Pipes & Cistern Questions and Answers

Answer: C

Let the capacity be \( C \) litres. The leak alone empties it in 6 hours, so it drains \( \tfrac{C}{6} \) litres an hour.

The inlet adds \( 4 \times 60 = 240 \) litres an hour, and the net effect empties the tank in 8 hours:

\[ \frac{C}{6} - 240 = \frac{C}{8} \]

\[ C\left(\frac{1}{6} - \frac{1}{8}\right) = 240 \quad\Longrightarrow\quad \frac{C}{24} = 240 \]

\[ C = 5760 \text{ litres} \]

The inlet rate must be converted from litres per minute to litres per hour to match the other terms.

Answer: B

Let the capacity of the cistern be LCM(12, 15, 20, 30) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Efficiency of pipe D = 60 / 30 = 2 units / minute
=> Combined efficiency of pipe A, pipe B, pipe C and pipe D = 10 units / minute
 
Therefore, time required to fill the cistern if all the pipes are opened simultaneously = 60 / 10 = 6 minutes

Answer: A

There are two important points to learn in this type of question,
First, if both will open then tank will be empty first.

Second most important thing is,
If we are calculating filling of tank then we will subtract as (filling-empting)

If we are calculating empting of thank then we will subtract as (empting-filling)

So lets come on the question now,

Part to emptied 2/5

Part emptied in 1 minute =

1/6−1/10

=1/15

=>1/15:2/5 :: 1:x

=>2/5∗15

= 6mins

Answer: B

Work out how much each pipe contributes in 3 minutes.

\[ P = \frac{3}{30} = \frac{1}{10}, \qquad Q = \frac{3}{20}, \qquad R = \frac{3}{10} \]

Over the common denominator 20:

\[ \frac{2}{20} + \frac{3}{20} + \frac{6}{20} = \frac{11}{20} \text{ total} \]

\[ \text{proportion of R} = \frac{6/20}{11/20} = \frac{6}{11} \]

The tank is only \( \tfrac{11}{20} \) full at that point, so the proportions are taken against that total rather than against a full tank.

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: 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: 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

"45% longer" means the 30 m pipe is 1.45 times the other.

\[ 30 = 1.45x \quad\Longrightarrow\quad x = \frac{30}{1.45} = 20.69 \]

Taking 45% off 30 would give 16.5, a different quantity: the percentage is measured against the shorter pipe, not the longer one.

Answer: D

Let the total time be x mins.

Part filled in first half means in x/2 = 1/40

Part filled in second half means in x/2 =

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.