Understanding the Pipe and Tank Filling Problem

This question involves two pipes, Pipe J and Pipe K, filling a tank. They work alternately, meaning one works for an hour, then the other works for an hour, and this cycle repeats. We are given the time each pipe takes to fill the tank individually and the order in which they start.

Step 1: Calculate Individual Filling Rates

The rate at which a pipe fills a tank is the reciprocal of the time it takes to fill the tank.

• Pipe J fills the tank in 48 hours. So, its rate is $\frac{1}{48}$ of the tank per hour.
• Pipe K fills the tank in 72 hours. So, its rate is $\frac{1}{72}$ of the tank per hour.

Step 2: Calculate Work Done in One Alternate Cycle

The pipes work alternately for one hour each, starting with Pipe K. A complete cycle consists of Pipe K working for 1 hour and then Pipe J working for 1 hour. This cycle takes a total of 2 hours.

In the first hour (Pipe K): Work done by K = $\frac{1}{72}$

In the second hour (Pipe J): Work done by J = $\frac{1}{48}$

Total work done in one 2-hour cycle = Work by K + Work by J = $\frac{1}{72} + \frac{1}{48}$

To add these fractions, we find a common denominator. The least common multiple (LCM) of 72 and 48 is 144.

Work done in one cycle = $\frac{1 \times 2}{72 \times 2} + \frac{1 \times 3}{48 \times 3} = \frac{2}{144} + \frac{3}{144} = \frac{2+3}{144} = \frac{5}{144}$ of the tank.

Step 3: Determine Number of Full Cycles to Fill Most of the Tank

Each 2-hour cycle fills $\frac{5}{144}$ of the tank. We want to find how many full cycles are needed to fill as much of the tank as possible without exceeding the total capacity (which is 1).

Let $n$ be the number of cycles. Total work after $n$ cycles = $n \times \frac{5}{144}$.

We need $n \times \frac{5}{144} \le 1$.

$5n \le 144$

$n \le \frac{144}{5} = 28.8$

So, 28 full cycles will complete the majority of the work.

Step 4: Calculate Work Done After Full Cycles and Remaining Work

Work done after 28 cycles = $28 \times \frac{5}{144} = \frac{140}{144}$ of the tank.

Time taken for 28 cycles = $28 \times 2$ hours = 56 hours.

Remaining work = Total capacity - Work done = $1 - \frac{140}{144} = \frac{144}{144} - \frac{140}{144} = \frac{4}{144} = \frac{1}{36}$ of the tank.

Step 5: Calculate Time to Complete the Remaining Work

After 28 cycles (56 hours), $\frac{1}{36}$ of the tank is remaining. The next hour is the 57th hour, and it's the start of a new cycle, so Pipe K will be working.

Pipe K's rate is $\frac{1}{72}$ per hour.

In the 57th hour, Pipe K works and fills $\frac{1}{72}$ of the tank.

Work done after 57 hours = Work after 56 hours + Work by K in 57th hour = $\frac{140}{144} + \frac{1}{72} = \frac{140}{144} + \frac{2}{144} = \frac{142}{144}$ of the tank.

Remaining work after 57 hours = $1 - \frac{142}{144} = \frac{2}{144} = \frac{1}{72}$ of the tank.

Now, the 58th hour starts, and it's Pipe J's turn. Pipe J's rate is $\frac{1}{48}$ per hour.

We need to find how long Pipe J takes to fill the remaining $\frac{1}{72}$ of the tank.

Time taken by J = $\frac{\text{Remaining Work}}{\text{Rate of J}} = \frac{1/72}{1/48} = \frac{1}{72} \times \frac{48}{1} = \frac{48}{72}$ hours.

Simplifying the fraction $\frac{48}{72}$: Divide both numerator and denominator by their greatest common divisor, 24.

$\frac{48}{72} = \frac{48 \div 24}{72 \div 24} = \frac{2}{3}$ hours.

So, Pipe J will take $\frac{2}{3}$ of the 58th hour to finish filling the tank.

Step 6: Calculate Total Time

Total time taken = Time for 28 full cycles + Time for K in 57th hour + Time for J in 58th hour

Total time = 56 hours + 1 hour + $\frac{2}{3}$ hours = 57 + $\frac{2}{3}$ hours = $57\frac{2}{3}$ hours.

The tank will be full in $57\frac{2}{3}$ hours.