If a pipe A can fill a tank 3 times faster than pipe B and takes 32 minutes less than pipe B to fill the tank. If both the pipes are opened simultaneously, then find the time taken to fill the tank?
3x – x = 32
x = 16
1/16 + 1/48 = 4/48
Time taken to fill the tank = 48/4 = 12 minutes
Two pipes P and Q can fill a tank in 24 minutes and 27 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 8 minutes?
Required time = y(1-(t/x)) = 27(1-(8/24))= 18 minutes
A full tank gets emptied in 8 minutes due to the presence of a leak in it. On opening a tap which can fill the tank at the rate of 9 L/min, the tank get emptied in 12 min. Find the capacity of a tank?
a = 8; b = 9; C = 12
Capacity of a tank = a*b*c/c-a = 8*9*12/4 = 216 Litre.
If a pipe A can fill a tank 3 times faster than pipe B. If both the pipes can fill the tank in 42 minutes, then the slower pipe alone will be able to fill the tank in?
Time is taken by pipe A = x
Time is taken by pipe B = x/3
1/x + 3/x = 1/42
x = 168 minutes
A large cistern can be filled by two pipes P and Q in 15 minutes and 10 minutes respectively. How many minutes will it take to fill the Cistern from an empty state if Q is used for half the time and P and Q fill it together for the other half?
Part filled by P and Q = 1/15 + 1/10 = 1/6
Part filled by Q = 1/10
x/2(1/6 + 1/10) = 2/15 = 15/2 = 7.5 minutes
A pipe can fill a cistern in 8 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely?
In One hour pipe can fill = 1/8
Time is taken to fill half of the tank = 1/2 * 8 = 4 hours
Part filled by four pipes in one hour = (4*1/8) = 1/2
Required Remaining Part = 1/2
Total time = 4 + 1 = 5
Two pipes P and Q are opened together to fill a tank. Both the pipes fill the tank in time “x” If Q separately took 16 minutes more time than “x” to fill the tank and Q took 36 minutes more time than “x” to fill the tank, then find out the value of x?
Time is taken to fill the tank by both Pipes x = √a*b
x = √16*36 = 4 * 6 = 24
A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged. If there is a leakage also which is capable of draining out the water from the tank at half of the rate of the outlet pipe, then what is the time taken to fill the empty tank when both the pipes are opened?
Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(6) = 16.66%
Efficiency of leakage = half of the rate of the outlet pipe = 8.33%
Net Efficiency = 75 – (16.66 + 8.33) = 50%
Required time = 100/50 = 2 hours
A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 4 hours when the inlet pipe is plugged. If both pipes are opened simultaneously at a time when the tank was one-third filled, when will the tank fill thereafter?
Inlet pipe Efficiency = 100/(8/6) = 75%
Outlet pipe Efficiency = 100/(4) = 25%
Net Efficiency = 75 – 25 = 50%(1/3)filled
2/3 filled = 100%
Required time = 100/50 = 2 hours
Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank. After how many hours Q was closed?
Pipe P Efficiency = 100/10 = 10%
Pipe Q Efficiency = 100/20 = 5%
Net Efficiency = 15%
15x + 10(8-x) = 100
x = 4
Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?
Pipe A’s work in % = 100/10 = 10%
Pipe B’s work in % = 100/20 = 5%
Pipe C’s work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours, tap C is closed + After the 4th hour, tap B is also closed = 100
⇒ (10+5+2.5)*2 + (10+5)*2 + X = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100-65 = 35%
A pipe can fill a tank 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 min in the beginning and then third pipe is also opened. Time taken to empty the water tank is?
x/6 – (x+5)/12 – (x+5)/15 = 0
x = 45 mins
Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern. When the cistern is full, in what time will the leak empty it?
Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108).
Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min.
Due to leakage, time taken = 7 hours 12 min + 48 min = 8 hours
Work done by two pipes and leak in 1 hour = 1/8.
Work done by the leak in 1 hour =(15/108)-(1/8)=(1/72).
Leak will empty the full cistern in 72 hours.
A tank is normally filled in 6 hours but takes two hours longer to fill because of a leak in the bottom of the tank. If the tank is full the leak will empty it in how many hours?
Work done by leak in 1 hr=(1/6-1/8)=1/24
Leak will empty the tank in 24 hours
Twelve pipes are connected to a Cistern. Some of them are inlet pipes and the others are outlet pipes. Each of the inlet pipes can fill the tank in 8 hours and each of the outlet pipes can empty the cistern completely in 6 hours. If all the pipes are kept open, the empty tank gets filled in 24 hours. How many inlet pipes are there?
(x/8)-[(12-x)/6] = 1/24
x = 7
A dam has four inlets – A, B, C and D. The dam can be filled in 12 minutes through the first three inlets and it can be filled in 15 minutes through the second, the third and fourth inlet also it can be filled through the first and the fourth inlet in 20 minutes. How much time required to fill up the dam by all the four inlets?
(1/A + 1/B + 1/C) = 1/12 …(i)
(1/B + 1/C + 1/D) = 1/15 …(ii)
(1/A + 1/D) = 1/20 …(iii)
From eqn (i) and (ii)
(1/A – 1/D) = 1/60…(iv)
From eqn (iii) and (iv)
Let the time taken to full the tank = T
T(1/A + 1/B +1/C +1/D)= 1
T(1/30 + 1/15) = 1
T = 10 mins
Three pipes P, Q and R connected to a Cistern. The first pipe (i.e) P can fill 1/2 part of the tank in one hour, second pipe, Q can fill 1/3 part of the cistern in one hour. R is connected to empty the cistern. After opening all the three pipes 7/12 part of the cistern. Then how much time required to empty the cistern completely?
In 1 hour, P can fill = 1/2 Part
Time taken to fill the Cistern by Pipe P = 2 hours
In 1 hour, Q can fill = 1/3 Part
Time taken to fill the Cistern by Pipe P = 3 hours
[1/2 + 1/3 – 1/R] = 7/12
1/R = 1/4
Time required to empty the Cistern = 4 hours
A Cistern can be filled by an inlet pipe at the rate of 4 litres per minute. A leak in the bottom of a cistern can empty the full tank in 8 hours. When the cistern is full, the inlet is opened and due to the leak, the cistern is empty in 40 hours. How many litres does the cistern hold?
Part emptied by the leak in 1 hour = 1/8
part filled by (leak & inlet open) in 1 hour = 1/40
Part filled by the inlet pipe in 1 hour = 1/8 – 1/40 = 1/10
Inlet pipe fills the tank in = 10 hours
Inlet pipe fills water at the rate of 4 litres a minute.
Capacity of Cistern = 10 * 60 * 4 = 2400 litre
In a tank there is a pipe which can be used for filling the tank as well as for emptying it. The capacity of the tank is 1200 m³. The emptying of the tank is 10 m³ per minute higher than its filling capacity and the pump needs 6 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pipe?
1200/x – 1200/(x+10) = 6
200/x – 200/(x+10) = 6
x2 + 10x – 2000 = 0
x = 40
Two pipes P and Q can fill a cistern in 12 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours will the tank be full?
Pipe P can fill = 1/12
Pipe Q can fill = 1/4
For every two hour, 1/12 + 1/4 = 1/3 Part filled
Total = 6 hours