Problem 53 - Intermediate Algebra

Question

A motel in a suburb of Chicago rents double rooms for $$\$ 120$$ per day and single rooms for $$\$ 90$$ per day. If a total of 55 rooms were rented for $$\$ 6150$$, how many of each kind were rented?

Step-by-Step Solution

Answer

40 double rooms and 15 single rooms were rented.

Step 1: Define Variables

Let \( x \) be the number of double rooms rented and \( y \) be the number of single rooms rented.

Step 2: Set Up Equations

We have two pieces of information: the total number of rooms rented is 55 and the total revenue is \$6150. Thus, we can set up the following system of equations: \( x + y = 55 \) (equation 1) and \( 120x + 90y = 6150 \) (equation 2).

Step 3: Solve for One Variable

From equation 1, express \( y \) in terms of \( x \): \( y = 55 - x \).

Step 4: Substitute and Solve

Substitute \( y = 55 - x \) into equation 2: \( 120x + 90(55 - x) = 6150 \). Simplifying, we get \( 120x + 4950 - 90x = 6150 \), which simplifies to \( 30x = 1200 \).

Step 5: Solve for \( x \)

Divide both sides by 30: \( x = 40 \). Hence, 40 double rooms were rented.

Step 6: Solve for \( y \)

Substitute \( x = 40 \) back into equation 1: \( y = 55 - 40 = 15 \). Hence, 15 single rooms were rented.

Key Concepts

Algebraic Problem Solving

Algebraic problem solving is a fundamental skill that helps us tackle a variety of mathematical challenges by using algebraic equations and expressions. In the context of our exercise, we are dealing with a real-world scenario involving money and rooms. To begin solving, it's important to first identify and define our variables clearly.

• In our problem, we defined the variables as follows: let \( x \) represent the double rooms rented and \( y \) represent the single rooms rented.
• Having these variables allows us to set up equations based on the problem's conditions, essentially translating words into math expressions.

This translation process is crucial because it forms the foundation for solving the problem by making it manageable through algebra. As we proceed, it's important to manipulate these equations, substituting one equation into another if needed, to find the values of unknown variables.

Equation Systems

Equation systems consist of two or more equations that are interrelated and share common variables. In solving these, there is a need to find the values of the variables that satisfy all the equations simultaneously. In our exercise, we encounter a system of two equations derived from the problem statements:

• Equation 1: \( x + y = 55 \)
• Equation 2: \( 120x + 90y = 6150 \)

These equations are set up from the total number of rooms rented and the total revenue. Solving such systems requires finding an effective method to reduce these equations to a simpler form.
A common technique used here is substitution or elimination, which helps in finding the solution efficiently. After isolating a variable from one equation, you can substitute it in the other, making it easier to solve. This technique allows you to handle practical problems involving multiple variables and conditions easily.

Word Problems

Word problems are a great way to apply mathematical theories to real-life situations. They require interpreting descriptive information and framing it as mathematical equations. In our exercise, we had a real-life scenario involving room rentals at a motel, with the challenge of figuring out how many of each type of room were rented.

• The first step in tackling word problems is proper understanding and reading to ascertain what is being asked.
• Next, break down the problem by identifying relevant quantities and relationships, which can often be translated into algebraic equations.
• Finally, once you have set up the equations properly, solving them with algebraic methods becomes efficient.

Word problems enhance critical thinking because they require piecing together data and navigating through numbers to derive a meaningful solution.