Pipes & Cistern Questions and Answers

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.

Answer: C

Run the hours in pairs, since A and B alternate with A first. Each pair of hours fills

\[ \frac{1}{6} + \frac{1}{4} = \frac{2 + 3}{12} = \frac{5}{12} \]

After two full pairs, that is 4 hours:

\[ 2 \times \frac{5}{12} = \frac{5}{6} \]

leaving \( \tfrac{1}{6} \). The fifth hour belongs to A, which fills exactly \( \tfrac{1}{6} \) in an hour, so the tank is full at the end of it:

\[ \text{total} = 5 \text{ hours} \]

The remainder happening to match A's hourly output is what makes the answer a whole number here.

Answer: B

Let the capacity of the tank be LCM (10, 15, 20) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 15 = 4 units / hour
=> Efficiency of pipe C = 60 / 20 = 3 units / hour
=> Combined efficiency of all three pipes = 13 units / hour
 
Till 9 AM, A works for 2 hours and B work for 1 hour.
=> Tank filled in 2 hours by A = 12 units
=> Tank filled in 1 hour by B = 4 units
=> Tank filled till 9 AM = 16 units
=> Tank still empty = 60 - 16 = 44 units
 
Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.
=> Tank filled in 3 more hours = 39 units
=> Tank filled till 12 PM = 16 + 39 units = 55 units
=> Tank empty = 60 - 55 = 5 units
 
Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units / hour.
=> Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours
 
Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM

Answer: B

Let the capacity of the pool be LCM (10, 20, 15) = 60 units.
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 15 = 4 units / hour
 
=> Efficiency of pipe A and pipe C working together = 10 units / hour
=> Efficiency of pipe B and pipe C working together = 7 units / hour
 
=> Pool filled in first hour = 10 units
=> Pool filled in second hour = 7 units
=> Pool filled in 2 hours = 10 + 7 = 17 units
 
We will have 3 cycles of 2 hours each such that A and C, and, B and C work alternatively.
=> Pool filled in 6 hours = 17 x 3 = 51 units
=> Pool empty = 60 - 51 = 9 units
Now, these 9 units would be filled by A and C working together with the efficiency of 10 units / hour.
=> Time required to fill these 9 units = 9/10 hour = 0.9 hours = 54 minutes
 
Therefore, total time required to fill the pool = 6 hours 54 minutes