Two pipes A and B can fill a tank in 10 hours and 15 hours respectively while a third pipe C can empty the full tank in 20 hours. All the pipes are opened for 5 hours and then C is closed. Find the time in which the tank is full?
(1/10 + 1/15 – 1/20)*5 + (1/10 + 1/15)*T = 1. We will get T = 2.5 hrs
so total time = 5 + 2.5 = 7.5 hrs
Three pipe P, Q and R can fill a tank in 12 minutes, 18 minutes and 24 minutes respectively. The pipe R is closed 12 minutes before the tank is filled. In what time the tank is full?
Let T is the time taken by the pipes to fill the tank
(1/12 + 1/18 + 1/24)*(T – 12) + (1/12 + 1/18)*12 = 1
We will get T = 108/13 = 8.(4/13) hrs
On pipe P is 4 times faster than pipe Q and takes 45 minutes less than pipe Q. In what time the cistern is full if both the pipes are opened together?
Let P takes x minutes to fill the tank alone, then Q will take 4x minutes to fill the tank
4x – x = 45, x = 15
So P will take 15 minutes and Q will take 60 minutes to fill the tank. Both will fill the tank in
(60*15)/(75) = 12 minutes
Two pipes can fill a tank in 15 and 20 hours respectively. The pipes are opened simultaneously and it is found that due to the leakage in the bottom, 17/7 hours extra are taken extra to fill the tank. If the tank is full, in what approximate time would the leak empty it?
Total time taken by both pipes before the leak was developed = 60/7 hours now, leaks is developed which will take T time to empty the tank so, (1/15 +1/20 – 1/T) = 1/11
solve for T, we will get 660/17 hours = 39 hours (approx.)
Two pipes A and B can fill a tank in 8 minutes and 12 minutes respectively. If both the pipes are openedsimultaneously, after what time should B be closed so that the tank is full in 6 minutes?
Let after x minutes pipe B is closed
(1/8 + 1/12)*x + (1/8)*(6 x) = 1
X= 3 minutes
In what time would a cistern be filled by three pipes whose diameters are 1cm, 2 cm and 3 cm running together, when the largest pipe alone can fill the tank in 21 minutes? The amount of water flowing through the pipe is directly proportional to the square of its diameter.
More the diameter more will be the water flowing through it and less will be the time taken.
Means bigger pipe will take less time to fill the tank
So, for 1 cm time, (1^{2})/(3^{2}) = 21/t, we get t = 189
For 2 cm time, (2^{2})/(3^{2}) = 21/t. We get t = 189/4
So total time = 1/21 + 1/189 + 4/189 = 2/27
So total time = 13.5 minutes
Two pipes P and Q can fill a tank in 10 min and 12 min respectively and a waste pipe can carry off 12 litres of water per minute. If all the pipes are opened when the tank is full and it takes one hour to empty the tank. Find the capacity of the tank.
Let the waste pipe take ‘T’ time to empty the tank.
(1/10 + 1/12 – 1/T)*60 = 1
We will get T = 5 min
So capacity = 5*12 = 60ltr
One pipe fill 1/4 of the tank in 4 minutes and another pipe fills 1/5 of the tank in 4 minutes. Find the time taken by both pipe together to fill half the tank?
First pipe will take 16 minutes to fill the tank alone. Similarly second pipe will take 20 minutes to fill the tank alone. Let T is the time in which both the pipes will fill half the tank
(1/16 + 1/20)*T = 1/2, we get T = 40/9 minutes
Two pipes can separately fill the tank in 15hrs and 30hrs respectively. Both the pipe are opened and when the tank is 1/3 full a leak is developed due to which 1/3 water supplied by the pipe leaks out. What is the total time to fill the tank?
(1/15 + 1/30)*T1 = 1/3, T1 = 10/3 hr
Now after leak is developed, [(1/15 + 1/30) – (1/3)*(1/15 + 1/30)]*T2 = 2/3
T2 = 10 hr. So total time = 10 + 10/3 = 40/3 hr
Three pipes A, B and C is attached to a cistern. A can fill it in 20 minutes and B can fill it in 30 minutes. C is a waste pipe. After opening both the pipes A and B, Riya leaves the cistern to fill and returns when the cistern is supposed to be filled. But she found that waste pipe C had been left open, she closes it and now the cistern takes 5 minutes more to fill. In how much time the pipe C can empty the full cistern?
The tank supposed to be filled in (30*20)/50 = 12 minutes
so, (1/20 + 1/30)*12 – 12/C + (1/20 + 1/30)*5 = 1 (A and B work for 12 minutes and also C work for 12 minutes and then A and B takes 5 more minutes to fill the tank)
solve for C, we will get C = 144/5 = 28.8
A pipe can empty a tank in 60 minutes alone. Another pipe whose diameter is twice the diameter of first pipe is also opened. Now find the time in which both pipe will empty the tank together.
Time taken by pipe to empty the tank is inversely proportional to cross sectional area.
So, time taken by second pipe will be = 60/4 = 15 min (πr^{2} = 1/60 and for second pipe 4πr^{2} = 1/T so we get T = 15 min)
Time taken by both to empty the pipe = (60*15)/75 = 12
Two pipes P and Q can fill a tank in 10 min and 12 min respectively and a waste pipe can carry off 12 litres of water per minute. If all the pipes are opened when the tank is full and it takes one hour to empty the tank. Find the capacity of the tank.
Let the waste pipe take ‘T’ time to empty the tank.
(1/10 + 1/12 – 1/T)*60 = 1
We will get T = 5 min
So capacity = 5*12 = 60ltr
Two pipes P and Q can fill a tank in 36 and 24 minutes respectively. If both the pipes are opened simultaneously, after how much time pipe Q should be closed so that tank is full in 30 minutes.
Let after T time, Q is closed, (1/36 + 1/24)*T + (1/36)*(30 – T) = 1
Two pipes A and B can fill a tank in 20 and 30 minutes respectively. Both the pipes are opened together but after 5 minutes pipe B is closed. What is the total time required to fill the tank
(1/20 + 1/30)*5 + (1/20)*T = 1
total time = T + 5 min
Three pipes P, Q and R can fill a tank in 12, 15 and 20 minutes respectively. If pipe P is opened all the time and pipe Q and R are opened for one hour alternatively. The tank will be full in
(1/12 + 1/15) + (1/12 + 1/20) = 17/60 (in 2 hrs this much tank is filled)
so in 6 hrs 51/60 is filled. Remaining, 9/60 = (1/12 + 1/15)*t,
so T = 1hr so total = 6 + 1 = 7 hr
A cistern can be filled by a pipe in 6 hours. A leak is developed at the bottom due to which it takes 2 hours more to fill the cistern. Find the time taken by the leak to empty the cistern when the cistern is full.
1/6 – 1/T = 1/8, solve for T
A pipe can fill a tank in 20 minutes and another pipe can fill the tank in 40 minutes. There is a waste pipe which can empty the tank in 15 minutes. First two pipes are opened for 5 minutes and then the third pipe is also opened. In what time the cistern is emptied after the third pipe also opened
(1/20 + 1/40)*5 + (1/20 + 1/40 – 1/15)*T = 1
Two pipes can separately fill the tank in 15hrs and 30hrs respectively. Both the pipe are opened and when the tank is 1/3 full a leak is developed due to which 1/3 water supplied by the pipe leaks out. What is the total time to fill the tank?
(1/15 + 1/30)*T1 = 1/3, T1 = 10/3 hr
now after leak is developed, [(1/15 + 1/30) – (1/3)*(1/15 + 1/30)]*T2 = 2/3
T2 = 10 hr. So total time = 10 + 10/3 = 40/3 hr
Pipe P is 4 times as fast as Q in filling a tank. If P takes 20 minutes to fill a tank, then what is the time taken by both the pipe P and Q to fill the tank?
P takes 20 minutes and it is 4 times faster than Q, it means Q will take 80 minutes to fill the tank.
(1/20 + 1/80)*t = 1. We get t = 16
In what time a cistern is filled by three pipes of diameter 2cm, 4cm and 6cm respectively. If the time taken by largest pipe to fill the tank is 40 minutes. Amount of water flowing through the pipe is proportional to the diameter of the pipe
Larger the crosssection area less will be time taken by pipe to fill the tank. 36/16 = T/40, T = 90min (for 4 cm pipe)
similarly for 2 cm pipe time taken will be = 360min
Total time = (1/360 + 1/90 + 1/40) = 1/p, so we get P = 25.5/7 minutes
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