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 $9878 dollars. How much did he invest at each rate?

Answer 1

• $89,800 at 6%
• $44,900 at 10%

Explanation:

You want to know the amounts invested if the total interest earned is $9878 and twice as much was invested at 6% as at 10%.

Setup

Let x represent the amount invested at 10%. Then 2x was invested at 6%, and the interest earned was ...

0.10x +0.06(2x) = 9878

Solution

Simplifying, we have ...

0.22x = 9878

x = 44900 . . . . . . . divide by 0.22

2x = 89800 . . . . . amount at 6%

The man invested $89,800 at 6% and $44,900 at 10%.

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