Let the annual interest rate for the $2000 investment be denoted as $r$ (in decimal form). Then, the annual interest rate for the $1000 investment is $r - 0.005$, as it is half a percent lower than $r$.

The total interest earned is the sum of the interest from both investments, and we know that the total interest is $190. The interest from an investment can be calculated using the formula:

$\text{Interest} = \text{Principal} \times \text{Rate}$

For the $2000 investment: \text{Interest from $2000} = 2000 \times r

For the $1000 investment: \text{Interest from $1000} = 1000 \times (r - 0.005)

The total interest is given as: $2000 \times r + 1000 \times (r - 0.005) = 190$

Now, solve the equation:

$2000r + 1000(r - 0.005) = 190$

Expanding the terms: $2000r + 1000r - 5 = 190$

Combine like terms: $3000r - 5 = 190$

Add 5 to both sides: $3000r = 195$

Solve for $r$: $r = \frac{195}{3000} = 0.065$

So, the annual interest rate for the $2000 investment is 6.5%. The rate for the $1000 investment, which is 0.5% lower, is: $r - 0.005 = 0.065 - 0.005 = 0.06$

Thus, Jason invested the $1000 at a rate of 6%.

Would you like further clarification or more details?

Here are 5 related questions for practice:

1. What would the total interest be if Jason invested $3000 at 6%?
2. How much interest would Jason earn on a $500 investment at 6% annually?
3. How would the total interest change if the second rate were 1% lower instead of 0.5% lower?
4. If Jason invested $1500 at 7%, how much interest would he earn in one year?
5. What is the relationship between the principal, rate, and time in calculating interest?

Tip: When solving word problems involving percentages, converting percentages to decimals (e.g., 6% to 0.06) helps simplify calculations.