Solving the Tank Filling Problem with Inlet and Outlet Pipes

This problem involves understanding the concept of work rates, specifically how quickly pipes can fill or drain a tank. The rate is usually expressed as the fraction of the tank filled or drained per unit of time (in this case, per hour).

Understanding Individual Pipe Rates

• An inlet pipe fills the tank in 140 hours. Its filling rate is $\frac{1}{140}$ of the tank per hour.
• An outlet pipe drains the tank in 63 hours. Its draining rate is $\frac{1}{63}$ of the tank per hour.

Calculating Combined Rates

When multiple pipes of the same type are working, their rates are added.

• There are 8 inlet pipes. Their combined filling rate is $8 \times \frac{1}{140} = \frac{8}{140}$. Simplifying this fraction: $\frac{8}{140} = \frac{2 \times 4}{35 \times 4} = \frac{2}{35}$ of the tank per hour.
• There are $y$ outlet pipes. Their combined draining rate is $y \times \frac{1}{63} = \frac{y}{63}$ of the tank per hour.

Determining the Net Rate

When inlet pipes (filling) and outlet pipes (draining) work simultaneously, the net rate is the difference between the filling rate and the draining rate. Since the tank gets filled, the combined filling rate must be greater than the combined draining rate.

Net filling rate = (Combined inlet rate) - (Combined outlet rate)

Net filling rate = $\frac{2}{35} - \frac{y}{63}$ of the tank per hour.

Relating Net Rate to Total Filling Time

We are given that the tank is completely filled in 105 hours when 8 inlet pipes and $y$ outlet pipes are open simultaneously. This means the net filling rate is $\frac{1}{105}$ of the tank per hour.

Setting up the Equation

We can now set the net filling rate equal to the rate derived from the total filling time:

$\frac{2}{35} - \frac{y}{63} = \frac{1}{105}$

Solving for y

To solve for $y$, we need to clear the denominators. We find the Least Common Multiple (LCM) of 35, 63, and 105.

• $35 = 5 \times 7$
• $63 = 9 \times 7 = 3^2 \times 7$
• $105 = 3 \times 35 = 3 \times 5 \times 7$
• LCM(35, 63, 105) = $3^2 \times 5 \times 7 = 9 \times 5 \times 7 = 315$.

Multiply the entire equation by 315:

$315 \times \left(\frac{2}{35} - \frac{y}{63}\right) = 315 \times \frac{1}{105}$

Distribute the multiplication:

$315 \times \frac{2}{35} - 315 \times \frac{y}{63} = 315 \times \frac{1}{105}$

Perform the divisions:

• $\frac{315}{35} = 9$
• $\frac{315}{63} = 5$
• $\frac{315}{105} = 3$

Substitute these values back into the equation:

$9 \times 2 - 5 \times y = 3 \times 1$

$18 - 5y = 3$

Now, isolate the term with $y$. Subtract 3 from both sides:

$18 - 3 = 5y$

$15 = 5y$

Finally, divide by 5 to find $y$:

$y = \frac{15}{5}$

$y = 3$

So, the value of $y$ is 3.

Checking the Answer

If $y=3$, the combined outlet rate is $3 \times \frac{1}{63} = \frac{3}{63} = \frac{1}{21}$ per hour.

The combined inlet rate is $\frac{2}{35}$ per hour.

Net rate = $\frac{2}{35} - \frac{1}{21}$.

LCM(35, 21) = $105$.

Net rate = $\frac{2 \times 3}{35 \times 3} - \frac{1 \times 5}{21 \times 5} = \frac{6}{105} - \frac{5}{105} = \frac{1}{105}$ per hour.

A net rate of $\frac{1}{105}$ per hour means the tank fills in 105 hours, which matches the problem statement. Thus, the value of $y=3$ is correct.

Net Rate = Combined Inlet Rate - Combined Outlet Rate = $\frac{2}{35} - \frac{y}{63}$

Given Filling Time = 105 hours, so Net Rate = $\frac{1}{105}$

Equation: $\frac{2}{35} - \frac{y}{63} = \frac{1}{105}$

Solution: $y=3$

Revision Table: Tank Filling Problem

Additional Information: Work and Time Problems

Work and time problems often involve calculating how long it takes to complete a task (like filling a tank, building a wall, etc.) when individuals or entities work at different rates, sometimes together and sometimes against each other. Key principles include:

• Work Rate: If someone can do a piece of work in $T$ days, their one day's work (rate) is $1/T$.
• Total Work: The total work is usually considered as 1 unit.
• Combined Work: If multiple people or pipes work together, their individual rates are added to find the combined rate. If one is working against the other (like a draining pipe vs. a filling pipe), the rates are subtracted to find the net rate.
• Work Done: Work Done = Rate $\times$ Time.

These principles are fundamental to solving problems involving pipes, people, or machines working to complete a task.