A Taxi Service Charges A Flat Fee Of $ 1.25 \$1.25 $1.25 And $ 0.75 \$0.75 $0.75 Per Mile. If Henri Has $ 14.00 \$14.00 $14.00 , Which Of The Following Shows The Number Of Miles He Can Afford To Ride In The Taxi?A. M ≤ 17 M \leq 17 M ≤ 17 B. M ≥ 17 M \geq 17 M ≥ 17

Introduction

When it comes to taxi services, understanding the pricing structure is crucial to determine the affordability of a ride. In this problem, we are given a taxi service that charges a flat fee of $\$1.25$ and an additional $\$0.75$ per mile. Henri has a budget of $\$14.00$, and we need to find out how many miles he can afford to ride in the taxi. This problem requires us to set up an equation and solve for the number of miles Henri can afford.

Setting Up the Equation

Let's denote the number of miles Henri can afford as $m$. The total cost of the ride will be the sum of the flat fee and the cost per mile. Since the flat fee is $\$1.25$, the cost per mile is $\$0.75$, and Henri has a budget of $\$14.00$, we can set up the following equation:

$$1.25 + 0.75m \leq 14.00$$

This equation represents the total cost of the ride, which must be less than or equal to Henri's budget of $\$14.00$.

Solving the Equation

To solve for $m$, we need to isolate the variable on one side of the equation. We can start by subtracting $1.25$ from both sides of the equation:

$$0.75m \leq 14.00 - 1.25$$

Simplifying the right-hand side of the equation, we get:

$$0.75m \leq 12.75$$

Next, we can divide both sides of the equation by $0.75$ to solve for $m$:

$$m \leq \frac{12.75}{0.75}$$

Calculating the Number of Miles

Now, we can calculate the number of miles Henri can afford by evaluating the expression on the right-hand side of the equation:

$$m \leq \frac{12.75}{0.75}$$

Using a calculator or performing the division manually, we get:

$$m \leq 17$$

Conclusion

Therefore, the correct answer is A. $m \leq 17$. This means that Henri can afford to ride in the taxi for up to $17$ miles, given his budget of $\$14.00$ and the pricing structure of the taxi service.

Discussion

This problem requires us to apply basic algebraic concepts, such as setting up and solving equations, to a real-world scenario. The pricing structure of the taxi service is a common example of a linear equation, where the cost per mile is a constant value. By solving for the number of miles Henri can afford, we can determine the maximum distance he can travel within his budget.

Additional Examples

Here are a few additional examples of how this problem can be applied to real-world scenarios:

• A taxi service charges a flat fee of $\$2.00$ and $\$1.00$ per mile. If Sarah has $\$15.00$, which of the following shows the number of miles she can afford to ride in the taxi?
• A taxi service charges a flat fee of $\$1.50$ and $\$0.50$ per mile. If John has $\$12.00$, which of the following shows the number of miles he can afford to ride in the taxi?

These examples require us to apply the same algebraic concepts and solve for the number of miles the individual can afford, given their budget and the pricing structure of the taxi service.

Final Thoughts

In conclusion, this problem requires us to apply basic algebraic concepts to a real-world scenario. By setting up and solving an equation, we can determine the number of miles Henri can afford to ride in the taxi, given his budget and the pricing structure of the taxi service. This problem is a great example of how algebra can be applied to real-world scenarios and is a useful tool for solving problems in mathematics and other fields.

Introduction

In our previous article, we explored a taxi service pricing problem where Henri had a budget of $\$14.00$ and the taxi service charged a flat fee of $\$1.25$ and $\$0.75$ per mile. We set up an equation and solved for the number of miles Henri could afford to ride in the taxi. In this article, we will provide a Q&A section to further clarify the concepts and provide additional examples.

Q: What is the pricing structure of the taxi service?

A: The taxi service charges a flat fee of $\$1.25$ and an additional $\$0.75$ per mile.

Q: How much does Henri have to spend on the taxi ride?

A: Henri has a budget of $\$14.00$.

Q: What is the equation that represents the total cost of the ride?

A: The equation is $1.25 + 0.75m \leq 14.00$, where $m$ is the number of miles Henri can afford.

Q: How do we solve for $m$ in the equation?

A: We can start by subtracting $1.25$ from both sides of the equation, then divide both sides by $0.75$ to solve for $m$.

Q: What is the value of $m$ that represents the number of miles Henri can afford?

A: The value of $m$ is $17$, which means Henri can afford to ride in the taxi for up to $17$ miles.

Q: What if the taxi service charged a flat fee of $\$2.00$ and $\$1.00$ per mile? How would we solve for the number of miles Sarah can afford?

A: We would set up a new equation, $2.00 + 1.00m \leq 15.00$, and solve for $m$ by subtracting $2.00$ from both sides and then dividing both sides by $1.00$.

Q: What if the taxi service charged a flat fee of $\$1.50$ and $\$0.50$ per mile? How would we solve for the number of miles John can afford?

A: We would set up a new equation, $1.50 + 0.50m \leq 12.00$, and solve for $m$ by subtracting $1.50$ from both sides and then dividing both sides by $0.50$.

Q: What is the main concept being applied in this problem?

A: The main concept being applied is algebra, specifically solving linear equations to represent real-world scenarios.

Q: Why is it important to understand the pricing structure of the taxi service?

A: Understanding the pricing structure is crucial to determine the affordability of a ride and to make informed decisions about transportation.

Q: Can this problem be applied to other real-world scenarios?

A: Yes, this problem can be applied to other real-world scenarios where pricing structures and budgets need to be considered.

Conclusion

In this Q&A article, we have provided additional examples and clarification on the concepts presented in our previous article. We have also highlighted the importance of understanding the pricing structure of the taxi service and the application of algebra to real-world scenarios.