Math Problem Statement

a. Complete an amortization schedule for a $17,000 loan to be repaid in equal installments at the end of each of the next three years. The interest rate is 10% compounded annually. Round all answers to the nearest cent.

Beginning

Repayment

Ending

Year

Balance

Payment

Interest

of Principal

Balance

1

$  fill in the blank 2

$  fill in the blank 3

$  fill in the blank 4

$  fill in the blank 5

$  fill in the blank 6

2

$  fill in the blank 7

$  fill in the blank 8

$  fill in the blank 9

$  fill in the blank 10

$  fill in the blank 11

3

$  fill in the blank 12

$  fill in the blank 13

$  fill in the blank 14

$  fill in the blank 15

$  

Solution

To solve for the amortization schedule, we can follow these steps:

Step 1: Determine the Annual Payment

The loan amount is $17,000, and the interest rate is 10% per year. The loan is to be repaid in equal installments at the end of each of the next three years. This is a typical case for an amortization schedule using the formula for the payment on an amortizing loan:

PMT=Pr1(1+r)nPMT = \frac{P \cdot r}{1 - (1 + r)^{-n}} Where:

  • PMTPMT = annual payment
  • PP = principal amount (loan balance), which is $17,000
  • rr = interest rate per period (10% or 0.10)
  • nn = number of periods (3 years)

Let's plug the values into the formula to find the annual payment:

PMT=17,000×0.101(1+0.10)3=1,7001(1.10)3PMT = \frac{17,000 \times 0.10}{1 - (1 + 0.10)^{-3}} = \frac{1,700}{1 - (1.10)^{-3}}

Now, we'll compute the exact value for the payment:

PMT=1,70010.7513=1,7000.24876,834.14PMT = \frac{1,700}{1 - 0.7513} = \frac{1,700}{0.2487} \approx 6,834.14

So, the annual payment is approximately $6,834.14.

Step 2: Set up the Amortization Table

Now we can set up the amortization table. For each year, we'll calculate:

  • Interest for the year: The interest for each year is calculated as the beginning balance multiplied by the interest rate (10%).
  • Principal repaid: The principal repayment is the total payment minus the interest for the year.
  • Ending balance: The ending balance is the beginning balance minus the principal repaid.

We'll now create the amortization schedule.


Amortization Schedule

Year 1:

  • Beginning Balance: $17,000.00
  • Payment: $6,834.14
  • Interest: $17,000 × 0.10 = $1,700.00
  • Principal Repayment: $6,834.14 - $1,700.00 = $5,134.14
  • Ending Balance: $17,000.00 - $5,134.14 = $11,865.86

Year 2:

  • Beginning Balance: $11,865.86
  • Payment: $6,834.14
  • Interest: $11,865.86 × 0.10 = $1,186.59
  • Principal Repayment: $6,834.14 - $1,186.59 = $5,647.55
  • Ending Balance: $11,865.86 - $5,647.55 = $6,218.31

Year 3:

  • Beginning Balance: $6,218.31
  • Payment: $6,834.14
  • Interest: $6,218.31 × 0.10 = $621.83
  • Principal Repayment: $6,834.14 - $621.83 = $6,212.31
  • Ending Balance: $6,218.31 - $6,212.31 = $0.00

Final Amortization Table

YearBeginning BalancePaymentInterestPrincipal RepaidEnding Balance
1$17,000.00$6,834.14$1,700.00$5,134.14$11,865.86
2$11,865.86$6,834.14$1,186.59$5,647.55$6,218.31
3$6,218.31$6,834.14$621.83$6,212.31$0.00

Let me know if you need further details or have any other questions!

Here are five related questions:

  1. How would the amortization schedule change if the loan term were 5 years instead of 3 years?
  2. What happens to the total interest paid over the life of the loan if the interest rate increases to 12%?
  3. How can the formula for calculating payments be adapted for loans with different compounding periods (e.g., monthly or quarterly)?
  4. What would happen if you made an extra payment during one of the years?
  5. How can you use the amortization schedule to calculate the total cost of a loan over its term?

Tip: The total interest paid on the loan can be found by subtracting the principal amount from the total of all payments made over the life of the loan.

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Math Problem Analysis

Mathematical Concepts

Amortization
Loan Repayment
Interest Calculation
Algebra

Formulas

PMT = P * r / (1 - (1 + r)^(-n))

Theorems

Amortization formula

Suitable Grade Level

Grades 11-12