Problem 61 - Algebra and Trigonometry

Question

Gas Station A gas station sells regular gas for \(\$ 2.20\) per gallon and premium gas for \(\$ 3.00\) allon. At the end of a business day 280 gallons of gas were sold, and receipts totaled S680. How many gallons of each type of gas were sold?

Step-by-Step Solution

Answer

200 gallons of regular gas and 80 gallons of premium gas were sold.

Step 1: Define Variables

Let \( x \) be the number of gallons of regular gas sold, and \( y \) be the number of gallons of premium gas sold.

Step 2: Set Up Equation for Total Gallons

The total number of gallons sold is 280. This gives us the equation: \( x + y = 280 \).

Step 3: Set Up Equation for Total Revenue

The total revenue from selling gas is \$680. The revenue from regular gas is \( 2.20x \), and from premium gas is \( 3.00y \). This gives us the second equation: \( 2.20x + 3.00y = 680 \).

Step 4: Solve the System of Equations

We now have two equations: \( x + y = 280 \) and \( 2.20x + 3.00y = 680 \). Solve this system by substitution or elimination. Substitute \( y = 280 - x \) into the second equation.

Step 5: Substitute and Simplify

Substitute \( y = 280 - x \) into the revenue equation: \( 2.20x + 3.00(280 - x) = 680 \). Simplify the equation: \( 2.20x + 840 - 3.00x = 680 \).

Step 6: Solve for x

Combine like terms: \( -0.80x = -160 \). Divide both sides by -0.80 to solve for \( x \): \( x = 200 \).

Step 7: Solve for y

Use \( y = 280 - x \) to find \( y \). Substitute \( x = 200 \): \( y = 280 - 200 = 80 \).

Step 8: Verify the Solution

Check that the solution satisfies both original equations. For the gallons equation: \( 200 + 80 = 280 \). For the revenue equation: \( 2.20(200) + 3.00(80) = 680 \). Both hold true.

Key Concepts

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In our gas station problem, algebra helps us create equations that represent real-world situations. We define variables such as \( x \) and \( y \) to represent the unknowns in the problem, like the number of gallons of gas sold. By using algebraic operations, we can form equations that reflect various conditions or constraints described in the problem. This allows us to methodically solve for these unknowns by expressing relationships using mathematical expressions.

Linear Equations

Linear equations are equations that make a straight line when graphed. They are used to describe a constant rate of change in mathematical problems. In the context of our word problem, two linear equations are set up based on the given information about gallons and prices. The first equation \( x + y = 280 \) represents the total gallons of gas sold. The second equation \( 2.20x + 3.00y = 680 \) represents the total revenue from the gas sold.
These equations are linear because they contain the variables \( x \) and \( y \) to the power of one. Solving these equations involves finding the values of the variables that satisfy both equations simultaneously, often using methods like substitution or elimination.

Word Problems

Word problems involve real-life situations translated into mathematical formulations. They require the reader to interpret text to identify the mathematical operations needed. In this gas station example, the problem gives us details about prices and sales, and we must express these verbally stated conditions in the form of equations. It's like solving a puzzle where different pieces of information combine to form a complete picture.
To tackle word problems effectively:

• Carefully read and understand what is being asked.
• Identify and define the unknown variables.
• Translate the wording into mathematical equations.
• Solve these equations using appropriate mathematical techniques.

Learning to interpret word problems can make math more applicable to everyday life and is a valuable skill in problem-solving scenarios.