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Pipe ans Cisterns

BASIC CONCEPTS 

There are only two concepts either part of ll or empty and time taken to ll or empty in an entire section of Pipes and Cisterns.

Note:
Time taken to ll the tank is considered as positive and noted with (+) sign.

Time taken to empty the tank is considered as negative noted with (-) sign.

Baic Concept 1:
If a pipe can ll or empty a tank in “x” hours, then the part of tank lled or emptied in 1 hour = 1/x.

Example: A pipe can ll the tank in 6 hours, then the part of tank lled in 1hour = 1/6th part.

Baic Concept 2:
If a pipe can ll or empty “1/x” part of a tank in 1hour, then it can ll or empty the whole tank in “x” hours.

Example: If a pipe can ll or empty 1/4 th part of a tank in 1hour, then it can ll the whole tank in 4 hour.

Baic Concept 3:
If a pipe lls a tank in “x” hours and another pipe lls the same tank in” y” hours.
Then total part lled by both pipes in 1 hour = 1/x +1/y

SOLVED EXAMPLES

Example:If a pipe can ll a tank in 2 hours and another pipe can ll the same tank in 6 hours, then what part of a tank will be lled by both the pipes in 1 hour, if they are opened at the same time?

Answer:
part lled by 1st pipe in 1 hour = 1/x = 1/2
part lled by 2nd pipe in 1 hour = 1/y = 1/6
total part lled by both pipes in 1 hour = (1/x +1/y)
= 1/2+1/6
=3+1/6
=4/6
=2/3 part.

Baic Concept 4:
If a pipe lls a tank in “x” hours and another pipe empties the same tank in” y” hours. Then total part lled by both pipes in 1 hour = 1/x – 1/y

Example:If a pipe can ll a tank in 5 hours and another pipe can empty the same tank in 10 hours, then what part of a tank will be lled by both the pipes in 1 hour, if they are opened at once?

Answer:
part lled by 1st pipe in 1 hour = 1/x = 1/5
part emptied by 2nd pipe in 1 hour = 1/y = 1/10
total part lled by both pipes in 1 hour = (1/x -1/y)
= 1/5-1/10
=2-1/10
=1/10th part.

Example:An outlet pipe can empty the cistern in 5, In what time will the pipe empty 2/5th part of the cistern?

Answer:
Time taken to empty full cistern = 5 hours
Time taken to empty 2/5th part of cistern = 2/5 x5 = 2 hours


MISCELLANEOUS CONCEPTS

Concept 1: If a pipe can ll or empty a tank in x hours and another pipe can ll or empty the same tank in y hours, then

(I) If both pipes either lls or empties the tank, then the time taken to ll or empty the tank when both pipes are opened is t = xy/x+y.

Example: Two pipes A and B can ll a tank in 18 hours and 12 hours respectively. If both the pipes are opened at once, how much time will be taken to ll the Tank?

Answer:
According to formula
Time taken by both pipes to ll the tank = xy/x+y
Here m = 18 and n=12
=18×12/18+12
=3×12/5
=36/5 hours.

(II) If the rst pipe lls the tank and the second pipe empties the tank, then the time taken to ll the tank when both pipes are opened is t= xy/x-y (if x>y)

Example:A pipe can ll a tank in 6 hours, while another pipe can empty it in 5 hours. If both the pipes are opened simultaneously, how much time will be taken
to ll the tank?

Answer: According to formula
Time taken by both pipes to ll the tank = t= xy/x-y (if x>y)
here m= 6, n=5;
=6×5/6-5
=30/1
=30 hours

(III) If the rst pipe lls the tank and the second pipe empties the tank, then the time taken to empty the tank when both pipes are opened is t= xy/y-x (if y>x)

Example:A pipe can ll a tank in 10 hours. Due to leak in the bottom, it lls the tank in 20 hours.If the tank is full, how much time will the leak take to empty the Tank?

Answer:
According to formula
Time taken by both pipes to ll the tank = xy/y-x (if y>x)
here m= 10, n=20;
=10×20/20-10
=200/10
=20 hours.

Concept 2: If three pipes can ll a tank separately in x,y and z hours, respectively,
then
(I) Part of tank lled in 1 hour by all the three pipes is given by (1/x+1/y+1/z)
(II)Total time taken to ll the tank is given byxyz/yz+xz+xy hours
Note:As mentioned earlier if any pipe is used to empty in the given three pipes, for that use negative (-) sign. and the formula was rewritten. For example assume that
in given data above, among three pipes, the 3rd pipe empty the tank then formula rewritten for

Example 1: Three pipes x,y,z can ll a tank separately in 4, 5 and 10 hours respectively. Find the time taken by all the three pipes to ll the tank when pipes are opened together?

Answer: According to formula
Total time taken to ll the tank is given byxyz/yz+xz+xy hours
Here values given as x = 4, y=5, z
Total time taken to ll the tank is given byxyz/yz+xz-xy hours
= 20x10x30/10×30 + 20×30 – 20×10
=6000/300+600-200
=6000/700
=60/7 hours.

Concept 3: Two pipes A and B together can ll a tank in time “.” If time taken by A alone is more than “t” by “a” and time taken by B alone is more than “t” by “b”, then t =√ab

Example: Two pipes A and B are opened together to ll a tank. Both the pipes ll the tank in time “t”. If A takes 4 minutes more time than “t” to ll the tank and B takes 64 minutes more time than “t” to ll the tank, nd the value of “t”?

Answer:
According to formula,t =√ab
here values are given as a=4, b=64
t= √4 x√64
=2 x8
=16 min.

Concept 4: A full tank gets emptied in “a” hours due to the presence of a leak in it. If a tap which lls it at a rate of “b” liters/hour, is opened, then it gets emptied in “C” hours. Then the volume of tank is abc/c-a

Example: A full tank get emptied in 6min due to the presence of an orice in it. On opening a tap which can ll the tank at the rate of 8L/min, the tank get emptied in 10 minutes. Find the volume of the tank?

Answer:
According to formula
volume of tank is = abc/c-a
= 6x8x10/10-6
=480/4
=120 litres.

Concept 5: If two taps A and B, which can ll a tank, such that eciency of A is “n” times of B and takes “t” minutes less or more than B to ll the tank, then
(I) Time taken to ll the tank by both pipes together = nt/n²-1 minutes.
(II) Time taken to ll the tank by faster tap = t/n-1 minutes.
(III) Time taken to ll the tank by slower tap = nt/n-1 minutes

Example: If tap A can ll a tank 3 times faster than tap B and takes 28 minutes less than tap B to ll the tank. If both the taps are opened simultaneously, then and the time taken to ll the tank?

Answer: According to the formula
Time taken to ll the tank by both pipes together = nt/n²-1 minutes.
here values are given n=3, t =28
=28 x 3/3²-1
=28 x3/8
=21/2 minutes.

Concept 6: Two pipes A and B can ll a tank in “x” minutes and “y” minutes respectively. If both the pipes are opened simultaneously, then the time after
which pipe B should be closed so that the tank is full in “t” minutes, is[y(1-t/x)] Minutes

Example: Two pipes A and B can ll a tank in 12 and 16 min, respectively. If both the pipes are opened simultaneously, after how much time should B be closed, so that the tank is full in 9 minutes?

Answer:
According to formula
Required time after which pipe B should be closed [y(1-t/x)] minutes,
Here values are given
x=12, y=16 and t =9
=16(1-9/12)
=16 x 3/12
= 4 minutes