Pipes and Cistern
- Three pipes 1, 2 and 3 together take 24 hours to fill a tank. The three pipes are opened for 6 hours after which Pipe 3 is closed. Pipe 1 and 2 take another 30 hours to fill the tank. In how much time would Pipe 3, alone fill the tank?
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According to question ,
Tank filled by all 3 in 6 hours = 
1 
6 = 1 24 4 Remaining = 1 - 1 = 3 is filled by, Pipe 1 and 2 in 30 hours 4 4 So, the entire tank would be filled by 1 and 2 in 40 hours 1 hour work of all 3 = 1 24 1 hour work of Pipe 1 & 2 = 1 40
Correct Option: D
According to question ,
Tank filled by all 3 in 6 hours = 1 6 = 1 24 4 Remaining = 1 - 1 = 3 is filled by, Pipe 1 and 2 in 30 hours 4 4 So, the entire tank would be filled by 1 and 2 in 40 hours 1 hour work of all 3 = 1 24 1 hour work of Pipe 1 & 2 = 1 40 Hence, 1 hour work of Pipe 3 = 1 - 1 = 2 = 1 24 40 120 60
∴ Pipe 3 alone would fill the tank in 60 hours.
- Two pipes x and y can fill a tank in 24 and 30 minutes respectively. Both the pipes are opened for 6 minutes, after which Pipe x is turned off. How much more time will it take to fill the tank?
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As per the given in question ,
When Pipe x is turned off (after 6 minutes),Work done by x and y in 6 minutes = 6 1 + 1 = 9 24 30 20 Remaining work = 1 - 9 = 11 20 20
which would be done by Pipe y alone.
1 work is done by Pipe y (alone) in 30 minutes .
Correct Option: A
As per the given in question ,
When Pipe x is turned off (after 6 minutes),Work done by x and y in 6 minutes = 6 1 + 1 = 9 24 30 20 Remaining work = 1 - 9 = 11 20 20
which would be done by Pipe y alone.
1 work is done by Pipe y (alone) in 30 minutes11 work is done by Pipe y (alone) in 30 × 11 = 33 minutes = 16.5 minutes. 20 20 2
- A tank is usually filled in 18 hours. But because of a leak in it’s bottom, it takes another 6 hours to fill. How much time will it take to the leak to empty the full tank?
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Let us consider the case when there is no leak –
Then in one hour, work done = 1 18 and in 6 hours = 6 = 1 18 3
This means 1/3 rd of the tank is emptied because of the leakage in 18 + 6 = 24 hours.
So, 1/3 rd is emptied in 24 hours, full tank would be emptied in 24 × 3 = 72 hours.
Second method to solve this question :
Using formula directly, we getTime required by leakage to empty the full tank = ab a - b
Here, a = 18, b = 24
Correct Option: A
Let us consider the case when there is no leak –
Then in one hour, work done = 1 18 and in 6 hours = 6 = 1 18 3
This means 1/3 rd of the tank is emptied because of the leakage in 18 + 6 = 24 hours.
So, 1/3 rd is emptied in 24 hours, full tank would be emptied in 24 × 3 = 72 hours.
Second method to solve this question :
Using formula directly, we getTime required by leakage to empty the full tank = ab a - b
Here, a = 18, b = 24Hence, t = 18 × 24 = 72 hours 24 - 18
- A pipe can fill a cistern in 12 minutes and another pipe can fill it in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then third pipe is also opened. In what time is the cistern emptied ?
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Part filled in 5 min. = 5 x (1/12 + 1/15) = 5 x 9/60 = 3/4
Part emptied in 1 min. (when all the pipes are opened) = 1/6 - (1/12 + 1/15) = (1/6 - 3/20) = 1/60Correct Option: D
Part filled in 5 min. = 5 x (1/12 + 1/15) = 5 x 9/60 = 3/4
Part emptied in 1 min. (when all the pipes are opened) = 1/6 - (1/12 + 1/15) = (1/6 - 3/20) = 1/60
Now, 1/60 part is emptied in 1 min. 3/4 part will be emptied in (60 x 3/4) = 45 min.
- Pipes 1 and 2 can fill a tank in 18 and 24 hours respectively. Both pipes work simultaneously for sometime after which Pipe 1 is
turned off. It takes 12 hours in all to fill the tank completely. Find the time for which Pipe 1 was turned on.
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Let the time for which Pipe 1 is turned on be ‘t’ hours, hence Pipe 1 has worked for ‘t’ hours and Pipe 2 has worked for 12 hours.
According to question ,∴ 1 (t) + 1 (12) = 1 18 24 ⇒ t + 1 = 1 or t = 1 ⇒ t = 9 18 2 18 2
∴ Pipe 1 was turned on for 9 hours.
Second method to solve this question :
For ‘t’ hours both pipes worked, and for (12 – t ) hours, only Pipe 2 worked, hence,t 1 + 1 + 1 (12 - t) = 1 18 24 24 7 t - t = 1 72 24 2
Correct Option: A
Let the time for which Pipe 1 is turned on be ‘t’ hours, hence Pipe 1 has worked for ‘t’ hours and Pipe 2 has worked for 12 hours.
According to question ,∴ 1 (t) + 1 (12) = 1 18 24 ⇒ t + 1 = 1 or t = 1 ⇒ t = 9 18 2 18 2
∴ Pipe 1 was turned on for 9 hours.
Second method to solve this question :
For ‘t’ hours both pipes worked, and for (12 – t ) hours, only Pipe 2 worked, hence,t 1 + 1 + 1 (12 - t) = 1 18 24 24 7 t - t = 1 72 24 2 ⇒ (14 - 6)t = 1 72 × 2 2 ⇒ t = 1 × 2 × 72 = 9 hours 2 8
