Pipe and Cistern(Concept of negative work)

Pipe and Cisterns:

Pipe and Cisterns are more similar to Time and work. But This topic has an additional concept of negative work. 

Negative work can be understood in the opposite direction of work. 

for example, some men are building while some are destroying the same wall. or some pipes are filling while some pipes are emptying the vessel. 

Negative work or opposite work is taken as  -ve in the equation to solve the problem. 

Pipes: Pipes are connected to a tank or cistern and are used to fill or empty the tank, they are of two types.

i. Inlet: A pipe connected with a tank or cistern that fills it is known as an inlet, which means the nature of the pipe is positive.

ii. Outlet: A pipe connected with a tank or cistern emptying it is known as an outlet, which means the nature of the pipe is negative. 

Concept of Negative work:

Suppose A and B are working to build a wall while C is working to break the wall. In such cases the wall is being built by A and B while it is broken by C. Here we consider the work as the building of the wall, We can say that C is doing negative work.

The net combined work per day here is :

A's work + B's work - C's work

Important points:

i.Pipe and Cisterns are more similar to Time and Work.

ii.If an inlet can completely fill the empty tank in x hours, the part of the tank filled in 1 hour is 1/x.

iii.If an outlet can empty the full tank in 1 hour, the part of the tank empty in 1 hour is 1/y.

iv.If both inlet and outlet are open, the net part of the tank filled in 1 hour = 1/x - 1/y.

v.If pipe A alone can fill the tank in x hours and pipe B can fill or empty the tank in y hours. If both pipes working simultaneously then tank to fill or empty by (XY/X±Y) hours,

for filling pipe we take +ve sign.

for empty or draining pipe we take - sign.

vi.Three pipes A, B, C can fill the tank in x,y, and z hours respectively. If all three pipes opened simultaneously the time taken to fill the cisterns is given by (XYZ)/(xy+yz+zx).

vii.Two pipes A and B can fill the tank in x,y respectively. There is also an outlet pipe C.if all three pipes opened simultaneously tank willful in z hours the time taken by C to empty the full tank is given by. xyz/(xz+yz-XY).

viii.One inlet pipe A is k times faster than the other inlet pipe B.

a. If B can fill a cistern in x hrs. then the time in which the cistern will be full if both the inlet pipes are opened together is x/(k+1) hrs.

b. If A can fill a cistern in y hrs. then the time in which the cistern will be full if both the inlet pipes are opened together is [k/(k+1)]y hrs.

xi.One pipe A is k times faster and takes x minute less time than the other inlet pipe B, then

a.The time taken to fill a cistern, if both the pipes are opened together is [kx/(k2-1)] min.

b.A will fill the cistern in (x/k-1)min.

c.B will finish the cistern in [kx/k-1]min.

Solved example based on Pipe and cisterns:

Ex.1 Two pipes A and B can fill a tank in 20 min and 30 min respectively. If both the pipes are opened simultaneously. How much time will be taken to till the tank?

Sol.Method 1

A = 20 min

B = 30 min

Part of the system filled by A and B in 1 min.  = 1/20 + 1/30 = 5/60 = 1/12

Both pipe A and B together fill the tank in 12 min.

Method 2.

LCM of 20 min and 30 min is 60

Tank filled in 1 min

A = 60/20 = 3

B = 60/30 = 2

A+B = 5

Time taken by A+B = 60/5 = 12 min

Method 3. By formula.

Both pipe A and B together fill the tank in

xy/(x+y) = 20*30 /(20+30) = 600/50 = 12 min. Ans.

Ex.2 If pipe 'A' can fill a tank in 8 hrs. and pipe B can empty a tank in 16 hr. When both pipes are opened simultaneously, How much time will be taken to fill the tank?

Sol. Method 1.

Here A = 8 hours, B = 16 hour

Part of cistern fill by A and B in 1 hour = 1/8 - 1/16 = (2-1)/16 = 1/16

Both pipes fill the tank in 16 hours.

Method 2.

LCM of 8 and 16 is 16

One hour work of

A = 16/8 = 2

B = 16/16 = 1

Time taken when A and B both are opened = A+B/1 = 16/1 = 16 hours.

Method 3.

By formula  xy/(x-y) 

Both pipe fill the tank = 16*8/(16-8) = 16 hours Ans.

Ex.3 Three taps A, B, C can fill an over a tank in 4,6 and 12 hours respectively. How long would these three taps take to fill the tank if all of them are opened together?

Sol. Method 1.

Here A = 4 hour, B = 6 hour and C = 12 hours

All together fill the tank in 1 hour = 1/4 + 1/6 +1/12  = (3+2+1)/12 = 1/2

All fill the tank in 2 hours

Method 2.

LCM  of 4,6 and 12 is 12

Tap fill tank in one hour is

A = 12/4 = 3

B = 12/6 = 2

C = 12/12 = 1

A+B+C = 6

Together they fill the tank in 12/6 = 2 hours

Method 3.

By formula:

xyz/(xy+yz+zx) = 4*6*12/(4*6+6*12+12*4) 

= 4*6*12/144 = 2 hours

Ex.4 Two taps A and B can fill an over a tank in 30 min and 60 min respectively. There is a third exhaust pipe C at the bottom of the tank. If all tapes are opened together then the tank will be full in 45 minutes, In what time can exhaust tap C empty the cistern when the tank is completely full.

Sol. Method 1.

Here A = 30 min, B= 60 min and A+B+C = 45 min.

C can empty the tank in one minute = 1/30 +1/60 - 1/45 = (6+3-4)/180

= 1/36 part

All fill the tank in 36 minutes

Method 2.

LCM of 30,60, and 45 is =180

Taps can fill or empty the tank in 1 minute.

A = 180/30 = 6

B = 180/60 = 3

A+B-C = 180/45 = 4

C = -5

C can emptied the tank = 180/5 min = 36 min

Method 3.

Formula = xyz/(xz+yz-xy)

30*60*45/(30*45+60*45-30*60) = 81000/(1350+2700-1800) = 36 min Ans.

Ex.5 One inlet pipe A is 3 times faster than the second inlet pipe B. If A can fill a cistern in 16 minutes, then find the time when the cistern will be full if both inlet pipes are opened together.

Sol. Here k = 3 and x = 16

Cistern will be full in = [x/(k+1)]y minutes

= [3/(3+1)]16 = 12 minutes.

Ex.6 One inlet pipe A is 3 times faster than the second inlet pipe B. If together can fill the tank in 36 min, then how many tie slower pipe can fill the tank.

Sol. Method 1

let time taken by faster pipe be x min then slower pipe take = 3x

1/x + 1/3x + 1/36

x = (36*4)/ 3 = 48 minutes 

Time taken by slower pipe to fill the tank = 3*48 = 144 min.

Method 2.

Both pipes fill the tank = x/(k+1)

Here both the pipes fill the tank = 36 mins

k=3, so slower pipe * fill the tank = 36* (3+1) = 144 min.

Previous year questions based on Pipes and Cisterns:

Ex.1 A tank can be filled by pipe A in 2 hours and pipe B in 6 hours at 10 AM. A was opened at what time will the tank be filled if pipe B is opened at 11 AM?

Sol.Part of tank filled by both pipe in 1 hours = 1/2 + 1/6 = 2/3

So time is taken to fill 2/3 part = 60 min.

Time take to fill 1/2 part = [(60*3)/2] * 1/2 = 45 min.

The tank will be filled at 11:45 AM 

Ex.2 If two pipes function simultaneously a tank is filled in 12 hours, one pipe fills the tank 10 hours faster than the others. How many hours does the faster pipe alone take to fill the tank?

Sol. If slower pipe fills the tank in x hours,

Then,   

1/x + 1/(x-10) = 1/12

LCM of x,x-10 is x(x-10)

(x-10+x)/x(x-10) = 1/12

x² -10x = 24x -120

x² - 34x + 120 = 0

x² -30x-4x+120=0

x(x-30)-4(x-30)=0

x=4,  x= 30

x=30 because x cannot be equal to 4 in this example.

Required time = 30-10 = 20 hours

Ex.3 Pipe A can fill an empty tank in 6 hours and pipe V in 8 hours if both the pipes are opened and after 2 hours pipe, A is closed, how much time 3 will take to fill the remaining tank?

Sol. Part of the tank is filled by pipe A and B in 2 hours. 

= 2(1/6 + 1/8) = 7/12

Remaining part = 1 - 7/12 = 5/12

This part is filled by pipe B, Time = 5/12 * 8 = 10/3.

Method 2.

LCM of 6 and 8 is 24 

One hour work of

A = 24/6 = 4

B= 24/8 = 3

A and B together can work in one hour = 7

they fill part of  tank in 2 hours = 7 * 2 = 14

Remaining part of tank is = 24-14 = 10

B will do it in = 10/3 hours Ans.

Ex.4 A tap drops at a rate of one drop per second. 60 drops make 100ml the number of liters water wasted in 300 days is?

Sol. 300 days = 300 * 24 hours = 300*24*60*60 seconds

Number of drops 600 drops = 100ml

Number of litres wasted is = [(300*24*60*60) /600] litres = 4620 litres

Ex.5 Having the same capacity 9 taps fill up a water tank in 20 minutes. How many taps of the same capacity are required to fill the same water tank in 15 minutes.

Sol. M1D1 = M2D2

9*20 = 15*x

x= 12

Here the relation of Man-Days can be applied.