Problem 25 - Algebra for College Students

Question

What rate of interest, to the nearest tenth of a percent, compounded quarterly is needed for an investment of \(\$ 1500\) to grow to \(\$ 2700\) in 10 years? \(\quad 5.9 \%\)

Step-by-Step Solution

Answer

The interest rate needed is approximately 19.5%.

Step 1: Identify the Formula

For compound interest, the formula to use is \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (initial investment). - \( r \) is the annual interest rate (as a decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years.

Step 2: Substitute Known Values

We substitute the following known values into the formula:- \( A = 2700 \)- \( P = 1500 \)- \( n = 4 \) (compounded quarterly)- \( t = 10 \) years.So, the formula becomes: \[ 2700 = 1500 \left(1 + \frac{r}{4}\right)^{40} \]

Step 3: Solve for the Rate of Interest

First, isolate the expression for the interest rate:\[ \left(1 + \frac{r}{4}\right)^{40} = \frac{2700}{1500} \] Simplify the fraction:\[ \left(1 + \frac{r}{4}\right)^{40} = 1.8 \] Take the 40th root on both sides to solve for \(1 + \frac{r}{4}\):\[ 1 + \frac{r}{4} = (1.8)^{\frac{1}{40}} \] Using a calculator, find \((1.8)^{\frac{1}{40}} \approx 1.048829\). Therefore,\[ 1 + \frac{r}{4} \approx 1.048829 \] Subtract 1 from both sides to solve for \( \frac{r}{4} \):\[ \frac{r}{4} \approx 0.048829 \]

Step 4: Calculate the Annual Interest Rate

Multiply both sides of the equation by 4 to solve for \( r \):\[ r \approx 4 \times 0.048829 \approx 0.195316 \] Convert the decimal to a percentage by multiplying by 100:\[ r \approx 19.53\% \]

Step 5: Round to the Nearest Tenth of a Percent

Round \( 19.53\% \) to the nearest tenth of a percent: \[ r \approx 19.5\% \]

Step 6: Conclusion

Thus, the interest rate needed is approximately \( 19.5\% \) when compounded quarterly to grow an investment from \\(1500 to \\)2700 in 10 years.

Key Concepts

Interest Rate Calculation

Understanding how to calculate the interest rate is key to determining how much an investment can grow over time. In this example, we were tasked with finding what interest rate is needed for a \\(1500 investment to grow to \\)2700 in 10 years, with interest compounded quarterly. The calculation begins by identifying the compound interest formula:

• \( A = P \left(1 + \frac{r}{n}\right)^{nt} \)

Here, \( A \) is the future value of the investment, \( P \) is the initial investment or principal, \( r \) is the annual interest rate in decimal, \( n \) is the frequency of compounding per year, and \( t \) is the number of years. Calculating the interest rate involves algebraically manipulating this formula to solve for \( r \). A thorough understanding of variables and algebra is essential as you use known data to isolate and solve for \( r \). This process enables investors to plan their potential returns based on a specified interest rate.

Quarterly Compounding

Quarterly compounding refers to the method of calculating interest payments on an investment four times a year. This approach is a common feature in compound interest calculations, offering a different growth trajectory than other compounding frequencies.

• In this problem, the investment sees its interest compounded quarterly, which means the total number of compounding periods over ten years is 40 (since \( n = 4 \) compounding periods per year, and \( t = 10 \) years).

When interest is compounded quarterly, as opposed to annually or semi-annually, interest on the principal amount is calculated more frequently, introducing a higher overall growth for the investment. Investors prefer quarterly compounding for its ability to maximize returns through more frequent growth.

Investment Growth

The growth of an investment is the additional amount gained over time, influenced by factors such as the rate of interest and compounding frequency. In our example, an initial investment of \\(1500 grows to \\)2700 after 10 years.

• This growth is the result of the compound interest calculated using the quarterly compounding formula.
• The difference between the initial and final amounts is a clear demonstration of how investments increase in value over a period of time with the right interest rate.

Compound interest impacts the growth rate, showcasing the power of letting investments mature. Through effective interest rate calculation and understanding compounding, one can strategize investments to maximize growth. Seeing how a \\(1500 investment turns into \\)2700 provides tangible evidence of well-managed financial planning.

Algebraic Equations

Algebraic equations are mathematical statements that show the equality between two expressions, heavily used in financial calculations to solve for unknown variables. In the compound interest formula, algebra helps us isolate \( r \), the interest rate.

• Transforming the original equation \( 2700 = 1500 \left(1 + \frac{r}{4}\right)^{40} \) allows us to simplify and focus on solving for \( r \).

This involves breaking down the equation by inverse operations, such as dividing, multiplying, and taking roots, demonstrating how algebra is a potent tool for making informed financial decisions. Mastering these calculations provides clarity and capability when working with different financial scenarios.