An inlet pipe can fill an empty tank in 140 hours while an outlet pipe drains a completely-filled tank in 63 hours. If 8 inlet pipes and y outlet pipes are opened simultaneously, when the tank is empty, then the tank gets completely filled in 105 hours. Find the value of y.
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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.
| Pipe Type | Individual Rate (per hour) | Number of Pipes | Combined Rate (per hour) |
|---|---|---|---|
| Inlet | $\frac{1}{140}$ | 8 | $8 \times \frac{1}{140} = \frac{2}{35}$ |
| Outlet | $\frac{1}{63}$ | $y$ | $y \times \frac{1}{63} = \frac{y}{63}$ |
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
| Concept | Description | Formula/Relation |
|---|---|---|
| Individual Rate | Fraction of work done by one unit in unit time. | If task takes T hours, rate is $\frac{1}{T}$ per hour. |
| Combined Rate (Same Type) | Sum of individual rates for pipes of the same type. | Rate$_{total}$ = Rate$_1$ + Rate$_2$ + ... |
| Net Rate (Filling & Draining) | Difference between filling rate and draining rate. | Net Rate = Filling Rate - Draining Rate |
| Time & Rate | Time taken is the reciprocal of the rate. | Time = $\frac{1}{\text{Rate}}$ or Rate = $\frac{1}{\text{Time}}$ |
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.
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