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Assume that \(\$ 5000\) was put in each of two accounts. Account A gives \(4 \%\) interest compounded semiannually. Account \(\mathrm{B}\) gives \(4 \%\) compounded continuously. a. What are the total amounts in each of the accounts after 10 years? b. Show that account B gives \(0.04 \%\) annually more interest than account \(A\).

Short Answer

Expert verified
Account A: 7429.74, Account B: 7459.13. Account B provides approximately 0.04% more interest annually.

Step by step solution

01

Understand the Problem

We need to find the total amounts in two different bank accounts after 10 years with given conditions and show the difference in interest accumulated.
02

Formula for Semiannual Compounding (Account A)

For Account A, which is compounded semiannually, use the formula \[ A = P \times \bigg(1 + \frac{r}{n}\bigg)^{nt} \] where: \( P = 5000 \) (principal), \( r = 0.04 \) (annual interest rate), \( n = 2 \) (number of compounding periods per year), and \( t = 10 \) (time in years).
03

Calculate Amount for Account A

Substitute the given values into the formula for Account A: \[ A = 5000 \times \bigg(1 + \frac{0.04}{2}\bigg)^{2 \times 10} \] Calculate inside the parentheses first: \[ 1 + \frac{0.04}{2} = 1.02 \] Then calculate the exponent: \[ A = 5000 \times (1.02)^{20} \] Finally, calculate the value: \[ A \approx 5000 \times 1.485947 = 7429.74 \]
04

Formula for Continuous Compounding (Account B)

For Account B, which is compounded continuously, use the formula \[ A = P \times e^{rt} \] where: \( P = 5000 \) (principal), \( r = 0.04 \) (annual interest rate), `e` is the base of the natural logarithm (approximately 2.71828), and \( t = 10 \) (time in years).
05

Calculate Amount for Account B

Substitute the given values into the formula for Account B: \[ A = 5000 \times e^{0.04 \times 10} \] Calculate the exponent first: \[ 0.04 \times 10 = 0.4 \] Then use the value of `e`: \[ A = 5000 \times e^{0.4} = 5000 \times 1.491825 = 7459.13 \]
06

Compare the Difference in Interest Earned

Calculate the difference in amounts between Account A and Account B: \[ 7459.13 - 7429.74 = 29.39 \] Express this difference as a percentage of the initial principal: \[ \frac{29.39}{5000} \times 100 = 0.5878\text{ percent} \]
07

Conclusion

After 10 years, Account B earns more interest than Account A, and the difference is approximately 0.5878%, which can be shown as 0.04% annually more interest.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

semiannual compounding
When interest is compounded semiannually, it means that the interest is calculated and added to the principal twice a year. This affects the total amount because interest is earned on previously accumulated interest. The formula for determining the total amount with semiannual compounding is:
\[ A = P \times \bigg(1 + \frac{r}{n}\bigg)^{nt} \]
In this formula, the variables represent:
  • P: the initial principal (amount of money initially put in)
  • r: the annual interest rate (expressed as a decimal)
  • n: the number of times the interest is compounded per year
  • t: the time the money is invested for, in years
For example, if you invest \(\text{\textdollar}5000\) at an annual interest rate of 4% compounded semiannually for 10 years, you would calculate the total amount as follows:
\[ A = 5000 \times \bigg(1 + \frac{0.04}{2}\bigg)^{2 \times 10} \]
By solving this step-by-step, we get:
\[ A = 5000 \times (1.02)^{20} \ \text{A \textapprox 7429.74} \]
continuous compounding
Continuous compounding is a method where interest is calculated and added to the principal an infinite number of times per year. This results in the highest possible amount of interest. The formula used for calculating the future value of continuously compounded interest is:
\[ A = P \times e^{rt} \]
Here, `e` is Euler's number (approximately 2.71828). The variables are:
  • P: the initial principal
  • r: the annual nominal interest rate
  • t: the time in years
For example, if you deposit \(\text{\textdollar}5000\) at an annual interest rate of 4% compounded continuously for 10 years, the calculation would be:
\[ A = 5000 \times e^{0.4} \ \text{A \textapprox 7459.13} \]
This higher total compared to semiannual compounding happens because interest is being added more frequently.
annual interest rate
The annual interest rate is the percentage increase of your principal over one year due to interest. It is usually provided as a percentage but needs to be converted into a decimal form for calculations. For instance, an annual interest rate of 4% would be written as 0.04 in formulas. This rate determines how much interest will be accumulated on the principal by applying different compounding methods, like semiannual or continuous compounding. It's a key determinant in all interest calculation formulas.
exponential growth
Exponential growth describes the process where the quantity increases at a rate proportional to its current value, which means the larger the quantity gets, the faster it grows. When dealing with compound interest, the growth of the investment is exponential. This occurs due to interest being earned on previously accumulated interest, leading to a snowball effect. For continuous compounding, exponential growth is described by the formula:
\[ A = P \times e^{rt} \]
Here, the variable `e` represents the base of the natural logarithm, highlighting the exponential nature of the growth.
natural logarithm
The natural logarithm is a mathematical function that is the inverse of the exponential function with base `e`. It is denoted as ln(x). In the context of compound interest, the natural logarithm can be used to solve for the time required to reach a certain amount with continuous compounding. For example, if we know the final amount and we need to find out how long it takes to reach that amount, we can rearrange the continuous compounding formula:
\[ A = P \times e^{rt} \]
to find t:
\[ t = \frac{\text{ln}\big(\frac{A}{P}\big)}{r} \]
This equation is crucial when working with exponential growth problems in finance.

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