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When water is flowing into a certain pool at the rate of \(\mathrm{r}\) gallons per hour, it takes \(9\) hours to fill the pool. In terms of \(\mathrm{r}\), how many hours will it take to fill the pool if water is flowing into the pool at the rate of \(\mathrm{r} + 5\) gallons per hour?
Let's start by understanding what's happening here in everyday terms. Think of filling a swimming pool with a garden hose. If you have a thin hose that flows slowly, it takes longer to fill the pool. If you get a thicker hose that flows faster, the same pool fills up quicker.
In our problem, we have the same pool being filled in two different scenarios:
The key insight is that the pool doesn't change size between scenarios - it has the same capacity in both cases. We need to find how the faster flow rate affects the time needed.
Process Skill: TRANSLATE
Let's think about this step by step. If water flows into the pool at r gallons per hour for 9 hours, how much water does the pool hold?
Using simple logic: if you pour water at a certain rate for a certain time, the total amount is rate × time.
So the pool's total capacity = \(\mathrm{r} \text{ gallons per hour} \times 9 \text{ hours} = 9\mathrm{r} \text{ gallons}\)
This makes perfect sense - if water flows at 10 gallons per hour for 9 hours, the pool holds 90 gallons total.
Pool Capacity = \(9\mathrm{r}\) gallons
Now we know the pool holds exactly \(9\mathrm{r}\) gallons. In the second scenario, water flows at \((\mathrm{r} + 5)\) gallons per hour. Since it's the same pool, it still needs to hold \(9\mathrm{r}\) gallons to be completely full.
Let's call the new time 't' hours. Using the same logic as before:
Rate × Time = Total Capacity
\((\mathrm{r} + 5) \text{ gallons per hour} \times \mathrm{t} \text{ hours} = 9\mathrm{r} \text{ gallons}\)
This gives us the equation: \((\mathrm{r} + 5) \times \mathrm{t} = 9\mathrm{r}\)
Process Skill: INFER
Now we solve for t by isolating it on one side of the equation:
\((\mathrm{r} + 5) \times \mathrm{t} = 9\mathrm{r}\)
To isolate t, we divide both sides by \((\mathrm{r} + 5)\):
\(\mathrm{t} = 9\mathrm{r} \div (\mathrm{r} + 5)\)
\(\mathrm{t} = \frac{9\mathrm{r}}{\mathrm{r} + 5}\)
Let's verify this makes sense: if r = 10, then originally it takes 9 hours at 10 gallons/hour. With the faster rate of 15 gallons/hour, it should take \(\frac{9(10)}{10+5} = \frac{90}{15} = 6\) hours. Indeed, 6 hours is less than 9 hours, which makes perfect sense!
Process Skill: MANIPULATE
The time required to fill the pool at the new rate is \(\frac{9\mathrm{r}}{\mathrm{r} + 5}\) hours.
Looking at our answer choices, this matches choice D: \(\frac{9\mathrm{r}}{\mathrm{r} + 5}\).
The answer is D.
Students often fail to recognize that the pool capacity remains the same in both scenarios. They might think they need to compare rates directly without establishing that the total amount of water (pool capacity) is the constant factor. This leads to setting up incorrect relationships or equations.
2. Confusing rate and time relationshipSome students incorrectly assume that if the rate increases by 5 gallons per hour, the time should also increase proportionally. They fail to understand the inverse relationship - when rate increases, time decreases for the same amount of work (filling the same pool).
When solving \((\mathrm{r} + 5) \times \mathrm{t} = 9\mathrm{r}\) for t, students might incorrectly divide or make algebraic errors. For example, they might write \(\mathrm{t} = 9\mathrm{r} - (\mathrm{r} + 5)\) instead of \(\mathrm{t} = 9\mathrm{r} \div (\mathrm{r} + 5)\), confusing subtraction with division.
2. Setting up the wrong equationEven if students understand the concept, they might set up the equation incorrectly. For instance, they might write \(\mathrm{r} \times \mathrm{t} = 9(\mathrm{r} + 5)\) instead of \((\mathrm{r} + 5) \times \mathrm{t} = 9\mathrm{r}\), mixing up which rate goes with which time scenario.
After correctly arriving at \(\mathrm{t} = \frac{9\mathrm{r}}{\mathrm{r} + 5}\), some students might get confused and select answer choice C: \(\frac{\mathrm{r}}{\mathrm{r} + 5}\), which is missing the factor of 9. This happens when students lose track of their work or misread their final expression.
Step 1: Choose a convenient value for r
Let's choose r = 10 gallons per hour. This is a logical choice because:
Step 2: Calculate the pool capacity
With r = 10 gallons per hour and time = 9 hours:
\(\text{Pool capacity} = \text{rate} \times \text{time} = 10 \times 9 = 90\) gallons
Step 3: Find the new time with increased rate
New rate = r + 5 = 10 + 5 = 15 gallons per hour
Since pool capacity remains 90 gallons:
\(\text{New time} = \frac{\text{Pool capacity}}{\text{New rate}} = \frac{90}{15} = 6\) hours
Step 4: Check against answer choices
We need to find which expression gives us 6 when r = 10:
Step 5: Verify with logic
The result makes intuitive sense: when the flow rate increases from 10 to 15 gallons per hour (a 50% increase), the time decreases from 9 to 6 hours (a 33% decrease), which follows the inverse relationship between rate and time.
Answer: D