Question

27) Cooley Landscaping Company needs to borrow $30,000 for a new front-end dirt loader. The bank is willing to loan the funds at 8.5% interest with annual payments at the end of the year for the next ten years. The loan is fully amortized, i.e., the loan will be fully paid off at the end of year ten. What is the annual payment on this loan for Cooley Landscaping?

Answer 1

The annual payment for Cooley Landscaping Company's $30,000 loan at 8.5% interest over 10 years is approximately $4,579.39.

Step-by-step explanation:

To calculate the annual payment for Cooley Landscaping Company's loan, we can use the formula for an amortizing loan which takes into account the principal amount, the interest rate, and the number of periods (years, in this case).

The formula for the annual payment (A) of an amortizing loan is:

A = P [i(1 + iⁿ)] / [(1 + i)ⁿ – 1]

Where:

• P = principal amount (the initial amount of the loan)
• i = annual interest rate (as a decimal)
• n = number of payments (years)

In this case, the principal amount P is $30,000, the annual interest rate i is 0.085, and the number of payments n is 10 years.

So,

A = $30,000 [0.085(1 + 0.085)^10] / [(1 + 0.085)¹⁰ – 1]

Using a calculator or spreadsheet, we find that the annual payment A is approximately $4,579.39.

Therefore, the annual payment for Cooley Landscaping's loan will be about $4,579.39.

Answer 2

The annual payment at the end of each year: $4,572.23

Step-by-step explanation:

The formular for calculating Present value of Annuity is applied in this case to help us find the equal annual payment.

Applying information in the question, we have the annuity that have:

n= 10 as there are 10 equal annual payments paid at the end of each year during 10 years;

i = 8.5% per annum compounded annually, as stated in the question;

PV = Borrowed amount = $30,000;

C = the equal annual payment.

The formular for PV of Annuity: PV = (C/i) x [ 1- (1+i)^(-n)] <=> C = (PV x i) / [ 1- (1+i)^(-n)]

Thus, C = (30,000 x 8.5%) / [ 1- 1.085^(-10) ] = $4,572.23