Question

Which is the better choice: $1000 deposited for a year at a rate of 4.8% compounded semiannually 19) or at a rate of 4.7% compounded quarterly?

Answer 1

$~~~~~~ \stackrel{\textit{\Large 4.8\% semiannually}}{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\ 1000\\ r=rate\to 4.8\%\to \frac{4.8}{100}\dotfill &0.048\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus twice} \end{array}\dotfill &2\\ t=years\dotfill &1 \end{cases}$

$A=1000\left(1+\frac{0.048}{2}\right)^{2\cdot 1}\implies \boxed{A\approx 1048.58}~~\textit{\Large \checkmark} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \stackrel{\textit{\Large 4.7\% quarterly}}{\textit{Compound Interest Earned Amount}}$

$A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\ 1000\\ r=rate\to 4.7\%\to \frac{4.7}{100}\dotfill &0.047\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &1 \end{cases} \\\\\\ A=1000\left(1+\frac{0.047}{4}\right)^{4\cdot 1}\implies \boxed{A\approx 1047.83}$