Question

June has a savings account with an annual simple interest rate of 2.6%. She hopes to gain $6,500 in interest over a period of eleven years. To accomplish this, June invested $17,801 in the account, but later realized that this was not enough money. To the nearest dollar, how much more money should June have initially invested to reach her goal? a. $6,335 b. $3,715 c. $1,409 d. $4,926

Answer 1

$\bf \qquad \textit{Simple Interest Earned}\\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\to &\ 6500\\ P=\textit{original amount deposited}\\ r=rate\to 2.6\%\to \frac{2.6}{100}\to &0.026\\ t=years\to &3 \end{cases}$

she should have invested P, in order to get I = 6500

but she instead invested 17,801, how much more did she need? well, P - 17801

solve for P

Answer 2

P=6500/(0.026*11)
P=22,727.27 to be invested
So how much more
22727.27-17801=4926.27