Calvin invested $7500 for one year, part 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%?

Let's assume Calvin invested "x" dollars at 12% annual interest.

Since the total amount invested is $7500, the rest of the money invested at 10% would be ($7500 - x) dollars.

Now, we can calculate the interest earned on each investment:

Interest earned on the amount invested at 12% = (x * 12%) = 0.12x

Interest earned on the amount invested at 10% = (($7500 - x) * 10%) = 0.10($7500 - x) = 750 - 0.10x

According to the given information, the total interest earned is $890.

So, we can set up the equation as follows:

0.12x + 750 - 0.10x = 890

Combining like terms, we get:

0.02x + 750 = 890

Subtracting 750 from both sides of the equation, we have:

0.02x = 140

Dividing both sides by 0.02, we get:

x = 7000

Therefore, Calvin invested $7000 at 12% annual interest.

To solve this problem, we can solve it by setting up a system of equations.

Let's assume Calvin invested x dollars at 12% and (7500 - x) dollars at 10%.

The amount of interest earned from the first investment at 12% is 0.12x.
The amount of interest earned from the second investment at 10% is 0.10(7500 - x) or (0.10)(7500) - (0.10)x.

The total interest earned is $890.

So, we can set up the following equation:
0.12x + 0.10(7500 - x) = 890

Now, let's solve for x:

0.12x + 0.10(7500 - x) = 890
0.12x + 750 - 0.10x = 890
0.02x = 890 - 750
0.02x = 140
x = 140 / 0.02
x = 7000

Therefore, Calvin invested $7000 at 12% interest.

part at 12% ---- x

part at 10% ---- 7500-x

solve for x ....

.12x + .10(7500-x) = 890