Question

Shoji invested $7000, part at 9% interest and part at 6% interest. The interest obtained from the 6% investment was half of the interest obtained from the 9% investment. How much was invested at each rate?

Answer 1

Shoji invested $4000 at 9% interest and $3000 at 6% interest.

Step-by-step explanation:

Let's assume that Shoji invested x dollars at 9% interest and (7000 - x) dollars at 6% interest

The interest obtained from the 9% investment is 0.09x dollars.

The interest obtained from the 6% investment is 0.06(7000 - x) dollars.

According to the problem, the interest obtained from the 6% investment is half of the interest obtained from the 9% investment. So, we can write the equation:

1. 0.06(7000 - x) = (1/2) * 0.09x

Solving the equation, we get x = 4000.

Therefore, $4000 was invested at 9% interest and $3000 was invested at 6% interest.