Problem 93

Question

Answer the question with an algebraic expression. The cost of a 5-pound box of candy is \(c\) cents. What is the price per pound?

Step-by-Step Solution

Verified
Answer
The price per pound is \( \frac{c}{5} \) cents.
1Step 1: Understand the Problem
We are given the total cost of a 5-pound box of candy, which is represented by the variable \( c \) (in cents). The task is to find out how much one pound of candy costs, requiring us to find the price per pound.
2Step 2: Setup the Division Statement
To find the price per pound, we need to divide the total cost of the box by the number of pounds in the box. The total number of pounds is 5, so the division statement is \( \frac{c}{5} \).
3Step 3: Formulate the Algebraic Expression
Based on the division setup, the algebraic expression for the price of one pound of candy is \( \frac{c}{5} \). This expression represents the cost for each single pound based on the total cost "\( c \)" cents.

Key Concepts

Division in AlgebraVariables in ExpressionsCost Calculation in Algebra
Division in Algebra
In algebra, division is an operation that helps us break down larger, complex quantities into smaller, manageable parts. It's a fundamental skill in various mathematical concepts. When dividing in algebra, we typically take a total amount and separate it into equal parts. For example, if you have a whole, like a total cost or quantity, you can find out the value for a single unit by dividing this total by the number of units you have.
This concept is vital in scenarios like calculating average, rates, or prices per unit.
For instance,
  • If you have a total cost represented by a variable, you can find the cost per unit by dividing this variable by the number of units.
  • This is expressed in algebraic form using fractions or division statements, typically like this: \( \frac{c}{5} \).
Hence, division in algebra simplifies understanding and computation of ratios, per-unit costs, and more.
Variables in Expressions
Variables are symbols used to represent numbers or values in algebra. They make expressions universal and flexible as they can stand for unknown quantities, allowing us to perform operations without knowing the exact numbers.
Variables are crucial in forming equations and expressions to solve mathematical problems dynamically. For example:
  • In the expression \( \frac{c}{5} \), "\(c\)" is a variable representing the total cost in cents.
  • This allows us to substitute different values of \( c \) to find what the cost of one pound of candy would be under different scenarios.
By using variables, we can generalize mathematical statements, making it easier to solve various related problems with a single formula or expression.
Cost Calculation in Algebra
Cost calculation in algebra often involves using expressions to determine unknown costs based on given information. This involves applying operations such as addition, subtraction, multiplication, and importantly, division to find per unit costs or total costs.
Using algebraic expressions for cost calculation allows us to develop formulas that can be applied in real-life financial situations, like budgeting, pricing, or expense analysis. For instance:
  • In the exercise, \( \frac{c}{5} \) is an expression that calculates the cost per pound of candy.
  • This is derived from dividing the total cost \( c \) by the total weight, which is an essential technique in pricing analysis.
These calculations are practical tools for making informed decisions about purchases, expenses, or savings.