krkan99
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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.

Respuesta :

[tex]\bf \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~A}{A=P\left(1+\frac{r}{n}\right)^{nt}} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 6\%\to \frac{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+\frac{0.06}{2}\right)^{2\cdot 1}\implies A=1000(1.03)^2\implies \boxed{A=1060.9}[/tex]

[tex]\bf -------------------------------\\\\ \qquad \textit{Compound Interest Earned Amount} \\\\ \stackrel{Bank~B}{A=P\left(1+\frac{r}{n}\right)^{nt}} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$1000\\ r=rate\to 6\%\to \frac{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} [/tex]

[tex]\bf A=1000\left(1+\frac{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[/tex]