Pipes & Cistern Questions and Answers

Answer: B

Let the capacity of the tank be LCM(10, 30) = 30 units.
=> Efficiency of pipe A = 30 / 10 = 3 units / minute
=> Efficiency of pipe A = 30 / 30 = 1 units / minute
=> Combined efficiency of pipe A and pipe B = 4 units / minute
 
Therefore, time required to empty the tank if both pipes work = 30 / 4 = 7 minutes 30 seconds

Answer: A

Pipe A can fill a tank in x hours and pipe B can fill a tank in y hours and pipe C can empty it in z hours. If all the pipes are operate simultaneously, 1 hours of work of each of the three pipes = 1/x +1/y - 1/z
1 minutes work of each of the three pipes = 1/10 + 1/12 - 1/20 = 15/2 hours
= 7 ½ hour = 7 hours 30 minutes.
 

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

A can do a piece of work in x minutes
B can do a piece of work in y minutes
C can do a piece of work in z minutes
Minute’s work of each of the three is 1/x +1/y + 1/z
1 minutes work of each of the three pipes = 1/20 + 1/15 + 1/12
= 1/5 or = 5 minutes

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

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

 Let the capacity of the tank be LCM (10, 30) = 30 units
=> Efficiency of pipe A = 30 / 10 = 3 units / hour
=> Efficiency of pipe B = 30 / 30 = 1 units / hour
=> Combined efficiency of both pipes = 4 units / hour
Now, total time taken by A and B working together to fill the tank if there was no leak = 30 / 4 = 7.5 hours
=> Actual time taken = 7.5 + 2.5 = 10 hours
 
The tank filled by A and B in these 2.5 hours is the extra work done to compensate the wastage by the leak in 10 hours.
=> 2.5 hours work of A and B together = 10 hours work of the leak
=> 2.5 x 4 = 10 x E, where ‘E’ is the efficiency of the leak.
=> E = 1 unit / hour
 
Therefore, time taken by the leak alone to empty the full tank = 30 / 1 = 30 hours

Answer: D

Let the time taken if both were working together be ‘n’ hours.
=> Time taken by A = n + 9
=> Time taken by B = n + 6.25
 
In such kind of problems, we apply the formula :
n2 = a x b, where ‘a’ and ‘b’ are the extra time taken if both work individually than if both work together.
Therefore, n2 = 9 x 6.25
=> n = 3 x 2.5 = 7.5
 
Thus, working together, pipes A and B require 7.5 hours.

Answer: C

Let the capacity of the tank be LCM (10, 20, 30) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 30 = 2 units / hour
Now, all three work for some time, say ‘t’ hours.
So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.
=> Combined efficiency of pipe A, pipe B and pipe C = 11 units / hour
=> Combined efficiency of pipe B and pipe C = 5 units / hour
 
So, we have 11 x t + 5 x 2 + 14 x 2 = 60
=> 11 t + 10 + 28 = 60
=> 11 t = 60 - 38
=> 11 t = 22
=> t = 2
 
Therefore, A was closed after 2 hours.

Answer: C

Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 - 48 = 12 units
Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 - 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
 
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds