Question

Mr. Ballew invested $3,000 more at14% than at 11% annual interest. Hisannual return is $1,670. How much didhe invest at each rate?

Answer 1

We have to transform this word problem into an equation.

• Let's call ,x, to the money invested at 11% of annual interest.

,

• The money invested at 14% of annual interest is x+3000.

,

• The addition of these two invested amounts is equivalent to 1670 because that's the return.

The equation is

$0.11x+0.14(x+3000)=1670$

Solve for x.

$\begin{gathered} 0.11x+0.14x+420=1670 \\ 0.25x=1670-420 \\ 0.25x=1250 \\ x=\frac{1250}{0.25} \\ x=5000 \end{gathered}$

Therefore, Mr. Ballew invested $5000 at 11% annual interest.

Add to obtain the other amount.

$5000+3000=8000$

Therefore, Mr. Ballew invested $8000 at 14% annual interest.