When it operates at its own constant rate, hose A can fill an empty tank twice as fast as hose...

1. Translate the problem requirements: Hose A fills the tank in x hours, hose B takes twice as long (2x hours) since A is twice as fast as B. We need to find how long both hoses together take to fill the same tank.
2. Express individual work rates: Convert the time information into rates of work per hour for each hose, thinking of rate as "fraction of tank filled per hour."
3. Combine the work rates: Add the individual rates to get the combined rate when both hoses work simultaneously.
4. Convert combined rate back to time: Use the relationship that time equals 1 divided by the combined rate to find the total time needed.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know in everyday language:

• Hose A is twice as fast as hose B at filling the tank
• Hose A can fill the entire tank in x hours
• We need to find how long both hoses take working together

Since hose A is twice as fast as hose B, this means hose B takes twice as long to do the same job. If hose A takes x hours, then hose B takes 2x hours to fill the same tank.

Think of it like this: if one worker can paint a fence in 2 hours, and another worker is half as fast, the slower worker would need 4 hours to paint the same fence.

Process Skill: TRANSLATE - Converting the relationship "twice as fast" into specific time values

2. Express individual work rates

Now let's think about how much work each hose does per hour. Instead of thinking about total time, let's think about "what fraction of the tank gets filled each hour?"

• Hose A fills the entire tank in x hours, so in 1 hour it fills \(\frac{1}{x}\) of the tank
• Hose B fills the entire tank in 2x hours, so in 1 hour it fills \(\frac{1}{2x}\) of the tank

This is like saying: if it takes you 4 hours to clean your room, then in 1 hour you clean 1/4 of your room.

So our work rates are:
• Hose A rate = \(\frac{1}{x}\) tanks per hour
• Hose B rate = \(\frac{1}{2x}\) tanks per hour

3. Combine the work rates

When both hoses work together, their efforts add up. It's like having two people working on the same task - they accomplish more together than either could alone.

Combined rate = Hose A rate + Hose B rate
Combined rate = \(\frac{1}{x} + \frac{1}{2x}\)

To add these fractions, we need a common denominator. The common denominator for x and 2x is 2x:
Combined rate = \(\frac{2}{2x} + \frac{1}{2x} = \frac{3}{2x}\)

This means together, the hoses fill \(\frac{3}{2x}\) of the tank each hour.

4. Convert combined rate back to time

Now we need to answer: "If the hoses fill \(\frac{3}{2x}\) of the tank each hour, how long does it take to fill the entire tank?"

This is like asking: "If you eat 3/4 of a pizza each hour, how long does it take to eat 1 whole pizza?"

Time needed = 1 ÷ (rate per hour)
Time needed = \(1 \div \frac{3}{2x}\)
Time needed = \(1 \times \frac{2x}{3} = \frac{2x}{3}\)

Therefore, working together, the two hoses will fill the tank in \(\frac{2x}{3}\) hours.

Final Answer

The answer is \(\frac{2x}{3}\) hours, which corresponds to choice A.

We can verify this makes sense: since we're using both hoses together, it should take less time than hose A alone (which takes x hours). Indeed, \(\frac{2x}{3} < x\), so this passes our reasonableness check.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "twice as fast" relationship
Students often confuse the speed relationship with the time relationship. When the problem states "hose A can fill a tank twice as fast as hose B," students might incorrectly think that if hose A takes x hours, then hose B takes \(\frac{x}{2}\) hours (half the time). However, being twice as fast means taking half the time, so if A takes x hours, B actually takes 2x hours. This is a critical conceptual error that affects the entire solution.

2. Forgetting to establish individual work rates
Some students jump directly to trying to combine the hoses without first determining each hose's individual rate of work. They might attempt shortcuts like "A is twice as fast, so together they work 3 times as fast" without properly defining what "rate" means in terms of fraction of tank filled per hour. This leads to incorrect setups and wrong answers.

3. Confusing time and rate concepts
Students may struggle with the fundamental concept that work rate is the reciprocal of time. They might try to add times directly (\(x + 2x = 3x\)) rather than understanding that rates add when working together, and that rate = \(\frac{1}{\text{time}}\). This conceptual gap prevents them from setting up the correct mathematical framework.

Errors while executing the approach

1. Fraction addition errors with different denominators
When adding \(\frac{1}{x} + \frac{1}{2x}\), students frequently make algebraic mistakes. Common errors include: adding denominators directly (getting \(\frac{2}{3x}\) instead of \(\frac{3}{2x}\)), forgetting to find a common denominator, or incorrectly converting \(\frac{1}{x}\) to \(\frac{2}{2x}\). These arithmetic errors lead to wrong combined rates.

2. Incorrect reciprocal calculation
After finding the combined rate of \(\frac{3}{2x}\), students must take the reciprocal to convert back to time. Many students struggle with dividing by a fraction, writing \(1 \div \frac{3}{2x}\) incorrectly. They might get \(\frac{3}{2x}\) instead of \(\frac{2x}{3}\), or make sign errors during the division process.

3. Lost track of units and what the variables represent
During calculations, students may lose sight of what their numbers represent. They might confuse whether they're working with "tanks per hour" or "hours per tank," leading to using the wrong formulas or taking reciprocals at inappropriate times.

Errors while selecting the answer

1. Failing the reasonableness check
Students might arrive at an answer but fail to verify it makes logical sense. For instance, if they get an answer larger than x (like \(\frac{3x}{2}\) or 3x), they should recognize this is unreasonable since two hoses working together must be faster than one hose alone. Missing this check means they don't catch computational errors.

2. Selecting the reciprocal of the correct answer
After calculating correctly and getting \(\frac{2x}{3}\), some students might second-guess themselves about whether they need the rate or the time. This confusion could lead them to select \(\frac{3}{2x}\) (choice D) instead of \(\frac{2x}{3}\) (choice A), essentially choosing the combined rate rather than the combined time.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a smart number for x

Let's choose x = 6 hours. This is a convenient number because it will make our rate calculations clean (6 is divisible by common denominators we'll encounter).

Step 2: Determine the time for hose B

Since hose A fills the tank twice as fast as hose B, hose B takes twice as long as hose A.

Time for hose B = 2x = 2(6) = 12 hours

Step 3: Calculate individual work rates

Work rate = fraction of tank filled per hour

• Hose A's rate = \(\frac{1}{6}\) of the tank per hour
• Hose B's rate = \(\frac{1}{12}\) of the tank per hour

Step 4: Find combined work rate

When working together: Combined rate = \(\frac{1}{6} + \frac{1}{12}\)

To add these fractions: \(\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\)

So together they fill \(\frac{1}{4}\) of the tank per hour.

Step 5: Calculate time for combined work

If they fill \(\frac{1}{4}\) of the tank per hour, then:

Time to fill entire tank = \(1 \div \frac{1}{4} = 4\) hours

Step 6: Express in terms of x

We found it takes 4 hours when x = 6

Since \(4 = \frac{2}{3} \times 6 = \frac{2}{3} \times x\)

Therefore, the answer is \(\frac{2x}{3}\)

This matches answer choice A, confirming our algebraic solution using concrete numbers.