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Home > Practice > Arithmetic Aptitude > Pipes and Cisterns > Miscellaneous

11. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

A. 81 minutes

B. 108 minutes

C. 144 minutes

D. 192 minutes

Discuss

Solution:

Let the slower pipe alone fill the tank in x minutes.
Then, faster pipe will fill it in

\frac{x}{3}

minutes.

\begin{aligned} ∴ \frac{1}{x} + \frac{3}{x} = \frac{1}{36} \\ \Rightarrow \frac{4}{x} = \frac{1}{36} \\ \Rightarrow x = 144 minutes \end{aligned}

12. 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 minutes

B. 20 minutes

C. 27.5 minutes

D. 30 minutes

Discuss

Solution:

\begin{aligned} Part filled by (A + B) in 1 minute \\ = \frac{1}{60} + \frac{1}{40} = \frac{1}{24} \\ Suppose the tank is filled in x minutes \\ Then, \frac{x}{2} (\frac{1}{24} + \frac{1}{40}) = 1 \\ \Rightarrow \frac{x}{2} \times \frac{1}{15} = 1 \\ \Rightarrow x = 30 minutes \end{aligned}

13. 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

Discuss

Solution:

Time taken by one tap to fill half of the the tank = 3 hours
Part filled by the four taps in 1 hour

\begin{aligned} = 4 \times \frac{1}{6} = \frac{2}{3} \\ Remaining part = 1 - \frac{1}{2} = \frac{1}{2} \\ ∴ \frac{2}{3} : \frac{1}{2} :: 1 : x \\ \Rightarrow x = \frac{1}{2} \times 1 \times \frac{3}{2} \\ = \frac{3}{4} h r s . i . e ., 45 mins . \\ So, total time taken = 3 h r s . 45 m i n s . \end{aligned}

14. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

A. 6 hours

B. $6 \frac{2}{3}$ hours

C. 7 hours

D. $7 \frac{1}{2}$ hours

Discuss

Solution:

\begin{aligned} (A + B)'s 1 hour work \\ = \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} \\ (A + C)'s 1 hour work \\ = \frac{1}{12} + \frac{1}{20} = \frac{8}{60} = \frac{2}{15} \\ Part filled in 2 hrs \\ = \frac{3}{20} + \frac{2}{15} = \frac{17}{60} \\ Part filled in 6 hrs \\ = 3 \times \frac{17}{60} = \frac{17}{20} \\ Remaining part \\ = 1 - \frac{17}{20} = \frac{3}{20} \end{aligned}

Now it is the turn of A and B and

\frac{3}{20}

part is filled by A and B in 1 hour
∴ Total time taken to fill tank
= (6 + 1) hrs
= 7 hrs

15. 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

Discuss

Solution:

Part filled in 2 hours =

\frac{2}{6}

\frac{1}{3}

Remaining part =

1 - \frac{1}{3}

\frac{2}{3}

∴ (A + B)'s 7 hour's work =

\frac{2}{3}

(A + B)'s 1 hour's work =

\frac{2}{21}

∴ C's 1 hour's work = {(A + B + C)'s 1 hour's work} - {(A + B's 1 hour's work}

\begin{aligned} = \frac{1}{6} - \frac{2}{21} \\ = \frac{1}{14} \end{aligned}

∴ C alone can fill the tank in 14 hours

16. Water is continuously supplied from a reservoir to a locality at the steady rate of 10000 liters per hours. When delivery exceeds demand the excess water is stored in a tank. If the demand for 8 consecutive three hour periods is 10000, 10000, 45000, 25000, 40000, 15000, 60000 and 35000 liters respectively, what will be the minimum capacity required of the water tank (in thousand liters)to meet the demand and avoid any wastage?

A. 10

B. 30

C. 40

D. 50

Discuss

Solution:

We have the following table:

Period	Supply	Demand	Calculation	Excess Qty. in tank
0 - 3 hrs	30000	10000	30000 - 10000	20000
3 - 6 hrs	30000	10000	20000 + 30000 - 10000	40000
6 - 9 hrs	30000	45000	40000 + 30000 - 45000	25000
9 - 12 hrs	30000	25000	25000 + 30000 - 25000	30000
12 - 15 hrs	30000	40000	30000 + 30000 - 40000	20000
15 - 18 hrs	30000	15000	20000 + 30000 - 15000	35000
18 - 21 hrs	30000	60000	35000 + 30000 - 60000	5000
21 - 24 hrs	30000	35000	5000 + 30000 - 35000	0

The excess quantity in tank at any time does not exceed 40000 litres,
Which is the required minimum capacity to avoid wastage.

17. A pipe can fill a tank in x hours and another pipe can empty it in y (y > x) hours. If both pipes are open, in how many hours will the tank is filled?

A. (x - y) hours

B. (y - x) hours

C. $\frac{x y}{x - y}$ hours

D. $\frac{x y}{y - x}$ hours

Discuss

Solution:

Net part filled in 1 hour

\begin{aligned} = (\frac{1}{x} - \frac{1}{y}) \\ = (\frac{y - x}{x y}) \end{aligned}

∴ The tank will be filled in

= (\frac{x y}{y - x}) hours

18. A pipe can fill a tank with water in 3 hours. Due to leakage in bottom, it takes $3 \frac{1}{2}$ hours to fill it. In what time the leak will empty the completely filled tank?

A. 12 hours

B. 21 hours

C. $6 \frac{1}{2}$ hours

D. $10 \frac{1}{2}$ hours

Discuss

Solution:

Let the total capacity of tank is 21 units
A's efficiency is 7 units/hr
A's efficiency after leakage 6 units/hr
leakage efficiency = 7 - 6 = 1 unit/hr
leakage will empty the filled tank

\frac{Total Capacity}{Efficiency} = \frac{21}{1}

= 21hours

19. Two pipe can fill a cistern in 3 hours and 4 hours respectively and a waste pipe can empty it in 2 hours. If all the three pipes are kept open, then the cistern will be filled in-

A. 5 hours

B. 8 hours

C. 10 hours

D. 12 hours

Discuss

Solution:

Part of the cistern filled in 1 hour

= \frac{1}{3} + \frac{1}{4} - \frac{1}{2}

(Cistern filled by 1^st pipe + Cistern filled by 2^nd pipe – Cistern emptied by 3^rd pipe )

\begin{aligned} \frac{4 + 3 - 6}{12} \\ = \frac{1}{12} \end{aligned}

Hence, the cistern will be filled in 12 hours.

20. $\frac{3}{4}$ part of the tank is full of water. When 30 litres of water is taken out, the tank becomes empty. The capacity of the tank is -

A. 36 liters

B. 42 liters

C. 40 liters

D. 38 liters

Discuss

Solution:

If tank has 4x liters of total capacity and its holds 3x liters of water and if 30 liters of water is taken out, then tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x = 4 × 10 = 40 liters

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