Question
$6000 are invested in a bank account at an interest rate of 4 percent per year.
Find the amount in the bank after 15 years if interest is compounded annually.
Find the amount in the bank after 15 years if interest is compounded quarterly.
Find the amount in the bank after 15 years if interest is compounded monthly.
Finally, find the amount in the bank after 15 years if interest is compounded continuously.
Find the amount in the bank after 15 years if interest is compounded annually.
Find the amount in the bank after 15 years if interest is compounded quarterly.
Find the amount in the bank after 15 years if interest is compounded monthly.
Finally, find the amount in the bank after 15 years if interest is compounded continuously.
Answers
GPT-4o mini
To solve the problem, we will use the formulas for compound interest based on the compounding frequency and compare the results.
### 1. Compounded Annually
The formula for compound interest compounded annually is:
\[
A = P(1 + r)^t
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount).
- \( r \) = annual interest rate (decimal).
- \( t \) = time the money is invested or borrowed for, in years.
Given:
- \( P = 6000 \)
- \( r = 0.04 \) (4%)
- \( t = 15 \)
Substituting the values into the formula:
\[
A = 6000(1 + 0.04)^{15}
\]
\[
A = 6000(1.04)^{15}
\]
\[
A = 6000 \times 1.800949 \approx 10805.69
\]
After 15 years, if interest is compounded annually, the amount in the bank will be approximately **$10,805.69**.
---
### 2. Compounded Quarterly
For quarterly compounding, the formula is:
\[
A = P\left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( n \) = number of times that interest is compounded per year.
Given:
- \( n = 4 \) (quarterly)
- The other variables remain the same.
Substituting the values into the formula:
\[
A = 6000\left(1 + \frac{0.04}{4}\right)^{4 \times 15}
\]
\[
A = 6000\left(1 + 0.01\right)^{60}
\]
\[
A = 6000(1.01)^{60}
\]
\[
A = 6000 \times 1.819396 \approx 10916.38
\]
After 15 years, if interest is compounded quarterly, the amount in the bank will be approximately **$10,916.38**.
---
### 3. Compounded Monthly
For monthly compounding, the formula is the same, with \( n = 12 \):
\[
A = P\left(1 + \frac{r}{n}\right)^{nt}
\]
Substituting the values into the formula:
\[
A = 6000\left(1 + \frac{0.04}{12}\right)^{12 \times 15}
\]
\[
A = 6000\left(1 + \frac{0.04}{12}\right)^{180}
\]
\[
A = 6000\left(1 + 0.0033333\right)^{180}
\]
\[
A = 6000(1.0033333)^{180}
\]
\[
A = 6000 \times 1.833148 \approx 10998.89
\]
After 15 years, if interest is compounded monthly, the amount in the bank will be approximately **$10,998.89**.
---
### 4. Compounded Continuously
For continuous compounding, the formula is:
\[
A = Pe^{rt}
\]
Where \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Substituting the values into the formula:
\[
A = 6000 \cdot e^{(0.04 \cdot 15)}
\]
\[
A = 6000 \cdot e^{0.6}
\]
\[
A = 6000 \cdot 1.822118 \approx 10932.71
\]
After 15 years, if interest is compounded continuously, the amount in the bank will be approximately **$10,932.71**.
### Summary
- Compounded Annually: **$10,805.69**
- Compounded Quarterly: **$10,916.38**
- Compounded Monthly: **$10,998.89**
- Compounded Continuously: **$10,932.71**
### 1. Compounded Annually
The formula for compound interest compounded annually is:
\[
A = P(1 + r)^t
\]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount).
- \( r \) = annual interest rate (decimal).
- \( t \) = time the money is invested or borrowed for, in years.
Given:
- \( P = 6000 \)
- \( r = 0.04 \) (4%)
- \( t = 15 \)
Substituting the values into the formula:
\[
A = 6000(1 + 0.04)^{15}
\]
\[
A = 6000(1.04)^{15}
\]
\[
A = 6000 \times 1.800949 \approx 10805.69
\]
After 15 years, if interest is compounded annually, the amount in the bank will be approximately **$10,805.69**.
---
### 2. Compounded Quarterly
For quarterly compounding, the formula is:
\[
A = P\left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( n \) = number of times that interest is compounded per year.
Given:
- \( n = 4 \) (quarterly)
- The other variables remain the same.
Substituting the values into the formula:
\[
A = 6000\left(1 + \frac{0.04}{4}\right)^{4 \times 15}
\]
\[
A = 6000\left(1 + 0.01\right)^{60}
\]
\[
A = 6000(1.01)^{60}
\]
\[
A = 6000 \times 1.819396 \approx 10916.38
\]
After 15 years, if interest is compounded quarterly, the amount in the bank will be approximately **$10,916.38**.
---
### 3. Compounded Monthly
For monthly compounding, the formula is the same, with \( n = 12 \):
\[
A = P\left(1 + \frac{r}{n}\right)^{nt}
\]
Substituting the values into the formula:
\[
A = 6000\left(1 + \frac{0.04}{12}\right)^{12 \times 15}
\]
\[
A = 6000\left(1 + \frac{0.04}{12}\right)^{180}
\]
\[
A = 6000\left(1 + 0.0033333\right)^{180}
\]
\[
A = 6000(1.0033333)^{180}
\]
\[
A = 6000 \times 1.833148 \approx 10998.89
\]
After 15 years, if interest is compounded monthly, the amount in the bank will be approximately **$10,998.89**.
---
### 4. Compounded Continuously
For continuous compounding, the formula is:
\[
A = Pe^{rt}
\]
Where \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
Substituting the values into the formula:
\[
A = 6000 \cdot e^{(0.04 \cdot 15)}
\]
\[
A = 6000 \cdot e^{0.6}
\]
\[
A = 6000 \cdot 1.822118 \approx 10932.71
\]
After 15 years, if interest is compounded continuously, the amount in the bank will be approximately **$10,932.71**.
### Summary
- Compounded Annually: **$10,805.69**
- Compounded Quarterly: **$10,916.38**
- Compounded Monthly: **$10,998.89**
- Compounded Continuously: **$10,932.71**