Solve the problem.
-Find the required annual interest rate, to the nearest tenth of a percent, for $5100 to grow to $6400 if interest is compounded quarterly for 7 years.
A) 0.8%
B) 3.3%
C) 4.1%
D) 1.6%

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. Here, $A = 6400$, $P = 5100$, $n = 4$ (since interest is compounded quarterly), and $t = 7$ years. We need to find $r$.Rearranging the formula to solve for $r$, we get:$$6400 = 5100(1 + \frac{r}{4})^{4*7}$$$$1.2549 = (1 + \frac{r}{4})^{28}$$Taking the 28th root of both sides:$$1 + \frac{r}{4} = (1.2549)^{\frac{1}{28}}$$$$\frac{r}{4} = (1.2549)^{\frac{1}{28}} - 1$$$$r = 4[(1.2549)^{\frac{1}{28}} - 1]$$Calculating this gives $r \approx 0.033$ or $3.3\%$, which is the annual interest rate to the nearest tenth of a percent.

Solve the Problem. -Find the Required Annual Interest Rate, to the Nearest Tenth