Problem 89 - Prealgebra and Introductory Algebra

Question

Suppose that you drive about \(12,000 \mathrm{mi}\) per year and that the cost of gasoline averages 3.70 dollar per gallon. a. Let \(x\) represent the number of miles per gallon your car gets. Write a variable expression for the amount you spend on gasoline in one year. b. Write and simplify a variable expression for the amount of money you will save each year if you increase your gas mileage by 5 miles per gallon. c. If you currently get 25 miles per gallon and you increase your gas mileage by 5 miles per gallon, how much will you save in one year?

Step-by-Step Solution

Answer

The yearly saving by improving the mileage by 5 mpg for a car originally getting 25 mpg are about $444.

Step 1: Formulating cost per year

The amount of gas needed in a year is calculated by dividing the total miles driven by the number of miles one can drive with a gallon of gas, \(x\). The cost per year in gas is then obtained by multiplying the number of used gallons with the price per gallon. This leads to the following expression for the cost per year: \((12,000 / x) * 3.7\)

Step 2: Formulating the yearly savings

The amount saved by driving more fuel efficiently by 5 miles per gallon is calculated by subtracting the cost per year for the improved efficiency from the cost per year for the original efficiency. The expression for the original cost was found in step 1, the expression for the improved efficiency is the same but using \(x+5\) instead of \(x\) for the miles per gallon. The expression for the yearly savings then becomes: \((12,000 / x) * 3.7 - (12,000 / (x+5)) * 3.7\)

Step 3: Applying the savings expression

Now the expression for the yearly savings can be applied to the specific situation where the car originally had a mileage of 25 miles per gallon. This is done by inserting 25 for \(x\) into the expression from step 2, and then doing the calculation: \((12,000 / 25) * 3.7 - (12,000 / (25+5)) * 3.7\) approximately equals 444

Key Concepts

miles per gallon

Miles per gallon (mpg) is a measure of how far a vehicle can travel with one gallon of gasoline. In algebraic terms, it represents the efficiency of a vehicle's fuel consumption. For instance, if your car gets 25 miles per gallon, it means you can drive 25 miles on just one gallon of gas. This is essential when calculating how much gasoline is needed for a given distance. Suppose you drive 12,000 miles per year; your mpg will help determine how many gallons you need for the entire year.
To figure out how much gasoline you use annually, you divide the total miles driven by the miles per gallon. The formula is:

• Gasoline needed = Total miles driven / Miles per gallon

This formula is foundational for budgeting fuel expenses and exploring how increased fuel efficiency can lead to cost savings.

cost of gasoline

The cost of gasoline is a critical factor when calculating yearly fuel expenses. It is expressed as the price you pay for each gallon of gas. In this exercise, the cost is given as \(3.70 per gallon. To find out how much you spend on gasoline annually, you multiply the number of gallons you use by the price per gallon.

• Annual gasoline cost = Number of gallons needed * Cost per gallon

Using the earlier example, if you need 600 gallons a year, at \)3.70 each, your yearly cost is computed as:\( 600 \times 3.7 = 2220 \)This figure gives you the total expense on gasoline per year, which is a significant part of your vehicle's operating costs. Knowing this cost helps in planning your budget effectively.

savings calculation

Understanding savings through improved mileage involves calculating how much money you can save if your car's fuel efficiency improves. In our example, increasing the mileage by 5 miles per gallon leads to direct savings. Here's how it works:First, compute the cost of driving at your current miles per gallon. Then, calculate the cost with the improved mileage.

• Current cost: \( (12,000 / x) \times 3.7 \) dollars
• Improved cost: \( (12,000 / (x + 5)) \times 3.7 \) dollars

The difference between these two calculations gives the potential savings. For instance, if a car moves from 25 mpg to 30 mpg,\[ \left( \frac{12,000}{25} \times 3.7 \right) - \left( \frac{12,000}{30} \times 3.7 \right) \approx 444 \]This means you would save approximately $444 annually by improving your efficiency.

variable expressions

Variable expressions are fundamental in algebra to represent dynamic quantities. Variables, like \(x\), stand in for unknown or varying values, allowing us to write flexible equations. In the context of this exercise, \(x\) represents the miles per gallon your car currently achieves.By using variable expressions, you can easily model real-life situations such as the annual cost of gasoline. For instance, the expression \((12,000 / x) \times 3.7\) effectively calculates total gasoline cost based on different fuel efficiencies without needing to plug in static numbers.Variable expressions enable us to explore different scenarios. They’re crucial for deriving formulas that model savings calculations or cost alterations prompted by changes in car efficiency, like increasing miles per gallon.