Pipes & Cistern Questions and Answers

Answer: D

The leak would empty the full cistern in 40 hours.

Without the leak the cistern fills in 8 hours; with it, in 10. The leak's rate is the difference between the two filling rates:

\[ \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40} \]

The leak empties 1/40 of the cistern per hour, so 40 hours to empty it. 16, 20 and 25 hours do not satisfy this; for example, a 20-hour leak would give a net rate of 1/8 - 1/20 = 3/40 and a filling time of 13.3 hours.

Answer: C

C alone takes 14 hours.

Working together, A, B and C fill 1/6 of the tank an hour, so in 2 hours they fill 1/3, leaving 2/3 for A and B:

\[ \begin{aligned} A + B &= \frac{2/3}{7} = \frac{2}{21} \text{ per hour} \\ C &= \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} \\ &= \frac{3}{42} = \frac{1}{14} \text{ per hour} \end{aligned} \]

So C alone fills the tank in 14 hours. Check: A and B alone would take 21/2 = 10.5 hours, and 1/10.5 + 1/14 = 1/6, the combined rate given. 10, 12 and 16 do not satisfy this.

Answer: C

Capacity of the tank = (12*13.5) litres

= 162 litres

Capacity of each bucket = 9 litres.

So we can get answer by dividing total capacity of tank by total capacity of bucket.

Number of buckets needed = (162/9) = 18 buckets

Answer: C

Let the capacity of the tank be LCM (10, 20, 30) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 30 = 2 units / hour
Now, all three work for some time, say ‘t’ hours.
So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.
=> Combined efficiency of pipe A, pipe B and pipe C = 11 units / hour
=> Combined efficiency of pipe B and pipe C = 5 units / hour
 
So, we have 11 x t + 5 x 2 + 14 x 2 = 60
=> 11 t + 10 + 28 = 60
=> 11 t = 60 - 38
=> 11 t = 22
=> t = 2
 
Therefore, A was closed after 2 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).

Without the leak the pump fills the tank at \( \tfrac{1}{2} \) per hour; with it, the effective rate is \( \tfrac{3}{7} \) per hour.

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

\[ \frac{1}{L} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} \]

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

Answer: A

If the whole tank takes 6 hours, then in one hour the pipe fills

\[ \frac{1}{6} \text{ of the tank} \]

This reciprocal is the basic move in every pipes-and-cisterns problem: times cannot be added, but rates can.

Answer: A

It is given that Pipe A,B and C work simultaneously.

Pipe A and B are filling the tank and Pipe C is emptying it.

In 1 hour volume of tank filled by Pipe A: 1/5th

In 1 hour volume of tank filled by Pipe B: 1/6th

In 1 hour volume of tank emptied by Pipe C: 1/12th

Working together Pipe A B and C fill x volume of tank.

=>1/x = (1/5)+(1/6)-(1/12)

=>x=60/17 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).

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: 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).

For the first 3 minutes both taps run:

\[ 3\left(\frac{1}{12} + \frac{1}{15}\right) = 3 \times \frac{5 + 4}{60} = \frac{27}{60} = \frac{9}{20} \]

B alone must supply the remaining \( \tfrac{11}{20} \):

\[ t = \frac{11}{20} \times 15 = 8.25 \text{ min} = 8 \text{ min } 15 \text{ s} \]

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.