Question number 8.




Fred purchased a new car.  His monthly payments are $ 350 per month.  He will be paying on his car for 5 years, and the loan costs 6% per year compounded monthly.  What was the original cost of the car?


 $ 18,102.95




 $ 18,101.95




 $ 18,105.95




 $ 18,103.95




 $ 18,099.95




 None of the above.



Question number 9.




Jackson deposits $ 100 each month into a savings account earning interest at the rate of 6% per year compounded monthly.  How much will he have in this account at the end of 4 years?


 $ 5,405.78




 $ 5,408.78




 $ 5,411.78




 $ 5,410.78




 $ 5,409.78




 None of the above.



Let's solve each problem step by step.

### Question 8: Fred's Car Loan

Given:
- Monthly payment: $ P = 350 $
- Loan duration: $ t = 5 $ years
- Interest rate: $ r = 6\% $ per year compounded monthly
- Compounded monthly: $ n = 12 $ times per year

We need to find the **present value** (or original cost of the car). The formula for the present value of an annuity (for loans) is:

\[
PV = P \times \left[ \frac{1 - (1 + r/n)^{-nt}}{r/n} \right]
\]

Where:
- $ PV $ = Present Value (the original loan amount or the car's cost)
- $ P $ = Monthly payment
- $ r = 0.06 $ (annual interest rate)
- $ n = 12 $ (compounded monthly)
- $ t = 5 $ years

First, convert the interest rate:
\[
r/n = \frac{0.06}{12} = 0.005 \quad \text{(monthly interest rate)}
\]

Calculate the exponent $ -nt $:
\[
-nt = -12 \times 5 = -60
\]

Now, plug the values into the present value formula:
\[
PV = 350 \times \left[ \frac{1 - (1 + 0.005)^{-60}}{0.005} \right]
\]

Solving this:

\[
PV = 350 \times \left[ \frac{1 - (1.005)^{-60}}{0.005} \right]
\]
\[
PV = 350 \times \left[ \frac{1 - 0.740818}{0.005} \right]
\]
\[
PV = 350 \times \left[ \frac{0.259182}{0.005} \right]
\]
\[
PV = 350 \times 51.8364 = 18,142.74
\]

Since the closest choice to this value is not provided, the correct answer is **None of the above**.

### Question 9: Jackson's Savings Account

Given:
- Monthly deposit: $ P = 100 $
- Duration: $ t = 4 $ years
- Interest rate: $ r = 6\% $ per year compounded monthly
- Compounded monthly: $ n = 12 $

We need to calculate the future value of this annuity (savings). The formula for the future value of an annuity is:

\[
FV = P \times \left[ \frac{(1 + r/n)^{nt} - 1}{r/n} \right]
\]

Where:
- $ FV $ = Future Value (total savings after 4 years)
- $ P = 100 $
- $ r = 0.06 $
- $ n = 12 $
- $ t = 4 $

First, calculate the monthly interest rate:
\[
r/n = \frac{0.06}{12} = 0.005
\]

Now, calculate the exponent $ nt $:
\[
nt = 12 \times 4 = 48
\]

Plug the values into the future value formula:
\[
FV = 100 \times \left[ \frac{(1 + 0.005)^{48} - 1}{0.005} \right]
\]

Solving this:

\[
FV = 100 \times \left[ \frac{(1.005)^{48} - 1}{0.005} \right]
\]
\[
FV = 100 \times \left[ \frac{1.27048 - 1}{0.005} \right]
\]
\[
FV = 100 \times \left[ \frac{0.27048}{0.005} \right]
\]
\[
FV = 100 \times 54.096 = 5,409.60
\]

Since this is closest to $ 5,409.78 $, the correct answer is **$5,409.78**.

---

Do you need further details or clarifications?

Here are 5 related questions:
1. How does compounding frequency affect the total amount paid for a loan?
2. What is the difference between simple interest and compound interest?
3. How does the present value of an annuity change if the interest rate is increased?
4. What factors determine the future value of a savings account?
5. How do you calculate the effective annual interest rate?

**Tip:** Always ensure to adjust the interest rate and time period to the compounding frequency when solving loan or savings problems!

Fred purchased a new car. His monthly payments are $350 per month. He will be paying on his car for 5 years, and the loan costs 6% per year compounded monthly. What was the original cost of the car? 

Jackson deposits $100 each month into a savings account earning interest at the rate of 6% per year compounded monthly. How much will he have in this account at the end of 4 years?

Learn how to solve problems involving the present and future value of annuities with a detailed step-by-step solution. This example includes calculating the original loan cost for a car with monthly payments and determining the future value of a savings account using compound interest formulas. Ideal for Grades 11-12 students.

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