Problem 76 - A Graphical Approach to College Algebra

Question

Investment problems such as those in Exercises \(75-80\) can be solved by using a method similar to the one explained in Example \(2,\) along with the simple- interest formula \(I=P R T\) where I is the interest earned, \(P\) is the initial amount of money deposited, \(R\) is the annual interest rate as a decimal, and \(T\) is the time the money is deposited in years. Solve each problem. Let \(T=1\) year for each exercise. Buying and Selling Land Bobby Aguillard bought two plots of land for a total of \(\$ 120,000 .\) When he sold the first plot, he made a profit of \(15 \% .\) When he sold the second, he lost \(10 \% .\) His total profit was \(\$ 5500 .\) How much did he pay for each piece of land?

Step-by-Step Solution

Answer

Bobby paid \$70,000 for the first plot and \$50,000 for the second plot.

Step 1: Define Variables

Let \(x\) be the price paid for the first plot, and hence \(120,000 - x\) is the price paid for the second plot.

Step 2: Express Profits and Losses as Equations

For the first plot: The selling price is \(x + 0.15x = 1.15x\) and the profit is \(0.15x\). For the second plot: The selling price is \( (120,000 - x) - 0.1(120,000 - x) = 0.9(120,000 - x)\) and the loss is \(-0.1(120,000 - x)\).

Step 3: Set Up the Total Profit Equation

According to the problem, his total profit equals \$5500. Therefore, the equation is \(0.15x - 0.1(120,000 - x) = 5500\).

Step 4: Simplify the Equation

Substitute and simplify the equation: \(0.15x - 12,000 + 0.1x = 5,500\), which simplifies to \(0.25x - 12,000 = 5,500\).

Step 5: Solve for x

Add 12,000 to both sides to isolate terms with \(x\): \(0.25x = 17,500\). Then, divide both sides by 0.25 to solve for \(x\): \(x = 70,000\).

Step 6: Find the Price of the Second Plot

Subtract the price of the first plot from the total: \(120,000 - x = 120,000 - 70,000 = 50,000\).

Key Concepts

Simple Interest Formula

The simple interest formula is a fundamental concept in finance that helps us calculate interest earned on investments. The formula is expressed as \( I = PRT \):

• \( I \) is the interest earned.
• \( P \) is the principal amount, or the initial sum of money invested.
• \( R \) represents the annual interest rate expressed as a decimal. For instance, an interest rate of 5% would be \( 0.05 \).
• \( T \) is the time duration in years for which the money is invested.

This formula makes it easy to compute the total interest over a set period, assuming a constant interest rate. In practical problems, it helps to calculate the additional money you gain when investing. So, using the right values for each variable allows solving real-life investment problems efficiently.

Profit and Loss Calculation

Understanding how to calculate profit and loss is crucial when analyzing the success of an investment or a sale. In a profit and loss scenario:

• A profit occurs when the selling price is higher than the purchase price. The profit can then be expressed as a percentage of the cost price.
• A loss occurs when the selling price is lower than the purchase price. Loss, too, can be expressed as a percentage of the cost price.

In the given exercise, Bobby Aguillard's total profit from two plots of land was \( \$5500 \). He made a 15% profit on the first plot and incurred a 10% loss on the second. These percentages allowed formulating equations that were key to finding how much he initially paid for each plot. Being able to convert these percentages into equations is essential in evaluating investment outcomes.

Algebraic Equations

Algebraic equations are powerful tools in solving investment problems like the one in the exercise. They allow us to represent complex real-world situations using mathematical expressions.

• The equation \( 0.15x - 0.1(120,000 - x) = 5500 \) helped determine the distribution of cost between the two plots.
• This equation was derived from the profit percentages and total profit combined.

Through the process of simplification and solving these equations, we can pinpoint exact numerical values, such as the price paid for each land plot. Manipulating such equations involves steps like distributing, combining like terms, adding, and dividing, all of which reveal the hidden relationships in the problem.

Variable Definition

Defining variables is a foundational skill for tackling mathematical problems involving multiple unknowns. In the exercise, we defined variables to represent the amounts spent on each plot:

• \( x \) was defined as the amount paid for the first plot.
• \( 120,000 - x \) was used for the second plot, reflecting that the total investment was \( \$120,000 \).

Variable definition allows for clear representation and understanding of what each part of the equation stands for. It simplifies the problem-solving process by making it possible to structure equations that can be solved systematically. Always think of variables as placeholders for unknown numbers that you aim to find through the problem-solving process.