Let's break down each question step by step.
Question 1: Account A vs. Account B
Account A (7.5% simple interest)
Using the formula for simple interest:
\[ \text{Total Interest} = P \times r \times t \]
Where:
- \( P = 10,000 \) (initial investment)
- \( r = 0.075 \) (interest rate)
- \( t = 10 \) (years)
Total Interest from Account A:
\[ \text{Total Interest} = 10,000 \times 0.075 \times 10 = 7,500 \]
Total amount after 10 years in Account A:
\[ 10,000 + 7,500 = 17,500 \]
Account B (5% compounded annually)
Using the formula for compound interest:
\[ A = P (1 + r)^t \]
Total amount from Account B:
\[ A = 10,000 \times (1 + 0.05)^{10} = 10,000 \times (1.62889) \approx 16,288.95 \]
Conclusion:
- Account A: $17,500 after 10 years
- Account B: $16,288.95 after 10 years
Recommendation: The business owner should choose Account A because it provides more money ($17,500) compared to Account B ($16,288.95).
Question 2: Earnings Comparison After 10 Years
You deposit $275 in a savings account that earns 4% simple interest. Your friend deposits $275 that earns 3% compounded yearly.
Your total earnings:
Your Account (4% Simple Interest)
\[ \text{Total Interest} = 275 \times 0.04 \times t \]
Your total amount after \( t \) years:
\[ A = 275 + (275 \times 0.04 \times t) = 275(1 + 0.04t) \]
Your friend's total amount (3% compounded annually):
Friend's Account (3% Compounded)
\[ A = 275 \times (1 + 0.03)^t = 275 \times (1.03)^t \]
To find when you and your friend have the same amount:
Set your amounts equal:
\[ 275(1 + 0.04t) = 275(1.03)^t \]
Remove \( 275 \):
\[ 1 + 0.04t = (1.03)^t \]
This requires numerical methods or trial and error to solve for \( t \). Testing small integers finds:
- For \( t = 10 \):
\[ 1 + 0.04(10) = 1.4 \]
\[ (1.03)^{10} \approx 1.3439 \] - For \( t = 9 \):
\[ 1 + 0.04(9) = 1.36 \]
\[ (1.03)^9 \approx 1.3677 \]
Find when they equal: By numerical trial, it happens around 9 years. Both will earn approximately 1.36 in this time frame.
Comparison after 5 and 10 years:
After 5 years:
- You: \( 275 + 275(0.20) = 330 \)
- Friend: \( 275(1.03^5) \approx 320.82 \) → You have more.
After 10 years:
- You: \( 275 + 275(0.40) = 385 \)
- Friend: \( 275(1.03^{10}) \approx 368.31 \) → You have more.
Question 3: Time to reach $7,000 from $2,000 at 5% compounded
Using the compound interest formula:
\[ 7000 = 2000 \times (1 + 0.05)^t \]
Divide through by 2000:
\[ 3.5 = (1.05)^t \]
Taking logarithm:
\( \log(3.5) = t \log(1.05) \)
\[ t \approx \frac{\log(3.5)}{\log(1.05)} \approx 25.08 \]
So it will take approximately 25 years.
Question 4: Total Amount Owed by Stan
Stan’s late payment for 2 months at a 3% per month interest rate.
Calculate the total amount for 2 months:
\[ \text{Amount after 1 month} = 300(1 + 0.03) = 309 \]
\[ \text{Amount after 2 months} = 309(1 + 0.03) = 318.27 \]
Total amount owed after two months:
\[ = 300 + 318.27 = 618.27 \]
Question 5: Simple Interest vs. Compounded Interest
If both accounts have the same initial amount and interest rate, compound interest will yield more over time. Compounded interest is calculated on the initial principal and the accumulated interest from previous periods. In contrast, simple interest is calculated only on the principal. After one year, interest earned with compound interest will be more than with simple interest due to earned interest being reinvested.