Problem 62 - A Graphical Approach to Precalculus with Limits

Question

Investment problems such as those in Exercises \(61-66\) 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 A man 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

The man paid $70,000 for the first plot and $50,000 for the second plot.

Step 1: Define Variables

Let's define variables for the amounts spent on each plot of land. Let \( x \) be the cost of the first plot, and \( y \) be the cost of the second plot. We know from the problem statement that the man bought these plots for a total of \( \$120,000 \). This gives us the first equation: \( x + y = 120,000 \).

Step 2: Formulate Equations from Profits and Losses

From the problem, the man gains a 15% profit on the first plot. Hence, the selling price of the first plot is \( 1.15x \). He loses 10% on the second plot, meaning the selling price is \( 0.90y \). The total profit is \( \$5500 \), and this gives the second equation: \( 1.15x + 0.90y = 120,000 + 5,500 \).

Step 3: Simplify and Solve the System of Equations

We have two equations: \( x + y = 120,000 \) and \( 1.15x + 0.90y = 125,500 \). We can solve this system using substitution or elimination. Let's use substitution. From the first equation, express \( y \) in terms of \( x \): \( y = 120,000 - x \). Substitute \( y \) in the second equation: \( 1.15x + 0.90(120,000 - x) = 125,500 \).

Step 4: Solve the Single Variable Equation

Substitute and simplify: \( 1.15x + 108,000 - 0.90x = 125,500 \). Combine like terms: \( 0.25x + 108,000 = 125,500 \). Solve for \( x \): \( 0.25x = 125,500 - 108,000 \) which simplifies to \( 0.25x = 17,500 \). Finally, \( x = \frac{17,500}{0.25} = 70,000 \).

Step 5: Determine the Cost of the Second Plot

Now that we know \( x \), substitute it back into the first equation to find \( y \): \( y = 120,000 - 70,000 = 50,000 \).

Key Concepts

System of Equations

In the context of investment problems, understanding how to solve a system of equations is crucial. A system of equations consists of two or more equations that share the same set of unknowns. Let's break it down into simple terms: imagine you have two puzzles that you need to solve with the same pieces.

To find out how much was paid for each plot of land, we set up two equations based on the given information:

• The sum of amounts paid for the two plots is $120,000, forming the equation: \( x + y = 120,000 \).
• Considering the profit and loss from selling the plots, we form the second equation: \( 1.15x + 0.90y = 125,500 \).

The solution to this system of equations will tell us the cost of each plot of land. Typically, substitution or elimination methods are used to find the values of the unknowns, \( x \) and \( y \). Substitution involves expressing one variable in terms of the other from one equation, and plugging it into the second equation, simplified as we see when finding \( y \) in terms of \( x \).When both equations yield the same values for \( x \) and \( y \), you have discovered the solution to the system of equations. This approach is vital in solving many real-world problems, particularly in business and finance.

Profit and Loss Calculation

Profit and loss calculations are a foundational concept for making sense of financial transactions, especially in investment problems. In this exercise, determining how much profit or loss was made from selling land required understanding percentages and how they affect the original amounts.

Let's break down what's happening:

• The first plot was sold at a 15% profit. This means that the selling price was 115% of the original price, represented as \( 1.15x \).
• The second plot, however, was sold at a 10% loss. Therefore, the selling price was only 90% of what was originally paid, or \( 0.90y \).

The total profit from selling both plots was \(5,500, which after adding to the initial investment sum (\)120,000), totals the selling prices \( 1.15x + 0.90y = 125,500 \). These calculations help us verify the precision in the financial transaction by ensuring the overall profitability analysis.
Understanding profits and losses not only assists in determining the selling prices but also enlightens other business activities like pricing strategies or investment decisions. In any business deal, knowing how percentages work can mean the difference between a successful venture and a loss.

Investment Problems

Investment problems focus on calculating and analyzing how money can grow over time or change through transactions. These problems often involve using equations based on profits and interests, such as the Simple Interest Formula: \( I = PRT \), where:

• \( I \) is the interest earned,
• \( P \) is the principal amount,
• \( R \) is the rate of interest expressed as a decimal,
• \( T \) is the time in years.

In this land purchase scenario, the ins and outs of the investment include the initial purchase cost and the profits/loss from sales. Although the Simple Interest Formula isn't applied directly here, understanding it builds foundational knowledge for similar problems.
Investment problems help in predicting future financial standing or assessing the viability of financial actions, like how much should be spent, saved, or the potential returns on investment. Importantly, applying these formulas and equations allows businesses to make informed decisions, mitigate risks, and maximize profitability, equipping you with the skills to solve real-world financial challenges effectively.