hours. When was A closed?
The temptation is to think in stages: work out how full the tank was when A shut, then hand the rest to B. That works, but it needs a variable in the middle and it is where sign errors creep in. There is a cleaner reading. B was never closed, so B ran for the full 12 hours, and B’s contribution is one multiplication.
Take the tank as LCM(25, 20) = 100 units, so A is 4 units an hour and B is 5. B ran 12 hours, so B delivered 5 × 12 = 60 units. The tank is full at the end, so A must have delivered the other 40. At 4 units an hour that is 10 hours, and A was closed after 10 hours.
Notice what never appeared: the fraction of the tank that was full at hour 10, and any equation with an unknown in it. The work split by pipe rather than by time, and that is the move that makes this model quick. Every question in it — “close A after how long”, “find X”, “how many more minutes” — is this same subtraction read from a different end.
02 Worked example
A takes 25 hours, B takes 20, and it is full in 12
This is the model’s standard shape. Pipe A alone fills a tank in 25 hours and pipe B alone in 20 hours. Both are opened together. After how long must A be closed so that the tank is exactly full at the 12-hour mark?
A ran for 10 of the 12 hours and was shut for the last 2. If you want the stage picture, it is there: at hour 10 the tank held 4×10 + 5×10 = 90 units, and B cleared the last 10 units in the remaining 2 hours. But that picture was the consequence of the answer, not the route to it — which is the point of the model.
03 The method
One equation, and the window it has to sit in
The method is a single line. The care goes into which pipe is which, and into checking that the stated total time is even possible.
| Wording | What stays open | What you compute |
|---|---|---|
| “Close A so it fills in T” | B, for all of T | A’s share, then A’s time |
| “After X hours Q is closed” | P, for all of T | P’s share, then Q’s share |
| “Both for k, then one off” | the survivor | leftover / survivor’s rate |
| “A and B closed, C finishes” | C, for all of T | the joint phase length |
| T below both-open time | impossible | no closing time exists |
| T at or above the stayer’s own time | degenerate | the leaver was never needed |
05 Cheat sheet
Model 2 on one page
Checked against A 25 hours, B 20 hours, tank full in 12 — the sum from section 02.
| Step | Rule | On A 25, B 20, T 12 |
|---|---|---|
| Tank | LCM of the solo times | LCM(25,20) = 100 units |
| Rates | tank / solo time | A = 4/h, B = 5/h |
| Stayer’s share | rate × T | 5 × 12 = 60 units |
| Leaver’s share | tank − stayer’s share | 100 − 60 = 40 units |
| Closing time | share / leaver’s rate | 40 / 4 = 10 hours |
| Lower bound on T | tank/(rA+rB) | 100/9 ≈ 11.11 hours |
| Upper bound on T | stayer’s solo time | 20 hours |
06 Where & why
Where this shows up
This is the model that separates candidates who have understood rates from candidates who have memorised ab/(a+b), which is why it appears so often at the harder end of a section.
The archetype. One multiplication, one subtraction, one division — and no algebra at all if you spot the staying pipe.
Same shape, worded so the pipe that closes is the one you would not expect. The wording is the difficulty; the arithmetic is identical.
Both pipes for a stated stretch, then one shuts, and the answer wanted is the extra time only. Reading “more” and returning the total is a standard way to lose the mark.
The same subtraction with C as the stayer. Its rate is small, so the leftover it has to clear is what fixes the joint phase.
07 Interview questions
What gets asked
Ten, from the method through to the two ways the question can be stated with no valid answer.
How do you handle a pipe that is closed before the tank is full?
A fills in 25 hours, B in 20, and the tank is full in 12. When is A closed?
Why not work out how full the tank was when the pipe closed?
P fills in 12 hours and Q in 36. Both open, Q is closed after X hours, total time 10 hours. Find X.
Two taps fill in 20 and 25 minutes. Both run for 5 minutes, then the second is shut. How much longer?
Which pipe do you call the staying pipe?
A, B and C fill in 4, 8 and 10 hours. All three open, then A and B are closed and C finishes. Total 2 hours 30 minutes. How long were all three open?
Can the stated total time be anything at all?
What happens if the stated total equals the staying pipe’s own time?
Is this model actually harder than the plain two-pipe one?
08 Practice problems
Six on the phase change
For each one, write down which pipe stays open before you touch a number. Two of these ask for extra time rather than total time — read the last line twice.