Pipes & Cistern Questions and Answers

Answer: C

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

A is five times as fast, so A's time is a fifth of B's.

\[ B - \frac{B}{5} = 32 \quad\Longrightarrow\quad \frac{4B}{5} = 32 \]

\[ B = 40, \qquad A = 8 \text{ minutes} \]

\[ \frac{1}{8} + \frac{1}{40} = \frac{5 + 1}{40} = \frac{3}{20} \]

\[ t = \frac{20}{3} \text{ minutes} \]

Answer: B

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

M runs for the whole 18 minutes:

\[ \frac{18}{22.5} = 0.8 \text{ of the tank} \]

N must supply the remaining 0.2 at a rate of \( \tfrac{1}{15} \) per minute:

\[ t = 0.2 \times 15 = 3 \text{ minutes} \]

So N is closed after 3 minutes. The question asks when N stops, not how long it is shut for.

Answer: A

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{X + 9} + \frac{1}{X + 16} = \frac{1}{X} \]

Clearing denominators:

\[ X(X + 16) + X(X + 9) = (X+9)(X+16) \]

\[ 2X^{2} + 25X = X^{2} + 25X + 144 \]

\[ X^{2} = 144 \quad\Longrightarrow\quad X = 12 \text{ minutes} \]

The result is the geometric mean of the two extra times: \( X = \sqrt{9 \times 16} = 12 \), a useful shortcut for this standard form.

Answer: D

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{38} + \frac{1}{19} - \frac{1}{133} \]

Over the common denominator 266:

\[ \frac{7 + 14 - 2}{266} = \frac{19}{266} = \frac{1}{14} \]

\[ t = 14 \text{ hours} \]

Answer: B

Take the pool as 120 units, the LCM of the three times.

\[ M = 3, \qquad N = 2, \qquad P = -4 \text{ units per minute} \]

A full three-minute cycle nets \( 3 + 2 - 4 = 1 \) unit.

After 116 cycles (348 minutes) the pool holds 116 units. Then:

\[ \text{minute } 349 \ (M): \ 116 + 3 = 119 \]

\[ \text{minute } 350 \ (N): \text{ needs 1 more unit at 2 per minute} \rightarrow 0.5 \text{ min} \]

\[ \text{total} = 349.5 \text{ minutes} \]

The pool fills part-way through a minute, so the answer is not a whole number of cycles - checking the level after each of the three phases is what catches that.

Answer: A

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{24} - \frac{1}{L} = \frac{1}{36} \]

\[ \frac{1}{L} = \frac{3 - 2}{72} = \frac{1}{72} \quad\Longrightarrow\quad L = 72 \text{ hours} \]

Answer: A

Flow varies as the square of the diameter, so the three pipes are in the ratio

\[ 2^{2} : 3^{2} : 4^{2} = 4 : 9 : 16 \]

The biggest carries 16 of the 29 parts and fills the pool in 126 minutes, so all three together work \( \tfrac{29}{16} \) times as fast:

\[ t = 126 \times \frac{16}{29} = \frac{2016}{29} \text{ minutes} \]

That is about 69.5 minutes - a little over half the 126, as expected when the flow nearly doubles.

Answer: D

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{22} + \frac{1}{33} = \frac{3 + 2}{66} = \frac{5}{66} \]

\[ t = \frac{66}{5} = 13.2 = 13\tfrac{1}{5} \text{ hours} \]

The joint time must be less than either pipe's own, which alone rules out the 18-hour option.

Answer: D

Let the time taken by the 1st pipe to fill the tank be x hours.
When both the pipes are open, it is given that:
1/x + 1/6 = 1/(4.2) of tank is filled in 1 hour.
1/x + 1/6 = 5/21
21(6 + x) = 30 x
9x = 126
x = 14 hours
Therefore, capacity of the tank = 100 * 14 = 1400 litres.

Answer: B

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6 + 3 + 1}{24} = \frac{10}{24} = \frac{5}{12} \]

\[ t = \frac{12}{5} = 2.4 \text{ hours} \]