Question

A continuously compounded savings account had an initial deposit of $10,000.00 and 10 years later has a balance of $13,125.87. At what interest rate was the savings account?

(A) 2.75%
(B) 3.25%
(C) 3.75%
(D) 4.25%

Answer 1

By using the formula for continuous compounding, the interest rate for the savings account that grew from $10,000 to $13,125.87 in 10 years is found to be 2.75%, matching option (A).

Step-by-step explanation:

To find the interest rate for a continuously compounded savings account with an initial deposit of $10,000 that grew to $13,125.87 in 10 years, we use the formula for continuous compounding, which is A = P * e^(rt). Here, A is the amount of money accumulated after n years, including interest, P is the principal amount (the initial amount of money), r is the annual interest rate (in decimal), t is the time the money is invested for in years, and e is the base of the natural logarithm.

To solve for the rate, r, we rearrange the formula:

r = (ln(A/P)) / t

Plugging the numbers into the formula gives us:

r = (ln(13125.87/10000)) / 10

r = 0.02775, or 2.775% when converted to a percentage.

Therefore, the interest rate was 2.75%, which corresponds to option (A).