__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