Understanding the Pipes and Cistern Leak Problem

This problem involves two pipes filling a cistern and a leak emptying it. We are given the time taken by each pipe to fill the cistern individually. We are also told the extra time taken to fill the cistern due to a leak. We need to find the time it takes for the leak alone to empty 70% of the cistern.

Calculating the Filling Rates of the Pipes

First, let's determine the rate at which each pipe fills the cistern per hour.

• Pipe A takes \(12\frac{1}{2}\) hours to fill the cistern. This is equal to \(12.5\) hours or \( \frac{25}{2} \) hours.
• The rate of Pipe A is \( \frac{1}{\text{Time taken by A}} = \frac{1}{12.5} = \frac{1}{\frac{25}{2}} = \frac{2}{25} \) of the cistern per hour.
• Pipe B takes \(25\) hours to fill the cistern.
• The rate of Pipe B is \( \frac{1}{\text{Time taken by B}} = \frac{1}{25} \) of the cistern per hour.

Determining the Combined Filling Rate Without Leakage

When both pipes A and B are open together, their rates add up to fill the cistern. The combined rate without any leak is:

Combined Rate (A+B) = Rate of A + Rate of B

Combined Rate (A+B) = \( \frac{2}{25} + \frac{1}{25} = \frac{3}{25} \) of the cistern per hour.

Calculating the Time Taken to Fill Without Leakage

The time taken to fill the cistern when only pipes A and B are working is the reciprocal of their combined rate:

Time (A+B) = \( \frac{1}{\text{Combined Rate (A+B)}} = \frac{1}{\frac{3}{25}} = \frac{25}{3} \) hours.

Finding the Actual Time Taken With the Leak

Due to the leakage, the pipes took one hour 40 minutes more to fill the cistern.

• Let's convert 1 hour 40 minutes into hours: \( 1 \text{ hour } + 40 \text{ minutes } = 1 + \frac{40}{60} \text{ hours } = 1 + \frac{2}{3} \text{ hours } = \frac{3+2}{3} = \frac{5}{3} \) hours.
• The actual time taken with the leak is the normal time plus the extra time:
• Actual Time (A+B - Leak) = Time (A+B) + Extra Time
• Actual Time (A+B - Leak) = \( \frac{25}{3} \text{ hours } + \frac{5}{3} \text{ hours } = \frac{25+5}{3} = \frac{30}{3} = 10 \) hours.

So, with the leak, it took 10 hours to fill the entire cistern.

Calculating the Effective Filling Rate With the Leak

In 10 hours, the cistern was filled (1 whole cistern). The effective filling rate when pipes A and B are working and the leak is active is:

Effective Rate (A+B - Leak) = \( \frac{\text{Amount filled}}{\text{Time taken}} = \frac{1 \text{ cistern}}{10 \text{ hours}} = \frac{1}{10} \) of the cistern per hour.

Determining the Rate of the Leak

The effective rate with the leak is the combined rate of A and B minus the rate of the leak (L).

Effective Rate (A+B - Leak) = Combined Rate (A+B) - Rate of Leak (L)

\( \frac{1}{10} = \frac{3}{25} - \text{Rate of Leak (L)} \)

Rearranging the equation to find the Rate of Leak:

Rate of Leak (L) = \( \frac{3}{25} - \frac{1}{10} \)

To subtract these fractions, we find a common denominator, which is 50.

Rate of Leak (L) = \( \frac{3 \times 2}{25 \times 2} - \frac{1 \times 5}{10 \times 5} = \frac{6}{50} - \frac{5}{50} = \frac{1}{50} \) of the cistern per hour.

This means the leak alone can empty \( \frac{1}{50} \) of the cistern in one hour.

Calculating Time for Leak to Empty Full Cistern

If the leak rate is \( \frac{1}{50} \) cistern per hour, the time taken for the leak alone to empty the entire cistern (1 whole cistern) is the reciprocal of the leak rate:

Time for Leak to Empty Full Cistern = \( \frac{1}{\text{Rate of Leak (L)}} = \frac{1}{\frac{1}{50}} = 50 \) hours.

Calculating Time for Leak to Empty 70% of Cistern

We need to find the time taken for the leak alone to empty 70% of the cistern.

70% of the cistern is \( \frac{70}{100} = 0.7 \) of the total capacity.

Time to empty 70% = \( 0.7 \times \) Time to empty 100%

Time to empty 70% = \( 0.7 \times 50 \) hours

Time to empty 70% = \( \frac{7}{10} \times 50 = 7 \times 5 = 35 \) hours.

Therefore, the leak alone will empty 70% of the cistern in 35 hours.

Step-by-Step Calculation Summary

1. Pipe A Rate: \( \frac{1}{12.5} = \frac{2}{25} \)/hour.
2. Pipe B Rate: \( \frac{1}{25} \)/hour.
3. Combined A+B Rate (No Leak): \( \frac{2}{25} + \frac{1}{25} = \frac{3}{25} \)/hour.
4. Time for A+B (No Leak): \( \frac{1}{\frac{3}{25}} = \frac{25}{3} \) hours.
5. Extra Time due to Leak: 1 hour 40 minutes = \( \frac{5}{3} \) hours.
6. Actual Time (A+B with Leak): \( \frac{25}{3} + \frac{5}{3} = \frac{30}{3} = 10 \) hours.
7. Effective Rate (A+B with Leak): \( \frac{1}{10} \)/hour.
8. Rate of Leak: \( \frac{3}{25} - \frac{1}{10} = \frac{6}{50} - \frac{5}{50} = \frac{1}{50} \)/hour.
9. Time for Leak to Empty Full Cistern: \( \frac{1}{\frac{1}{50}} = 50 \) hours.
10. Time for Leak to Empty 70% of Cistern: \( 0.70 \times 50 = 35 \) hours.

Revision Table: Key Concepts in Pipes and Cisterns

Additional Information: Solving Time and Work Problems

Pipes and cistern problems are a type of time and work problem. The core idea is that the rate of work is inversely proportional to the time taken to complete the work. If a pipe fills a cistern in 'T' hours, its filling rate is \( \frac{1}{T} \) of the cistern per hour. Similarly, if a leak empties a cistern in 'T' hours, its emptying rate is \( \frac{1}{T} \) of the cistern per hour.

When multiple pipes or leaks are working simultaneously, their rates are combined:

• Pipes filling the cistern have positive rates.
• Leaks or pipes emptying the cistern have negative rates.
• The net rate is the sum of the individual rates, considering their signs.

Total work is usually considered as '1' unit (representing the full cistern). The formula connecting Work, Rate, and Time is:

Work = Rate \( \times \) Time

In this problem, the total work is filling 1 cistern. When calculating the time to empty a percentage of the cistern, the 'Work' becomes that percentage of 1 (e.g., 70% = 0.7).

Understanding these basic principles allows you to solve various problems involving pipes filling or emptying tanks, often complicated by different rates, start times, or leaks.