Problem 51
Question
During a vacation, the Mulligan family drove 63 miles in 1 hour. If they averaged the same speed during their trip, which equation can be used to find how far the Mulligan family drove in 6 hours? \(\mathbf{A} \frac{63}{x}=6\) B \(\frac{x}{6}=63\) C \(6 x=63\) D \(63 x=6\)
Step-by-Step Solution
Verified Answer
The correct option is B: \(\frac{x}{6} = 63\).
1Step 1: Understanding Average Speed
The Mulligan family drove 63 miles in 1 hour. This means their speed was a rate of 63 miles per hour. Our goal is to find out how far they would travel if they maintained this speed for 6 hours.
2Step 2: Defining the Formula for Distance
We know that distance can be calculated using the formula: \(Distance = Speed \times Time\). Given the speed is 63 miles per hour and the time is 6 hours, we want to find the distance for 6 hours.
3Step 3: Setting Up the Equation
Using the formula \(Distance = Speed \times Time\) and substituting the known values, we set up the equation as \(Distance = 63 \times 6\). Here, Distance is represented by a variable, say \(x\), so it can be written as \(x = 63 \times 6\).
4Step 4: Identifying the Equation
The problem asks us to identify the equation that represents this situation from given options. The equation \(x = 63 \times 6\) matches option B's format when rewritten as \(\frac{x}{6} = 63\).
5Step 5: Concluding the Correct Option
Comparing our formulated equation to the given options, it's evident that option B, \(\frac{x}{6} = 63\), is correctly structured to solve for the distance based on the constant speed and travel time.
Key Concepts
Distance CalculationSpeed and Time RelationshipEquations for Word Problems
Distance Calculation
Distance calculation is a fundamental concept in prealgebra problems. It involves determining how far an object has traveled, given a certain speed and time. The core equation for distance calculation is:
To apply this equation, you simply multiply the speed by the time traveled. For instance, if a car travels at a speed of 63 miles per hour for 6 hours, you can calculate the distance by multiplying:
This straightforward calculation helps you determine the total distance covered with uniform speed over a given timeframe. By mastering this formula, you can solve various real-world problems involving travel distances.
- \( \text{Distance} = \text{Speed} \times \text{Time} \)
To apply this equation, you simply multiply the speed by the time traveled. For instance, if a car travels at a speed of 63 miles per hour for 6 hours, you can calculate the distance by multiplying:
- \( 63 \text{ (speed)} \times 6 \text{ (time)} = 378 \text{ miles} \)
This straightforward calculation helps you determine the total distance covered with uniform speed over a given timeframe. By mastering this formula, you can solve various real-world problems involving travel distances.
Speed and Time Relationship
Let's delve into the relationship between speed and time. Speed can be defined as the rate at which an object covers a distance. It is typically expressed in units such as miles per hour (mph) or kilometers per hour (kph). The relationship between speed and time is crucial in calculating how far an object travels. Here’s how it works:
Understanding this relationship is key to solving many word problems in prealgebra, as it allows you to determine one of these variables if the other two are known. For example, knowing that a car travels 63 miles in one hour helps you determine how far it will travel in 6 hours at the same speed.
- Faster speeds result in greater distances covered in the same amount of time, while slower speeds cover less distance.
- If time increases while the speed stays constant, distance will also increase proportionally.
Understanding this relationship is key to solving many word problems in prealgebra, as it allows you to determine one of these variables if the other two are known. For example, knowing that a car travels 63 miles in one hour helps you determine how far it will travel in 6 hours at the same speed.
Equations for Word Problems
Solving word problems using equations is a common challenge in prealgebra. These problems require translating a real-world situation into a mathematical equation, which can then be solved to find the answer. In this context, equations related to distance, speed, and time are particularly useful. Here’s a structured approach:
By following these steps, you can effectively handle word problems. Selecting the right equation from multiple options, as in the exercise above, ensures you correctly model the situation with mathematical terms.
- Identify the quantities involved: In our problem, these are speed, time, and distance.
- Find the relationship: Use the formula \( \text{Distance} = \text{Speed} \times \text{Time} \) to relate these quantities.
- Set up the equation: Substitute the known values into the formula to create an equation. For example, if the speed is 63 miles/hour and the time is 6 hours, the equation is \( x = 63 \times 6 \).
- Solve the equation: To find the unknown, solve the equation using appropriate algebraic methods.
By following these steps, you can effectively handle word problems. Selecting the right equation from multiple options, as in the exercise above, ensures you correctly model the situation with mathematical terms.
Other exercises in this chapter
Problem 51
Simplify to make the calculation as easy as possible. \(72(38)+(-72)(18)\)
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The width of a beach is changing at a rate of \(-9\) inches per year. How long will it take for the width of the beach to change \(-4.5\) feet?
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Use the Distributive Property to write each expression as an equivalent algebraic expression. $$10(r-s)$$
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Solve each equation. Check your solution $$25 z=175$$
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