Solving the Pipe and Tank Filling Problem

This problem involves calculating the time taken to fill a tank when different pipes with varying rates of filling and emptying work together for specific durations.

Understanding the Individual Pipe Rates

First, let's determine the fraction of the tank each pipe can fill or empty in one hour. This is the rate of each pipe.

• Pipe S1 fills the tank in 15 hours. Its filling rate is $\frac{1}{15}$ of the tank per hour.
• Pipe S2 fills the tank in 20 hours. Its filling rate is $\frac{1}{20}$ of the tank per hour.
• Pipe S3 empties the tank in 40 hours. Its emptying rate is $\frac{1}{40}$ of the tank per hour. We consider this a negative rate when calculating combined work.

Work Done in the First 2 Hours

Initially, only pipes S1 and S2 are open for 2 hours. Let's find their combined filling rate and the amount of the tank filled in this period.

Combined rate of S1 and S2 = Rate of S1 + Rate of S2

Combined rate = $\frac{1}{15} + \frac{1}{20}$

To add these fractions, we find a common denominator, which is 60.

Combined rate = $\frac{4}{60} + \frac{3}{60} = \frac{4+3}{60} = \frac{7}{60}$ of the tank per hour.

Amount of tank filled in the first 2 hours = Combined rate $\times$ Time

Amount filled = $\frac{7}{60} \times 2 = \frac{14}{60} = \frac{7}{30}$ of the tank.

Calculating Remaining Capacity

The tank has a total capacity of 1 (representing a full tank). After the first 2 hours, $\frac{7}{30}$ of the tank is filled. The remaining capacity that needs to be filled is:

Remaining capacity = Total capacity - Amount filled

Remaining capacity = $1 - \frac{7}{30} = \frac{30}{30} - \frac{7}{30} = \frac{30-7}{30} = \frac{23}{30}$ of the tank.

Combined Rate When S3 is Opened

After 2 hours, pipe S3 is also opened. Now, pipes S1, S2 (filling), and S3 (emptying) are working together. Let's find their combined rate.

Combined rate of S1, S2, and S3 = Rate of S1 + Rate of S2 - Rate of S3

Combined rate = $\frac{1}{15} + \frac{1}{20} - \frac{1}{40}$

To combine these fractions, we find a common denominator, which is 120.

Combined rate = $\frac{8}{120} + \frac{6}{120} - \frac{3}{120} = \frac{8+6-3}{120} = \frac{14-3}{120} = \frac{11}{120}$ of the tank per hour.

This is the effective filling rate when all three pipes are open.

Time to Fill the Remaining Capacity

Now we need to find how much time it will take to fill the remaining $\frac{23}{30}$ of the tank at the combined rate of $\frac{11}{120}$ tank per hour.

Time = $\frac{\text{Remaining Capacity}}{\text{Combined Rate}}$

Time = $\frac{23/30}{11/120}$

To divide by a fraction, we multiply by its reciprocal.

Time = $\frac{23}{30} \times \frac{120}{11}$

We can simplify the expression: $\frac{120}{30} = 4$.

Time = $\frac{23}{1} \times \frac{4}{11} = \frac{23 \times 4}{11} = \frac{92}{11}$ hours.

This is the time taken to fill the tank completely after pipe S3 is opened.

Summary of Calculation

Therefore, the tank will be completely filled in $\frac{92}{11}$ hours after pipe S3 is opened.

Revision Table: Pipe and Cistern Concepts

Additional Information on Work Rate Problems

Pipe and cistern problems are a type of work rate problem. The key idea is to calculate the amount of work done per unit of time (the rate). Here are some related points:

• Work is often represented as '1 unit' (a full tank, completing a job).
• Rates are usually expressed as 'fraction of work per unit time'.
• If a pipe empties, its rate is considered negative for combined calculations.
• Pay close attention to the time intervals when different sets of pipes are working. Calculate the work done in each interval separately.
• Determine the remaining work (or capacity) after each interval.
• Calculate the combined rate for the next interval based on which units (pipes) are working.
• Finally, calculate the time needed to complete the remaining work at the new combined rate.