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

November 8, 2016
Updated 2016/11/08 at 8:33 AM

### 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), 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)
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
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
=> (11x/225) + [(60 – 2x)/75] = 1
=> 11x +
180 – 6x = 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
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.
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
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
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
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
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
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)
=> (2x –
5)(x – 9) = x(x – 5)
=> x² – 18x +
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
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
=> x2 – 2x –
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
Option C
Explanation:
Suppose pipe
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)

=> (7/x)
= (1/5)

=> x
= 35 hrs.

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