Question

A man invests his savings in two accounts, one paying 6 percent and the other paying 10 percent simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $9570. How much did he invest at each rate?

Answer 1

To solve the investment problem, we set up a system of equations using the amounts invested at 6 percent and 10 percent interest rates, resulting in investments of $17000 at 6 percent and $8500 at 10 percent to achieve an annual interest of $9570.

Step-by-step explanation:

The question is a classic problem in simple interest and involves setting up a system of equations to resolve the investments made in two different accounts with differing interest rates.

Let's call the amount invested at 6 percent x, and hence the amount invested at 10 percent will be x/2 (since he puts twice as much in the 6 percent account). The total interest from these two investments is $9570 per year.

The interest earned from the 6 percent account in one year is 0.06x, and the interest from the 10 percent account is 0.10(x/2).

So, we have the system of equations:
0.06x + 0.10(x/2) = $9570

After solving, we get that x, the amount invested at 6 percent, is $17000, and x/2, the amount invested at 10 percent, is $8500.