Question

Sang invested some money at 11% interest. Sang also invested $191 more than 5 times that amount at 9%. How much is invested at each rate if Sang receives $2534.39 in interest after one year? (Round to two decimal places if necessary.)

Answer 1

Let 'x' be the amount invested at 11% interest and let 'y' be the amount invested at 9%. Then, since Sang investedn $191 more than 5 times the amount 'y' at 9%, we have:

$y=5x+191$

Also, since the total interest is the sum of the interest of both accounts, we have the following equation:

$0.11x+0.9y=2534.39$

if we substitute the first equation on the second, we get the following:

$\begin{gathered} 0.11x+0.9(5x+191)=2534.39 \\ \Rightarrow0.11x+0.45x+171.9=2534.39 \\ \Rightarrow0.56x=2534.39-171.9=2362.49 \\ \Rightarrow x=\frac{2362.49}{0.56}=4218.73 \\ x=4218.73 \end{gathered}$

now that we know the amount y, we can use it to find the amount y:

$\begin{gathered} y=5(4218.73)+191=21093.65+191=21284.65 \\ \Rightarrow y=21284.65 \end{gathered}$

therefore, Sang invested $4218.73 at 11% interest and $21,284.64 at 9% interest