Wednesday, January 29, 2014
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.
