Understanding the Tank Filling Problem

This problem involves calculating the amount of work done by different taps filling a tank over specific time periods. We are given the time each tap takes to fill the tank individually and how the taps are opened and closed at different stages. The goal is to find the percentage of the total work (filling the tank) that was done by tap A alone.

Calculating Individual Tap Work Rates

The work rate of a tap is the reciprocal of the time it takes to fill the tank. If a tap fills the tank in \(t\) hours, its rate is \(1/t\) of the tank per hour.

• Tap A fills the tank in 10 hours.
• Tap B fills the tank in 20 hours.
• Tap C fills the tank in 25 hours.

Their individual work rates per hour are:

• Rate of A = \(\frac{1}{10}\) tank/hour
• Rate of B = \(\frac{1}{20}\) tank/hour
• Rate of C = \(\frac{1}{25}\) tank/hour

Analyzing the Stages of Tank Filling

The filling process occurs in three distinct stages:

1. Stage 1: Taps A, B, and C are open simultaneously for the first 2 hours.
2. Stage 2: Tap C is closed, and taps A and B remain open from hour 2 to hour 4 (for 2 hours).
3. Stage 3: Tap B is also closed, and tap A fills the remaining tank from hour 4 onwards.

Calculating Work Done in Each Stage

Let's calculate the fraction of the tank filled in each stage.

Stage 1: Taps A, B, and C open (0 to 2 hours)

Combined rate of A, B, and C = Rate of A + Rate of B + Rate of C

Combined rate = \(\frac{1}{10} + \frac{1}{20} + \frac{1}{25}\)

To add these fractions, find a common denominator, which is 100.

Combined rate = \(\frac{1 \times 10}{10 \times 10} + \frac{1 \times 5}{20 \times 5} + \frac{1 \times 4}{25 \times 4} = \frac{10}{100} + \frac{5}{100} + \frac{4}{100} = \frac{10 + 5 + 4}{100} = \frac{19}{100}\) tank/hour.

Work done in Stage 1 (2 hours) = Combined rate \(\times\) Time = \(\frac{19}{100} \times 2 = \frac{38}{100} = \frac{19}{50}\) of the tank.

Stage 2: Taps A and B open (2 to 4 hours)

This stage lasts for \(4 - 2 = 2\) hours.

Combined rate of A and B = Rate of A + Rate of B

Combined rate = \(\frac{1}{10} + \frac{1}{20}\)

Common denominator is 20.

Combined rate = \(\frac{1 \times 2}{10 \times 2} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{2 + 1}{20} = \frac{3}{20}\) tank/hour.

Work done in Stage 2 (2 hours) = Combined rate \(\times\) Time = \(\frac{3}{20} \times 2 = \frac{6}{20} = \frac{3}{10}\) of the tank.

Total Work Done in the first 4 hours

Total work done after Stage 1 and Stage 2 = Work in Stage 1 + Work in Stage 2

Total work = \(\frac{19}{50} + \frac{3}{10}\)

Common denominator is 50.

Total work = \(\frac{19}{50} + \frac{3 \times 5}{10 \times 5} = \frac{19}{50} + \frac{15}{50} = \frac{19 + 15}{50} = \frac{34}{50} = \frac{17}{25}\) of the tank.

Stage 3: Tap A open alone (from 4 hours onwards)

The remaining fraction of the tank to be filled is \(1 - \text{Total work done in first 4 hours}\).

Remaining work = \(1 - \frac{17}{25} = \frac{25}{25} - \frac{17}{25} = \frac{25 - 17}{25} = \frac{8}{25}\) of the tank.

Tap A fills the remaining tank alone at its rate of \(\frac{1}{10}\) tank/hour.

Time taken by A to fill the remaining work = \(\frac{\text{Remaining work}}{\text{Rate of A}}\) = \(\frac{8/25}{1/10} = \frac{8}{25} \times \frac{10}{1} = \frac{8 \times 10}{25} = \frac{80}{25}\) hours.

This time can be simplified: \(\frac{80}{25} = \frac{16 \times 5}{5 \times 5} = \frac{16}{5}\) hours.

Calculating Total Work Done by Tap A

Tap A was open during all three stages:

• Stage 1 (0 to 2 hours): A worked for 2 hours.
• Stage 2 (2 to 4 hours): A worked for 2 hours.
• Stage 3 (from 4 hours onwards): A worked for \(\frac{16}{5}\) hours.

Total work done by A = (Rate of A \(\times\) Time in Stage 1) + (Rate of A \(\times\) Time in Stage 2) + (Rate of A \(\times\) Time in Stage 3)

Work by A in Stage 1 = \(\frac{1}{10} \times 2 = \frac{2}{10} = \frac{1}{5}\) of the tank.

Work by A in Stage 2 = \(\frac{1}{10} \times 2 = \frac{2}{10} = \frac{1}{5}\) of the tank.

Work by A in Stage 3 = \(\frac{1}{10} \times \frac{16}{5} = \frac{16}{50} = \frac{8}{25}\) of the tank.

Total work done by A = \(\frac{1}{5} + \frac{1}{5} + \frac{8}{25}\)

To add these fractions, find a common denominator, which is 25.

Total work by A = \(\frac{1 \times 5}{5 \times 5} + \frac{1 \times 5}{5 \times 5} + \frac{8}{25} = \frac{5}{25} + \frac{5}{25} + \frac{8}{25} = \frac{5 + 5 + 8}{25} = \frac{18}{25}\) of the tank.

Finding the Percentage of Work Done by Tap A

To express the fraction of work done by A as a percentage, multiply the fraction by 100%.

Percentage of work by A = \(\left(\frac{18}{25}\right) \times 100\%\)

Percentage of work by A = \(18 \times \frac{100}{25}\%\)

Percentage of work by A = \(18 \times 4\%\)

Percentage of work by A = \(72\%\)

Therefore, tap A itself did 72% of the total work to fill the tank.

Revision Table: Key Concepts for Tank and Tap Problems

Additional Information: Time and Work Principles

Tank and tap problems are a type of time and work problem. The basic principle is that the amount of work done is directly proportional to the rate of work and the time spent working (Work = Rate × Time). In these problems:

• The 'work' is filling (or emptying) the tank, often represented as 1 unit.
• 'Rate' is how much of the tank is filled (or emptied) per hour or minute.
• Filling is considered positive work, and emptying (by a leak or outlet tap) is considered negative work.
• When taps work together, their rates are combined. Filling rates are added, and emptying rates are subtracted from filling rates.

Understanding these fundamental principles helps break down complex problems with multiple taps and changing conditions into manageable steps, calculating the work done in each phase and then summing it up or finding the remaining work.