Question

A city utility department charges residential customers \(\$ 2.50\) per \(1,000\) gallons of water and \(\$ 16.00\) per month for trash pickup. Which of the following expressions gives a residential customer's total monthly charges, in dollars, for use of \(g\) thousand gallons of water and trash pickup? F. \(\quad 2.50 g+\quad 16.00\) G. \(\quad 2.50 g+1,016.00\) H. \(16.00 g+\quad 2.50\) J. \(\quad 18.50 g\) K. \(2,500.00 g+16.00\)

Short answer

Answer: The correct expression is 2.50g + 16.00.

Step-by-step solution

Determine water charges

To determine the charge for water usage, we need to multiply the cost per \(1,000\) gallons by the number of \(g\) thousand gallons used. This gives us: \(2.50 * g\) or \(2.50g\).

Include trash pickup charges

To include the cost of trash pickup, we add the fixed monthly charge to the water usage charge. Thus, the total cost becomes: \(2.50g + 16.00\).

Match the expression with the answer choices

We compare our derived expression, \(2.50g + 16.00\), with the answer choices provided. We find that it matches answer choice F: \(\quad 2.50 g+\quad 16.00\). The correct expression for the total monthly charges for a residential customer is therefore given by F.

Key concepts

Utility Charges Calculation

Understanding utility charges calculation is essential for managing personal finances and for solving ACT Math Problems related to real-life scenarios. In the context of the exercise provided, the utility charges involve two components: a variable component based on water consumption and a fixed component for trash pickup.

Here's how to calculate such utility charges:

• Identify the variable charge rate, which in this case is \(\$ 2.50\) per \(1,000\) gallons of water.
• Determine the fixed charge, which here is \(\$ 16.00\) per month for trash pickup.
• Multiply the variable rate by the number of units consumed (thousands of gallons used) to find the total variable charge.
• Add the fixed charge to the variable charge to determine the total monthly utility charges.

This method can be applied to any utility charges that include both variable and fixed components. It's crucial to keep the units consistent when making these calculations and to be able to translate these processes into algebraic expressions, which brings us to our next concept.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, symbols, and variables that represent unknown values. They are an integral part of solving word problems, particularly in ACT Math, as they provide a framework for representing real-world scenarios algebraically.

In our example, the expression for calculating a customer's monthly utility charges is represented by \(2.50g + 16.00\), where \(g\) is the number of thousands of gallons of water used. A few important aspects of working with algebraic expressions include:

• Identifying variables that represent unknown quantities, which allows us to solve for them.
• Understanding how to translate worded scenarios into algebraic terms, such as turning 'per \(1,000\) gallons of water' into the multiplication \(2.50 \times g\).
• Applying the correct mathematical operations, like addition or multiplication, to accurately depict the situation being modeled.

The clarity of an algebraic expression is pivotal as it directly impacts one's ability to solve complex problems effectively.

Word Problem Solving

Solving word problems involves several key steps and skills that encompass reading comprehension, translation of text into mathematical language, and critical thinking.

To tackle word problems effectively:

• Read the problem carefully to understand what is being asked.
• Identify the relevant information and what quantities represent, just as \(g\) represents the thousand gallons of water usage in the utility charges problem.
• Translate the written description into mathematical expressions or equations. This involves recognizing key words that indicate mathematical operations. For instance, 'per' usually suggests multiplication, whereas 'total' implies addition.
• Solve the equation using appropriate algebraic methods.
• Check your answer back in the context of the problem to ensure it makes sense.

When applying these steps to our example, it becomes clear how each piece of information leads to the creation of an algebraic expression that models the situation. Practice and familiarity with the language of word problems are two tools that will enhance a student's ability to solve them quickly and accurately.