Final answer:
The interest rate on the $3000 investment is 6%, and the interest rate on the $15,000 investment is 6.6%. This is determined by setting up and solving an algebraic equation using the given information that the annual interest on a $15,000 investment exceeds that of a $3000 investment by $810 and that the $15,000 is invested at a 0.6% higher interest rate.
Step-by-step explanation:
The problem at hand is a typical algebra question involving interest rates and investments.
Let r represent the interest rate (in decimal form) on the $3000 investment. Then, the interest rate on the $15,000 investment would be r + 0.006 (0.6% higher than the rate of the $3000 investment).
The annual interest on the $3000 investment is $3000 ⋅ r, and the annual interest on the $15,000 investment is $15,000 ⋅ (r + 0.006).
We are given that the difference in interest between the two investments is $810, so:
$15,000 ⋅ (r + 0.006) - $3000 ⋅ r = $810
By simplifying this equation, we find:
$15,000r + $90 - $3000r = $810
$12,000r = $720
r = 0.06
Therefore, the interest rate on the $3000 investment is 6%, and the interest rate on the $15,000 investment is 6.6%.