**Important Facts:**

1.INLET:A pipe connected with a tank or cistern or a reservoir,

that fills it, it is known as Inlet.

OUTLET:A pipe connected with a tank or a cistern or a reservoir,

emptying it, is known as 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 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 filled in 1 hour=(1/y -1/x).

v) If two pipes A and B can fill a tank in x hours and y hours

respectively. If both the pipes are opened simultaneously, part

filled by A+B in 1 hour= 1/x +1/y.

Simple Problems

1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.

If both the pipes are opened simultaneously, how much 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)’s in 1 hour=1/36 +1/45= 9/180 =1/20

Hence, both the pipes together will fill the tank in 20 hours.

2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While

3rd pipe empties the full tank n 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 be filled in 15/2hours= 7 hrs 30 min

3)A cistern can be filled by a tap in 4 hours while it can be emptied

by another tap in 9 hours. If both the taps are opened simultaneously,

then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36

Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

4)If two pipes function simultaneously, the reservoir will be filled in

12 days.One pipe fills the reservoir 10 hours faster than the other.

How many hours does it take the second pipe to fill the reservoir.

Sol: Let the reservoir be filled by the 1st pipe in x hours.

The second pipe will fill it in (x+10) hours

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

=> (2x+10)/((x)*(x+10))= 1/12

=> x=20

So, the second pipe will take 30 hours to fill the reservoir.

5)A cistern has two taps which fill it in 12 min and 15 min respectively.

There is also a waste pipe in the cistern. When all the three are opened,

the empty cistern is full in 20 min. How long will the waste pipe take to

empty the full cistern?

Sol: Work done by a waste pipe in 1 min

=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)

6)A tap can fill a tank in 6 hours. After half the tank is filled, three

more similar taps are opened. What is the total time taken to fill the

tank completely?

Sol: Time taken by one tap to fill the half of the tank =3 hours

Part filled by the four taps in 1 hour=4/6=2/3

Remaining part=1 -1/2=1/2

Therefore, 2/3:1/2::1:x

or x=(1/2)*1*(3/2)=3/4 hours.

i.e 45 min

So, total time taken= 3hrs 45min.

7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B

can empty it in 6 min. If both pipes are open, how long will it take to

empty or fill the tank completely ?

Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.

Part to be emptied=2/5.

Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15

Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.

So, the tank be emptied in 6 min.

8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for

Bucket P to fill the empty drum. How many turns it will take for both the

buckets P&Q, having each turn together to fill the empty drum?

Sol: Let the capacity of P be x lit.

Then capacity of Q=x/3 lit

Capacity of the drum=60x lit

Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

Complex Problems

1)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 took 32min 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 two pipes to fill the tank=112/15 hrs.

Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours

Therefore, work done by (two pipes + leak) in 1 hr= 1/8

work done by leak n 1 hour=15/112 -1/8=1/112

Leak will empty full cistern n 112 hours.

2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After

7 min, C also opened. In how much time, the tank s full.

Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20

Remaining part= 1-7/20=13/20

Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60

Now, 1/60 part is filled in 1 min.

13/20 part is filled n (60*13/20)=39 min

Total time taken to fill the tank=39+7=46 min

3)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

x(1/24+1/32) +(18-x)1/24 =1

=> x=8

Hence B must be closed after 8 min.

4)Two pipes A& B together can fill a cistern in 4 hours. Had they been

opened separately, then B would have taken 6 hours more than A to fill

the cistern. How much time will be taken by A to fill the cistern

separately?

Sol: Let the cistern be filled by pipe A alone in x hours.

Pipe B will fill it in x+6 hours

1/x + 1/x+6=1/4

Solving this we get x=6.

Hence, A takes 6 hours to fill the cistern separately.

5)A tank is filled by 3 pipes with uniform flow. The first two pipes

operating simultaneously fill the tan in the same time during which

the tank is filled by the third pipe alone. The 2nd pipe fills the tank

5 hours faster than first pipe and 4 hours slower than third pipe. The

time required by first pipe is :

Sol: Suppose, first pipe take x hours to fill the tank then

B & C will take (x-5) and (x-9) hours respectively.

Therefore, 1/x +1/(x-5) =1/(x-9)

On solving, x=15

Hence, time required by first pipe is 15 hours.

6)A large tanker can be filled by two pipes A& B in 60min and 40 min

respectively. How many minutes will it take to fill the tanker from

empty state if B is used for half the time & A and B fill it together for

the other half?

Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24

Suppose the tank is filled in x minutes

Then, x/2(1/24+1/40)=1

=> (x/2)*(1/15)=1

=> x=30 min.

7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.

If they are opened on alternate hours and if pipe A s opened first, in

how many hours, the tank shall be full.

Sol: (A+B)’s 2 hours work when opened alternatively =1/6+1/4 =5/12

(A+B)’s 4 hours work when opened alternatively=10/12=5/6

Remaining part=1 -5/6=1/6.

Now, it is A’s turn and 1/6 part is filled by A in 1 hour.

So, total time taken to fill the tank=(4+1)= 5 hours.

8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.

If A is open all the time and B and C are open for one hour each

alternatively, the tank will be full in.

Sol: (A+B)’s 1 hour’s work=1/12+1/15=9/60=3/20

(A+C)’s 1 hour’s work=1/20+1/12=8/60=2/15

Part filled in 2 hours=3/20+2/15=17/60

Part filled in 2 hours=3/20+2/15= 17/60

Part filled in 6 hours=3*17/60 =17/20

Remaining part=1 -17/20=3/20

Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.

Therefore, total time taken to fill the tank=6+1=7 hours

9)A Booster pump can be used for filling as well as for emptying a tank.

The capacity of the tank is 2400 m3. The emptying capacity of the tank is

10 m3 per minute higher than its filling capacity and the pump needs 8

minutes lesser to empty the tank than it needs to fill it. What is the

filling capacity of the pump?

Sol: Let, the filling capacity of the pump be x m3/min

Then, emptying capacity of the pump=(x+10) m3/min.

So,2400/x – 2400/(x+10) = 8

on solving x=50.

1.INLET:A pipe connected with a tank or cistern or a reservoir,

that fills it, it is known as Inlet.

OUTLET:A pipe connected with a tank or a cistern or a reservoir,

emptying it, is known as 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 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 filled in 1 hour=(1/y -1/x).

v) If two pipes A and B can fill a tank in x hours and y hours

respectively. If both the pipes are opened simultaneously, part

filled by A+B in 1 hour= 1/x +1/y.

Simple Problems

1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.

If both the pipes are opened simultaneously, how much 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)’s in 1 hour=1/36 +1/45= 9/180 =1/20

Hence, both the pipes together will fill the tank in 20 hours.

2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While

3rd pipe empties the full tank n 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 be filled in 15/2hours= 7 hrs 30 min

3)A cistern can be filled by a tap in 4 hours while it can be emptied

by another tap in 9 hours. If both the taps are opened simultaneously,

then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36

Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

4)If two pipes function simultaneously, the reservoir will be filled in

12 days.One pipe fills the reservoir 10 hours faster than the other.

How many hours does it take the second pipe to fill the reservoir.

Sol: Let the reservoir be filled by the 1st pipe in x hours.

The second pipe will fill it in (x+10) hours

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

=> (2x+10)/((x)*(x+10))= 1/12

=> x=20

So, the second pipe will take 30 hours to fill the reservoir.

5)A cistern has two taps which fill it in 12 min and 15 min respectively.

There is also a waste pipe in the cistern. When all the three are opened,

the empty cistern is full in 20 min. How long will the waste pipe take to

empty the full cistern?

Sol: Work done by a waste pipe in 1 min

=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)

6)A tap can fill a tank in 6 hours. After half the tank is filled, three

more similar taps are opened. What is the total time taken to fill the

tank completely?

Sol: Time taken by one tap to fill the half of the tank =3 hours

Part filled by the four taps in 1 hour=4/6=2/3

Remaining part=1 -1/2=1/2

Therefore, 2/3:1/2::1:x

or x=(1/2)*1*(3/2)=3/4 hours.

i.e 45 min

So, total time taken= 3hrs 45min.

7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B

can empty it in 6 min. If both pipes are open, how long will it take to

empty or fill the tank completely ?

Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.

Part to be emptied=2/5.

Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15

Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.

So, the tank be emptied in 6 min.

8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for

Bucket P to fill the empty drum. How many turns it will take for both the

buckets P&Q, having each turn together to fill the empty drum?

Sol: Let the capacity of P be x lit.

Then capacity of Q=x/3 lit

Capacity of the drum=60x lit

Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

Complex Problems

1)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 took 32min 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 two pipes to fill the tank=112/15 hrs.

Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours

Therefore, work done by (two pipes + leak) in 1 hr= 1/8

work done by leak n 1 hour=15/112 -1/8=1/112

Leak will empty full cistern n 112 hours.

2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After

7 min, C also opened. In how much time, the tank s full.

Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20

Remaining part= 1-7/20=13/20

Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60

Now, 1/60 part is filled in 1 min.

13/20 part is filled n (60*13/20)=39 min

Total time taken to fill the tank=39+7=46 min

3)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

x(1/24+1/32) +(18-x)1/24 =1

=> x=8

Hence B must be closed after 8 min.

4)Two pipes A& B together can fill a cistern in 4 hours. Had they been

opened separately, then B would have taken 6 hours more than A to fill

the cistern. How much time will be taken by A to fill the cistern

separately?

Sol: Let the cistern be filled by pipe A alone in x hours.

Pipe B will fill it in x+6 hours

1/x + 1/x+6=1/4

Solving this we get x=6.

Hence, A takes 6 hours to fill the cistern separately.

5)A tank is filled by 3 pipes with uniform flow. The first two pipes

operating simultaneously fill the tan in the same time during which

the tank is filled by the third pipe alone. The 2nd pipe fills the tank

5 hours faster than first pipe and 4 hours slower than third pipe. The

time required by first pipe is :

Sol: Suppose, first pipe take x hours to fill the tank then

B & C will take (x-5) and (x-9) hours respectively.

Therefore, 1/x +1/(x-5) =1/(x-9)

On solving, x=15

Hence, time required by first pipe is 15 hours.

6)A large tanker can be filled by two pipes A& B in 60min and 40 min

respectively. How many minutes will it take to fill the tanker from

empty state if B is used for half the time & A and B fill it together for

the other half?

Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24

Suppose the tank is filled in x minutes

Then, x/2(1/24+1/40)=1

=> (x/2)*(1/15)=1

=> x=30 min.

7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.

If they are opened on alternate hours and if pipe A s opened first, in

how many hours, the tank shall be full.

Sol: (A+B)’s 2 hours work when opened alternatively =1/6+1/4 =5/12

(A+B)’s 4 hours work when opened alternatively=10/12=5/6

Remaining part=1 -5/6=1/6.

Now, it is A’s turn and 1/6 part is filled by A in 1 hour.

So, total time taken to fill the tank=(4+1)= 5 hours.

8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.

If A is open all the time and B and C are open for one hour each

alternatively, the tank will be full in.

Sol: (A+B)’s 1 hour’s work=1/12+1/15=9/60=3/20

(A+C)’s 1 hour’s work=1/20+1/12=8/60=2/15

Part filled in 2 hours=3/20+2/15=17/60

Part filled in 2 hours=3/20+2/15= 17/60

Part filled in 6 hours=3*17/60 =17/20

Remaining part=1 -17/20=3/20

Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.

Therefore, total time taken to fill the tank=6+1=7 hours

9)A Booster pump can be used for filling as well as for emptying a tank.

The capacity of the tank is 2400 m3. The emptying capacity of the tank is

10 m3 per minute higher than its filling capacity and the pump needs 8

minutes lesser to empty the tank than it needs to fill it. What is the

filling capacity of the pump?

Sol: Let, the filling capacity of the pump be x m3/min

Then, emptying capacity of the pump=(x+10) m3/min.

So,2400/x – 2400/(x+10) = 8

on solving x=50.

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