16. PIPES AND CISTERNS
IMPORTANT FACTS AND FORMULAE
1.
Inlet: A pipe connected with a tank or a cistern
or a reservoir, that fills it, is known as an inlet.
Outlet: A
pipe connected with a tank or a cistern or a reservoir, emptying it, is
known as an outlet.
2. (i) If a pipe can fill a tank in x
hours, then : part filled in 1 hour = 1/x
(ii) If a pipe
can empty a full tank in y hours, then : part emptied in 1 hour = 1/y
(iii) If a pipe can .fill a tank
in x hours and another pipe can empty the full tank in y hours
(where y> x), then on opening both the pipes, the net part
filled in 1 hour =
(1/x)-(1/y)
(iv) If a pipe can
fill a tank in x hours and another pipe can empty the full tank in y hours
(where x > y), then on opening both the pipes, the net part emptied
in 1 hour =
(1/y)-(1/x)
SOLVED EXAMPLES
Ex. 1:Two pipes A and B can fill a tank in
36 bours and 46 bours respectively. If both the pipes are opened
simultaneously, bow mucb time will be taken to fill the
tank?
Sol: Part filled by A in 1 hour = (1/36);
Part filled by B in 1 hour = (1/45);
Part filled by
(A + B) In 1 hour =(1/36)+(1/45)=(9/180)=(1/20)
Hence, both the pipes
together will fill the tank in 20 hours.
Ex.
2: Two pipes can fill a tank in 10hours 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?
Sol: Net
part filled In 1 hour =(1/10)+(1/12)-(1/20)=(8/60)=(2/15).
The tank will be full in 15/2 hrs = 7 hrs 30 min.
Ex. 3: If two pipes function
simultaneously, tbe reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than tbe otber. How many hours does it take the second
pipe to fill the reservoir?
Sol:let the reservoir be
filled by first pipe in x hours.
Then ,second pipe fill it in (x+10)hrs.
Therefore (1/x)+(1/x+10)=(1/12) ó(x+10+x)/(x(x+10))=(1/12).
ó x^2 –14x-120=0 ó (x-20)(x+6)=0
óx=20 [neglecting the negative value
of x]
so, the second pipe will take (20+10)hrs.
(i.e) 30 hours to fill the reservoir
Ex. 4: A
cistern has two taps which fill it in 12 minutes and 15minutes respectively.
There is also a waste pipe in the cistern. When all the 3 are opened , the
empty cistern is full in 20 minutes. How long will the waste pipe take to empty
the full cistern?
Sol: Workdone by the waste
pipe in 1min
=(1/20)-(1/12)+(1/15) = -1/10 [negative sign means
emptying]
therefore the waste pipe will empty the
full cistern in 10min
Ex. 5: An electric pump can fill a tank in 3 hours. Because of a leak in ,the tank it took 3(1/2)
hours to fill the tank. If the tank is full, how much time will the leak take
to empty it ?
Sol:
work done by the leak in 1 hour=(1/3)-(1/(7/2))=(1/3)-(2/7)=(1/21).
The leak will empty .the tank in 21 hours.
Ex. 6. Two pipes can fill a cistern in 14
hours and 16 hours respectively. The pipes
are opened simultaneously and it is
found that due to leakage in the bottom
it tooki 32 minutes more to fill the cistern.When the
cistern is full, in what time will the leak
empty it?
Sol: Work done by the two
pipes in 1 hour =(1/14)+(1/16)=(15/112).
Time taken
by these pipes to fill the tank = (112/15) hrs = 7 hrs 28 min.
Due to
leakage, time taken = 7 hrs 28 min + 32 min = 8 hrs
Work done by (two pipes + leak) in 1 hour
= (1/8).
Work done by the leak m 1 hour =(15/112)-(1/8)=(1/112).
Leak will empty the full cistern in 112
hours.
Ex. 7: Two pipes A and B can fill a tank in 36 min. and 45 min.
respectively. A water pipe C can empty
the tank in 30 min. First A and B are opened. after 7 min,C is also opened. In
how much time, the tank is full?
Sol:Part filled in 7 min.
= 7*((1/36)+(1/45))=(7/20).
Remaining
part=(1-(7/20))=(13/20).
Net part
filled in 1min. when A,B and C are opened=(1/36)+(1/45)-(1/30)=(1/60).
Now,(1/60)
part is filled in one minute.
(13/20) part
is filled in (60*(13/20))=39 minutes.
Ex.8: Two pipes A,B can fill a tank in 24 min. and 32 min. respectively.
If both the pipes are opened simultaneously, after how much time B should be
closed so that the tank is full in 18 min.?
Sol: let B be closed
after x min. then ,
Part filled
by (A+B) in x min. +part filled by A in
(18-x)min.=1
Therefore
x*((1/24)+(1/32))+(18-x)*(1/24)=1 ó (7x/96) + ((18-x)/24)=1.
ó 7x +4*(18-x)=96.
Hence, be
must be closed after 8 min.
