Solving for Peters Salary Using Proportions

Solving for Peters Salary Using Proportions

Understanding how to solve salary-related problems using proportions is a valuable skill, especially in a professional setting. This article will walk you through a specific problem where we need to find Peters salary based on the given proportions and Mary's salary. We will demonstrate several methods to arrive at the solution and explore the underlying mathematical principles.

Problem Statement

The question at hand asks, 'If 3/4 of Peters salary is equal to 2/4 of Marys salary, and Marys salary is 120000, what is Peters salary?' Let's break this down step by step.

Mathematical Formulation

Using the information given, we can set up the following premise and assumptions:

Premise

3/4P2/4M

Assumptions

Mary's salary, M, is 120000 Peter's salary, P, in terms of M is 2/4M

Calculation

Let's solve for Peter's salary with the given values. We start by substituting Mary's salary into the equation:

3/4P 2/4M

3/4P 2/4 * 120000

3/4P 60000

To find P, we can multiply both sides of the equation by 4/3:

P 60000 * 4/3

P 80000

Proof of Solution

To verify our solution, we can substitute P 80000 back into the original equation:

3/4 * 80000 2/4 * 120000

60000 60000

The equation checks out, confirming that Peter's salary is indeed 80000.

Alternative Methods

There are several ways to approach this problem. Here are a couple of additional methods:

Method 1

Since 3/4P 2/4M, and M 120000, we can equate:

3/4P 2/4 * 120000

3/4P 60000

P 60000 * 4/3

P 80000

Method 2

Representing the problem with symbols:

P Peter’s salary

M Mary’s salary 120000

3/4P 2/4M

3/4P 60000

P 60000 * 4/3

P 80000

Conclusion

In conclusion, by using the principles of proportion and algebra, we have found that Peter’s salary is 80000. This method not only helps in solving the given problem but also has broader applications in financial and economic contexts.

Key Takeaways

Using proportions to solve for unknown values in salary problems Applying algebraic equations to find solutions Ensuring consistency and accuracy in calculations