Understanding the Pipe Filling Problem

This problem involves two pipes, A and B, filling a tank at different rates and operating alternately. We need to figure out the total time it takes for the tank to be completely filled under these conditions, starting with pipe B.

Calculating Individual Pipe Rates

First, let's determine the rate at which each pipe fills the tank per hour. The rate is the reciprocal of the time taken to fill the entire tank.

• Pipe A fills the tank in 10 hours. So, Pipe A's rate is \(\frac{1}{10}\) of the tank per hour.
• Pipe B fills the tank in 16 hours. So, Pipe B's rate is \(\frac{1}{16}\) of the tank per hour.

Work Done in One Alternate Cycle

The pipes are opened alternately for 1 hour each, with pipe B starting first. This means the sequence is B for 1 hour, then A for 1 hour, then B for 1 hour, and so on. A full cycle consists of Pipe B working for 1 hour and Pipe A working for the next 1 hour, totaling 2 hours.

In one cycle (2 hours), the amount of tank filled is the sum of the work done by B in 1 hour and A in 1 hour.

Work done in one cycle = (Rate of B \(\times\) 1 hour) + (Rate of A \(\times\) 1 hour)

Work done in one cycle = \(\frac{1}{16} + \frac{1}{10}\)

To add these fractions, we find a common denominator for 16 and 10. The least common multiple (LCM) of 16 and 10 is 80.

• \(\frac{1}{16} = \frac{1 \times 5}{16 \times 5} = \frac{5}{80}\)
• \(\frac{1}{10} = \frac{1 \times 8}{10 \times 8} = \frac{8}{80}\)

Work done in one cycle = \(\frac{5}{80} + \frac{8}{80} = \frac{13}{80}\) of the tank.

So, in every 2-hour cycle, \(\frac{13}{80}\) of the tank is filled.

Calculating Number of Full Cycles

The total work required is to fill 1 full tank (which is equivalent to \(\frac{80}{80}\)). We want to find out how many full 2-hour cycles are needed to fill a significant portion of the tank without exceeding the total capacity.

Let's see how many times the work done in one cycle (\(\frac{13}{80}\)) fits into the total work (1 or \(\frac{80}{80}\)).

Number of cycles \(\approx \frac{\text{Total Work}}{\text{Work per Cycle}} = \frac{1}{\frac{13}{80}} = \frac{80}{13}\)

\(\frac{80}{13} \approx 6.15\). This means there will be 6 full cycles.

Work done in 6 cycles = \(6 \times \frac{13}{80} = \frac{78}{80}\) of the tank.

Time taken for 6 cycles = \(6 \text{ cycles} \times 2 \text{ hours/cycle} = 12 \text{ hours}\).

Calculating Remaining Work

After 6 full cycles (12 hours), a portion of the tank is still empty. The remaining work is:

Remaining Work = Total Work - Work Done in 6 Cycles

Remaining Work = \(1 - \frac{78}{80} = \frac{80}{80} - \frac{78}{80} = \frac{2}{80} = \frac{1}{40}\) of the tank.

Completing the Remaining Work

After 6 full cycles (12 hours), the last pipe to operate in the 6th cycle was A (since B started, cycle 1 ends with A, cycle 2 with A, ..., cycle 6 with A). The next pipe to open is B.

Pipe B needs to fill the remaining \(\frac{1}{40}\) of the tank. Pipe B's rate is \(\frac{1}{16}\) of the tank per hour.

Time taken by B to fill remaining work = \(\frac{\text{Remaining Work}}{\text{Rate of B}}\)

Time taken by B = \(\frac{\frac{1}{40}}{\frac{1}{16}} = \frac{1}{40} \times \frac{16}{1} = \frac{16}{40}\) hours.

Simplifying the fraction \(\frac{16}{40}\) by dividing both numerator and denominator by their greatest common divisor, 8:

\(\frac{16 \div 8}{40 \div 8} = \frac{2}{5}\) hours.

Total Time to Fill the Tank

The total time taken to fill the tank is the sum of the time taken for the full cycles and the time taken for the remaining work.

Total Time = Time for 6 cycles + Time for remaining work by B

Total Time = \(12 \text{ hours} + \frac{2}{5} \text{ hours}\)

Total Time = \(12\frac{2}{5}\) hours.

Revision Table: Pipes and Cisterns Concepts

Understanding key terms and concepts is crucial for solving pipes and cisterns problems.

• Work Rate: If a pipe fills a tank in 't' hours, its work rate is \(1/t\) of the tank filled per hour.
• Total Work: Filling the entire tank is considered 1 unit of work.
• Work Done: Work Done = Rate \(\times\) Time.
• Time Taken: Time Taken = Work Done / Rate.
• Alternate Working: When pipes work alternately, calculate the work done in one complete cycle (usually consisting of one hour for each pipe involved).

Additional Information: Alternate Working Problems

Problems involving alternate working pipes require careful tracking of who works during which hour and how much work is completed in each cycle. It's important to determine how many full cycles can occur before the remaining work is small enough to be completed by just one pipe. Always identify which pipe's turn it is to work on the remaining fraction of the job.

For instance, if pipes A and B work alternately starting with A, the first hour is A, the second is B, the third is A, and so on. If they start with B, the first hour is B, the second is A, the third is B, and so on. The pattern of who works last in a cycle depends on the number of pipes and the duration of the cycle, but in a simple two-pipe alternating system, the turn after an even number of hours belongs to the pipe that started second (Pipe A in this case, if B started first).