How long must $4400 be in a bank at 3% compounded annually to become $5913.23? (Round to the nearest year.)
A) 12 yr
B) 9 yr
C) 11 yr
D) 10 yr

Question

Quizplus · Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ 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 (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. In this case, since the interest is compounded annually, $n = 1$, and we are given $A = 5913.23$, $P = 4400$, and $r = 0.03$ (3% as a decimal). We need to solve for $t$.Rearranging the formula to solve for $t$, we get:$$t = \frac{\log(A/P)}{\log(1 + r)}$$Substituting the given values:$$t = \frac{\log(5913.23/4400)}{\log(1 + 0.03)}$$$$t = \frac{\log(1.344375)}{\log(1.03)}$$$$t \approx 10$$Therefore, it takes approximately 10 years for $4400 to grow to $5913.23 at an annual compound interest rate of 3%.

How Long Must $4400 Be in a Bank at 3