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).

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.

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 =

So all pipes will fill the tank in 3 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: B

Tank can be filled by pipe A in 12 minutes and pipe B in 24minutes.

=> Part filled by A in 1 minute= 1/12

=> Part filled by B in 1 minute= 1/24

=> Part filled by both in 1 minute= 1/12 + 1/24

= 3/24=1/8

Time taken to fill the entire tank = 8 minutes

Answer: B

If A can do a piece of work in x hours and B can do a piece of work in y hours.
Then A and B together will do the work in = 1/x+1/y

Time taken by A = 10 hours
Time taken by B = 15 hours
Time taken by A and B = (15×10 / 15+10) = 6 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).

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

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

Answer: A

Part filled by (A + B) in 1 min =		1	+	1		=	1	.
		20		30			12

 Both pipes can fill the tank in 12 minutes.

Answer: B

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =		4 x	1		=	2	.
			6			3

Remaining part =		1 -	1		=	1	.
			2			2

	2	:	1	:: 1 : x
	3		2

x =		1	x 1 x	3		=	3	hours i.e., 45 mins.
		2		2			4

So, total time taken = 3 hrs. 45 mins.