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 4202 dollars. How much did he invest at each rate?

Answer 1

• Amount at 6% = $38,200
• Amount at 10% = $19,100

Step-by-step explanation:

Assume x is the amount in the 10% account.

The formula to solve would be;

(2x * 6%) + x * 10% = 4,202

0.12x + 0.1x = 4,202

0.22x = 4,202

x = 4,202/0.22

x = $19,100

The amount he invested at 6% is therefore;

= 19,100 * 2

= $38,200