One pipe fills a tank in 6 hours and another in 4 hours. If both are opened together, how long will the tank take to fill?

2 h 24 min. Combined rate = 16 + 14 = 512 tank per hour, so the tank fills in 125 = 2 25 hours = 2 h 24 min.

A pipe fills a tank in 5 hours, but a leak at the bottom empties a full tank in 20 hours. With both working, how long will an empty tank take to fill?

6 h 40 min. Net rate = 15 - 120 = 320 per hour, so it fills in 203 = 6 23 hours = 6 h 40 min.

Two pipes fill a tank in 20 and 30 minutes. Both are opened together, but after 5 minutes the first pipe is closed. How much longer does the second pipe take to fill the rest?

17.5 min. In 5 minutes both pipes fill 520 + 530 = 512. The remaining 712 at the second pipe's rate of 130 per minute takes 712 30 = 17.5 minutes.

Pipe A fills a tank in 12 minutes and pipe B empties it in 18 minutes. If both are opened when the tank is empty, how long will it take to fill?

36 min. Net rate = 112 - 118 = 136 per minute, so the tank fills in 36 minutes.

Pipes A and B fill a tank in 24 and 32 minutes. Both are opened together. After how many minutes should B be closed so that the tank is full in exactly 18 minutes?

8 min. A runs the full 18 minutes, filling 1824 = 34. B must supply the other 14 at 132 per minute, needing 14 32 = 8 minutes, so B is closed after 8 minutes.

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Quantitative Aptitude · Time, Work & Distance

Pipes and Cisterns — Questions & Answers

Pipes and cisterns is time-and-work with a tank: an inlet fills at a positive rate, an outlet or leak empties at a negative rate. Combine the rates and invert to get the time.

5 solved questions
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Key formulas

Inlet filling in $x$ hours: rate $+\dfrac{1}{x}$ per hour. Outlet emptying in $y$ hours: rate $-\dfrac{1}{y}$
Both open: net rate $= \dfrac{1}{x} - \dfrac{1}{y}$ ; time to fill $= \dfrac{xy}{y-x}$ (for $y > x$ )
If a full tank empties in $y$ hours because of a leak, the leak's rate is $\dfrac{1}{y}$

5 solved Pipes and Cisterns questions

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01
One pipe fills a tank in 6 hours and another in 4 hours. If both are opened together, how long will the tank take to fill?
5 steps · reveal one at a time
1. Step 1: Find individual rates
  First, let's figure out how much of the tank each pipe fills in one hour. The first pipe fills the tank in 6 hours, so its rate is $\frac{1}{6}$ of the tank per hour. The second pipe fills the tank in 4 hours, so its rate is $\frac{1}{4}$ of the tank per hour.
2. Step 2: Calculate combined rate
  When both pipes are open, their rates add up. So, the combined filling rate is $\frac{1}{6} + \frac{1}{4}$ . To add these fractions, find a common denominator, which is 12. This gives us $\frac{2}{12} + \frac{3}{12} = \frac{5}{12}$ of the tank per hour.
3. Step 3: Determine total time
  If the pipes together fill $\frac{5}{12}$ of the tank in one hour, then the time it takes to fill the entire tank (1 whole tank) is the reciprocal of this rate. So, the total time is $\frac{12}{5}$ hours.
4. Step 4: Convert to hours and minutes
  The time $\frac{12}{5}$ hours can be written as $2\frac{2}{5}$ hours. To convert the fraction of an hour to minutes, multiply by 60: $\frac{2}{5} \times 60 = 24$ minutes. So, the total time is 2 hours and 24 minutes.
5. Step 5: The answer
  So the answer is (B) 2 h 24 min, as the combined rate of filling results in the tank being full in exactly that amount of time.
02
A pipe fills a tank in 5 hours, but a leak at the bottom empties a full tank in 20 hours. With both working, how long will an empty tank take to fill?
5 steps · reveal one at a time
1. Step 1: Find filling rate
  First, let's find the rate at which the pipe fills the tank. If it fills the tank in $5$ hours, its filling rate is $1/5$ of the tank per hour.
2. Step 2: Determine net rate
  The leak empties the tank in $20$ hours, so its emptying rate is $1/20$ per hour. To find the net rate at which the tank fills, we subtract the leak's rate from the pipe's rate: $1/5 - 1/20$ .
3. Step 3: Calculate total time
  To subtract, find a common denominator: $4/20 - 1/20 = 3/20$ . So, the tank fills at a net rate of $3/20$ per hour. To find the total time, we take the reciprocal: $1 \div (3/20) = 20/3$ hours.
4. Step 4: Convert to minutes
  The time is $20/3$ hours, which is $6$ and $2/3$ hours. To convert the fraction of an hour to minutes, multiply by $60$ : $(2/3) \times 60 = 40$ minutes. So, the total time is $6$ hours and $40$ minutes.
5. Step 5: The answer
  So the answer is (B) 6 h 40 min, as this is the total time it takes for the tank to fill with both the pipe and the leak working.
03
Two pipes fill a tank in 20 and 30 minutes. Both are opened together, but after 5 minutes the first pipe is closed. How much longer does the second pipe take to fill the rest?
5 steps · reveal one at a time
1. Step 1: Calculate combined work
  First, let's find out how much of the tank both pipes fill together in the initial 5 minutes. The first pipe fills $1/20$ of the tank per minute, and the second fills $1/30$ . So, in 5 minutes, they fill $5/20 + 5/30$ of the tank.
2. Step 2: Simplify filled portion
  To simplify $5/20 + 5/30$ , we can find a common denominator. This becomes $1/4 + 1/6$ . With a common denominator of 12, this is $3/12 + 2/12 = 5/12$ . So, $5/12$ of the tank is filled.
3. Step 3: Find remaining work
  If $5/12$ of the tank is filled, the remaining portion is $1 - 5/12 = 7/12$ . This is the amount the second pipe needs to fill alone after the first pipe is closed.
4. Step 4: Calculate time needed
  The second pipe fills $1/30$ of the tank per minute. To find how long it takes to fill the remaining $7/12$ , we divide the remaining work by its rate: $(7/12) \div (1/30)$ . This simplifies to $(7/12) \times 30$ minutes.
5. Step 5: The answer
  So the answer is (C) 17.5 min. Calculating $(7/12) \times 30$ gives $7 \times (30/12) = 7 \times (5/2) = 35/2 = 17.5$ minutes.
04
Pipe A fills a tank in 12 minutes and pipe B empties it in 18 minutes. If both are opened when the tank is empty, how long will it take to fill?
4 steps · reveal one at a time
1. Step 1: Understand individual rates
  First, let's find the rate at which each pipe operates. Pipe A fills the tank, so its rate is positive: $1/12$ of the tank per minute. Pipe B empties the tank, so its rate is negative: $-1/18$ of the tank per minute.
2. Step 2: Calculate net filling rate
  When both pipes are open, the net rate at which the tank fills is the sum of their individual rates. Since one fills and one empties, we subtract the emptying rate from the filling rate. Net rate $= \text{Rate A} + \text{Rate B}$ .
3. Step 3: Perform the calculation
  Substitute the rates into the equation: Net rate $= \frac{1}{12} - \frac{1}{18}$ . To subtract these fractions, find a common denominator, which is 36. So, Net rate $= \frac{3}{36} - \frac{2}{36} = \frac{1}{36}$ of the tank per minute.
4. Step 4: The answer
  So the answer is (D) 36 min. If the tank fills at a net rate of $1/36$ of the tank per minute, it will take the reciprocal of this rate to fill the entire tank, which is 36 minutes.
05
Pipes A and B fill a tank in 24 and 32 minutes. Both are opened together. After how many minutes should B be closed so that the tank is full in exactly 18 minutes?
3 steps · reveal one at a time
1. Step 1: Pipe A's contribution
  Pipe A fills the tank in 24 minutes. Since it runs for the entire 18 minutes, it fills a fraction of the tank equal to $\frac{18}{24}$ . This simplifies to $\frac{3}{4}$ of the tank.
2. Step 2: Pipe B's task
  If Pipe A fills $\frac{3}{4}$ of the tank, the remaining portion that Pipe B must fill is $1 - \frac{3}{4} = \frac{1}{4}$ . Pipe B fills the tank completely in 32 minutes.
3. Step 3: The answer
  So the answer is (B) 8 min. To fill the remaining $\frac{1}{4}$ of the tank, Pipe B needs $\frac{1}{4} \times 32 = 8$ minutes. Thus, Pipe B is closed after 8 minutes.

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