The problem describes a scenario involving an inlet pipe filling a tank and an outlet pipe draining it, with both pipes working alternately for one hour each, starting with the inlet pipe. We need to find the total time taken to fill the tank.

Calculating Pipe Rates

First, let's determine the filling rate of the inlet pipe and the draining rate of the outlet pipe.

• The inlet pipe fills the tank in \(4\frac{1}{2}\) hours.
  • Convert the mixed number to an improper fraction: \(4\frac{1}{2} = \frac{(4 \times 2) + 1}{2} = \frac{8+1}{2} = \frac{9}{2}\) hours.
  • The rate of the inlet pipe is the reciprocal of the time taken: Rate\(_\text{inlet}\) = \(1 \div \frac{9}{2} = \frac{2}{9}\) tank per hour.
• The outlet pipe drains the tank in \(7\frac{1}{5}\) hours.
  • Convert the mixed number to an improper fraction: \(7\frac{1}{5} = \frac{(7 \times 5) + 1}{5} = \frac{35+1}{5} = \frac{36}{5}\) hours.
  • The rate of the outlet pipe is the reciprocal of the time taken: Rate\(_\text{outlet}\) = \(1 \div \frac{36}{5} = \frac{5}{36}\) tank per hour. Since this is a draining pipe, its effect is negative.

Work Done in One Cycle

The pipes work alternately for one hour each, starting with the inlet pipe. One cycle consists of 1 hour of the inlet pipe working followed by 1 hour of the outlet pipe working. A full cycle takes 2 hours.

In the first hour (inlet pipe): Amount filled = \(1 \times \text{Rate}_\text{inlet} = 1 \times \frac{2}{9} = \frac{2}{9}\) tank.

In the second hour (outlet pipe): Amount drained = \(1 \times \text{Rate}_\text{outlet} = 1 \times \frac{5}{36} = \frac{5}{36}\) tank.

Net amount filled in one 2-hour cycle = Amount filled - Amount drained \( = \frac{2}{9} - \frac{5}{36} \).

To subtract these fractions, we find a common denominator, which is 36.

\(\frac{2}{9} = \frac{2 \times 4}{9 \times 4} = \frac{8}{36}\)

Net amount filled in one cycle = \(\frac{8}{36} - \frac{5}{36} = \frac{8-5}{36} = \frac{3}{36} = \frac{1}{12}\) tank.

So, in every 2-hour cycle, \(\frac{1}{12}\) of the tank is filled.

Calculating Time to Fill the Tank

The pipes work alternately until the tank is completely filled. Since the inlet pipe is the one filling the tank, it will be the pipe that finishes the job in its turn. The outlet pipe will not drain a full tank.

We need to find the number of full 2-hour cycles that occur before the tank is almost full, such that the remaining amount can be filled by the inlet pipe in its turn.

Let's find how many cycles fill the tank to a point where the remaining capacity is less than or equal to the amount the inlet pipe can fill in one hour (\(\frac{2}{9}\)).

Let N be the number of full cycles. After N cycles (2N hours), the amount filled is \(N \times \frac{1}{12} = \frac{N}{12}\).

Consider the amount filled after a certain number of cycles:

• After 9 cycles (18 hours): Amount filled = \(9 \times \frac{1}{12} = \frac{9}{12} = \frac{3}{4}\). Remaining capacity = \(1 - \frac{3}{4} = \frac{1}{4}\). In the 19th hour, the inlet pipe works. The inlet pipe fills \(\frac{2}{9}\) in one hour. Since \(\frac{1}{4} > \frac{2}{9}\) (because \(\frac{9}{36} > \frac{8}{36}\)), the inlet pipe cannot fill the remaining \(\frac{1}{4}\) in one hour. So, more than 9 cycles are needed before the final filling stage.
• After 10 cycles (20 hours): Amount filled = \(10 \times \frac{1}{12} = \frac{10}{12} = \frac{5}{6}\). Remaining capacity = \(1 - \frac{5}{6} = \frac{1}{6}\). In the 21st hour, the inlet pipe works. The inlet pipe fills \(\frac{2}{9}\) in one hour. Since \(\frac{1}{6} < \frac{2}{9}\) (because \(\frac{3}{18} < \frac{4}{18}\)), the inlet pipe can fill the remaining \(\frac{1}{6}\) in less than one hour. This means that after 10 full cycles, the next turn of the inlet pipe will finish filling the tank.

So, 10 full cycles occur, taking \(10 \times 2 = 20\) hours. After these 20 hours, the tank is \(\frac{5}{6}\) full.

The remaining amount to be filled is \(1 - \frac{5}{6} = \frac{1}{6}\) of the tank.

This remaining \(\frac{1}{6}\) of the tank is filled by the inlet pipe in the next turn (which is the 21st hour).

Time taken by the inlet pipe to fill the remaining \(\frac{1}{6}\) tank = \(\frac{\text{Remaining capacity}}{\text{Rate}_\text{inlet}} = \frac{1/6}{2/9}\) hours.

Time taken = \(\frac{1}{6} \div \frac{2}{9} = \frac{1}{6} \times \frac{9}{2} = \frac{9}{12} = \frac{3}{4}\) hours.

The total time taken to fill the tank is the time for 10 full cycles plus the time taken by the inlet pipe to fill the remaining amount.

Total time = 20 hours + \(\frac{3}{4}\) hours = \(20\frac{3}{4}\) hours.

Therefore, the tank will be completely filled in \(20\frac{3}{4}\) hours.

Revision Table: Pipe and Tank Concepts

Understanding the basic concepts is key to solving pipe and tank problems.

• Rate of work: If a pipe fills or drains a tank in 'T' hours, its rate is \(1/T\) of the tank per hour.
• Inlet pipe: Has a positive work rate (fills the tank).
• Outlet pipe: Has a negative work rate (drains the tank).
• Combined rate: When multiple pipes work together, their rates are added (inlet rates are positive, outlet rates are negative).
• Alternating work: When pipes work alternately, calculate the net work done in one cycle (one turn of each pipe) and the duration of one cycle.
• Finishing pipe: In alternating problems involving filling and draining, the filling pipe usually finishes the job. Calculate cycles up to a point where the remaining work can be completed by the filling pipe in its turn.

Additional Information: Time and Work Analogy

Pipe and tank problems are similar to time and work problems. Filling a tank is like completing a job, and the pipes are like workers. An inlet pipe is like a worker doing positive work, and an outlet pipe is like a worker doing negative work (undoing the work).

• Rate = Work / Time.
• Work = Rate \(\times\) Time.
• Time = Work / Rate.

In this problem, the 'work' is filling 1 full tank. The 'rates' are the fractions of the tank filled or drained per hour.

When pipes work alternately, we look at the progress made over a full cycle (inlet + outlet) to find the combined effect, as the net change repeats with each cycle.