Problem 8 - Algebra for College Students

Question

8\. Solve \(i=P r t\) for \(P\), given that \(r=8 \frac{1}{2} \%, t=2\) years, and \(i=\$ 204\).

Step-by-Step Solution

Answer

The principal amount, \( P \), is \$1200.

Step 1: Understand the formula

The formula given is the interest formula: \( i = P r t \). In this formula, \( i \) represents the interest earned, \( P \) is the principal amount, \( r \) is the rate of interest per year in decimal form, and \( t \) is the time in years. We need to solve for \( P \).

Step 2: Convert percentage to decimal

The given interest rate \( r = 8 \frac{1}{2} \% \) needs to be converted into decimal form. First, express the mixed number as an improper fraction: \( 8 \frac{1}{2} = \frac{17}{2} \). Then convert it to decimal: \( r = \frac{17}{2} \times \frac{1}{100} = 0.085 \).

Step 3: Substitute known values into the equation

Substitute \( i = 204 \), \( r = 0.085 \), and \( t = 2 \) into the formula \( i = P r t \). This gives us the equation: \( 204 = P \times 0.085 \times 2 \).

Step 4: Simplify the equation

Calculate the product of \( r \) and \( t \): \( 0.085 \times 2 = 0.17 \). So, the equation becomes: \( 204 = P \times 0.17 \).

Step 5: Solve for P

To solve for \( P \), divide both sides of the equation by 0.17: \( P = \frac{204}{0.17} \). Calculating gives \( P = 1200 \).

Key Concepts

Interest Calculation

Interest calculation is a fundamental concept in algebraic formulas and is widely used in finance. At its core, this concept helps you understand how much extra money, known as interest, you earn or pay on a principal amount over a set period. The formula for calculating simple interest is:

• \( i = P r t \)

where:

• \( i \) is the interest earned or paid.
• \( P \) is the principal amount or initial investment.
• \( r \) is the annual interest rate in decimal form.
• \( t \) is the time the money is invested or borrowed for, in years.

To accurately find any of these components, understanding the others and how they interact is crucial. For instance, knowing how to manipulate this formula allows you to solve for \( P \), given the interest, rate, and time.

Rate Conversion

In financial calculations, rates are typically presented as percentages but must be converted to decimals for use in formulas. This conversion is essential for accurate calculations. The conversion is simple:

• Express the percentage as a fraction of 100: \( 8 \frac{1}{2} \% \) becomes \( \frac{17}{2} \times \frac{1}{100} \).

This ensures that you can accurately use the rate in mathematical equations. It is crucial to perform this step correctly to prevent errors in the calculation. The resulting decimal \( r = 0.085 \) can then be used in the interest formula to find the principal or check the correctness of given data.

Solving Equations

Solving equations involves finding the value of an unknown in an equation. This requires a systematic approach. For example, in the equation \( i = P r t \), if we need to find \( P \), follow these steps:

• Substitute known values into the equation: \( 204 = P \times 0.085 \times 2 \).
• Simplify by multiplying the known factors: \( 0.17 \times P = 204 \).
• Isolate \( P \) by dividing both sides by the product of \( r \) and \( t \): \( P = \frac{204}{0.17} \).

Then, solve for \( P \). This method of isolating the variable is fundamental in algebraic problem-solving, ensuring you understand and document each step properly.

Mathematical Problem Solving

Mathematical problem-solving is a skill that combines understanding concepts with executing the right operations. Approaching a problem, break it down into manageable parts:

• Identify what is being asked. In our case, find \( P \).
• Gather and convert all given data, such as rates and times.
• Substitute these values into a relevant formula.
• Simplify and solve the equation step by step.

By following methodical steps, you ensure a logical flow of action, which minimizes errors and enhances comprehension. This approach not only solves the current problem but also builds strong foundations for tackling more complex algebraic problems in the future.