approximately how much principal would need to be placed into an account earning 3.575% interest compounded…

approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of $68,000 at the end of 30 years?\na. $23,706\nb. $23,377\nc. $52,069\nd. $58,944\nplease select the best answer from the choices provided\no a\no b\no c\no d

approximately how much principal would need to be placed into an account earning 3.575% interest compounded quarterly so that it has an accumulated value of $68,000 at the end of 30 years?\na. $23,706\nb. $23,377\nc. $52,069\nd. $58,944\nplease select the best answer from the choices provided\no a\no b\no c\no d

Answer

Explanation:

Step1: Identify compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the accumulated amount, $P$ is the principal, $r$ is the annual interest rate (in decimal form), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. We want to solve for $P$, so we can rewrite the formula as $P=\frac{A}{(1 +\frac{r}{n})^{nt}}$.

Step2: Convert given values to appropriate form

The annual interest rate $r = 3.575%=0.03575$, the number of times compounded per year $n = 4$ (quarterly compounding), the number of years $t = 30$, and the accumulated amount $A=68000$.

Step3: Substitute values into the formula for $P$

$P=\frac{68000}{(1+\frac{0.03575}{4})^{4\times30}}$. First, calculate the value inside the parentheses: $\frac{0.03575}{4}=0.0089375$, then $1+\frac{0.03575}{4}=1.0089375$. Next, calculate the exponent: $4\times30 = 120$. So, $(1.0089375)^{120}\approx2.867$. Then, $P=\frac{68000}{2.867}\approx23716$, which is closest to $$23706$.

Answer:

A. $$23,706$