If $10,000 is locked in a savings account offering 6.5% per annum, compounded monthly. How fast is the balance growing three years later? Round to the nearest dollar. Ensure that your answer intuitively makes sense to you.

Answer 1

First, we need to calculate the balance in the account after 3 years. We can use the formula for compound interest:

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

Where A is the final amount, P is the principal amount ($10,000 in this case), r is the annual interest rate (6.5% or 0.065 as a decimal), n is the number of times interest is compounded per year (monthly means 12 times), and t is the number of years (3 years in this case).

A = 10000 * (1 + 0.065/12)^(12*3)
A = 10000 * (1.005417)^(36)
A ≈ 12394.84

The balance after 3 years is approximately $12,394.84.

Now we need to find out how fast it's growing at this point. We can do this by finding the interest earned in one month and annualizing it (since the interest rate is given per annum).

First, let's calculate the balance in the account 1 month before the three-year mark:

A_prev_month = P(1 + r/n)^(nt - 1)
A_prev_month = 10000 * (1 + 0.065/12)^(12*3 - 1)
A_prev_month = 10000 * (1.005417)^(35)
A_prev_month ≈ 12355.18

Now we subtract the balance at the month before from the balance at the three-year mark.

Interest earned in 1 month = A - A_prev_month
Interest_earned_1_month ≈ 12394.84 - 12355.18
Interest_earned_1_month ≈ 39.66

Now, we can calculate the annualized growth rate:

Annualized_growth_rate = Interest_earned_1_month * 12
Annualized_growth_rate ≈ 39.66 * 12
Annualized_growth_rate ≈ 475.92

So, the balance is growing at a rate of approximately $476 per year 3 years after the $10,000 was first placed into the account.