Pipes & Cistern Questions and Answers

Answer: D

Work in tanks per hour. The pump alone fills \( \tfrac{1}{2} \); with the leak draining at \( \tfrac{1}{L} \) the net rate fills the tank in \( 2\tfrac{1}{3} = \tfrac{7}{3} \) hours:

\[ \frac{1}{2} - \frac{1}{L} = \frac{3}{7} \]

\[ \frac{1}{L} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} \quad\Longrightarrow\quad L = 14 \text{ hours} \]

A sanity check: the leak costs only 20 minutes over the fill, so it must be far weaker than the pump - a leak emptying the tank in 4 or even 7 hours would be far too strong.

Answer: B

Put both rates over a common unit of minutes. The filling tap does \( \tfrac{1}{48} \) of the tank each minute and the emptying tap removes \( \tfrac{1}{120} \), since 2 hours is 120 minutes:

\[ \frac{1}{48} - \frac{1}{120} = \frac{5 - 2}{240} = \frac{3}{240} = \frac{1}{80} \]

So the tank fills in 80 minutes, that is 1 hour 20 minutes.

\[ 11{:}40\text{ AM} + 1\text{ h }20\text{ min} = 1{:}00\text{ PM} \]

Answer: C

The tap fills \( \tfrac{1}{8} \) of the tank per hour and the hole empties \( \tfrac{1}{12} \), so the net rate is the difference:

\[ \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24} \]

\[ \therefore\; \text{time} = 24 \text{ hours} \]

Adding the two times, or averaging them, has no meaning here: rates add, times do not.

Answer: A

Rates add when pipes work together:

\[ \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \]

\[ \therefore\; \text{time} = 12 \text{ minutes} \]

The combined time must be shorter than either pipe alone, so anything of 20 minutes or more can be discarded before any arithmetic is done.

Answer: C

Let the slower pipe take \( x \) minutes; the faster one is three times as quick and so takes \( x/3 \). Their rates add:

\[ \frac{1}{x} + \frac{3}{x} = \frac{4}{x} = \frac{1}{36} \quad\Longrightarrow\quad x = 144 \text{ minutes} \]

Seen another way, the two pipes contribute in the ratio 1 : 3, so the slower one does a quarter of the work and needs four times the joint time: \( 4 \times 36 = 144 \). The faster pipe alone would take 48 minutes.

Answer: C

Two pipes fill and one empties, so the outflow is subtracted:

\[ \frac{1}{20} + \frac{1}{30} - \frac{1}{40} \]

Over a common denominator of 120:

\[ \frac{6 + 4 - 3}{120} = \frac{7}{120} \]

\[ \therefore\; \text{time} = \frac{120}{7} = 17\tfrac{1}{7} \text{ hours} \]

The net rate is positive, so the tank does fill; had pipe C been faster than A and B combined the tank would never fill at all.

Answer: D

Deal with the two stages separately. In the first 3 minutes both pipes run:

\[ 3\left(\frac{1}{12} + \frac{1}{15}\right) = 3 \times \frac{5 + 4}{60} = 3 \times \frac{9}{60} = \frac{9}{20} \]

That leaves \( 1 - \tfrac{9}{20} = \tfrac{11}{20} \) for B alone at \( \tfrac{1}{15} \) per minute:

\[ t = \frac{11/20}{1/15} = \frac{11}{20} \times 15 = \frac{33}{4} = 8\tfrac{1}{4} \text{ minutes} \]

which is 8 minutes 15 seconds, since a quarter of a minute is 15 seconds.

Answer: B

Pipe A runs for the whole 30 minutes; pipe B runs for some shorter time \( t \) before being turned off. Note \( 37\tfrac{1}{2} = \tfrac{75}{2} \), so A's rate is \( \tfrac{2}{75} \):

\[ 30 \times \frac{2}{75} + \frac{t}{45} = 1 \]

\[ \frac{60}{75} + \frac{t}{45} = 1 \quad\Longrightarrow\quad \frac{t}{45} = 1 - \frac{4}{5} = \frac{1}{5} \]

\[ t = \frac{45}{5} = 9 \text{ minutes} \]

The question asks after how many minutes B is turned off, which is this 9 - not the 21 minutes for which B is then idle.

Answer: C

Run the hours in pairs, since A and B alternate with A first. Each pair of hours fills

\[ \frac{1}{6} + \frac{1}{4} = \frac{2 + 3}{12} = \frac{5}{12} \]

After two full pairs, that is 4 hours:

\[ 2 \times \frac{5}{12} = \frac{5}{6} \]

leaving \( \tfrac{1}{6} \). The fifth hour belongs to A, which fills exactly \( \tfrac{1}{6} \) in an hour, so the tank is full at the end of it:

\[ \text{total} = 5 \text{ hours} \]

The remainder happening to match A's hourly output is what makes the answer a whole number here.

Answer: D

Let the total time be \( t \). For the first half of it B works alone at \( \tfrac{1}{40} \); for the second half both work, at \( \tfrac{1}{60} + \tfrac{1}{40} \):

\[ \frac{t}{2}\cdot\frac{1}{40} + \frac{t}{2}\left(\frac{1}{60} + \frac{1}{40}\right) = 1 \]

\[ \frac{t}{80} + \frac{t}{2}\cdot\frac{5}{120} = 1 \quad\Longrightarrow\quad \frac{t}{80} + \frac{t}{48} = 1 \]

Over a denominator of 240:

\[ \frac{3t + 5t}{240} = 1 \quad\Longrightarrow\quad \frac{8t}{240} = 1 \quad\Longrightarrow\quad t = 30 \text{ minutes} \]

The phrase half the time means half of the unknown answer, which is why the equation has to be set up in \( t \) rather than solved forwards.