Problem 33
Question
Solve each problem. Patrick drives 40 miles to work, and Guy drives 60 miles to work. Guy claims that he drives at the same speed as Patrick, but it takes him only 12 minutes longer to get to work. If this is true, then how long does it take each of them to get to work? What are their speeds? Do you think that Guy's claim is correct?
Step-by-Step Solution
Verified Answer
Patrick takes 24 minutes and Guy takes 36 minutes. Their speed is 100 mph. Guy's claim is correct.
1Step 1 - Define Variables
Let the speed at which both Patrick and Guy drive be denoted by \( s \) miles per hour. Let the time it takes for Patrick to get to work be \( t \) hours.
2Step 2 - Write Equations for Time
Patrick's time to get to work is \( t = \frac{40}{s} \). Guy's time to get to work is \( t + \frac{12}{60} = \frac{60}{s} \).
3Step 3 - Simplify Guy's Time Expression
Convert 12 minutes to hours by dividing by 60: \( \frac{12}{60} = 0.2 \). So, Guy's time is \( t + 0.2 \).
4Step 4 - Set Up an Equation
Set up the equation using the times: \( t + 0.2 = \frac{60}{s} \).
5Step 5 - Substitute Patrick's Time
Substitute \( t \) from Patrick's equation into Guy's equation: \( \frac{40}{s} + 0.2 = \frac{60}{s} \).
6Step 6 - Solve for Speed \( s \)
Isolate \( s \):\[ \frac{60}{s} - \frac{40}{s} = 0.2 \]\[ \frac{20}{s} = 0.2 \]\[ s = \frac{20}{0.2} \]\[ s = 100 \text{ mph} \]
7Step 7 - Calculate Patrick's Time
Using \( s = 100 \):\[ t = \frac{40}{100} = 0.4 \text{ hours} \]Convert to minutes: \( 0.4 \times 60 = 24 \text{ minutes} \)
8Step 8 - Calculate Guy's Time
Guy takes 12 minutes longer: \( 24 + 12 = 36 \text{ minutes} \).
9Step 9 - Verify Consistency
If Guy drives at 100 mph: \( \frac{60}{100} = 0.6 \text{ hours} = 36 \text{ minutes} \). The claim appears to be correct.
Key Concepts
Linear EquationsDistance-Speed-Time RelationshipVariable IsolationUnit Conversion
Linear Equations
In algebra, linear equations are foundational tools used to model relationships where the change between variables is consistent.
For example, when solving the problem given, we formulate linear equations to represent the time it takes for Patrick and Guy to travel to work.
Patrick's time to travel 40 miles is given by the equation: \( t = \frac{40}{s} \), where \( t\) is time in hours, and \( s\) is speed in miles per hour.
Similarly, Guy's time, considering he drives 60 miles and adds an extra 12 minutes or 0.2 hours, can be expressed as: \( t + 0.2 = \frac{60}{s}\).
Such equations help us represent their travel times and solve for unknowns like speed using straightforward algebraic manipulations.
For example, when solving the problem given, we formulate linear equations to represent the time it takes for Patrick and Guy to travel to work.
Patrick's time to travel 40 miles is given by the equation: \( t = \frac{40}{s} \), where \( t\) is time in hours, and \( s\) is speed in miles per hour.
Similarly, Guy's time, considering he drives 60 miles and adds an extra 12 minutes or 0.2 hours, can be expressed as: \( t + 0.2 = \frac{60}{s}\).
Such equations help us represent their travel times and solve for unknowns like speed using straightforward algebraic manipulations.
Distance-Speed-Time Relationship
The distance-speed-time relationship is a crucial concept in algebra and real-life situations like traveling.
This relationship is usually expressed in the formula: \( \text{Distance} = \text{Speed} \times \text{Time}\), or rearranging it: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
In the example exercise, we used this relationship to find how long it takes for Patrick and Guy to get to work.
Patrick's travel time is based on his speed \( s\) and distance 40 miles: \( \text{Time} = \frac{40}{s}\).
For Guy, to account for his 60-mile trip: \( \text{Time} = \frac{60}{s}\), but since he takes 12 minutes (0.2 hours) longer: \( \text{Time} = \frac{60}{s} + 0.2\). Understanding and using this relationship allows us to set up equations to solve for the speed.
This relationship is usually expressed in the formula: \( \text{Distance} = \text{Speed} \times \text{Time}\), or rearranging it: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
In the example exercise, we used this relationship to find how long it takes for Patrick and Guy to get to work.
Patrick's travel time is based on his speed \( s\) and distance 40 miles: \( \text{Time} = \frac{40}{s}\).
For Guy, to account for his 60-mile trip: \( \text{Time} = \frac{60}{s}\), but since he takes 12 minutes (0.2 hours) longer: \( \text{Time} = \frac{60}{s} + 0.2\). Understanding and using this relationship allows us to set up equations to solve for the speed.
Variable Isolation
In algebra, we often need to isolate a variable to solve an equation.
This process involves rearranging the equation so that the variable we are solving for is on one side of the equation, and everything else is on the other side.
In our exercise, we needed to find the common speed \( s\) for Patrick and Guy.
Starting with the equation: \( t + 0.2 = \frac{60}{s} \), we substituted \( t \) with \( \frac{40}{s} \), resulting in: \( \frac{40}{s} + 0.2 = \frac{60}{s} \).
The goal is to isolate \( s \). Subtracting \( \frac{40}{s} \) from both sides: \( 0.2 = \frac{60}{s} - \frac{40}{s} \).
Simplifying this: \( 0.2 = \frac{20}{s} \), and multiplying both sides by \( s \) to isolate it: \( s \times 0.2 = 20 \).
Finally, solving for \( s \): \( s = \frac{20}{0.2} = 100 \text{ mph} \). Isolating variables allows us to solve for them effectively.
This process involves rearranging the equation so that the variable we are solving for is on one side of the equation, and everything else is on the other side.
In our exercise, we needed to find the common speed \( s\) for Patrick and Guy.
Starting with the equation: \( t + 0.2 = \frac{60}{s} \), we substituted \( t \) with \( \frac{40}{s} \), resulting in: \( \frac{40}{s} + 0.2 = \frac{60}{s} \).
The goal is to isolate \( s \). Subtracting \( \frac{40}{s} \) from both sides: \( 0.2 = \frac{60}{s} - \frac{40}{s} \).
Simplifying this: \( 0.2 = \frac{20}{s} \), and multiplying both sides by \( s \) to isolate it: \( s \times 0.2 = 20 \).
Finally, solving for \( s \): \( s = \frac{20}{0.2} = 100 \text{ mph} \). Isolating variables allows us to solve for them effectively.
Unit Conversion
Unit conversion is the process of changing a measure to a different unit while maintaining the same quantity.
In many problems, consistent units are essential for accuracy.
In our exercise, we needed to convert 12 minutes into hours to match other time calculations.
Since 1 hour = 60 minutes, we converted 12 minutes to hours by dividing: \( \frac{12 \text{ minutes}}{60} = 0.2 \text{ hours} \).
Without consistent units, our equations would be incorrect, making solving the problem difficult or impossible.
Similarly, converting Patrick's time from hours to minutes involved multiplying by 60: \( 0.4 \text{ hours} \times 60 = 24 \text{ minutes} \). Mastering unit conversion is vital to solving algebra problems effectively.
In many problems, consistent units are essential for accuracy.
In our exercise, we needed to convert 12 minutes into hours to match other time calculations.
Since 1 hour = 60 minutes, we converted 12 minutes to hours by dividing: \( \frac{12 \text{ minutes}}{60} = 0.2 \text{ hours} \).
Without consistent units, our equations would be incorrect, making solving the problem difficult or impossible.
Similarly, converting Patrick's time from hours to minutes involved multiplying by 60: \( 0.4 \text{ hours} \times 60 = 24 \text{ minutes} \). Mastering unit conversion is vital to solving algebra problems effectively.
Other exercises in this chapter
Problem 32
Reduce each rational expression to its lowest terms. $$\frac{b^{8}-a b^{5}}{a b^{5}}$$
View solution Problem 33
Find the solution set to each equation. $$\frac{2}{x+1}=\frac{x-1}{4}$$
View solution Problem 33
Simplify each complex fraction. $$\frac{\frac{4}{y}-\frac{y+4}{y-3}}{\frac{2}{y-3}+\frac{y+1}{y}}$$
View solution Problem 33
Reduce each rational expression to its lowest terms. $$\frac{a-b}{2 b-2 a}$$
View solution