Question

Suppose that a loan of $6000 is given at an interest rate of 9% compounded each year.

Answer 1

Answer: a) $6649

b) $7128.6

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 6000

r = 9% = 9/100 = 0.09

n = 1 because it was compounded once in a year.

a)

t = 1 year

Therefore,

A = 6000(1 + 0.09/1)^1 × 1

A = 6000(1.09)

A = $6649

b) t = 2 years

A = 6000(1.09)^2

A = $7128.6

Answer 2

Answer:

• $6540.00
• $7128.60

Step-by-step explanation:

Each year, the balance is multiplied by (1+r) where r is the annual interest rate.

(a) Balance at the end of year 1:

  $6000×1.09 = $6540

(b) Balance at the end of year 2:

  $6540×1.09 = $7128.60