Question

Bank a offers a savings account with a 6% APR compounded semiannually. Bank B offers the same rate but compounded monthly. If $1000 is invested in both banks, find the different in interest earned at the end of each year.

Answer 1

`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~A}{A=P\left(1+(r)/(n)\right)^(nt)} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 6\%\to (6)/(100)\to &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semi-annually, thus twice} \end{array}\to &2\\ t=years\to &1 \end{cases} \\\\\\ A=1000\left(1+(0.06)/(2)\right)^(2\cdot 1)\implies A=1000(1.03)^2\implies \boxed{A=1060.9}`


`\bf -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~B}{A=P\left(1+(r)/(n)\right)^(nt)} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 6\%\to (6)/(100)\to &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve}\\ \end{array}\to &12\\ t=years\to &1 \end{cases}`


`\bf A=1000\left(1+(0.06)/(12)\right)^(2\cdot 1)\implies A=1000(1.005)^(12)~~\approx~~ \boxed{A=1061.67781}\\\\ -------------------------------\\\\ thus\qquad \qquad 1061.67781~-~1060.9`