Pipes and
Cisterns Practice Quiz – Set 2
1. A container of capacity 25 L has an inlet and
an outlet tap. If both are opened simultaneously, the container is filled in 5
minutes. But if the outlet flow rate is doubled and taps opened the container
never gets filled up. Which of the following can be outlet flow rate?
a) 6 lit/min
b) 4 lit/min
c) 3 lit/min
d) 5lit/min
Answer: D)
Explanation: Let, the inlet pipe fills x l/min
And outlet pipe empties y
l/min
Then x –y = 25/5 =
5
… (1)
When outflow rate is
doubled, the tank never gets filled up, which means
x – 2y ≤ 0
x ≤ 2y
putting x =2y in Eq. (1)
y = 5
Any value of y greater or
equal to 5 will satisfy the given condition.
2. Three pipes P1, P2 and P3 can fill a glass tub
from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When
the glass tub is empty, all the three pipes are opened. P1,P2 and P3 discharge
chemical solutions C, H and T respectively. What is the proportion of the
solution T in the liquid in the pot after 3 minutes?
a) 11/6
b) 5/11
c) 6/11
d) 1/11
e) None of these
Answer: C)
Explanation: Part of the glass tub filled by P1 in
1 minute = 1 ⁄30
Part of the glass tub
filled by P2 in 1 minute = 1 ⁄20
Part of the glass tub
filled by P3 in 1 minute = 1 ⁄10
Here we have to find the
proportion of the solution T.
Pipe P3 discharges
chemical solution T
Part of the glass tub
filled by P3 in 3 minutes = 3 *1/10 = 3/10
Part of the glass tub
filled by pipe P1, P2 and P3 together in 1 minute = 1 /30 + 1/20 + 1/10 = 11/60
Part of the glass tub
filled by pipe P1, P2 and P3 together in 3 minute = 3 * 11/60 = 11/20
Required proportion =
(3/10)/(11/12)= 6/11
3. A tank is filled by three pipes P1, P2, P3 with
uniform flow. The P1 and P2 operating simultaneously fill the tank in the same
time during which the tank is filled by the P3 alone. The P2 fills the tank 5
hrs faster than the P1 and 4 hrs slower than the P3. The time required by the
first pipe, P1 is:
a) 6 hrs
b) 10 hrs
c) 15 hrs
d) 30 hrs
e) None of these
Answer: C)
Explanation: P1 = (x + 5) hr
P2 = x hr
P3 = (x – 4) hr
According to question
1/(x + 5) + (1/x) + 1/(x
– 4)
x^2- 8x – 20 = 0
x = 10 hr
Time required by first
pipe = 15 hr
4. Two water taps T1 and T2 can fill a tank in 900
seconds and 2400 seconds respectively. Both the taps are opened together but
after 240 seconds, tap T1 is turned off. What is the total time required to
fill the tank?
a) 30 min 10 sec
b) 25 min 20 sec
c) 14 min 40 sec
d) 20 min 10 sec
e) None of these
Answer: B)
Explanation: 900 sec = 15 min , 2400 sec = 40 min
and 240 sec = 4 min
Part filled by tap T1 in
1 minute = 1/15
Part filled by tap T2 in
1 minute = 1/40
Part filled by tap T1 and
tap T2 in 1 minute = 1/15 + 1/40 = 11/120
Tap T1 and tap T2 were
open for 4 minutes
Part filled by tap T1 and
tap T2 in these 4 minutes = 4 * 11/120 = 11/30
Remaining part to be
filled = 1 – 11/30 = 19/30
Time taken by tap T2 to fill
this remaining part = (19/30)/(1/40) = 76/3 = 25(1/3)
i.e. 25 min 20 sec
5. Two pipes P and Q can fill a container in 25
and 30 minutes respectively and a waste pipe R can empty 3 gallons per minute.
All the three pipes P,Q and R working together can fill the container in 15
minutes. The capacity of the container is:
a) 350 gallons
b) 450 gallons
c) 420 gallons
d) 550 gallons
e) None of these
Answer: B)
Explanation: Part filled by pipe P in 1
minute=1/25
Part filled by pipe Q in
1 minute=1/30
Let the waste pipe R can
empty the full container in x minutes
Then, part emptied by
waste pipe R in 1 minute=1/x
All the three pipes P,Q
and R can fill the container in 15 minutes
i.e. part filled by all
the three pipes in 1 minute = 1/15
1/25 + 1/30 – 1/x = 1/15
x = 150
i.e, the waste pipe R can
empty the full container in 150 minutes
Given that waste pipe R
can empty 3 gallons per minute
ie, in 150 minutes, it
can empty 150 * 3 = 450 gallons
Hence, the volume of the
container = 450 gallons
6. Two taps T1 and T2 together can fill the tank
in 2160 seconds. If the tap T1 can fill a tank four times as fast as another
tap T2, then what is the time taken by the slower tap to fill the tank alone?
a) 188 min
b) 184 min
c) 160 min
d) 180 min
e) None of these
Answer: D)
Explanation: 2160 seconds = 36 min
Let the slower tap T2
alone can fill the tank in x minutes
Then the faster tap T2
can fill the tank in minutes = x/4
Part filled by the slower
tap in 1 minute =1/x
Part filled by the faster
tap in 1 minute =4/x
Part filled by both the
taps in 1 minute = 1/x + 4/x
It is given that both the
taps together can fill the tank in 2160 seconds i.e. 36 minutes
Part filled by both the
taps in 1 minute =1/36
1/x +4/x =1/36
x = 5 × 36 = 180
i.e. the slower tap T2
alone fill the tank in 180 minutes
7. A large tanker can be filled by two pipes P1
and P2 in 3600 seconds and 40 minutes respectively. How many seconds will it
take to fill the tanker from empty state if P2 is used for half the time and P1
and P2 fill it together for the other half?
a) 30 min
b) 1200 sec
c) 25 min
d) 1800 sec
e) 35 min
Answer: D)
Explanation: Part filled by pipe P1 in 1 minute
=1/60
Part filled by pipe P2 in
1 minute =1/40
Part filled by pipes P1
and pipe P2 in 1 minute = 1/60 + 1/40 = 1/24
Suppose the tank is
filled in x minutes
Then, to fill the tanker
from empty state, P2 is used for x/2 minutes
And P1 and P2 is used for
the rest x/2 minutes
x/2 * 1/40 + x/2 * 1/24 =
1
x/2(1/40 + 1/24) =1
x = 30 min = 1800 sec.
8. A small hole in the bottom of a cistern can
empty the full container in 360 minutes. An inlet pipe fills water at the rate
of 4 liters a minute. When the container is full, the inlet is opened and due
to the leak, the cistern is empty in 24 hours. How many liters does the cistern
hold?
a) 1290 litre
b) 2120 litre
c) 1920 litre
d) 2020 litre
e) None of these
Answer: C)
Explanation: 360 min = 6 hr
Part emptied by the leak
in 1 hour =1/6
Net part emptied by the
leak and the inlet pipe in 1 hour =1/24
Part filled by the inlet
pipe in 1 hour =1/6 – 1/24 =1/8
i.e., inlet pipe fills
the cistern in 8 hours = (8 × 60) minutes = 480 minutes
Given that the inlet pipe
fills water at the rate of 4 liters a minute
Hence, water filled in
480 minutes = 480 × 4 = 1920 litre
i.e. The cistern can hold
1920 litre
9. Three taps T1, T2 and T3 can fill a tank in
720, 900 and 1200 minutes respectively. If T1 is open all the time and T2 and
T3 are open for one hour each alternately, the tank will be full in:
a) 6 2/3 hrs
b) 9 hrs
c) 4 hrs
d) 7 hrs
e) 8 hrs
Answer: D)
Explanation: 720 min =12 hr, 900 min = 15 hr and
1200 min = 20 hr
Part filled by pipe T1 in
1 hour =1/12
Part filled by pipe T2 in
1 hour =1/15
Part filled by pipe T3 in
1 hour =1/20
In first hour, T1 and T2
is open
In second hour, T1 and T3
is open
then this pattern goes on
till the tank fills
Part filled by pipe T1
and pipe T2 in 1 hour =1/12 +1/15 = 3/20
Part filled by pipe T1
and pipe T3 in 1 hour =1/12 + 1/20 = 2/15
Part filled in 2 hour =
3/20 +2/15 =17/60
Part filled in 6 hour
=17/60 * 3 = 17/20
Remaining part = (1 –
17/20) = 3/20
Now, 6 hours are over and
only 3/20 part needed to be filled.
At this 7th hour, T1 and
T2 is open
Time taken by pipe T1 and
T2 to fill this 3/20 part = (3/20)/(3/20) = 1 hr
Total time taken = 6 hr +
1 hr = 7 hr
10. A booster pump can be used for filling as well
as for emptying a container. The capacity of the container is 2400. The
emptying of the container is 10 per
minute higher than its filling capacity and the pump needs 8 minutes lesser to
empty the container than it needs to fill it. What is the filling capacity of
the pump?
a) 40 m^3 / min
b) 50 m^3 / min
c) 60 m^3 / min
d) 70 m^3 / min
e) None of these
Answer: B)
Explanation: Let the filling capacity of the pump
= x m^3/ min.
Then the emptying
capacity of the pump = (x + 10) m^3 / min.
Time required for filling
the container = 2400/x min.
Time required for
emptying the container = 2400/(x+10) min.
Pump needs 8 minutes
lesser to empty the container than it needs to fill it
2400/x – 2400/(x+10)=8
+ 10x – 3000 = 0
x = 50 or -60
i.e. filling capacity of
the pump = 50 m^3 / min
