- Translate the problem requirements: We have two teams racing 300 miles. Team A finished 3 hours faster than Team B, and Team A's speed was 5 mph faster than Team B's. We need to find Team B's average speed.
- Set up the relationship using the distance formula: Since \(\mathrm{distance} = \mathrm{speed} \times \mathrm{time}\), and both teams traveled 300 miles, we can express their times in terms of their speeds.
- Create equations based on the given constraints: Use the fact that Team A took 3 fewer hours and had 5 mph greater speed to set up equations relating Team A's and Team B's performance.
- Solve the resulting equation: Substitute the relationships to get a single equation in terms of Team B's speed and solve for the answer.
Execution of Strategic Approach
1. Translate the problem requirements
Let's start by understanding what we know in plain English:
- Both teams traveled the same distance: 300 miles
- Team A finished 3 hours faster than Team B
- Team A's speed was 5 mph faster than Team B's speed
- We need to find Team B's average speed
This is a classic relationship problem involving distance, speed, and time. Think of it like two friends driving to the same destination - one drives faster and arrives earlier.
Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships
2. Set up the relationship using the distance formula
Since both teams traveled 300 miles, we can use the basic relationship: \(\mathrm{Time} = \frac{\mathrm{Distance}}{\mathrm{Speed}}\)
Let's call Team B's speed 'x' mph. This means:
- Team B's speed: x mph
- Team A's speed: (x + 5) mph (since it's 5 mph faster)
- Team B's time: \(\frac{300}{x}\) hours
- Team A's time: \(\frac{300}{x + 5}\) hours
The key insight is that Team A took 3 fewer hours than Team B to complete the same 300-mile course.
3. Create equations based on the given constraints
Since Team A finished 3 hours faster than Team B, we can write:
Team B's time - Team A's time = 3 hours
Substituting our expressions:
\(\frac{300}{x} - \frac{300}{x + 5} = 3\)
This equation captures the entire problem: the difference in their times equals 3 hours.
Process Skill: INFER - Recognizing that the time difference gives us the key equation to solve
4. Solve the resulting equation
Now we solve: \(\frac{300}{x} - \frac{300}{x + 5} = 3\)
To solve this, let's find a common denominator for the fractions:
\(\frac{300(x + 5)}{x(x + 5)} - \frac{300x}{x(x + 5)} = 3\)
Simplifying the numerator:
\(\frac{300(x + 5) - 300x}{x(x + 5)} = 3\)
\(\frac{300x + 1500 - 300x}{x(x + 5)} = 3\)
\(\frac{1500}{x(x + 5)} = 3\)
Cross multiply:
\(1500 = 3x(x + 5)\)
\(1500 = 3x^2 + 15x\)
\(500 = x^2 + 5x\)
\(x^2 + 5x - 500 = 0\)
Using the quadratic formula or factoring:
\((x + 25)(x - 20) = 0\)
So x = -25 or x = 20
Since speed cannot be negative, x = 20 mph.
Process Skill: MANIPULATE - Carefully handling fractions and quadratic equations to reach the solution
5. Final Answer
Team B's average speed was 20 mph.
Let's verify: If Team B traveled at 20 mph, it took \(\frac{300}{20} = 15\) hours.
Team A traveled at 25 mph, taking \(\frac{300}{25} = 12\) hours.
The difference is \(15 - 12 = 3\) hours. ✓
The answer is D. 20
Common Faltering Points
Errors while devising the approach
- Misinterpreting the time relationship: Students often confuse which team finished faster and set up the equation as Team A's time - Team B's time = 3, instead of the correct Team B's time - Team A's time = 3. This happens because they don't carefully process that "Team A finished 3 fewer hours" means Team A took less time.
- Incorrect variable assignment: Some students assign variables inconsistently, such as letting x represent Team A's speed instead of Team B's speed, then forgetting this assignment when setting up equations. This creates confusion throughout the solution process.
- Missing the distance formula connection: Students may recognize this as a speed/time problem but fail to connect that since both teams travel the same distance (300 miles), they can use \(\mathrm{Time} = \frac{\mathrm{Distance}}{\mathrm{Speed}}\) to create the necessary relationships between the variables.
Errors while executing the approach
- Common denominator errors: When solving \(\frac{300}{x} - \frac{300}{x + 5} = 3\), students frequently make mistakes finding the common denominator \(x(x + 5)\), either by incorrectly combining fractions or making sign errors when subtracting the numerators.
- Algebraic manipulation mistakes: Students often lose track of steps when cross-multiplying and simplifying, particularly when going from \(1500 = 3x(x + 5)\) to the standard quadratic form \(x^2 + 5x - 500 = 0\). Division errors and sign mistakes are common here.
- Quadratic factoring errors: Students may struggle to factor \(x^2 + 5x - 500 = 0\) correctly, either by using incorrect factor pairs or making arithmetic errors when checking that \((-25) \times 20 = -500\) and \((-25) + 20 = -5\).
Errors while selecting the answer
- Choosing the wrong team's speed: After finding x = 20, students sometimes select this as Team A's speed instead of Team B's speed, even though the question specifically asks for Team B's average mph. This happens when they lose track of their variable definitions.
- Accepting negative solutions: Some students might consider x = -25 as a valid answer choice if they see a similar number among the options, forgetting that speed cannot be negative in real-world contexts.
