Question

Sam invested $3200, some at 6% interest and the rest at 9.2%. How much did he invest at each rate if he received $249.60 in interest in one year?

Answer 1

Okay, here we have this:

Considering the provided information, we are going to calculate investment at each interest rate, then we have the following:

We will first take x to be the amount of money invested at 6 percent interest.

Therefore the amount invested at 9.2 percent will be: (3200-x) dollars.

In this way we obtain the following equation of total interest:

$0.06x+0.092(3200-x)=249.60$

Let's solve for x:

$\begin{gathered} 0.06x+294.4-0.092x=249.6 \\ -0.032x+294.4=249.6 \\ 44.8=0.032x \\ x=\frac{44.8}{0.032} \\ x=1400 \end{gathered}$

Finally we obtain that the amount of money invested at 6 percent interest is $1400. And the amount of money invested at 9.2 percent interest is ($3200-$1400)=$1800