Question

A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $6600 dollars. How much did he invest at each rate?

Your answer is:
Amount invested at 6% equals $
Amount invested at 10% equals $

Answer 1

To find out how much the man invested at each interest rate, we set up a system of equations and solved them to find that the man invested $60,000 at 6% and $30,000 at 10%.

Step-by-step explanation:

The student is asking for help to determine the amount invested at two different rates of simple interest given the total annual interest earned from both investments. To solve this, we can set up a system of equations. Let's call the amount invested at 6% x and the amount at 10% y.

Since the man invests twice as much in the 6% account, then x = 2y. The total interest from both accounts is $6600, so the simple interest formula, which is interest (I) equals principal (P) times rate (r) times time (t), can be used to set up another equation: 0.06x + 0.10y = $6600.

Using substitution, we can replace x with 2y in the equation to find y. By solving the system of equations, we can determine how much was invested at each rate:

1. 0.06x + 0.10y = $6600 becomes 0.06(2y) + 0.10y = $6600.
2. Simplify to 0.12y + 0.10y = $6600, which gives us 0.22y = $6600.
3. Divide both sides by 0.22 to find y = $30,000.
4. Since x = 2y, we have x = 2($30,000) which gives us x = $60,000.

Therefore, the man invested $60,000 at 6% and $30,000 at 10%.