Solving the Pipe and Tank Filling Problem

This problem involves calculating the time taken to fill a tank when multiple pipes are working simultaneously, some filling and some emptying. To solve this, we first determine the work done by each pipe in one hour (their rate), and then combine these rates to find the net work done per hour when all pipes are open.

Understanding the Rate of Each Pipe

The rate of a pipe is the fraction of the tank it can fill or empty in one hour. If a pipe can fill a tank in \(T\) hours, its filling rate is \(1/T\) tank per hour. If a pipe can empty a tank in \(T\) hours, its emptying rate is \(1/T\) tank per hour (and this rate is subtracted from the filling rates).

• Pipe P is an inlet pipe and fills the tank in 10 hours.
• Rate of Pipe P in one hour = \( \frac{1}{10} \) of the tank.

• Pipe Q is an inlet pipe and fills the tank in 12 hours.
• Rate of Pipe Q in one hour = \( \frac{1}{12} \) of the tank.

• Pipe R is an outlet pipe and empties the tank in 30 hours.
• Rate of Pipe R in one hour = \( \frac{1}{30} \) of the tank. Since it's an outlet, this rate is negative when considering the net filling.

Calculating the Combined Filling Rate

When all three pipes P, Q, and R are opened simultaneously, the net amount of tank filled in one hour is the sum of the filling rates minus the emptying rate.

Combined Rate per hour = (Rate of P) + (Rate of Q) - (Rate of R)

Combined Rate per hour = \( \frac{1}{10} + \frac{1}{12} - \frac{1}{30} \)

To add and subtract these fractions, we find a common denominator for 10, 12, and 30. The least common multiple (LCM) of 10, 12, and 30 is 60.

Combined Rate per hour = \( \frac{6}{60} + \frac{5}{60} - \frac{2}{60} \)

Combined Rate per hour = \( \frac{6 + 5 - 2}{60} \)

Combined Rate per hour = \( \frac{11 - 2}{60} \)

Combined Rate per hour = \( \frac{9}{60} \)

This fraction can be simplified by dividing the numerator and denominator by their greatest common divisor, which is 3.

Combined Rate per hour = \( \frac{9 \div 3}{60 \div 3} = \frac{3}{20} \)

So, when all pipes are open, \( \frac{3}{20} \) of the tank is filled in one hour.

Calculating the Total Time to Fill the Tank

If \( \frac{3}{20} \) of the tank is filled in 1 hour, the total time taken to fill the entire tank (which is 1 whole) is the reciprocal of the combined rate.

Total Time = \( \frac{1}{\text{Combined Rate per hour}} \)

Total Time = \( \frac{1}{\frac{3}{20}} \)

Total Time = \( \frac{20}{3} \) hours

Converting Hours to Hours and Minutes

The total time is \( \frac{20}{3} \) hours. We can convert this into hours and minutes.

\( \frac{20}{3} \) hours = \( 6 \frac{2}{3} \) hours

This means 6 full hours and \( \frac{2}{3} \) of an hour.

To convert the fraction of an hour to minutes, we multiply by 60 (since there are 60 minutes in an hour).

\( \frac{2}{3} \) hours = \( \frac{2}{3} \times 60 \) minutes

\( \frac{2}{3} \times 60 = 2 \times \frac{60}{3} = 2 \times 20 = 40 \) minutes

So, the total time taken to fill the tank is 6 hours and 40 minutes.

Therefore, if all the pipes are opened simultaneously, it will take 6 hours and 40 minutes to fill the tank completely.

Combined Rate = \( \frac{1}{10} + \frac{1}{12} - \frac{1}{30} = \frac{6+5-2}{60} = \frac{9}{60} = \frac{3}{20} \) tank/hour.

Time to fill = \( \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{3}{20}} = \frac{20}{3} \) hours.

\( \frac{20}{3} \) hours = 6 hours and \( \frac{2}{3} \times 60 \) minutes = 6 hours and 40 minutes.

Revision Table: Pipe and Tank Concepts

Additional Information: Solving Work and Time Problems

Problems involving pipes filling or emptying tanks are classic examples of 'work and time' problems. The core idea is to figure out how much of the total 'work' (filling the tank) is done in a unit of time (usually an hour or a minute).

• LCM Method: An alternative approach is to assume the tank capacity is the LCM of the times taken by individual pipes. In this case, LCM(10, 12, 30) = 60 units. Then, find the work done by each pipe per hour (efficiency):
  • P fills \( \frac{60}{10} = 6 \) units/hour.
  • Q fills \( \frac{60}{12} = 5 \) units/hour.
  • R empties \( \frac{60}{30} = 2 \) units/hour.
  Net filling per hour = \( 6 + 5 - 2 = 9 \) units/hour. Total time = \( \frac{\text{Total Capacity}}{\text{Net Rate}} = \frac{60}{9} = \frac{20}{3} \) hours, which is 6 hours 40 minutes. This method often simplifies calculations by avoiding fractions initially.
• Different Scenarios: Problems can become more complex if pipes are opened or closed at different times, or if the tank is partially filled initially. The rate concept remains fundamental. You calculate the work done in specific time intervals and adjust the remaining work accordingly.
• Consistency of Units: Always ensure that the time units (hours, minutes) are consistent throughout the calculation. Convert everything to the same unit before calculating rates.