Question

A New York City taxi charges an initial fee of \(\$ 2.50\) plus \(\$ 0.50\) for every \(\frac{1}{5}\) of a mile traveled. If Jim's total fare (not including tip) equals \(\$ 14.50\), how many miles did Jim travel? (A) \(4.8\) miles (B) \(5.8\) miles (C) 24 miles (D) 29 miles

Short answer

The short answer based on the provided step-by-step solution is: Jim traveled 24 miles.

Step-by-step solution

Write down the given information

We know that the initial fee is \(2.50\), and every \(\frac{1}{5}\) of a mile costs an additional \(0.50\). Let's denote the distance Jim traveled by \(x\). Therefore, we can represent the total cost as: \(2.50 + 0.50(\frac{1}{5}x) = 14.50\)

Solve the equation for \(x\)

Now, we want to solve the equation for the distance \(x\). First, we can rewrite the equation as: \((2.50 + 0.50(\frac{1}{5}x) = 14.50)

Simplify the equation

Now, we can simplify the equation by multiplying the distance by the rate: \(2.50 + 0.10x = 14.50\)

Subtract the initial fee from both sides of the equation

To isolate the term with the distance, subtract the initial fee from both sides of the equation: \(0.10x = 12\)

Divide both sides by the rate

Now, to find the distance in miles, divide both sides of the equation by the rate (\(0.10\)): \(x = 120\)

Convert the distance to miles

We know that \(x\) represents the distance in \(\frac{1}{5}\) of a mile. To convert the value of \(x\) to miles, divide it by \(5\): \(miles = \frac{x}{5}\) \(miles = \frac{120}{5}\) \(miles = 24\) So, the distance Jim traveled is 24 miles, which corresponds to answer choice (C).

Key concepts

Algebraic Equations

Understanding algebraic equations is essential for solving a variety of math problems, including setting up and solving problems related to distance and cost calculations. An algebraic equation is a mathematical statement that shows the relationship between different variables and constants. It typically involves equalities and can include addition, subtraction, multiplication, and division.

In our exercise example, the equation \(2.50 + 0.50(\frac{1}{5}x) = 14.50\) uses algebra to represent a real-world situation involving a taxi fare. The initial fee and rate per distance unit are both constants, while the distance \(x\) is the variable we aim to find. To solve algebraic equations, one often simplifies the equation, isolates the variable, and then solves for the variable through arithmetic operations.

Linear Equations

Linear equations are a subset of algebraic equations and are characterized by variables that have a power of one. These equations form straight lines when graphed on a coordinate axis. They generally have the standard form \(ax + b = c\), where \(x\) is the variable, and \(a\), \(b\), and \(c\) are constants with \(a\) not equal to zero.

In the taxi fare problem, we can see the equation \(2.50 + 0.10x = 14.50\) as a simple linear equation once simplified. This simplicity allows for easy manipulation to solve for \(x\), representing the distance traveled. In this case, solving a linear equation involves subtracting the initial fee from both sides and then dividing by the rate to find the value of \(x\).

Math Word Problems

Math word problems require reading comprehension, translation of words into an algebraic representation, and then using math skills to solve the problem. They often depict realistic scenarios that involve everyday contexts, such as shopping or travel, to help students apply mathematical concepts to life.

To approach them effectively:

• Identify the quantities and their relationships.
• Translate text into mathematical expressions.
• Carefully define the variable and construct an equation that models the scenario.
• Solve the equation for the unknown.

For instance, in our GED practice question, we process the information about the taxi fare and convert it into the equation \(2.50 + 0.50(\frac{1}{5}x) = 14.50\). The process of translation from text to equations is a critical skill in tackling word problems.

Distance Calculation

Distance calculation is a practical application of algebra. It involves determining the extent of movement from one point to another, and it's a common type of problem in GED math practice tests. In these problems, distance is often the variable we need to solve for, and the equation given usually relates distance to other factors such as speed and time or, as in our example, the rate and total cost.

To calculate distance in the problem we're discussing:

• Understand the relationship of the cost per distance unit.
• Set up an equation with distance as the variable.
• Perform algebraic operations to solve for the distance.

After isolating the distance variable and solving for it, always remember to convert the value into actionable units, miles in this case, if necessary. The answer not only serves as a numerical solution but also helps in understanding the relationship between linear movement and its associated cost.