Pipes & Cistern Questions and Answers

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).

For the first 30 minutes both taps run:

\[ \frac{1}{30} - \frac{1}{45} = \frac{3 - 2}{90} = \frac{1}{90} \text{ per minute} \]

\[ 30 \times \frac{1}{90} = \frac{1}{3} \text{ filled} \]

With Tap B closed, the remaining \( \tfrac{2}{3} \) is filled at \( \tfrac{1}{30} \) per minute:

\[ \frac{2}{3} \times 30 = 20 \text{ minutes} \]

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).

With all three open:

\[ \frac{1}{15} + \frac{1}{10} - \frac{1}{30} = \frac{2 + 3 - 1}{30} = \frac{2}{15} \text{ per hour} \]

\[ 2 \text{ hours} \rightarrow \frac{4}{15} \text{ filled} \]

After R is closed, P and Q fill at \( \tfrac{1}{15} + \tfrac{1}{10} = \tfrac{1}{6} \):

\[ \frac{11}{15} \div \frac{1}{6} = 4.4 \text{ hours} \]

\[ \text{total} = 2 + 4.4 = 6.4 \text{ h}, \qquad \text{without the mistake} = 6 \text{ h} \]

\[ \text{extra} = 0.4 \text{ h} = 24 \text{ minutes} \]

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).

\[ \text{A's rate} = \frac{1}{2.4} - \frac{1}{4} = \frac{5 - 3}{12} = \frac{1}{6} \]

So A alone would fill the cistern in 6 hours. At 100 litres an hour:

\[ \text{capacity} = 6 \times 100 = 600 \text{ litres} \]

Pipe B's own time is only needed to isolate A's rate; the capacity then follows from A's known flow.

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).

Without the leak the cistern fills in \( 8 - 1 = 7 \) hours.

\[ \frac{1}{7} - \frac{1}{L} = \frac{1}{8} \]

\[ \frac{1}{L} = \frac{1}{7} - \frac{1}{8} = \frac{8 - 7}{56} = \frac{1}{56} \]

\[ L = 56 \text{ hours} \]

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).

\[ \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3 + 2 - 1}{6} = \frac{2}{3} \text{ per hour} \]

\[ t = \frac{1}{2/3} = 1.5 \text{ hours} \]

\[ 9{:}00 + 1{:}30 = \textbf{10{:}30 a.m.} \]

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).

P and Q together fill in 6 hours, and P works at twice Q's rate:

\[ 3q = \frac{1}{6} \quad\Longrightarrow\quad q = \frac{1}{18}, \qquad p = \frac{1}{9} \]

Adding R brings the time to 5 hours:

\[ r = \frac{1}{5} - \frac{1}{6} = \frac{1}{30} \]

\[ q + r = \frac{1}{18} + \frac{1}{30} = \frac{5 + 3}{90} = \frac{4}{45} \]

\[ t = \frac{45}{4} = 11.25 \text{ hours} \]

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).

With the leak the cistern takes \( 10 + 5 = 15 \) hours.

\[ \frac{1}{10} - \frac{1}{L} = \frac{1}{15} \]

\[ \frac{1}{L} = \frac{3 - 2}{30} = \frac{1}{30} \quad\Longrightarrow\quad L = 30 \text{ hours} \]

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).

Per hour: \( P = \tfrac{2}{3} \), \( Q = \tfrac{1}{2} \), \( R = -2 \).

By 10 a.m. P has run 2 hours and Q one hour:

\[ 2 \times \frac{2}{3} + 1 \times \frac{1}{2} = \frac{4}{3} + \frac{1}{2} = \frac{11}{6} \]

From 10 a.m. the net rate turns negative:

\[ \frac{2}{3} + \frac{1}{2} - 2 = -\frac{5}{6} \text{ per hour} \]

\[ t = \frac{11/6}{5/6} = 2.2 \text{ hours} = 2 \text{ h } 12 \text{ min} \]

\[ 10{:}00 + 2{:}12 = \textbf{12{:}12 p.m.} \]

The calculation treats the inflow as accumulating without an overflow, which is the standard convention for these problems.

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).

\[ \frac{1}{6} - \frac{1}{12} = \frac{2 - 1}{12} = \frac{1}{12} \]

So one twelfth of the tank is filled in an hour, and the tank fills in 12 hours overall.

Answer: A

Flow through a pipe is proportional to its cross-sectional area, which varies as the square of the diameter.

\[ \frac{(2d)^{2}}{d^{2}} = 4 \]

Four times the flow means a quarter of the time:

\[ \frac{84}{4} = 21 \text{ minutes} \]

Halving the time to 42 minutes would assume flow proportional to diameter rather than to area.