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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
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Explanation :
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+25/30 =1
x/45 = 5/30 =1/6
x=45/6
x=7.5m
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
3x=42
X=14
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
P+Q= 1/8
Q= 1/8 – 1/12 = 3-2/24 = 1/24
Q= 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
5x/60+16-x/20 =1
5x+48-3x/60 =1
2x+48 = 60
2x=12
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
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