# Pipes and Cisterns PRACTICE SET 1 FOR RRVUNL AE & JE,UPPCL AE,DMRC/NMRC,SSC JE,BSPHCL JE

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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

2. If a pipe
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),

The net part filled in 1 hour = (1/α) - (1/β)

*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)

The net part emptied in 1 hour = (1/β) - (1/α)

__Also, these shortcut formulas can be 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),

*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

Time taken
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

Time taken
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)

__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?**

__Solution:__

__By 1st method;__
A fills the
tank in 1 hr = 1/30 parts

B fills the
tank in 1 hr = 1/45 parts

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

So, time
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?**

__Solution:__
The tank
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.**

__Solution:__
B fills the
tank in 1 hr = 1/32 parts

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

Given,

A fill the full
tank in 24 hrs

Therefore, A
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:**

(A) 5 min

(B) 9 min

(C) 10 min

(D) 15 min

Answer:
Option B

__Explanation:__
Let B be
turned off after

*x*minutes. Then,
Part filled
by (A + B) in

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

*x*(^{2}/_{75}+^{1}/_{45}) + [(30 -*x*).^{2}/_{75}= 1
=> (11

*x*/225) + [(60 - 2*x*)/75] = 1
=> 11

*x*+ 180 - 6*x*= 225
=>

*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:**

(A) 60
gallons

(B) 100
gallons

(C) 120
gallons

(D) 180
gallons

Answer:
Option C

__Explanation:__
Work done by
the waste pipe in 1 minute =

^{1}/_{15 }- (^{1}/_{20 }+^{1}/_{24})
= (

^{1}/_{15}-^{11}/_{120})
= - (

^{1}/_{40}) [-ve sign means emptying]
∴ Volume
of

^{1}/_{40}part = 3 gallons.
Volume of
whole = (3 × 40) gallons = 120 gallons.

__Question No. 03__**Two pipes ‘**

*A*’ and ‘*B*’ can fill a tank in 15 minutes and 20 minutes respectively. Both the 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.

(B) 11 min.
45 sec.

(C) 12 min.
30 sec.

(D) 14 min.
40 sec.

Answer:
Option D

__Explanation:__
Part filled
in 4 minutes = 4 {(1/15) + (1/20)} = (7/15)

Remaining
part = {1 - (7/15)} = (8/15)

Part filled
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.

__Question No. 04__**A large tanker can be filled by two pipes ‘**

*A*’ and ‘*B*’ in 60 minutes and 40 minutes respectively. 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

__Explanation:__
Part filled
by (A + B) in 1 minute = {(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.

__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?**

(A) 3 hrs 15
min

(B) 3 hrs 45
min

(C) 4 hrs

(D) 4 hrs 15
min

Answer:
Option B

__Explanation:__
Time taken
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

Remaining
part = {1 - (1/2)} = 1/2

∴ (2/3) : (1/2)

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

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

__Question No. 06__**Three pipes**

*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 7 hours. The number of hours taken by ‘*C*’ alone to fill the tank is:
(A) 10

(B) 12

(C) 14

(D) 16

Answer:
Option C

__Explanation:__
Part filled
in 2 hours = 2/6 = 1/3

Remaining
part = [1 - (1/3)] = 2/3

∴ (A
+ 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}

= {(1/6) -
(2/21)} = 1/14

∴ C
alone can fill the tank in 14 hours.

__Question No. 07__**Three pipes**

*A*,*B*and*C*can fill a tank from empty to full 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 3 minutes?
(A) 5/11

(B) 6/11

(C) 7/11

(D) 8/11

Answer:
Option B

__Explanation:__
Part filled
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

∴ Required
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:**

(A) 6 hours

(B) 10 hours

(C) 15 hours

(D) 30 hours

Answer:
Option C

__Explanation:__
Suppose, first
pipe alone takes

*x*hours to fill the tank
Then, second
and third pipes will take (

*x*-5) and (*x*- 9) hours respectively to fill the tank.
∴

^{1}/_{x}+^{1}/_{(x - 5)}=^{1}/_{(x - 9)}
=>

^{(x - 5 + x)}/_{x (x - 5)}=^{1}/_{(x - 9)}
=> (2

*x*- 5)(*x*- 9) =*x*(*x*- 5)
=>

*x*² - 18*x*+ 45 = 0
=> (

*x*- 15)(*x*- 3) = 0
=>

*x*= 15. [Neglecting*x*= 3]

__Question No. 09__**Two pipes**

*A*and*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?
(A) 1 hour

(B) 2 hours

(C) 6 hours

(D) 8 hours

Answer:
Option C

__Explanation:__
Let the
cistern be filled by pipe A alone in

*x*hours.
Then, pipe B
will fill it in (

*x*+ 6) hours.
∴ (1/

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

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

*x*^{2}- 2*x*- 24 = 0
=> (

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

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

__Question No. 10__**A tank is filled in 5 hours by three pipes**

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

Answer:
Option C

__Explanation:__
Suppose pipe
A alone takes

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

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

*x*) + (4/

*x*) = (1/5)

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