Problem 23 - Precalculus Mathematics for Calculus

Question

What annual rate of interest would you have to earn on an investment of \(\$ 3500\) to ensure receiving \(\$ 262.50\) interest after 1 year?

Step-by-Step Solution

Answer

The annual interest rate required is 7.5%.

Step 1: Identify the Formula Needed

To find the annual interest rate, we can use the simple interest formula: \[ I = P \times r \times t \] where \( I \) is the interest earned, \( P \) is the principal amount, \( r \) is the rate of interest, and \( t \) is the time in years.

Step 2: Setup Known Values

In this problem, we know the principal \( P = 3500 \), the interest \( I = 262.50 \), and the time \( t = 1 \) year. We need to find the rate of interest \( r \).

Step 3: Rearrange the Formula to Solve for Interest Rate

Rearrange the formula to solve for \( r \):\[ r = \frac{I}{P \times t} \]

Step 4: Substitute Known Values into the Formula

Substitute the values into the formula: \[ r = \frac{262.50}{3500 \times 1} \]

Step 5: Calculate the Interest Rate

Perform the calculation: \[ r = \frac{262.50}{3500} \] which evaluates to \( r = 0.075 \) or 7.5%.

Key Concepts

Simple Interest Formula

Interest calculation is straightforward when using the simple interest formula. This formula helps you to determine either the interest earned, the principal amount, or the rate of interest over a specific time period. The formula is: \[ I = P \times r \times t \] where:

• \( I \) is the interest earned
• \( P \) is the principal amount
• \( r \) is the rate of interest (usually given as a decimal)
• \( t \) is the time, typically in years

The simple interest doesn't take into account the compounding of interest, which makes it suitable for short-term investments or loans. Each year's interest is calculated on the principal, ensuring consistent and predictable outcomes. For example, a principal of \( \\(3500 \) at a rate of \( 7.5\% \) per year will generate a fixed \( \\)262.50 \) annually.
To effectively use the formula, make sure all units are consistent: time should align with the rate's unit of time (per year) to avoid calculation errors.

Principal Amount

The principal amount is the initial sum of money that you invest or borrow before any interest is added. It's crucial in determining how much interest you can earn or have to pay over a given time. In the context of simple interest, this amount remains unchanged throughout the period. For instance, if you start with a principal amount of \( \\(3500 \), it remains \( \\)3500 \) regardless of how much interest you earn on it.
This stability makes the calculation of simple interest much more straightforward than compound interest, which adds each period's interest to the principal. The principal is simply your starting point for calculating potential financial growth or obligations. By knowing your principal, you can predict the exact amount of interest you will earn or owe at the end of a specific period. Understanding the principal helps you make informed financial decisions based on clear expectations of potential outcomes.

Step-by-Step Solution

Breaking down problems into simple steps can make tackling complex calculations much easier. When calculating the interest rate in our exercise, a systematic approach was taken.

Identifying the Appropriate Formula

Begin with recognizing that the problem requires the simple interest formula: \[ I = P \times r \times t \]. Selecting the correct formula is crucial since it dictates the entire process.

Application and Substitution

Next, inputs were identified: a principal of \( \\(3500 \), an interest of \( \\)262.50 \), and a time of 1 year. Knowing these values allows us to isolate the unknown – the interest rate \( r \). Rearranging gives: \[ r = \frac{I}{P \times t} \] which leads to substituting known values into the equation, simplifying the process.

Solve and Conclude

Inputting these values results in: \[ r = \frac{262.50}{3500 \times 1} \] leading to a rate of \( 7.5\% \). By breaking down the problem, complex calculations become less daunting, allowing for accurate and efficient problem-solving.