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