Pipes and Cisterns — Extra Practice

A pipe that fills in $n$ hours fills $\frac{1}{n}$ of the tank in one hour.

• Inlet: fills an empty tank. Outlet: empties a full one. Both work on $\frac{1}{\text{time}}$ — only the sign differs: filling $+$, emptying $-$.
• A pipe that fills $\frac{p}{q}$ in one hour fills the whole tank in $\frac{q}{p}$ hours. The reciprocal of the part is the time.
• For a fraction $\frac{p}{q}$ of the tank the time is $\frac{p}{q}\times$ (time for the whole).
• Effect of diameter: the flow is proportional to the area of the opening, and area $\propto d^2$. Double the diameter ⟹ four times the flow ⟹ a quarter of the time.

Two inlets: add the rates. One inlet, one outlet: subtract.

1. Write each pipe’s rate: $+\frac{1}{\text{time}}$ for an inlet, $-\frac{1}{\text{time}}$ for an outlet.
2. Add them — this is the net rate. If positive the tank fills; if negative it empties.
3. The reciprocal of the net rate (ignoring the sign) is the time.
4. Check: when the outlet is faster ($n<m$), the tank never fills — it empties.

Two inlets and an outlet, or three inlets.

Three pipes use the same method — add the three rates with a minus sign for the outlet. The fastest way is to take the LCM of all the times and write each pipe’s rate as “units per hour”.

The leak’s rate from the filling time with and without the leak.

A leak is just an outlet whose time is unknown. The tap’s rate $\frac{1}{m}$ minus the leak’s rate leaves the combined $\frac{1}{n}$, so the leak’s rate $=\frac{1}{m}-\frac{1}{n}$. Its reciprocal is the time the leak alone takes to empty a full tank.

When the total and two pipes are known find the third, plus $t=\sqrt{ab}$ and “$n$ times faster” questions.

• Third pipe unknown: its rate $=$ total rate $-$ the known rates. If it comes out negative the pipe is an outlet.
• Formula $t=\sqrt{ab}$: if two pipes fill together in $t$ and alone take $t+a$ and $t+b$, then $\frac{1}{t}=\frac{1}{t+a}+\frac{1}{t+b}$ gives $t^2=ab$.
• $n$ times faster and $d$ sooner: time ratio $1:n$; make the difference of the parts equal $d$ to get $k$.

Add net rate $\times$ time stage by stage until the sum reaches $1$.

1. Cut the time into stages (“both open”, “only one open”, “all three open”).
2. Filling in a stage $=$ that stage’s net rate $\times$ its duration.
3. When the total filling is $1$ the tank is full; solve the equation for the unknown.
4. If the total time is asked, add the durations of all stages.

Pipes open for one minute each in turns, or are opened at different hours.

1. In turns: find the net filling of one cycle (inlets $+$, outlets $-$).
2. Count how many whole cycles bring the tank close to full, then finish the rest with the next inlet — it may not need a full minute.
3. Opened at different hours: write each pipe’s running duration and add rate $\times$ duration.
4. If a clock time is asked, add the duration to the start time at the end.

When a pipe’s rate is given in litres, equate the difference in the tank’s rates with it.

• A pipe given in litres per minute has a direct rate; the others are in “part of the tank per minute”. To combine them take the capacity as $C$ and write every rate in litres.
• Total in terms of $C$: $\frac{C}{\text{time}}$ litres per minute.
• $p\%$ more capacity: rate ratio $100:(100+p)$; B alone $=T\times \frac{200+p}{100+p}$.
• Counting pipes: take $x$ inlets and $N-x$ outlets and equate the total rate.

The slips that cost marks.

Habits for speed.

• Take the LCM as the tank’s capacity (litres): each pipe’s litres per hour becomes a whole number.
• Two together: $\frac{mn}{m+n}$; one fills, one empties: $\frac{mn}{n-m}$ (if the inlet is the faster).
• Check: two inlets together take less than either alone; an outlet makes the filling take longer.
• Remember $t=\sqrt{ab}$ — “so many minutes more” questions finish in one line.

One-rule questions.

Exam-level questions.

Questions at the hard end of competitive exams.