Sue invested $10,000, part at 3.6% annual interest and the balance 7.8% annual interest. How much is the invested at each rate if a 1-year interest payment is $667.02?

Question

The answers are

$7,310 at 7.8%

$2,690 at 3.6%

I just don’t know how to get to the answers.

Answer 1

Total available to invest is $10,000. Total interest I earned after 1 year is $667.02.

Using this information, let P1 be the amount invested at 3.6%, which is the same as 0.036. Then 10,000-P1 is the amount left to invest at 7.8%, which is 0.078.

Thus, I = .036 x P1 + (10,000-P1) x 0.078 ==> I = .036P1 + 780 -0.078P1 ==> I = 780 - 0.042P1

We know I=667.02, so plugging it into the formula we get 667.02 = 780 - 0.042P1

Subtract 780 from both sides ==> -112.98 = -0.042P1.

Solve for P1 by dividing both sides by -0.042 to get P1 = $2690 invested at 3.6% and 10,000-2690 = $7310 invested at 7.8%.

The solutions are $2690 and $7310.