Pipes & Cistern Questions and Answers

Answer: D

We are given that A fills the pool at the rate of 4 liters every 3 minutes and B fills the pool at the rate of 3 liters every 4 minutes.
=> Pool filled by A in 12 minutes = 4 x 4 = 16 liters
=> Pool filled by B in 12 minutes = 3 x 3 = 9 liters
=> Pool filled in 12 minutes = 16 + 9 = 25 liters

or we can say to fill 25 liters water, they will take 12 minutes.
Therefore to fill 1000 liters they will take (12*1000)/25 minutes i.e. 8 hours.
=> Since they will put a bucket at the same instant every 12 minutes. (LCM of 3 and 4 minutes = 12 minutes).
=> After every 12 minutes they will take rest for 3 minutes.
=> Overall  (8*60)/12 i.e 40 times they take rest while working .
=> So every 12 minutes should be considered 15 (12+3) minutes for 40 times.
=> But after having taken 39 times . The pool of capacity 1000 liters will be filled.
=> Last 3 minutes must not be considered.

Therefore, to fill 1000 liters (25 x 40 liters), it would take (12*1000)/25 minutes i.e. 8 hours. + (39*3) minutes = 8 hrs. + 117 minutes
So ans is  9 hrs 57 minutes.

Answer: A

Lets suppose tank got filled by first pipe in X hours,

So, second pipe will fill the tank in X + 10 hours.

Answer: B

Pipe A runs for the whole 30 minutes; pipe B runs for some shorter time \( t \) before being turned off. Note \( 37\tfrac{1}{2} = \tfrac{75}{2} \), so A's rate is \( \tfrac{2}{75} \):

\[ 30 \times \frac{2}{75} + \frac{t}{45} = 1 \]

\[ \frac{60}{75} + \frac{t}{45} = 1 \quad\Longrightarrow\quad \frac{t}{45} = 1 - \frac{4}{5} = \frac{1}{5} \]

\[ t = \frac{45}{5} = 9 \text{ minutes} \]

The question asks after how many minutes B is turned off, which is this 9 - not the 21 minutes for which B is then idle.

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: D

Part filled by (A + B) in 1 minute =		1	+	1		=	1	.
		60		40			24

Suppose the tank is filled in x minutes.

Then,	x		1	+	1		= 1
	2		24		40

	x	x	1	= 1
	2		15

 x = 30 min.

Answer: D

As the inlet can fill the tank in 12 hours, in one hour it will fill 1/12th of the tank.
Similarly, the outlet pipe in one hour can empty 1/16th part of the tank.
If both are opened simultaneously, the part of the tank filled in one hour is:
1/12 - 1/16 = 1/48
Hence, the tank gets filled in 48 hours.

Answer: C

In 1 minute pipe A can fill 1/45th part of the tank and pipe B can fill 1/30th part of the tank. If they are opened simultaneously then in 1 minute they can fill (1/45 + 1/30) part of the tank = 1/18th part of the tank.
Hence, in 18 minutes the tank gets filled if pipes A & B are opened together.

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: D

Part filled in 4 minutes = 4		1	+	1		=	7	.
		15		20			15

Remaining part =		1 -	7		=	8	.
			15			15

	1	:	8	:: 1 : x
	20		15

x =		8	x 1 x 20		= 10	2	min = 10 min. 40 sec.
		15				3

 The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Answer: D

Let B be turned off after X minutes.

Part of water will be filled by A + B in x minutes + After tap B is closed rest filled by A in (30-x) minutes = 1

=> x/(time taken to fill the container by each separately) + 30 - x/(time taken to fill the container by A alone) = 1
=> x(2/75 + 1/45) + (30-x)2/75 = 1
=> 11x + 180 - 6x = 225

Therefore, x = 9 minutes