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**Important formulas – Pipes and Cisterns:**

**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.

1. If a pipe

can fill a tank in α hours, then: part filled in 1 hour = 1/α

can fill a tank in α hours, then: part filled in 1 hour = 1/α

2. If a pipe

can empty a full tank in β

then: part emptied in 1 hour = 1/β

can empty a full tank in β

*hours,*then: part emptied in 1 hour = 1/β

3. If a pipe

can fill a tank in α hours and another pipe can empty the full tank in β hours (where

β> α),

opening both the pipes,

The net part

filled in 1 hour = (1/α) – (1/β)

can fill a tank in α hours and another pipe can empty the full tank in β hours (where

β> α),

*then on*opening both the pipes,

The net part

filled in 1 hour = (1/α) – (1/β)

4. If a pipe

can fill a tank in α hours and another pipe can empty the full tank in β hours (where α > β), then on

opening both the pipes,

The net part

emptied in 1 hour = (1/β) – (1/α)

can fill a tank in α hours and another pipe can empty the full tank in β hours (where α > β), then on

opening both the pipes,

The net part

emptied in 1 hour = (1/β) – (1/α)

__Also, these shortcut formulas can be__

used:used:

1. If a pipe

can fill a tank in α hours and another pipe can empty the full tank in β hours (where

β> α),

taken to fill the tank, when both the pipe are opened, αβ/(β – α)

can fill a tank in α hours and another pipe can empty the full tank in β hours (where

β> α),

*then time*taken to fill the tank, when both the pipe are opened, αβ/(β – α)

2. If a pipe

can fill a tank in α hours and another pipe can fill the same tank in β hours,

then the net part the time taken to fill, when both the pipes are open

can fill a tank in α hours and another pipe can fill the same tank in β hours,

then the net part the time taken to fill, when both the pipes are open

Time taken

to fill the tank = αβ/(α + β)

to fill the tank = αβ/(α + β)

3. If a pipe

fills a tank in α hour and another fills the same tank in β hours, but a third

one empties the full tank in ɣ hours, and all of them are opened together, then

fills a tank in α hour and another fills the same tank in β hours, but a third

one empties the full tank in ɣ hours, and all of them are opened together, then

Time taken

to fill the tank = αβɣ/(βɣ +αɣ – αβ)

to fill the tank = αβɣ/(βɣ +αɣ – αβ)

4. A pipe

can fill a tank in α hours. Due to a leak in the bottom, it is filled in β

hours. The time taken by the leak to empty the tank is, αβ/(β – α)

can fill a tank in α hours. Due to a leak in the bottom, it is filled in β

hours. The time taken by the leak to empty the tank is, αβ/(β – α)

__Example: 01__**Two pipes A and B can fill the tank in 30**

hrs and 45 hrs respectively. If both the pipes are opened simultaneously, how

much time will be taken to fill the tank?

hrs and 45 hrs respectively. If both the pipes are opened simultaneously, how

much time will be taken to fill the tank?

__Solution:__

__By 1st method;__
A fills the

tank in 1 hr = 1/30 parts

tank in 1 hr = 1/30 parts

B fills the

tank in 1 hr = 1/45 parts

tank in 1 hr = 1/45 parts

A and B

together fills the tank in 1 hr = 1/30 + 1/45 = 1/18 parts

together fills the tank in 1 hr = 1/30 + 1/45 = 1/18 parts

So, time

required to fill the tank is 18 hrs.

required to fill the tank is 18 hrs.

**By 2nd method;**Time taken = αβ/(α + β)

= (30 × 45)/(30 + 45) = 18 hrs.

__Example: 02__**Pipe A can fill a tank in 25 hrs while B**

alone can fill it in 30 hrs and C can empty the full tank in 45 hrs. If all the

pipes are opened together, how much time will be needed to make the tank full?

alone can fill it in 30 hrs and C can empty the full tank in 45 hrs. If all the

pipes are opened together, how much time will be needed to make the tank full?

__Solution:__
The tank

will be full in = (25 × 30 × 45)/[(30 × 45) + (25 × 45) – (25 × 30)] = 19.56

hours.

will be full in = (25 × 30 × 45)/[(30 × 45) + (25 × 45) – (25 × 30)] = 19.56

hours.

__Example: 03__**Two pipes A and B would fill a cistern in**

24 hrs and 32 hrs respectively. If both the pipes are opened together; find

when the first pipe must be turned off so that the cistern may be just filled

in 16 hrs.

24 hrs and 32 hrs respectively. If both the pipes are opened together; find

when the first pipe must be turned off so that the cistern may be just filled

in 16 hrs.

__Solution:__
B fills the

tank in 1 hr = 1/32 parts

tank in 1 hr = 1/32 parts

B fills the

tank in 16 hrs = 16/32 parts = 1/2 part.

tank in 16 hrs = 16/32 parts = 1/2 part.

Given,

A fill the full

tank in 24 hrs

tank in 24 hrs

Therefore, A

fills the 1/2 part in just 12 hrs.

fills the 1/2 part in just 12 hrs.

So, the

first pipe A should work for 12 hrs.

**Alternate Method:**The first should work

for = [1 – (16/32)] × 24 = 12 hrs.

**SOLVED EXAMPLES**

__Question No. 01__**Two pipes A**

and B can fill a cistern in 37½ minutes

and 45 minutes respectively. Both pipes are opened. The cistern will be filled

in just half an hour, if the B is turned off after:

and B can fill a cistern in 37½ minutes

and 45 minutes respectively. Both pipes are opened. The cistern will be filled

in just half an hour, if the B is turned off after:

(A) 5 min

(B) 9 min

(C) 10 min

(D) 15 min

Answer:

Option B

Option B

__Explanation:__
Let B be

turned off after

turned off after

*x*minutes. Then,
Part filled

by (A + B) in

min. = 1.

by (A + B) in

*x*min. + Part filled by A in (30 –*x*)min. = 1.

∴

+ 1/45) + [(30 –

*x*

(2/75+ 1/45) + [(30 –

*x*). 2/75 = 1
=> (11

*x*/225) + [(60 – 2*x*)/75] = 1
=> 11

180 – 6

*x*+180 – 6

*x*= 225
=>

9

*x*=9

__Question No. 02__**Two pipes**

can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3

gallons per minute. All the three pipes working together can fill the tank in

15 minutes. The capacity of the tank is:

can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3

gallons per minute. All the three pipes working together can fill the tank in

15 minutes. The capacity of the tank is:

(A) 60

gallons

gallons

(B) 100

gallons

gallons

(C) 120

gallons

gallons

(D) 180

gallons

gallons

Answer:

Option C

Option C

__Explanation:__
Work done by

the waste pipe in 1 minute = 1/15 – (1/20

+ 1/24)

the waste pipe in 1 minute = 1/15 – (1/20

+ 1/24)

= (1/15

– 11/120)

– 11/120)

= – (1/40) [-ve sign means emptying]

∴ Volume

of 1/40 part = 3 gallons.

of 1/40 part = 3 gallons.

Volume of

whole = (3 × 40) gallons = 120 gallons.

whole = (3 × 40) gallons = 120 gallons.

__Question No. 03__**Two pipes ‘**

the pipes are opened together but after 4 minutes, pipe ‘

*A*’ and ‘*B*’ can fill a tank in 15 minutes and 20 minutes respectively. Boththe pipes are opened together but after 4 minutes, pipe ‘

*A*’ is turned off. What is the total time required to fill the tank?
(A) 10 min.

20 sec.

20 sec.

(B) 11 min.

45 sec.

45 sec.

(C) 12 min.

30 sec.

30 sec.

(D) 14 min.

40 sec.

40 sec.

Answer:

Option D

Option D

__Explanation:__
Part filled

in 4 minutes = 4 {(1/15) + (1/20)} = (7/15)

in 4 minutes = 4 {(1/15) + (1/20)} = (7/15)

Remaining

part = {1 – (7/15)} = (8/15)

part = {1 – (7/15)} = (8/15)

Part filled

by B in 1 minute = (1/20)

by B in 1 minute = (1/20)

∴ (1/20) : (8/15)

**::**1 :*x**x*=

{(8/15) × 1 × 20} = 10⅔ min = 10 min. 40 sec.

∴ The

tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

__Question No. 04__**A large**

tanker can be filled by two pipes ‘

and ‘

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

if ‘

tanker can be filled by two pipes ‘

*A*’and ‘

*B*’ in 60 minutes and 40 minutesrespectively. How many minutes will it take to fill the tanker from empty state

if ‘

*B*’ is used for half the time and ‘*A*’ and ‘*B*’ fill it together for the other half?
(A) 15 min

(B) 20 min

(C) 27.5 min

(D) 30 min

Answer:

Option D

Option D

__Explanation:__
Part filled

by (A + B) in 1 minute = {(1/60) + (1/40)} = (1/24)

by (A + B) in 1 minute = {(1/60) + (1/40)} = (1/24)

Suppose the

tank is filled in

tank is filled in

*x*minutes.
Then, (

*x*/2) × {(1/24) + (1/40)} = 1
=> (

*x*/2) × (1/15) = 1
=>

30 min.

*x*=30 min.

__Question No. 05__**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?

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?

(A) 3 hrs 15

min

min

(B) 3 hrs 45

min

min

(C) 4 hrs

(D) 4 hrs 15

min

min

Answer:

Option B

Option B

__Explanation:__
Time taken

by one tap to fill half of the

tank = 3 hrs.

by one tap to fill half of the

tank = 3 hrs.

Part filled

by the four taps in 1 hour = {4 × (1/6)} = 2/3

by the four taps in 1 hour = {4 × (1/6)} = 2/3

Remaining

part = {1 – (1/2)} = 1/2

part = {1 – (1/2)} = 1/2

∴ (2/3) : (1/2)

**::**1 :*x*
=>

mins.

*x*= [(1/2) × 1 × (3/2)] = (3/4) hours*i.e.,*45mins.

So, total

time taken = 3 hrs. 45 mins.

time taken = 3 hrs. 45 mins.

__Question No. 06__**Three pipes**

and

After working at it together for 2 hours, ‘

is closed and ‘

hours. The number of hours taken by ‘

alone to fill the tank is:

*A*,*B*and

*C*can fill a tank in 6 hours.After working at it together for 2 hours, ‘

*C*’is closed and ‘

*A*’ and ‘*B*’ can fill the remaining part in 7hours. The number of hours taken by ‘

*C*’alone to fill the tank is:

(A) 10

(B) 12

(C) 14

(D) 16

Answer:

Option C

Option C

__Explanation:__
Part filled

in 2 hours = 2/6 = 1/3

in 2 hours = 2/6 = 1/3

Remaining

part = [1 – (1/3)] = 2/3

part = [1 – (1/3)] = 2/3

∴ (A

+ B)’s 7 hour’s work = 2/3

+ B)’s 7 hour’s work = 2/3

∴ C’s

1 hour’s work = {(A + B + C)’s 1 hour’s work} – {(A + B)’s 1 hour’s work}

1 hour’s work = {(A + B + C)’s 1 hour’s work} – {(A + B)’s 1 hour’s work}

= {(1/6) –

(2/21)} = 1/14

(2/21)} = 1/14

∴ C

alone can fill the tank in 14 hours.

alone can fill the tank in 14 hours.

__Question No. 07__**Three pipes**

and

full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is

empty, all the three pipes are opened.

minutes?

*A*,*B*and

*C*can fill a tank from empty tofull in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is

empty, all the three pipes are opened.

*A*,*B*and*C*discharge chemical solutions*P*,*Q*and*R*respectively. What is the proportion of the solution ‘*R*’ in the liquid in the tank after 3minutes?

(A) 5/11

(B) 6/11

(C) 7/11

(D) 8/11

Answer:

Option B

Option B

__Explanation:__
Part filled

by (A + B + C) in 3 minutes = 3 [(1/30) + (1/20) + (1/10)] = 3 × (11/60) =

11/20

by (A + B + C) in 3 minutes = 3 [(1/30) + (1/20) + (1/10)] = 3 × (11/60) =

11/20

Part filled

by C in 3 minutes = 3/10

by C in 3 minutes = 3/10

∴ Required

ratio = (3/10) × (20/11) = 6/11

ratio = (3/10) × (20/11) = 6/11

__Question No. 08__**A tank is**

filled by three pipes with uniform flow. The first two pipes operating

simultaneously fill the tank in the same time during which the tank is filled

by the third pipe alone. The second pipe fills the tank 5 hours faster than the

first pipe and 4 hours slower than the third pipe. The time required by the

first pipe is:

filled by three pipes with uniform flow. The first two pipes operating

simultaneously fill the tank in the same time during which the tank is filled

by the third pipe alone. The second pipe fills the tank 5 hours faster than the

first pipe and 4 hours slower than the third pipe. The time required by the

first pipe is:

(A) 6 hours

(B) 10 hours

(C) 15 hours

(D) 30 hours

Answer:

Option C

Option C

__Explanation:__
Suppose, first

pipe alone takes

pipe alone takes

*x*hours to fill the tank
Then, second

and third pipes will take (

respectively to fill the tank.

and third pipes will take (

*x*-5) and (*x*– 9) hoursrespectively to fill the tank.

∴ 1/x

+ 1/(x – 5) = 1/(x – 9)

+ 1/(x – 5) = 1/(x – 9)

=> (x

– 5 + x)/x (x – 5) = 1/(x – 9)

– 5 + x)/x (x – 5) = 1/(x – 9)

=> (2

5)(

*x*–5)(

*x*– 9) =*x*(*x*– 5)
=>

45 = 0

*x*² – 18*x*+45 = 0

=> (

15)(

*x*–15)(

*x*– 3) = 0
=>

[Neglecting

*x*= 15.[Neglecting

*x*= 3]

__Question No. 09__**Two pipes**

separately, then ‘

6 hours more than ‘

cistern. How much time will be taken by ‘

to fill the cistern separately?

*A*and*B*together can fill a cistern in 4 hours. Had they been openedseparately, then ‘

*B*’ would have taken6 hours more than ‘

*A*’ to fill thecistern. How much time will be taken by ‘

*A*’to fill the cistern separately?

(A) 1 hour

(B) 2 hours

(C) 6 hours

(D) 8 hours

Answer:

Option C

Option C

__Explanation:__
Let the

cistern be filled by pipe A alone in

cistern be filled by pipe A alone in

*x*hours.
Then, pipe B

will fill it in (

will fill it in (

*x*+ 6) hours.
∴ (1/

+ 1/(

*x*)+ 1/(

*x*+6) = 1/4
=> [(

*x*+ 6 +*x*)/*x*(*x*+ 6)]*= 1/4*
=>

24 = 0

*x*2 – 2*x*–24 = 0

=> (

4) = 0

*x*-6)(*x*+4) = 0

=>

6. [Neglecting the negative value of

*x*=6. [Neglecting the negative value of

*x*]

__Question No. 10__**A tank is**

filled in 5 hours by three pipes

as fast as ‘

filled in 5 hours by three pipes

*A*,*B*and*C*. The pipe ‘*C’*is twiceas fast as ‘

*B*’ and ‘*B*’ is twice as fast as ‘*A*’. How much time will pipe ‘*A*’ alone take to fill the tank?
(A) 20 hours

(B) 25 hours

(C) 35 hours

(D) None of

these

these

Answer:

Option C

Option C

__Explanation:__
Suppose pipe

A alone takes

A alone takes

*x*hours to fill the tank.
Then, pipes

*B*and*C*will take*x*/2 and*x*/4 hours respectively to fill the tank∴ (1/*x*)

+ (2/*x*) + (4/*x*) = (1/5)