If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles,...

1. Translate the problem requirements: We need to find the time for Car C to travel 5 miles, where Car C travels at the average speed of Car A (2 miles in n hours) and Car B (3 miles in m hours).
2. Calculate individual rates: Determine the speed (rate) of each car in miles per hour using distance/time.
3. Find the average rate: Calculate the arithmetic mean of the two rates by adding them and dividing by 2.
4. Apply time formula: Use the relationship time = distance/rate to find how long Car C takes to travel 5 miles at the average rate.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in simple terms:

• Car A travels 2 miles and takes n hours to do it
• Car B travels 3 miles and takes m hours to do it
• Car C will travel at the average speed of Car A and Car B
• We need to find how long Car C takes to travel 5 miles

Think of this like comparing how fast different people walk. If one person walks 2 miles in n hours and another walks 3 miles in m hours, we want to know how long it takes someone walking at the average of those two speeds to walk 5 miles.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical relationships

2. Calculate individual rates

Now let's figure out how fast each car actually goes. Speed is simply how much distance you cover per unit of time.

For Car A: It goes 2 miles in n hours, so its speed is 2 miles ÷ n hours = \(\frac{2}{\mathrm{n}}\) miles per hour

For Car B: It goes 3 miles in m hours, so its speed is 3 miles ÷ m hours = \(\frac{3}{\mathrm{m}}\) miles per hour

Think of it this way: if you walk 2 miles in 4 hours, you're walking at 2÷4 = 0.5 miles per hour. Same logic applies here, just with variables instead of numbers.

3. Find the average rate

To find the average of two speeds, we add them up and divide by 2 (since we have 2 cars).

Average speed = (Speed of Car A + Speed of Car B) ÷ 2
Average speed = \(\left(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}}\right) ÷ 2\)
Average speed = \(\left(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}}\right) \times \frac{1}{2}\)
Average speed = \(\frac{\left(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}}\right)}{2}\)

To make this easier to work with, let's get a common denominator for the fractions inside:
\(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}} = \frac{2\mathrm{m} + 3\mathrm{n}}{\mathrm{nm}}\)

So our average speed becomes: \(\frac{2\mathrm{m} + 3\mathrm{n}}{\mathrm{nm}} ÷ 2 = \frac{2\mathrm{m} + 3\mathrm{n}}{2\mathrm{nm}}\)

4. Apply time formula

Now we use the basic relationship: Time = Distance ÷ Speed

Car C needs to travel 5 miles at the average speed we just calculated.

Time for Car C = 5 miles ÷ Average speed
Time for Car C = \(5 ÷ \left[\frac{2\mathrm{m} + 3\mathrm{n}}{2\mathrm{nm}}\right]\)

When we divide by a fraction, it's the same as multiplying by its reciprocal:
Time for Car C = \(5 \times \frac{2\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}}\)
Time for Car C = \(\frac{10\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}}\)

Wait, let me double-check this against the answer choices. Looking at choice A, we have \(\frac{10\mathrm{nm}}{3\mathrm{n} + 2\mathrm{m}}\). Let me verify our average calculation:

Actually, let me recalculate the average more carefully:
Average speed = \(\frac{\left(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}}\right)}{2}\)

Getting common denominator: \(\frac{2}{\mathrm{n}} = \frac{2\mathrm{m}}{\mathrm{nm}}\) and \(\frac{3}{\mathrm{m}} = \frac{3\mathrm{n}}{\mathrm{nm}}\)
So: \(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}} = \frac{2\mathrm{m} + 3\mathrm{n}}{\mathrm{nm}}\)

Therefore: Average speed = \(\frac{2\mathrm{m} + 3\mathrm{n}}{\mathrm{nm}} ÷ 2 = \frac{2\mathrm{m} + 3\mathrm{n}}{2\mathrm{nm}}\)

Time = Distance ÷ Speed = \(5 ÷ \left[\frac{2\mathrm{m} + 3\mathrm{n}}{2\mathrm{nm}}\right] = 5 \times \frac{2\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}} = \frac{10\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}}\)

This matches choice A if we note that \(2\mathrm{m} + 3\mathrm{n} = 3\mathrm{n} + 2\mathrm{m}\) (same expression, terms reordered).

5. Final Answer

The time it takes Car C to travel 5 miles is \(\frac{10\mathrm{nm}}{3\mathrm{n} + 2\mathrm{m}}\) hours.

This corresponds to Answer Choice A.

To verify: Car C travels at the average of \(\frac{2}{\mathrm{n}}\) and \(\frac{3}{\mathrm{m}}\) miles per hour, which is \(\frac{3\mathrm{n} + 2\mathrm{m}}{2\mathrm{nm}}\) miles per hour. To travel 5 miles at this speed takes \(5 ÷ \left[\frac{3\mathrm{n} + 2\mathrm{m}}{2\mathrm{nm}}\right] = \frac{10\mathrm{nm}}{3\mathrm{n} + 2\mathrm{m}}\) hours.

Common Faltering Points

Errors while devising the approach

1. Confusing rate vs. time relationships: Students often mix up the formula for rate calculations. They might think Car A's rate is \(\frac{\mathrm{n}}{2}\) instead of \(\frac{2}{\mathrm{n}}\), forgetting that rate = distance ÷ time, not time ÷ distance.

2. Misunderstanding what "average of rates" means: Some students might try to find the average time instead of average speed, or attempt to average the given times (n and m) directly rather than first calculating individual rates.

3. Setting up the wrong final equation: Students may correctly find the average rate but then use the wrong relationship to find time, such as multiplying distance by rate instead of dividing distance by rate.

Errors while executing the approach

1. Common denominator errors: When adding fractions \(\frac{2}{\mathrm{n}} + \frac{3}{\mathrm{m}}\), students frequently make algebraic mistakes. They might write the common denominator result as \(\frac{2\mathrm{n} + 3\mathrm{m}}{\mathrm{nm}}\) instead of the correct \(\frac{2\mathrm{m} + 3\mathrm{n}}{\mathrm{nm}}\), mixing up which coefficient goes with which variable.

2. Dividing by fractions incorrectly: When calculating time = \(5 ÷ \text{[average rate]}\), students often struggle with dividing by a complex fraction. They may forget to flip the fraction when dividing, or make errors in the reciprocal multiplication step.

3. Arithmetic errors in the final calculation: Students may correctly set up \(5 \times \frac{2\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}}\) but then make simple multiplication errors, forgetting to multiply \(5 \times 2\mathrm{nm} = 10\mathrm{nm}\) in the numerator.

Errors while selecting the answer

1. Not recognizing equivalent expressions: Students might arrive at \(\frac{10\mathrm{nm}}{2\mathrm{m} + 3\mathrm{n}}\) but fail to recognize that this equals choice A: \(\frac{10\mathrm{nm}}{3\mathrm{n} + 2\mathrm{m}}\), since addition is commutative. They may think their answer doesn't match any of the choices.

2. Selecting reciprocal answers: If students made an error in the time vs. rate relationship, they might get \(\frac{3\mathrm{n} + 2\mathrm{m}}{10\mathrm{nm}}\) and incorrectly select a choice that looks similar, such as choice C: \(\frac{2\mathrm{n} + 3\mathrm{m}}{5\mathrm{nm}}\), not noticing the coefficient differences.

Alternate Solutions

Smart Numbers Approach

We can solve this problem by choosing convenient values for n and m that make our calculations cleaner while preserving the relationships in the problem.

Step 1: Choose Smart Values

Let's choose values that will make rate calculations simple:

• Let n = 2 hours (time for Car A to travel 2 miles)
• Let m = 3 hours (time for Car B to travel 3 miles)

These values are chosen because they create clean rate calculations and avoid complex fractions.

Step 2: Calculate Individual Rates

Using Rate = Distance ÷ Time:

• Car A's rate = 2 miles ÷ 2 hours = 1 mile per hour
• Car B's rate = 3 miles ÷ 3 hours = 1 mile per hour

Step 3: Find Average Rate

Average rate of Car C = (1 + 1) ÷ 2 = 1 mile per hour

Step 4: Calculate Time for Car C

Time = Distance ÷ Rate = 5 miles ÷ 1 mph = 5 hours

Step 5: Verify with Answer Choice A

Using our smart numbers in choice A: \(\frac{10\mathrm{nm}}{3\mathrm{n} + 2\mathrm{m}}\)

= \(\frac{10 \times 2 \times 3}{3 \times 2 + 2 \times 3} = \frac{60}{6 + 6} = \frac{60}{12} = 5\) hours ✓

This confirms that answer choice A gives us the correct result with our smart numbers.