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: A

The tank will be filled in 18 hours.

Work out how much of the tank each pipe handles in one hour. The filling pipe adds 1/6 of the tank and the emptying pipe removes 1/9, so the net filling per hour is:

\[ \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18} \]

One-eighteenth of the tank fills each hour, so the whole tank takes 18 hours.

The trap is to add the rates (which gives 3.6 hours) or to subtract the times (which gives 3 hours); neither is offered. 12, 10 and 15 hours would each need a faster net rate than 1/18.

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).

\[ \frac{1}{10} + \frac{1}{15} - \frac{1}{w} = \frac{1}{18} \]

\[ \frac{3 + 2}{30} - \frac{1}{w} = \frac{1}{18} \quad\Longrightarrow\quad \frac{1}{6} - \frac{1}{w} = \frac{1}{18} \]

\[ \frac{1}{w} = \frac{3 - 1}{18} = \frac{1}{9} \quad\Longrightarrow\quad w = 9 \text{ minutes} \]

Answer: C

When we have question like one is filling the tank and other is empting it, then we subtraction as,

Filled in 1 hour = 1/4
Empties in 1 hour = 1/9

Net filled in 1 hour = 1/4 - 1/9
= 5/36

So cistern will be filled in 36/5 hours i.e. 7.2 hours

Answer: D

One tap fills the tank in 6 hours, so the first half takes 3 hours. Three more taps then join it, making four in all and quadrupling the rate:

\[ \text{time for the second half} = \frac{3 \text{ hours}}{4} = 45 \text{ minutes} \]

\[ \text{total} = 3\text{ h} + 45\text{ min} = 3\text{ h }45\text{ min} \]

The three extra taps are added to the original one, not substituted for it - reading it as three taps in total would give an hour for the second half.

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: D

Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30

Part filled by (A+B+C) in 1 hour =

1/5+1/10+1/30

=1/3

So all pipes will fill the tank in 3 hours.

Answer: C

Work done by the inlet in 1 hour =

1/6−1/8=1/24

Work done by inlet in 1 min=1/24∗1/60=1/1440

=>Volume of 1/1440 part

= 4 liters
Volume of whole = (1440 * 4) litres = 5760 litres.

Answer: D

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =

3 1/2=7/2

Work done with leak in 1 hr = 2/7

Work done by leak in 1 hr = 1/3−2/7 = 1/21

So tank will be empty by the leak in 21 hours.