Question

Omar invested $68,000 in an account paying an interest rate of 7 1/2% compounded annually. Madeline invested $68,000 in an account paying an interest rate of 7 1/4% compounded daily. After 8 years, how much more money would Madeline have in her account than Omar, to the nearest dollar?

Answer 1

Explanation:

Omar: first find 7 1/2 in decimal form .then multiply the decimal 7.5 by the amount invested 68,000. then multiply amount a year in interest 510 by number of years 8 then you get 4,080.

Madeline: first find 7 1/4 as decimal .then multiply the decimal 7.25 by the amount invested 68,000. then multiply amount a year in interest

Answer 2

ok, let's say Omar invested at 7.5% and Madeline invested at 7.25%, and let's assume a year has 365 days, so daily compounding will have 365 periods in a year.

`~~~~~~ \stackrel{ \textit{\LARGE Omar} }{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$68000\\ r=rate\to 7.5\%\to (7.5)/(100)\dotfill &0.075\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &8 \end{cases}`

`A = 68000\left(1+(0.075)/(1)\right)^(1\cdot 8) \implies \boxed{A \approx 121276.49} \\\\[-0.35em] ~\dotfill`

`~~~~~~ \stackrel{ \textit{\LARGE Madeline} }{\textit{Compound Interest Earned Amount}} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$68000\\ r=rate\to 7.25\%\to (7.25)/(100)\dotfill &0.0725\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &8 \end{cases}`

`A = 68000\left(1+(0.0725)/(365)\right)^(365\cdot 8) \implies \boxed{A \approx 121443.62} \\\\[-0.35em] ~\dotfill\\\\ 121443.62~~ - ~~121276.49 ~~ \approx ~~ \text{\LARGE 167}`