Question

kevin is going to invest in an account paying an interest rate of 2% compounded monthly. how much would kevin need to invest, to the nearest hundred dollars, for the value of the account to reach $84,000 in 10 years?

Explanation:

Step1: Recall compound interest formula

The present value formula for compound interest is $P = \frac{A}{(1 + \frac{r}{n})^{nt}}$, where $A = 84000$, $r = 0.02$, $n = 12$, $t = 10$.

Step2: Calculate the denominator

First compute $\frac{r}{n} = \frac{0.02}{12} \approx 0.0016667$, then $nt = 12 \times 10 = 120$. So $(1 + 0.0016667)^{120} \approx e^{120 \times \ln(1.0016667)} \approx e^{0.198026} \approx 1.2190$.

Step3: Compute present value

$P = \frac{84000}{1.2190} \approx 68910$, rounded to nearest hundred is 68900.

Answer:

68900