Solving the Pipe Filling Tank Alternately Problem

This problem involves two pipes, A and B, filling a tank at different rates and working alternately for one hour each. We need to find the total time taken to fill the tank.

Understanding Work Rate

The work rate of a pipe is the fraction of the tank it can fill in one hour. If a pipe can fill a tank in $T$ hours, its work rate is $1/T$ tank per hour.

• Pipe A fills the tank in 32 hours.
• Pipe B fills the tank in 28 hours.

So, their individual work rates are:

• Rate of A = $\frac{1}{32}$ tank/hour
• Rate of B = $\frac{1}{28}$ tank/hour

Work Done in One Cycle

The pipes work alternately for 1 hour each, starting with A. A complete cycle consists of Pipe A working for 1 hour and then Pipe B working for 1 hour. The duration of one cycle is $1 + 1 = 2$ hours.

In the first hour (starting with A), Pipe A fills $\frac{1}{32}$ of the tank.

In the second hour (Pipe B's turn), Pipe B fills $\frac{1}{28}$ of the tank.

Total work done in one cycle (2 hours) = Work done by A in 1 hour + Work done by B in 1 hour

Work done in one cycle = $\frac{1}{32} + \frac{1}{28}$

To add these fractions, we find a common denominator, which is the Least Common Multiple (LCM) of 32 and 28. LCM(32, 28) = 224.

$\frac{1}{32} = \frac{1 \times 7}{32 \times 7} = \frac{7}{224}$

$\frac{1}{28} = \frac{1 \times 8}{28 \times 8} = \frac{8}{224}$

Work done in one cycle = $\frac{7}{224} + \frac{8}{224} = \frac{7+8}{224} = \frac{15}{224}$

So, in every 2-hour cycle, $\frac{15}{224}$ of the tank is filled.

Calculating Full Cycles

We need to find how many full cycles are required to fill most of the tank without exceeding the total capacity (which is 1 whole tank). Let $N$ be the number of full cycles.

$N \times \frac{15}{224} < 1$

$N < \frac{224}{15}$

$\frac{224}{15} \approx 14.93$

We can complete 14 full cycles.

Time taken for 14 cycles = $14 \text{ cycles} \times 2 \text{ hours/cycle} = 28$ hours.

Amount of tank filled after 14 cycles = $14 \times \frac{15}{224} = \frac{210}{224}$

Handling the Remaining Work

After 28 hours, the fraction of the tank filled is $\frac{210}{224}$.

Remaining fraction to be filled = $1 - \frac{210}{224} = \frac{224 - 210}{224} = \frac{14}{224}$.

After 14 full cycles, it is Pipe A's turn to work (since A started). Pipe A works for the next hour (the 29th hour).

In 1 hour, Pipe A fills $\frac{1}{32}$ of the tank. We can express this with the common denominator: $\frac{1}{32} = \frac{7}{224}$.

Amount of tank filled by A in the 29th hour = $\frac{7}{224}$.

Total tank filled after 29 hours = Amount filled after 14 cycles + Amount filled by A in the 29th hour

$\text{Total filled} = \frac{210}{224} + \frac{7}{224} = \frac{217}{224}$

Remaining fraction to be filled after 29 hours = $1 - \frac{217}{224} = \frac{224 - 217}{224} = \frac{7}{224}$.

Now, it is Pipe B's turn to work (the 30th hour onwards if needed).

Pipe B fills $\frac{1}{28}$ of the tank in 1 hour. B's rate is $\frac{1}{28} = \frac{8}{224}$ tank per hour.

We need to fill $\frac{7}{224}$ of the tank. Since B's rate ($\frac{8}{224}$) is greater than the remaining amount ($\frac{7}{224}$), B will take less than 1 hour to finish the job.

Time taken by Pipe B to fill the remaining $\frac{7}{224}$ of the tank = $\frac{\text{Remaining work}}{\text{Rate of B}}$

$\text{Time by B} = \frac{\frac{7}{224}}{\frac{8}{224}} = \frac{7}{8}$ hours.

Total Time to Fill the Tank

Total time = Time for 14 full cycles + Time taken by A in the 29th hour + Time taken by B for the remaining work

$\text{Total time} = 28 \text{ hours} + 1 \text{ hour} + \frac{7}{8} \text{ hours}$

$\text{Total time} = 29 + \frac{7}{8} \text{ hours}$

$\frac{7}{8} = 0.875$

Total time = $29 + 0.875 = 29.875$ hours.

Rounding to the Nearest Integer

The question asks for the time in hours, rounded to the nearest integer. $29.875$ rounded to the nearest integer is 30.

Therefore, the tank will be filled in approximately 30 hours.

Revision Table: Key Concepts for Pipe and Cistern Problems

Additional Information: Alternate Work Problems

Problems involving pipes or people working alternately are common in time and work. The key steps are:

• Calculate the individual work rates.
• Identify the pattern or cycle of work.
• Calculate the total work done in one complete cycle and the time taken for one cycle.
• Determine how many full cycles are needed to complete most of the work without exceeding the total amount.
• Calculate the work done and time taken for these full cycles.
• Calculate the remaining work.
• Determine which worker/pipe starts after the full cycles and calculate the time they take to complete the remaining work (or part of it if it takes more than their turn time).
• Continue this process hour by hour if necessary until the tank is filled or work is complete.
• Sum up the time taken for full cycles and the subsequent hours/fractions of hours.

This method systematically breaks down the problem into manageable steps, especially useful when the total time is not an exact number of cycles.