Question

If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested?

Answer 1

`~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$132.90\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 5.5\%\to (5.5)/(100)\dotfill &0.055\\ t=years \end{cases} \\\\\\ 132.90 = (6000)(0.055)(t)\implies \cfrac{132.90}{(6000)(0.055)}=t\implies \cfrac{443}{1100}=t \\\\\\ \stackrel{\textit{converting that to days}}{\cfrac{443}{1100}\cdot 365} ~~ \approx ~~ 147~days`

now, if we move back from August 14th by 147 days backwards, that'd put us on March 20th.