Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 6 hours. The number of hours taken by C alone to fill the cistern is
A+B+C in 1h = 1/6
A+B+C in 2h = 2/6 = 1/3
Remaining = 1-1/3 = 2/3
A+B in 6hrs = 2/3
A+B in 1hr = 2/18
C alone to fill the cistern = 1/6 – 2/18 = 3-2/18 = 1/18
Pipes A and B can fill a tank in 5 and 3 hrs respectively. Pipe C can empty empty it in 15 h. The tank is half full. All the three pipes are in operation simultaneously. After how much time the tank will be full ?
In 1 hr = 1/5+1/3 – 1/15 = 3+5-1/15 = 7/15
½ tank filled by 3 pipes = 15/7*1/2 = 15/14 =1(1/14)
Two pipes A and B can fill a tank in 10 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, Pipe A is turned off. What is the total time required to fill the tank ?
A + B in 4 minute = 4 (1 / 10 + 1 / 20) = 4(2+1/20) = 12/20 = 3/5
Part remaning = 1 – (3 / 5) = 2 / 5
1 / 20 part is filled by B in 1 minute
2 / 5 part will be filled in = (20)* (2 / 5) = 8 minutes
Total = 8+4 = 12m
Two pipes A and B can fill a tank in 6 hours and 5 hours respectively. If they are turned on alternatively for 1 hour each, find the time in which the tank is full. (Assume pipe A is opened first)
Total= 30, A = 30/6 =1/5, B = 30/5 =1/6
In 2 hrs = 5+6 =11
In 4hrs = 22
Remaining = 30-22 =8
1hr Pipe A = 8-5= 3,Remaining B = 3*1/6 = 30min
Total = 5hrs 30min
Pipes A, B and C can fill a tank in 3, 4 and 6 hours respectively. If all the pipes are opened together and after 30 minutes pipes B and C are turned off, find the total time in which the tank is full.
In 1 hr A, B, C = 1/3+1/4+1/6 = 8+6+4/24 = 18/24 = 6/8 = ¾
Filled in 30m = 3/8
Remaining = 1-3/8 =5/8
Pipe A = 3*5/8 = 15/8
Total = 15/8+1/2 = 15+4/8 = 19/8 = 2(3/8) hrs
Two pipes M and N can fill a tank in 30 and 45 minutes respectively. If both the pipes were open for few minutes after N was closed and the tank was full in 25 minutes, find the time for pipe N was open.
X(1/30+1/45) + 1/30(25-x) = 1
x/45 = 5/30 =1/6
A cistern is filled by 3 pipes A, B and C with uniform flow. The second pipe B takes3/2 times the time taken by A to fill the tank, while C takes twice the time taken by B to fill the tank. If all the three pipes can fill the tank in 7 hours, find the time required by pipe A alone to fill the tank.
1/x + 1/ (3/2x) + ½(3x/2) = 1/7
6/3x = 1/7
3x/6 = 7
Two pipes P and Q can fill a tank in 8 hours. If only pipe P is open then it would take 4 hours longer to fill the tank. Find how much longer would it take if only pipe Q is open.
P= 8+4 = 12
Q= 1/8 – 1/12 = 3-2/24 = 1/24
Q alone= 24-8 = 16
Two pipes P and Q can fill a tank in 20m and 30m respectively. If both the pipes are opened simultaneously, after how much time should Q be closed so that the tank is full in 16minutes ?
X(1/20+1/30) +(16-x)1/20 = 1
2x+48 = 60
X=12/2 = 6
A tap can fill a tank in 12 minutes and another tap can empty the tank in 6 minutes.If the tank is already full and then both the taps are opened the tank will be
1/12 – 1/6 = 1-2/12 = -1/12
Two taps can separately fill the tank in 18min and 12min respectively and when the waste pipe is open, they can together fill the tank in 9 minutes.The waste pipe can empty the tank in
1/18 + 1/12 = 3+2/36 = 5/36
5/36 – 1/9 = 5-4/36 = 1/36 ⇒ 36min
Two pipes can fill the tank in 4hrs 5hrs respectively while the third pipe can empty the tank in 20hrs, if all the pipes are opened together, then the tank will be filled in
1/4+1/5 -1/20 = 5+4-1/20 = 8/20
20/8 ⇒ 2(1/2)hrs
10 buckets of water fill a taken when the capacity of each bucked is 14 liter. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 7litres ?
10*14 = x*7
X = 10*14/7 = 20
A leak in the bottom of a tank can empty the full tank in 7 hours. An inlet pipe fills water at the rate of 2 litres a minute. When the tank is full the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank in litres is
In 1 hr = 1/7 – 1/8 = 8-7/56 = 1/56
In 1 min = 1/(56*60) = 1/3360
Inlet pipe fill water at the rate of 2 liters a minute = 2*3360 = 6720litres
Two pipes P and Q can fill a tank in 6 hours and 8 hours respectively. If they are opened on alternate hours and if pipe P is opened first, in how many hours, the tank shall be full ?
1/6+ 1/8 = 8+6/48 = 14/48 = 7/24…………in 2 hr
P ⇒ 6* 3/24 ⇒ 3/4 = 45min
Total = 6hrs 45min
Two pipes can fill a tank in 10 hours and 12 hours respectively while a third pipe empties the full tank in 20 hours.If all the three pipes operate simultaneously, in how much time will the tank be filled?
Net apart filled in 1 hour = (1/10+1/12-1/20)
= 8/60 = 2/15.
Tank will be full in = 15/2 hours ⇒ 7 hrs 30 min.
One pipe can fill a tank 4 times as fast as another pipe. If together the two pipes can fill the tank in 15 minutes, then the slower pipe alone will be able to fill the tank in:
1/x+4/x = 1/15
5/x = 1/15
X = 75 minutes
Bucket A has thrice the capacity as bucket B. It takes 20 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets A and B having each turn together to fill the empty drum
A = 3B
A = 60, B = 20
No of turns = xy/x+y
No of turns = 60*20/20+60 = 1200/80 = 15 turns
Two pipes A and B would fill a cistern in 20min and 30min respectively, both pipes are kept open for 10min and the first pipe be turned off after that the cistern may be filled in
10(1/20+1/30) = 10[3+2/60] = 50/60 = 5/6
Remaining part = 1-5/6 = 6-5/6 = 1/6
Second pipe = 30*1/6 = 5min
Tap A can fill the empty tank in 12hrs but due to leak in the bottom it is filled in 15hrs.If the tank is full ,then tap is closed.In how many hours the leak can empty the tank ?
1/12 – 1/15 = 5-4/60 = 1/60