Question
You deposit $1,000 in a savings account that earns 6% interest per year.
Simple Interest
t Principal Annual Interest End of year Balance
1 $1,000.00 $60.00 $1,060.00
2 $1,000.00 $60.00 $1,120.00
3
4
5
6
7
8
9
10
Compound Interest
(compounded annually)
t Principal Annual Interest End of year Balance
1 $1,000.00
$60.00 $1,060.00
2 $1,060.00
$63.60
$1,123.00
3
4
5
6
7
8
9
10
Complete each table. Which type of interest gives the greater balance? Use the simple interest formula and the given values to find each missing value.
Graph the end-of-year balances for each type of interest in part A on the same graph. Label each.
You deposit $400 in an account that earns 3.5% interest compounded annually. What is the account balance after 2 years? After 10 years? After 25 years?
A firefighter invests $40,000 in a retirement account for 2 years. The interest rate is 6%. The interest is compounded monthly. What will the final balance be?
P=$2,000, R=5%, t=5
. Find I.
I=$2,566.72, R=12%, t=3
. Find P.
I=$842.96, P=500, t=2
. Find 𝑅.
I=$1,738.91, P=1,500, R=3%
. Find 𝑡.
Answers
GPT-4o mini
Let's start by completing the tables for simple interest and compound interest:
### Simple Interest
To calculate simple interest using the formula
\[ I = P \cdot r \cdot t \]
where:
- \( I \) = interest earned
- \( P \) = principal amount (initial deposit)
- \( r \) = interest rate (as a decimal)
- \( t \) = time (in years)
Using your values:
- \( P = 1000 \)
- \( r = 0.06 \)
We can fill out the table as follows:
| t | Principal | Annual Interest | End of Year Balance |
|---|---------------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,000.00 | $60.00 | $1,120.00 |
| 3 | $1,000.00 | $60.00 | $1,180.00 |
| 4 | $1,000.00 | $60.00 | $1,240.00 |
| 5 | $1,000.00 | $60.00 | $1,300.00 |
| 6 | $1,000.00 | $60.00 | $1,360.00 |
| 7 | $1,000.00 | $60.00 | $1,420.00 |
| 8 | $1,000.00 | $60.00 | $1,480.00 |
| 9 | $1,000.00 | $60.00 | $1,540.00 |
| 10| $1,000.00 | $60.00 | $1,600.00 |
### Compound Interest
To calculate compound interest using the formula
\[ A = P(1 + r)^t \]
where:
- \( A \) = the amount of money accumulated after n years, including interest
- \( P = 1,000 \) (initial deposit)
- \( r = 0.06 \) (interest rate)
- \( t \) = time (in years)
We can fill out the compound interest table as follows:
| t | Principal | Annual Interest | End of Year Balance |
|---|----------------|-------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,060.00 | $63.60 | $1,123.60 |
| 3 | $1,123.60 | $67.42 | $1,191.02 |
| 4 | $1,191.02 | $71.46 | $1,262.48 |
| 5 | $1,262.48 | $75.75 | $1,338.23 |
| 6 | $1,338.23 | $80.29 | $1,418.52 |
| 7 | $1,418.52 | $85.11 | $1,503.63 |
| 8 | $1,503.63 | $90.22 | $1,593.85 |
| 9 | $1,593.85 | $95.63 | $1,689.48 |
| 10| $1,689.48 | $101.37 | $1,790.85 |
### Comparison
From the tables, we see that the balances at the end of 10 years are:
- Simple Interest: $1,600.00
- Compound Interest: $1,790.85
**Conclusion**: Compound interest gives a greater balance than simple interest.
### Graph the Balances
When graphing, the X-axis can represent the number of years (0 to 10) and the Y-axis represents the account balance. You will plot two lines, one for simple interest that increases linearly and one for compound interest that curves upward.
### Additional Calculations
1. **For the account that earns 3.5% interest compounded annually:**
- \( P = 400 \)
- \( r = 0.035 \)
**After 2 years:**
\[ A = 400(1 + 0.035)^2 = 400(1.071225) \approx 428.49 \]
**After 10 years:**
\[ A = 400(1 + 0.035)^{10} = 400(1.410802) \approx 564.32 \]
**After 25 years:**
\[ A = 400(1 + 0.035)^{25} = 400(2.314783) \approx 925.91 \]
2. **For the firefighter's investment:**
- $40,000 invested at 6% for 2 years compounded monthly.
- The formula is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where \( n = 12 \) (monthly compounding).
\[ A = 40000 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} = 40000 \left(1.005)^{24} \approx 40000(1.12749) \approx 45099.60 \]
3. **Calculating Interest (I):**
- For \( P = 2000, R = 5\% (0.05), t = 5 \):
\[ I = P \cdot r \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
4. **Finding Principal (P):**
- \( I = 842.96, R = 12\% (0.12), t = 3 \):
\[ P = \frac{I}{r \cdot t} = \frac{842.96}{0.12 \cdot 3} \approx 2,341.67 \]
5. **Finding Rate (R):**
- \( I = 842.96, P = 500, t = 2 \):
\[ R = \frac{I}{P \cdot t} = \frac{842.96}{500 \cdot 2} \approx 0.16859 \approx 16.86\% \]
6. **Finding Time (t):**
- \( I = 1738.91, P = 1500, R = 3\% (0.03) \):
\[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} \approx 38.63 \text{ years} \]
This should cover all the calculations and questions you posed! Let me know if you need any further assistance.
### Simple Interest
To calculate simple interest using the formula
\[ I = P \cdot r \cdot t \]
where:
- \( I \) = interest earned
- \( P \) = principal amount (initial deposit)
- \( r \) = interest rate (as a decimal)
- \( t \) = time (in years)
Using your values:
- \( P = 1000 \)
- \( r = 0.06 \)
We can fill out the table as follows:
| t | Principal | Annual Interest | End of Year Balance |
|---|---------------|------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,000.00 | $60.00 | $1,120.00 |
| 3 | $1,000.00 | $60.00 | $1,180.00 |
| 4 | $1,000.00 | $60.00 | $1,240.00 |
| 5 | $1,000.00 | $60.00 | $1,300.00 |
| 6 | $1,000.00 | $60.00 | $1,360.00 |
| 7 | $1,000.00 | $60.00 | $1,420.00 |
| 8 | $1,000.00 | $60.00 | $1,480.00 |
| 9 | $1,000.00 | $60.00 | $1,540.00 |
| 10| $1,000.00 | $60.00 | $1,600.00 |
### Compound Interest
To calculate compound interest using the formula
\[ A = P(1 + r)^t \]
where:
- \( A \) = the amount of money accumulated after n years, including interest
- \( P = 1,000 \) (initial deposit)
- \( r = 0.06 \) (interest rate)
- \( t \) = time (in years)
We can fill out the compound interest table as follows:
| t | Principal | Annual Interest | End of Year Balance |
|---|----------------|-------------------|---------------------|
| 1 | $1,000.00 | $60.00 | $1,060.00 |
| 2 | $1,060.00 | $63.60 | $1,123.60 |
| 3 | $1,123.60 | $67.42 | $1,191.02 |
| 4 | $1,191.02 | $71.46 | $1,262.48 |
| 5 | $1,262.48 | $75.75 | $1,338.23 |
| 6 | $1,338.23 | $80.29 | $1,418.52 |
| 7 | $1,418.52 | $85.11 | $1,503.63 |
| 8 | $1,503.63 | $90.22 | $1,593.85 |
| 9 | $1,593.85 | $95.63 | $1,689.48 |
| 10| $1,689.48 | $101.37 | $1,790.85 |
### Comparison
From the tables, we see that the balances at the end of 10 years are:
- Simple Interest: $1,600.00
- Compound Interest: $1,790.85
**Conclusion**: Compound interest gives a greater balance than simple interest.
### Graph the Balances
When graphing, the X-axis can represent the number of years (0 to 10) and the Y-axis represents the account balance. You will plot two lines, one for simple interest that increases linearly and one for compound interest that curves upward.
### Additional Calculations
1. **For the account that earns 3.5% interest compounded annually:**
- \( P = 400 \)
- \( r = 0.035 \)
**After 2 years:**
\[ A = 400(1 + 0.035)^2 = 400(1.071225) \approx 428.49 \]
**After 10 years:**
\[ A = 400(1 + 0.035)^{10} = 400(1.410802) \approx 564.32 \]
**After 25 years:**
\[ A = 400(1 + 0.035)^{25} = 400(2.314783) \approx 925.91 \]
2. **For the firefighter's investment:**
- $40,000 invested at 6% for 2 years compounded monthly.
- The formula is:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where \( n = 12 \) (monthly compounding).
\[ A = 40000 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} = 40000 \left(1.005)^{24} \approx 40000(1.12749) \approx 45099.60 \]
3. **Calculating Interest (I):**
- For \( P = 2000, R = 5\% (0.05), t = 5 \):
\[ I = P \cdot r \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
4. **Finding Principal (P):**
- \( I = 842.96, R = 12\% (0.12), t = 3 \):
\[ P = \frac{I}{r \cdot t} = \frac{842.96}{0.12 \cdot 3} \approx 2,341.67 \]
5. **Finding Rate (R):**
- \( I = 842.96, P = 500, t = 2 \):
\[ R = \frac{I}{P \cdot t} = \frac{842.96}{500 \cdot 2} \approx 0.16859 \approx 16.86\% \]
6. **Finding Time (t):**
- \( I = 1738.91, P = 1500, R = 3\% (0.03) \):
\[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} \approx 38.63 \text{ years} \]
This should cover all the calculations and questions you posed! Let me know if you need any further assistance.