Pipes and Cisterns

Pipes as Work with Signs

A tank (or cistern) is filled by an inlet and emptied by an outlet. A waste pipe, a drain and a leak are all outlets. This is time and work with one change: some of the "workers" undo the work. The full tank is the whole job, and the total-work (LCM units) method from the Time and Work basics section works here exactly as it is.

The only new rule is the sign.

• An inlet puts water in, so its rate is +.
• An outlet or a leak takes water out, so its rate is −.
• Add the rates of every pipe that is open, with their signs. That is the net rate, and time = (water to be moved) ÷ (net rate).

"Pipe A fills a tank in 34 hours and pipe B empties the full tank in 51 hours. Both are opened when the tank is empty. When will it be full?"

Take the tank as 102 units, the LCM of 34 and 51.

The tank gains 1 unit an hour, so it is full in 102 hours. In fractions: $\frac{1}{34} - \frac{1}{51} = \frac{3 - 2}{102} = \frac{1}{102}$ of the tank an hour.

For an inlet that fills in $x$ hours and an outlet that empties in $y$ hours ($y$ more than $x$), the filling time is

$$\frac{xy}{y - x}$$

Here $\frac{34 \times 51}{51 - 34} = \frac{1734}{17} = 102$.

Which pipe wins? If the outlet is the faster pipe (it empties in less time than the inlet fills), the net rate is negative and an empty tank never fills. A full tank slowly empties instead. An inlet that fills in 38 hours and an outlet that empties in 19 hours, on a full tank: 38 units, +1 − 2 = −1 unit an hour, so the tank is empty in 38 hours.

Change to one unit of time first. An inlet fills a tank in 50 minutes and an outlet empties it in 1 hour 15 minutes. Use 75 minutes for the outlet. The LCM of 50 and 75 is 150 units: +3 − 2 = +1 unit a minute, so the tank fills in 150 minutes = 2 hours 30 minutes.

[SSC Tip] The usual slip is to add the outlet as if it were a second inlet. For the 34-hour and 51-hour pipes that gives $\frac{34 \times 51}{85}$ = 20.4 hours, and SSC puts it in the options. A quick check: with an outlet working against it, the tank must take longer than the inlet alone, so here any answer under 34 hours is wrong.

Two or Three Pipes Together

Two inlets. Pipe A fills a tank in 18 hours and pipe B in 90 hours. Take the tank as 90 units: A puts in 5 and B puts in 1 unit an hour, 6 together, so the tank fills in 90 ÷ 6 = 15 hours. The two-inlet formula gives the same: $\frac{xy}{x + y} = \frac{18 \times 90}{108} = 15$.

Two inlets and an outlet. A and B fill a tank in 6 hours and 27 hours, and C empties it in 54 hours. All three are opened together.

Time = 54 ÷ 10 = 5.4 hours. Turn the decimal part into minutes: 0.4 hour = 0.4 × 60 = 24 minutes, so the answer is 5 hours 24 minutes. Reading 5.4 hours as "5 hours 4 minutes" or "5 hours 40 minutes" is a classic slip.

Three inlets. A, B and C fill a tank in 10, 14 and 35 hours. With 70 units: 7 + 5 + 2 = 14 units an hour, so together they take 5 hours.

What part did each pipe fill? Pipes that run together for the whole time share the tank in the ratio of their rates. In the three-inlet tank above the rates are 7 : 5 : 2, so A fills $\frac{1}{2}$ of the tank, B fills $\frac{5}{14}$ and C fills $\frac{1}{7}$.

Part filled in a given time. In the 6, 27 and 54-hour tank, after 3 hours the tank has 3 × 10 = 30 of its 54 units, so it is $\frac{5}{9}$ full.

[Tier 2] Pair data with an outlet. The A + B, B + C, C + A method from the Time and Work basics works here too, as long as you keep the signs. "Inlets A and B together fill a tank in 8 hours. A with outlet C open fills it in 16 hours, and B with C open fills it in 48 hours." Take 48 units: A + B = 6, A − C = 3 and B − C = 1. Adding the last two gives A + B − 2C = 4, so 2C = 6 − 4 = 2 and C = 1. So C empties the full tank in 48 hours, A = 3 + 1 = 4 units an hour (12 hours) and B = 1 + 1 = 2 units an hour (24 hours).

With only inlets open, the time together must be less than the fastest pipe's time; with an outlet added, it must be more than the inlets would take on their own. Those two checks clear most wrong options.

The Leak Found Late

A favourite SSC story: a tank should fill in a certain time but takes longer, and the reason turns out to be a leak. You are asked how long the leak alone would take to empty a full tank.

"A pipe fills a tank in 25 hours. Because of a leak, it takes 30 hours. In how much time can the leak empty the full tank?"

Take the tank as 150 units (LCM of 25 and 30).

• The pipe alone puts in 150 ÷ 25 = 6 units an hour.
• With the leak open, the tank gains only 150 ÷ 30 = 5 units an hour.
• So the leak takes out 6 − 5 = 1 unit an hour, and it empties the full tank in 150 hours.

It is the same formula as before, with $x$ as the usual time and $y$ as the time with the leak: $\frac{xy}{y - x} = \frac{25 \times 30}{5} = 150$.

"Half an hour more." A tap fills a cistern in 3 hours, but with a leak it takes $3\frac{1}{2} = \frac{7}{2}$ hours. The leak's rate is $\frac{1}{3} - \frac{2}{7} = \frac{7 - 6}{21} = \frac{1}{21}$, so the leak empties the cistern in 21 hours. Write the mixed time as a fraction before you turn it into a rate.

Two pipes and a leak. Pipes A and B fill a tank in 18 and 36 hours, which is 12 hours together. Because of a leak the tank takes 15 hours. Now compare 12 and 15 hours: with 60 units, the pipes alone give 5 units an hour and with the leak 4, so the leak takes out 1 unit an hour and alone empties the tank in 60 hours. Find the usual time of the pipes together first, then treat them like a single pipe.

The leak starts midway. A pipe fills a tank in 14 hours. When the tank is half full, a leak starts that would empty a full tank in 42 hours. The first half takes 7 hours. After that the net rate is $\frac{1}{14} - \frac{1}{42} = \frac{2}{42} = \frac{1}{21}$, so the second half takes 10.5 hours. Total: 17 hours 30 minutes, not 21 hours (21 hours is the answer only if the leak was there from the start).

Finding the inlet from the leak. "A leak empties a full tank in 56 hours. With an inlet also open, the empty tank fills in 8 hours. How long would the inlet take alone?" The inlet has to beat the leak, so its rate is $\frac{1}{8} + \frac{1}{56} = \frac{8}{56} = \frac{1}{7}$, and it takes 7 hours alone. Subtracting instead gives $9\frac{1}{3}$ hours, a pipe slower than the pipe and leak together, which cannot be right.

[SSC Tip] The leak's time is always more than the delayed filling time. For "25 hours became 30 hours", options like 5 hours (30 − 25) or 55 hours (25 + 30) are traps; the leak takes 150 hours, longer than both.

Pipes Opened or Closed at Different Times

When pipes are opened or closed at different moments, draw a short time line. For each pipe, find how long it was open, multiply by its units per hour, and add with signs. The total has to equal the tank.

One pipe closed after some hours. A and B fill a tank in 12 and 36 hours. Both are opened, and A is closed after 5 hours. Take 36 units: A gives 3 and B gives 1 an hour. In 5 hours together they put in 20 units. B fills the other 16 at 1 an hour, in 16 hours. So B takes 16 hours more, and the tank is full 21 hours after the start. Read which one the question wants: "how much more time" or "total time".

One pipe opened later. Pipe A fills a tank in 27 hours and pipe B in 18 hours. A is opened at 7 a.m. and B at 9 a.m. With 54 units, A gives 2 and B gives 3 an hour.

After 9 a.m. the other 50 units take 50 ÷ 5 = 10 hours, so the tank is full at 7 p.m.

Closed before the end. A and B fill a tank in 35 and 70 minutes (70 units: A gives 2 and B gives 1 a minute). Both are opened, but B is closed 5 minutes before the tank is full. If the tank fills in T minutes, A runs T minutes and B runs T − 5, so 2T + (T − 5) = 70, which gives 3T = 75 and T = 25 minutes. Another way to see it: had B stayed on for those last 5 minutes, it would have added 5 more units, so both pipes together would have done 75 units in T minutes at 3 a minute.

When should the outlet be closed? Inlet A fills a tank in 16 minutes and outlet B empties it in 40 minutes (80 units: A +5, B −2 a minute). Both are opened together. After how many minutes should B be closed so that the tank is full in 18 minutes? A runs all 18 minutes and puts in 90 units. The tank holds only 80, so B must have taken out 10 units, which takes 10 ÷ 2 = 5 minutes.

Note that "B was closed after 5 minutes" and "B was closed 13 minutes before the tank was full" describe the same thing here. Questions are worded both ways, and the options usually carry both numbers, so check which one is asked.

All Opened, Then Some Closed

All open, then one closed. A and B fill a tank in 12 and 28 hours, and C empties it in 21 hours. All three are opened together, and C is closed after 4 hours. Take 84 units: A +7, B +3 and C −4, so all three give 6 units an hour. In 4 hours the tank has 24 units. A and B fill the other 60 at 10 an hour, in 6 more hours, so the tank is full 10 hours after the start.

Finding a pipe from "together, then one closed". "A, B and C together fill a tank in 16 hours. After 4 hours C is closed, and A and B fill the rest in 18 hours. How long would C take alone?" Take 48 units, so all three give 3 an hour. In 4 hours they put in 12 units. A and B then fill 36 units in 18 hours, which is 2 units an hour. So C gives 3 − 2 = 1 unit an hour and alone takes 48 ÷ 1 = 48 hours.

Half full, then more pipes. "A tap fills a tank in 10 hours. When the tank is half full, three more taps of the same size are opened. How long does it take in all?" The first half takes 5 hours. Then 4 taps work together, so the second half takes 5 ÷ 4 = $1\frac{1}{4}$ hours. Total: 6 hours 15 minutes. The usual slip is to make the 4 taps fill a whole tank (10 ÷ 4 = 2.5 hours) instead of the half that is left.

Half the time each, or half the tank each. Pipe A fills a tank in 10 hours and pipe B in 90 hours, so together they take 9 hours ($\frac{1}{10} + \frac{1}{90} = \frac{10}{90} = \frac{1}{9}$).

• A for the first half of the time, B for the second half. If the tank takes T hours, $\frac{T}{2} \times \frac{1}{10} + \frac{T}{2} \times \frac{1}{90} = 1$, so $\frac{T}{2} \times \frac{1}{9} = 1$ and T = 18 hours. This is always twice the time the two pipes take together.
• A fills half the tank, then B fills the other half. A takes 5 hours for its half and B takes 45 hours for its half, so the total is 50 hours. This is always the average of the two times.

[SSC Tip] In "all opened, then some closed" questions, make a two-row table: the pipes open in the first stage and for how long, then the pipes open in the second stage. The units from the two rows add up to the tank. Most mistakes come from keeping an outlet in the second stage after it has been closed, or leaving it out of the first stage. And "half the time" and "half the tank" give different answers, so SSC likes to put both in the options.

Alternate Hours: Two Pipes in Turns

Here the pipes take turns: A runs for an hour, then B, then A again. Group the hours into rounds (one turn each), but finish the job hour by hour, because the tank can fill in the middle of a round.

"Pipes A and B fill a tank in 4 hours and 9 hours. They are opened alternately for 1 hour each, starting with A. When will the empty tank be full?"

Take 36 units: A gives 9 units in its hour and B gives 4. One round (2 hours) adds 13 units.

Two rounds (4 hours) give 26 units, and A's next hour makes it 35. The 1 unit left falls in B's turn, and B needs $\frac{1}{4}$ of an hour for it. So the tank is full after $5\frac{1}{4}$ hours = 5 hours 15 minutes.

Starting with B changes the last part. Two rounds still give 26 units in 4 hours, but now B's hour comes next and makes it 30. A then needs 6 of its 9 units, which is $\frac{2}{3}$ of an hour. The tank is full after 5 hours 40 minutes.

The quick way. Take as many whole rounds as you can without filling the tank, then go hour by hour with what is left, in the right order of turns.

Three pipes in turn. A, B and C fill a tank in 4, 8 and 16 hours and run one hour each in the order A, B, C. With 16 units they give 4, 2 and 1 in their turns, so a 3-hour round adds 7. Two rounds (6 hours) give 14 units, and A's next turn needs only 2 units, which is half of its hour: $6\frac{1}{2}$ hours.

A and B together, then A alone. Some questions pair the pipes differently, say A and B together in the first hour and A alone in the next. The method does not change: one round is those two hours, and the last part goes hour by hour.

Don't divide the tank by the units per round and multiply by the round length. That gives 36 ÷ 13 × 2 = $5\frac{7}{13}$ hours, the usual wrong option, because it treats both pipes as if they ran at an average speed when the last bit is done by one pipe at its own speed.

Inlet and Outlet in Turns: The Last-Hour Trap

When an inlet and an outlet take turns, each round gains only a little, so it is tempting to divide the tank by that small gain. That misses the fact that the tank can become full during an inlet hour, before the outlet gets its next turn.

"Pipe A fills a tank in 4 hours and pipe B empties it in 7 hours. Starting with A, they are opened alternately for 1 hour each. When will the empty tank be full?"

Take 28 units: A adds 7 in its hour and B takes out 4. One round (2 hours) gains just 3 units.

The trap: 28 ÷ 3 rounds × 2 hours = $18\frac{2}{3}$ hours.

The right way, in three steps.

1. Net gain per round: +3 units.
2. The level from which one A-hour fills the tank: 28 − 7 = 21 units.
3. Rounds needed to reach that level: 21 ÷ 3 = 7 rounds = 14 hours. Then A's next hour fills it.

So the tank is full after 15 hours. Here is how the last few hours go.

After hour 13 the tank has 25 units, close but not full, and the outlet takes it back down to 21 before A finishes the job in hour 15.

If the inlet's hourly units are more than what is left at the start of its last hour, take only the part of the hour that is needed.

Emptying a full tank in turns. The same trap works the other way. "A full tank has an outlet that empties it in 7 hours and an inlet that fills it in 14 hours. Starting with the outlet, they are opened alternately for 1 hour each. When will the tank be empty?" With 14 units, the outlet takes out 2 in its hour and the inlet puts back 1, so each round loses 1 unit. The outlet can finish the job in one hour once 2 units or fewer are left. After 12 rounds (24 hours) the tank has 2 units, and the outlet's next hour empties it: 25 hours, not 28.

Part-Full Tanks and Emptying

Write the starting level and the target level in units before anything else. Then time = (units to be moved) ÷ (net rate).

A tank already part full. A tank is $\frac{1}{4}$ full, and an inlet fills the whole tank in 16 hours. Only $\frac{3}{4}$ needs filling, so it takes $\frac{3}{4}$ × 16 = 12 hours.

Inlet and outlet on a part-full tank. First decide whether the tank will fill or empty: look at the sign of the net rate. "A tank is $\frac{2}{5}$ full. Pipe A fills it in 30 hours and pipe B empties it in 24 hours. Both are opened." With 120 units, A gives +4 and B −5, so the tank loses 1 unit an hour. It holds 48 units now, so it is empty in 48 hours. A student who assumes it fills would work on the empty $\frac{3}{5}$ and say 72 hours, but this tank never fills.

Emptying with two outlets. Outlets A and B empty a full tank in 7 and 28 hours. With 28 units: 4 + 1 = 5 units out an hour, so together they empty it in 28 ÷ 5 = 5.6 hours = 5 hours 36 minutes.

Outlet first, then an inlet. "A full tank is emptied by an outlet in 9 hours. After 1 hour, an inlet that fills the tank in 27 hours is also opened. When will the tank be empty?" With 27 units, the outlet takes out 3 an hour, so after 1 hour 24 units are left. Then the tank loses 3 − 1 = 2 units an hour, so it needs 12 more hours.

The tank is empty 13 hours after the start.

Inlet first, outlet opened later. An inlet fills a tank in 3 hours, and an outlet that empties it in 6 hours is opened when the tank is half full. The first half takes $1\frac{1}{2}$ hours. After that the tank fills at $\frac{1}{3} - \frac{1}{6} = \frac{1}{6}$ of the tank an hour, so the second half takes 3 hours: 4 hours 30 minutes in all.

Capacity in Litres

When rates are given in litres, the litres are the units. Keep one unit of time all the way through, usually minutes.

Straight from rates. Two inlets let in 25 and 18 litres a minute and an outlet drains 13 litres a minute. Net flow = 25 + 18 − 13 = 30 litres a minute, so a 2,400-litre tank fills in 2,400 ÷ 30 = 80 minutes.

A pipe's time plus a litre rate. "Pipe A fills a tank in 40 minutes. An outlet removes 9 litres a minute. With both open, the tank fills in 1 hour. Find the capacity." Call the capacity C litres. A puts in $\frac{C}{40}$ litres a minute, and with the outlet open the tank gains $\frac{C}{60}$ litres a minute:

$$\frac{C}{40} - 9 = \frac{C}{60} \Rightarrow \frac{3C - 2C}{120} = 9 \Rightarrow C = 1080$$

So the tank holds 1,080 litres. Check: A gives 27 litres a minute, the tank gains 18 a minute, and 1,080 ÷ 18 = 60 minutes.

A leak, then an inlet as well. "A leak empties a full tank in 2 hours. An inlet that lets in 12 litres a minute is opened, and now the full tank empties in 3 hours." In minutes, the leak takes out $\frac{C}{120}$ litres a minute, and with the inlet on the tank loses $\frac{C}{180}$ litres a minute. The difference is the inlet:

$$\frac{C}{120} - \frac{C}{180} = 12 \Rightarrow \frac{C}{360} = 12 \Rightarrow C = 4320$$

The tank holds 4,320 litres. Check: the leak takes out 36 litres a minute, the inlet puts back 12, and 4,320 ÷ 24 = 180 minutes = 3 hours.

From a rate difference. "Pipe A fills a tank in 21 minutes and pipe B in 35 minutes. A lets in 4 litres a minute more than B." Each minute A fills $\frac{1}{21}$ and B fills $\frac{1}{35}$ of the tank, a difference of $\frac{5 - 3}{105} = \frac{2}{105}$ of the tank, and that is 4 litres. So $\frac{1}{105}$ of the tank is 2 litres, and the tank holds 210 litres (A gives 10 and B gives 6 litres a minute).

From a time difference. "A pipe fills a tank in 4 hours. If it let in 50 litres a minute more, it would take only 3 hours." In minutes, $\frac{C}{180} - \frac{C}{240} = 50$, so $\frac{C}{720} = 50$ and C = 36,000 litres. Check: 150 litres a minute for 240 minutes and 200 litres a minute for 180 minutes both make 36,000.

Units to keep handy: 1 hour = 60 minutes, 1 cubic metre = 1,000 litres, 1 kilolitre = 1,000 litres.

[SSC Tip] Mixing hours with litres per minute gives answers 60 times too big or too small, and SSC puts exactly those numbers in the options. Change every time to minutes before you write the equation.

Relative Rates, Identical Pipes and Time Saved

"Twice as fast." If A fills twice as fast as B, A takes half of B's time, and their rates are 2 : 1. "Together they fill a tank in 10 hours." Give A 2 units and B 1 unit an hour. Together they put in 3 units an hour for 10 hours, so the tank is 30 units. A alone takes 15 hours and B alone 30 hours.

Efficiency in per cent. "A is 25% more efficient than B" means the rates are 125 : 100 = 5 : 4, so the times are 4 : 5. Rates and times always go opposite ways.

A chain of rates. "A is twice as fast as B, and B is three times as fast as C. All three together fill a tank in 6 hours." Rates A : B : C = 6 : 3 : 1, 10 units an hour in all, so the tank is 60 units. C alone takes 60 hours, B 20 hours and A 10 hours.

Identical pipes. Count pipe-minutes the way you count man-days in work. "12 identical pipes fill a tank in 35 minutes. If 5 of them are shut, how long will the rest take?" The tank needs 12 × 35 = 420 pipe-minutes, and 7 pipes take 420 ÷ 7 = 60 minutes.

Time saved or lost. Work out the time before and after, then subtract.

• Adding an inlet: A and B fill a tank in 24 and 120 hours. With 120 units they give 5 + 1 = 6 an hour, so 20 hours together. Add pipe C, which fills it in 30 hours (4 units an hour): 10 an hour, so 12 hours. 8 hours saved.
• Closing an outlet: inlets that together fill a tank in 3 hours, with an outlet open that empties it in 6 hours, take $\frac{3 \times 6}{6 - 3}$ = 6 hours. Closing the outlet brings it back to 3 hours: 3 hours saved.

[Tier 2] Times that differ by a fixed amount. "A takes 8 hours less than B, and together they fill a tank in $7\frac{1}{2}$ hours." Testing the options is fastest: A = 12 and B = 20 work, because $\frac{1}{12} + \frac{1}{20} = \frac{8}{60} = \frac{2}{15}$, which is the rate for $7\frac{1}{2}$ hours. (In full: with B = $x$, $x^2 - 23x + 60 = 0$, so $x$ = 20.)

[Tier 2] The square-root rule. If A alone takes $a$ hours more than A and B together, and B alone takes $b$ hours more, then the time together is $\sqrt{ab}$. With $a$ = 6 and $b$ = 54, together = $\sqrt{324}$ = 18 hours, A = 24 and B = 72. Check: $\frac{1}{24} + \frac{1}{72} = \frac{3 + 1}{72} = \frac{1}{18}$.

[Tier 2] Pipe size. At the same flow speed, the water a pipe carries goes up with the square of its diameter. If a 2 cm pipe fills a tank in 36 minutes, a 3 cm pipe fills it in 36 × $\frac{4}{9}$ = 16 minutes.

Solved Examples

Question 1

Pipe A can fill a tank in 24 hours and pipe B can empty the full tank in 56 hours. If both pipes are opened together when the tank is empty, in how many hours will the tank be full? (a) 16.8 hours (b) 32 hours (c) 42 hours (d) 80 hours

Answer. Take the tank as 168 units (LCM of 24 and 56). A puts in 7 units an hour and B takes out 3, so the tank gains 4 units an hour. It is full in 168 ÷ 4 = (c) 42 hours. If you add B's 3 units instead of taking them away, you get 168 ÷ 10 = 16.8 hours, which is the time for two inlets. Any answer below 24 hours cannot be right, because B is working against A.

Question 2

Two pipes A and B can fill a cistern in 20 minutes and 25 minutes respectively, and a third pipe C can empty the full cistern in 50 minutes. If all three pipes are opened together, in how much time will the empty cistern be filled? (a) $14\frac{2}{7}$ minutes (b) $11\frac{1}{9}$ minutes (c) $9\frac{1}{11}$ minutes (d) $33\frac{1}{3}$ minutes

Answer. Take the cistern as 100 units (LCM of 20, 25 and 50). A adds 5, B adds 4 and C takes out 2 units a minute, so the net is 5 + 4 − 2 = 7 units a minute. Time = $\frac{100}{7}$ = (a) $14\frac{2}{7}$ minutes.

Question 3

Pipes A, B and C can fill a tank in 14 hours, 35 hours and 70 hours respectively. All three are opened together when the tank is empty. What fraction of the tank will have been filled by pipe B when the tank is full? (a) $\frac{5}{8}$ (b) $\frac{1}{35}$ (c) $\frac{1}{3}$ (d) $\frac{1}{4}$

Answer. Take the tank as 70 units. A adds 5, B adds 2 and C adds 1 unit an hour, 8 units in all, so the tank is full in 70 ÷ 8 = 8.75 hours. In that time B puts in 2 × 8.75 = 17.5 units, which is $\frac{17.5}{70}$ of the tank. So the answer is (d) $\frac{1}{4}$. Quicker: pipes that run together the whole time share the tank in the ratio of their rates, 5 : 2 : 1, and B's share is $\frac{2}{8}$.

Question 4

A pipe can fill a tank in 7 hours. Because of a leak in the bottom, it takes 9 hours to fill the tank. If the tank is full, in how much time will the leak alone empty it? (a) 2 hours (b) 31.5 hours (c) 63 hours (d) 16 hours

Answer. Take the tank as 63 units (LCM of 7 and 9). The pipe alone puts in 9 units an hour, but with the leak the tank gains only 7 units an hour. So the leak takes out 9 − 7 = 2 units an hour and empties the full tank in 63 ÷ 2 = (b) 31.5 hours. The formula $\frac{xy}{y - x} = \frac{7 \times 9}{2}$ gives the same.

Question 5

Two taps can fill a cistern in 45 minutes and 90 minutes. Both taps are opened together, but because of a leak the cistern takes 10 minutes longer than usual to fill. In how much time can the leak alone empty the full cistern? (a) 2 hours (b) 5 hours (c) 1 hour 30 minutes (d) 1 hour 10 minutes

Answer. First find the usual time. Taking the cistern as 90 units, the taps put in 2 + 1 = 3 units a minute, so it usually fills in 30 minutes. With the leak it takes 40 minutes. Now compare 30 and 40 minutes: with a 120-unit cistern, the taps alone give 4 units a minute and the taps with the leak give 3, so the leak takes out 1 unit a minute. It empties the full cistern in 120 minutes = (a) 2 hours.

Question 6

A pipe can fill an empty tank in 24 hours. When the tank is one-fourth full, a leak develops which alone can empty the full tank in 96 hours. How long does it take in all to fill the tank? (a) 32 hours (b) 24 hours (c) 30 hours (d) 38 hours

Answer. Up to one-fourth full there is no leak, so that part takes $\frac{1}{4}$ × 24 = 6 hours. After that the net rate is $\frac{1}{24} - \frac{1}{96} = \frac{3}{96} = \frac{1}{32}$ of the tank an hour, so the remaining $\frac{3}{4}$ takes $\frac{3}{4}$ × 32 = 24 hours. Total = 6 + 24 = (c) 30 hours. 32 hours treats the leak as open from the start.

Question 7

Pipes A and B can fill a tank in 12 hours and 42 hours respectively. Both are opened together, and after 6 hours pipe A is closed. How much more time will pipe B take to fill the tank? (a) 21 hours (b) $4\frac{2}{7}$ hours (c) 12 hours (d) 15 hours

Answer. Take the tank as 84 units. A gives 7 and B gives 2 units an hour, so in 6 hours together they put in 54 units. That leaves 30 units for B alone at 2 an hour, which takes (d) 15 hours.

Question 8

Pipes A, B and C can fill a tank in 8 hours, 36 hours and 72 hours respectively. Pipe A is opened at 6 a.m., pipe B at 8 a.m. and pipe C at 10 a.m. At what time will the empty tank be full? (a) 12 noon (b) 12:40 p.m. (c) 1:40 p.m. (d) 12:20 p.m.

Answer. Take the tank as 72 units: A gives 9, B gives 2 and C gives 1 unit an hour.

After 10 a.m., 32 units are left at 12 units an hour, which takes $2\frac{2}{3}$ hours = 2 hours 40 minutes. So the tank is full at (b) 12:40 p.m.

Question 9

Pipes A and B can fill a tank in 40 minutes and 60 minutes respectively. Both are opened together, but pipe B is turned off 10 minutes before the tank is full. In how much time is the tank filled? (a) 28 minutes (b) 24 minutes (c) 34 minutes (d) 30 minutes

Answer. Take the tank as 120 units: A gives 3 and B gives 2 units a minute. Say the tank fills in T minutes. A runs for all T minutes and B for T − 10 minutes, so 3T + 2(T − 10) = 120, which gives 5T = 140 and T = (a) 28 minutes. Check: A gives 3 × 28 = 84 and B gives 2 × 18 = 36, and 84 + 36 = 120.

Question 10

Pipe A can fill a tank in 20 minutes and pipe B can empty the full tank in 35 minutes. Both are opened together when the tank is empty. After how many minutes should pipe B be closed so that the tank is full in exactly 24 minutes? (a) 4 minutes (b) 17 minutes (c) 7 minutes (d) 12 minutes

Answer. Take the tank as 140 units: A puts in 7 units a minute and B takes out 4. A runs for all 24 minutes and puts in 168 units. The tank holds only 140, so B must have taken out 168 − 140 = 28 units, which takes 28 ÷ 4 = 7 minutes. So B is closed after (c) 7 minutes. Check: in the first 7 minutes the tank gains 3 units a minute (21 units), and A alone adds 119 more in the next 17 minutes. Don't confuse this with the 17 minutes A runs alone after B is closed.

Question 11

Pipes A and B can fill a tank in 12 hours and 27 hours respectively, and pipe C can empty it in 36 hours. All three are opened together when the tank is empty. After 3 hours, pipe C is closed. In how much more time will the tank be full? (a) 9 hours (b) 6 hours (c) 7 hours 48 minutes (d) $8\frac{4}{13}$ hours

Answer. Take the tank as 108 units: A gives 9, B gives 4 and C takes out 3 units an hour. With all three open the tank gains 10 units an hour, so in 3 hours it has 30 units. A and B then fill the other 78 units at 13 an hour, in 78 ÷ 13 = (b) 6 hours.

Question 12

Three pipes A, B and C together can fill a tank in 14 hours. After all three have been open for 4 hours, pipe C is closed, and A and B fill the rest of the tank in 15 hours. In how many hours can pipe C alone fill the tank? (a) 21 hours (b) 35 hours (c) 63 hours (d) 42 hours

Answer. In 4 hours all three fill $\frac{4}{14} = \frac{2}{7}$ of the tank. A and B then fill the other $\frac{5}{7}$ in 15 hours, so together they would fill the whole tank in 21 hours. C's rate = $\frac{1}{14} - \frac{1}{21} = \frac{3 - 2}{42} = \frac{1}{42}$, so C alone takes (d) 42 hours. In units: with 42 units, all three give 3 an hour and A and B give 2, so C gives 1.

Question 13

A tap can fill a tank in 12 hours. After half the tank is filled, two more taps of the same size are opened. How long does it take in all to fill the tank? (a) 8 hours (b) 10 hours (c) 4 hours (d) 6 hours

Answer. The first half takes 6 hours. Then three taps work together, so the second half takes 6 ÷ 3 = 2 hours. Total = 6 + 2 = (a) 8 hours. 10 hours would mean the three taps fill a whole tank after the first half, not just the half that is left.

Question 14

Pipes A and B can fill a tank in 4 hours and 10 hours respectively. They are opened alternately for 1 hour each, starting with A. In how much time will the empty tank be filled? (a) $5\frac{5}{7}$ hours (b) 6 hours (c) $5\frac{1}{2}$ hours (d) $2\frac{6}{7}$ hours

Answer. Take the tank as 20 units: A gives 5 units in its hour and B gives 2, so each 2-hour round adds 7 units. 2 rounds (4 hours) give 14 units. Next is A's hour, which makes it 19 units after 5 hours. B then needs only 1 more unit, which takes $\frac{1}{2}$ of its hour. So the tank is full after (c) $5\frac{1}{2}$ hours.

Question 15

Pipe A can fill a tank in 6 hours and pipe B can empty the full tank in 11 hours. They are opened alternately for 1 hour each, starting with A. After how many hours will the empty tank be full? (a) $26\frac{2}{5}$ hours (b) 23 hours (c) 22 hours (d) 24 hours

Answer. Take the tank as 66 units: A adds 11 units in its hour and B takes out 6, so each 2-hour round gains only 5 units. Do not just divide 66 by 5. Once the tank has 55 units at the start of an A-hour, A's 11 units finish it. 11 rounds (22 hours) give 55 units, and A's next hour takes it to 66. So the tank is full after (b) 23 hours. Near the end it goes 50 units after hour 20, 61 after hour 21, 55 after hour 22 and 66 after hour 23.

Question 16

Pipes A and B can fill a tank in 5 hours and 10 hours, and pipe C can empty it in 20 hours. The three pipes are opened one at a time for 1 hour each in the order A, B, C, and this is repeated. In how much time will the empty tank be filled? (a) 12 hours (b) 11 hours (c) 10 hours (d) $10\frac{1}{2}$ hours

Answer. Take the tank as 20 units: A adds 4, B adds 2 and C takes out 1, so each 3-hour round gains 5 units. 3 rounds (9 hours) give 15 units. A's next hour makes it 19, and B needs only 1 more unit, which is half of its hour. So the tank is full after (d) $10\frac{1}{2}$ hours.

Question 17

A tank is one-third full. Pipe A can fill the tank in 42 minutes and pipe B can empty it in 30 minutes. If both pipes are opened together, what will happen? (a) It will be empty in 35 minutes (b) It will be full in 70 minutes (c) It will be empty in 105 minutes (d) It will be full in 35 minutes

Answer. B empties faster than A fills, so the water level goes down. Net rate = $\frac{1}{30} - \frac{1}{42} = \frac{7 - 5}{210} = \frac{1}{105}$ of the tank lost each minute. Only $\frac{1}{3}$ of the tank is there, so it empties in $\frac{1}{3}$ × 105 = 35 minutes. The answer is (a) It will be empty in 35 minutes. 105 minutes is the time to empty a full tank, but only a third of it is there.

Question 18

A full tank has two outlet pipes that can empty it in 4 hours and 15 hours, and an inlet pipe that can fill it in 20 hours. If all three pipes are opened together, in how much time will the tank be empty? (a) $3\frac{3}{19}$ hours (b) 3 hours 45 minutes (c) $2\frac{8}{11}$ hours (d) 5 hours

Answer. Take the tank as 60 units. The outlets take out 15 and 4 units an hour and the inlet puts in 3, so the tank loses 15 + 4 − 3 = 16 units an hour. The full tank is empty in 60 ÷ 16 = 3.75 hours = (b) 3 hours 45 minutes.

Question 19

Two inlet pipes fill a tank at 17 litres and 13 litres per minute, and an outlet pipe drains it at 8 litres per minute. With all three pipes open, the empty tank is filled in 1 hour 15 minutes. What is the capacity of the tank? (a) 2,850 litres (b) 2,250 litres (c) 1,650 litres (d) 1,320 litres

Answer. Net flow = 17 + 13 − 8 = 22 litres a minute. 1 hour 15 minutes = 75 minutes, so the tank holds 22 × 75 = (c) 1,650 litres.

Question 20

A leak in the bottom of a tank can empty the full tank in 7 hours. An inlet pipe that lets in 5 litres of water a minute is opened, and now the full tank is emptied in 12 hours. How many litres does the tank hold? (a) 1,500 litres (b) 2,520 litres (c) 10,080 litres (d) 5,040 litres

Answer. Work per minute: 7 hours = 420 minutes and 12 hours = 720 minutes. If the tank holds C litres, the leak takes out $\frac{C}{420}$ litres a minute, and with the inlet open the tank loses $\frac{C}{720}$ litres a minute. The difference is the inlet: $\frac{C}{420} - \frac{C}{720} = 5$. The LCM of 420 and 720 is 5,040, so $\frac{12C - 7C}{5040} = 5$, which gives C = (d) 5,040 litres. Check: the leak takes out 12 litres a minute, the inlet puts back 5, and 5,040 ÷ 7 = 720 minutes = 12 hours.

Question 21

Pipe A can fill a tank in 36 minutes and pipe B can fill it in 60 minutes. Pipe A lets in 4 litres more water per minute than pipe B. What is the capacity of the tank? (a) 96 litres (b) 360 litres (c) 180 litres (d) 720 litres

Answer. Each minute A fills $\frac{1}{36}$ of the tank and B fills $\frac{1}{60}$. The difference is $\frac{1}{36} - \frac{1}{60} = \frac{5 - 3}{180} = \frac{1}{90}$ of the tank, and that is 4 litres. So the tank holds 4 × 90 = (b) 360 litres. Check: A gives 360 ÷ 36 = 10 litres a minute and B gives 360 ÷ 60 = 6, which are 4 litres apart.

Question 22

Pipe A fills a tank three times as fast as pipe B. Together, the two pipes fill the tank in 18 minutes. How long would pipe B alone take to fill it? (a) 72 minutes (b) 24 minutes (c) 54 minutes (d) 36 minutes

Answer. The rates are 3 : 1, so give A 3 units a minute and B 1 unit. Together they put in 4 units a minute for 18 minutes, so the tank is 72 units. B alone takes 72 ÷ 1 = (a) 72 minutes, and A alone takes 72 ÷ 3 = 24 minutes. 54 minutes, 18 × 3, ignores that A was doing most of the work.

Question 23

Eight pipes of the same size together fill a tank in 45 minutes. If two of the pipes stop working, how long will the remaining pipes take to fill the tank? (a) $33\frac{3}{4}$ minutes (b) 54 minutes (c) 90 minutes (d) 60 minutes

Answer. The tank needs 8 × 45 = 360 pipe-minutes, just like man-days in work. With 6 pipes it takes 360 ÷ 6 = (d) 60 minutes.

Question 24

[Tier 2] Pipe A takes 21 hours less than pipe B to fill a tank. Working together, they fill it in 10 hours. How long does pipe A alone take? (a) 35 hours (b) 21 hours (c) 14 hours (d) 15 hours

Answer. If B takes $x$ hours, A takes $x - 21$, and $\frac{1}{x} + \frac{1}{x - 21} = \frac{1}{10}$. In an MCQ it is faster to test the options: if A takes 14 hours then B takes 35, and $\frac{1}{14} + \frac{1}{35} = \frac{5 + 2}{70} = \frac{1}{10}$, which works. So A takes (c) 14 hours. Solved fully, the equation is $x^2 - 41x + 210 = 0$, so $x$ = 35 or 6; $x$ = 6 is impossible because A would then take a negative time.