Question

A person invested $7500 for 1 year apart at 7%, part at 9%, and the remainder at 12% . The total annual income from these investments was $772. The amount of money invested at 12% was $700 more than the amounts invested at 7% and 9% combined. find the amount invested at each rate.

Answer 1

The amount invested at 7% was $3000, the amount invested at 9% was $3800, and the amount invested at 12% was $7000.

Step-by-step explanation:

Let's denote the amount invested at 7% as x, the amount invested at 9% as y, and the amount invested at 12% as x + y + 700.

We can set up the following system of equations based on the given information:

• x + y + (x + y + 700)

= 7500 (Total amount invested)

• x(0.07) + y(0.09) + (x + y + 700)(0.12)

= 772 (Total annual income)

Simplifying and solving this system of equations, we can find that :

x = 3000, y = 3800, and x + y + 700 = 7500.

Therefore, $3000 was invested at 7%, $3800 was invested at 9%, and $7000 was invested at 12%.