# CRE 6 - Pipes & Cisterns | Time and Work

**CRE 6 - Pipes & Cisterns | Time and Work**

Two pipes can fill a cistern in 18 hrs and 24 hrs respectively. In how much time will they fill the cistern if both of them are opened together?

- A.
$10\frac{5}{7}$

- B.
$10\frac{5}{8}$

- C.
$10\frac{2}{7}$

- D.
$10\frac{6}{7}$

Answer: Option C

**Explanation** :

Work done by 1st pipe in 1 hour = 1/18

Work done by 2nd pipe in 1 hour = 1/24

∴ Work done by both pipes in 1 hour = $\frac{1}{8}+\frac{1}{24}=\frac{7}{72}$ 1/18+1/24)=7/72

∴ Both pipes together can fill in 72/7 hrs i.e., $10\frac{2}{7}$ hours

Hence, option (c).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

Pipe A can fill a tank in 8 hrs and pipe B can empty it in 16 hrs. If both the pipes are opened simultaneously, how many hours will they take to fill up an empty tank?

- A.
16 hours

- B.
20 hours

- C.
18 hours

- D.
14 hours

Answer: Option A

**Explanation** :

Part filled by A in 1 hour = 1/8

Part emptied by B in 1 hour = 1/16

∴ Net filling in 1 hour = (1/8 - 1/16) = 1/16

∴ Tank full in 16 hours.

Hence, option (a).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

One tap can fill a tank in 16 hours and another can empty the tank in 40 hours. A third tap is also, connected to the tank. If all the three taps are opened together, the tank is filled in 30 hours. At what rate does the third pipe work?

- A.
Fills the tank in 24 hours

- B.
Empties the tank in 80 hours

- C.
Fills the tank in 120 hours

- D.
Empties the tank in 120 hours

- E.
Empties the tank in 240 hours

Answer: Option E

**Explanation** :

Let the third tap fill the tank in x hours.

When the three taps are opened together, part of the tank filled in 1 hour = 1/30

∴ 1/16 - 1/40 + 1/x = 1/30

∴ 1/x = 1/30 - 1/16 + 1/40

∴ 1/x = (8 - 15 + 6)/240

∴ 1/x = -1/240

∴ The third pipe empties the tank in 240 hours.

Hence, option (e).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

Two pipes A & B can fill a cistern in 16 and 24 minutes respectively. If both the pipes are opened simultaneously, then after how much time should B be closed so that the tank is filled in 12 minutes?

- A.
8 mins

- B.
6 mins

- C.
10 mins

- D.
14 mins

Answer: Option B

**Explanation** :

Let B closed after x mins, then part filled by (A + B) in x mins + part filled by A in (12 - x) mins = 1]

∴ $x\left(\frac{1}{16}+\frac{1}{24}\right)+\left(12-x\right)\frac{1}{16}=1$ or $\frac{5x}{48}+\frac{12-x}{16}=1$

⇒ 5x + 36 - 3x = 48

⇒ 2x = 12 ⇒ x = 6

So, B must be closed after 6 mins.

Hence, option (b).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

If the first pipe takes 60 mins to fill a tank and second takes 90 mins to empty it, then in how much time will the tank be full, if the second pipe is opened 30 mins earlier than the first pipe and the tank is empty at first?

- A.
150 mins

- B.
180 mins

- C.
130 mins

- D.
Never

Answer: Option B

**Explanation** :

The time required to fill the tank is 180 minutes

1/60 - 1/90 = 1/180

The fact that the second pipe was opened earlier makes no difference as the tank is empty.

Hence, option (b).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

There are three pipes A, B and C in a tank. If all the pipes are open together, a tank can be filled in 12 hours. If pipe A can fill the tank in 30 hours and pipe B can fill the tank in 24 hours then at what rate does pipe C work?

- A.
Fills the tank in 24 hours

- B.
Empties the tank in 80 hours

- C.
Fills the tank in 120 hours

- D.
Empties the tank in 120 hours

- E.
Fills the tank in 180 hours

Answer: Option C

**Explanation** :

Let pipe C can fill the tank in x hours.

When the three pipes are opened together, part of the tank filled in 1 hour = 1/12.

$\frac{1}{30}+\frac{1}{24}+\frac{1}{x}=\frac{1}{12}$

∴ $\frac{1}{x}=\frac{1}{12}-\frac{1}{30}-\frac{1}{24}$

∴ $\frac{1}{x}=\frac{1}{120}$

∴ Pipe C fills the tank in 120 hours.

Hence, option (c).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

Pipe A can fill a tank at the rate of 10 liters/minute while B can fill the same tank in 15 minutes. A and B together can fill the empty tank in 12 minutes. Find the capacity of the tank (in liters).

Answer: 600

**Explanation** :

Let the volume of the tank be V liters.

Efficiency of Pipe A = 10 liters/minute.

Efficiency of Pipe B = V/15 liters/minute.

Combined efficiency of Pipes A and B = V/12 liters/minute.

⇒ 10 + V/15 = V/12

⇒ 10 = V/60

⇒ V = 600 liters

Hence, 600.

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

Three pipes A, B and C can fill a cistern in 12 hours. After working at it together for 4 hours, C is closed and A and B can fill the remaining part in 14 hours. The number of hours taken by C alone to fill the cistern is:

- A.
24

- B.
28

- C.
32

- D.
36

- E.
30

Answer: Option B

**Explanation** :

A, B and C together can fill 1/12th of cistern in one hour.

A, B and C together worked for two hours = 4 × (1/12) = 1/3 is filled.

A and B together can fill remaining 2/3 in 14 hours

⇒ A and B together can fill complete cistern in 14 × (3/2) = 21 hours.

A and B can together fill 1/21 of tank in one hour where A, B and C can fill 1/12th in one hour.

⇒ C alone can (1/12) - (1/21) of the cistern in one hour = 1/28th of cistern.

So, C alone can fill the cistern in 28 hours.

Hence, option (b).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

Two pipes of circular cross-section fill a tank. Radius of bigger pipe is twice the radius of smaller pipe. If bigger pipe can fill the tank in 60 minutes, how long will it take for both the pipes to fill the tank together?

- A.
1 hour 12 minutes

- B.
1 hour

- C.
40 minutes

- D.
48 minutes

Answer: Option D

**Explanation** :

We know, efficiency of a pipe ∝ cross-section

and, cross-section ∝ (radius)^{2}

⇒ efficiency ∝ (radius)^{2}

∴ $\frac{efficiencyofbiggerpipe}{efficiencyofsmallerpipe}$ = $\frac{radiusofbiggerpipe}{radiusofsmallerpipe}$ = $\frac{4}{1}$

Also, Time ∝ 1/efficiency

⇒ $\frac{Timetakenbybiggerpipe}{Timetakenbysmallerpipe}$ = $\frac{efficiencyofsmallerpipe}{efficiencyofbiggerpipe}$

⇒ 60/x = 1/4

⇒ x = 240 minutes

Now, when they work together time taken to fill the tank = $\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$60$}\right.}+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$240$}\right.}}$ = 48 minutes.

Hence, option (d).

Workspace:

**CRE 6 - Pipes & Cisterns | Time and Work**

An inlet tap can fill a tank in 20 hours while an outlet tap situated at the bottom of the tank can empty a filled tank in 30 hours. How long will it take to fill an empty tank if the outlet tap is located at one-fourth of the height of the tank from bottom.

- A.
60 hours

- B.
50 hours

- C.
40 hours

- D.
30 hours

- E.
None of these

Answer: Option B

**Explanation** :

Let the work to be done = LCM (20, 30) = 60 units

∴ Efficiency of inlet tap = 60/20 = 3 units/hour

Efficiency of outlet tap = 60/30 = -2 units/hour [-ve indicates that the pipe is an outlet pipe]

Since the tap is located at one-fourth of the height from the bottom, it means that till one-fourth of the tank is filled only inlet pipe would be functioning. After that both inlet and outlet pipes would be functioning.

Time to fill one-fourth of the tank = 15/3 = 5 hours [only inlet pipe is functioning]

Time to fill remaining three-fourth of the tank = 45/(3 - 2) = 45 hours [both inlet and outlet pipes are functioning]

∴ Total time taken to fill the tank = 5 + 45 = 50 hours.

Hence, option (b).

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.