When water is flowing into a certain pool at the rate of r gallons per hour, it takes 9 hours...

1. Translate the problem requirements: We need to understand that pool capacity remains constant, and we're comparing two scenarios with different flow rates to find the new time required
2. Establish the pool capacity relationship: Use the first scenario (rate r, time 9 hours) to express the total pool capacity in terms of r
3. Apply the same capacity to the new scenario: Set up an equation where the same pool capacity is filled at the new rate (r + 5) to find the new time
4. Solve for the new time algebraically: Manipulate the equation to isolate the time variable and express it in terms of r

Execution of Strategic Approach

1. Translate the problem requirements

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:

• Scenario 1: Water flows at rate r gallons per hour, takes 9 hours
• Scenario 2: Water flows at rate (r + 5) gallons per hour, takes ? hours

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

2. Establish the pool capacity relationship

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

3. Apply the same capacity to the new scenario

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

4. Solve for the new time algebraically

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

4. Final Answer

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.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what stays constant

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 relationship

Some 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).

Errors while executing the approach

1. Incorrect algebraic manipulation

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 equation

Even 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.

Errors while selecting the answer

1. Choosing the reciprocal

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.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for r

Let's choose r = 10 gallons per hour. This is a logical choice because:

• It makes calculations clean and simple
• It satisfies the constraint that r must be positive
• The value r + 5 = 15 is also easy to work with

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:

• Choice A: \(\frac{9(10 + 5)}{10} = \frac{9(15)}{10} = \frac{135}{10} = 13.5\) ✗
• Choice B: \(\frac{10 + 5}{9 \times 10} = \frac{15}{90} = \frac{1}{6}\) ✗
• Choice C: \(\frac{10}{10 + 5} = \frac{10}{15} = \frac{2}{3}\) ✗
• Choice D: \(\frac{9 \times 10}{10 + 5} = \frac{90}{15} = 6\) ✓
• Choice E: \(\frac{10}{9 \times 15} = \frac{10}{135} = \frac{2}{27}\) ✗

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