Problem 87 - Intermediate Algebra

Question

Solve each problem by writing a variation model. Finding Distance. The distance that a car can go varies directly as the number of gallons of gasoline it consumes. If a car can go 288 miles on 12 gallons of gasoline, how far can it go on a full tank of 18 gallons?

Step-by-Step Solution

Answer

The car can go 432 miles on 18 gallons.

Step 1: Understand the Variation Model

We are dealing with a direct variation model where the distance \( d \) varies directly as the number of gallons \( g \). This relationship can be expressed as \( d = k \, g \), where \( k \) is the constant of variation.

Step 2: Identify Known Values

According to the problem, a car travels 288 miles on 12 gallons of gasoline. Thus, \( d = 288 \) and \( g = 12 \).

Step 3: Solve for the Constant of Variation

Using the direct variation formula \( d = k \, g \), substitute \( d = 288 \) and \( g = 12 \). Thus, \( 288 = k \, 12 \). Solve for \( k \) to find \( k = \frac{288}{12} = 24 \).

Step 4: Use the Constant to Find the New Distance

We now want to find the distance \( d \) traveled on 18 gallons. Using the formula \( d = k \, g \) with \( k = 24 \) and \( g = 18 \), compute \( d = 24 \, 18 = 432 \).

Step 5: Conclusion

The car can travel 432 miles on a full tank of 18 gallons of gasoline using the direct variation model.

Key Concepts

Constant of Variation

When solving problems of direct variation, the constant of variation is a pivotal concept. It serves as the constant ratio that links two variables together. In our case of distance and gasoline consumption, it offers a way to describe how the distance a car can travel is dependent on the gas it uses.

The formula representing direct variation is expressed as:

• \( d = k \cdot g \)

This formula reflects the relationship where \( d \) (distance) directly varies with \( g \) (gallons of gasoline), and \( k \) is the constant of variation. To find \( k \), rearrange the equation and divide the distance by the gallons of gasoline:

• \( k = \frac{d}{g} \)

In the exercise, with 288 miles for 12 gallons, the calculation reveals \( k = 24 \). This means that for every gallon, the car travels 24 miles. This constant provides a foundation to predict the car's travel based on any quantity of gasoline it consumes.

Distance Calculation

Distance calculation in the context of direct variation involves using the constant of variation you previously found. The process becomes straightforward once \( k \) is determined. Let's see how this works for a different scenario.

With \( k = 24 \), if you wish to calculate how far a car travels with 18 gallons of gasoline, you apply the direct variation formula:

• \( d = k \cdot g \)

With the known values, substitute:\

• \( d = 24 \times 18 = 432 \)

This result indicates that the car can travel 432 miles with 18 gallons. By incorporating the constant of variation, calculations become predictions based on consistent measurement, ensuring you know how far you can drive based on the fuel you use.

Algebraic Problem Solving

Algebraic problem solving facilitates finding unknowns through a structured approach using equations. For direct variation problems, it's particularly useful as it provides a clear pathway to discovering relationships between variables.

Start by clearly identifying what you know and what you need to find. Building an equation with variables makes managing the problem methodical. Here's how it applied to our situation:

• First, express the known variation model: \( d = k \cdot g \)
• Plug in known values to solve for the constant \( k \): \( 288 = k \cdot 12 \)
• Rearrange and solve for \( k \): \( k = 24 \)
• Use \( k \) to solve for any unknown distance (\( d \)) given a different amount of fuel.

Through algebra, you're adding flexibility and power to how you solve and predict outcomes for various practical scenarios. Taking the time to set up your equations correctly ensures you're able to tackle similar problems with confidence in the future.