I invested $42,000 in three funds paying 5%, 7%, and 9% simple interest. The total annual interest from these investments was $2,600. The amount of money invested at 5% was $200 less than the amount invested at 7% and 9% combined. How much was invested in each fund?
To solve this problem, let's denote the amounts invested in the three funds as follows:
Let x be the amount invested at 5%.
The amount invested at 7% is (x + $200) since it is $200 more than the amount invested at 5%.
The amount invested at 9% is (x + $200) since it is $200 more than the amount invested at 5%.
Now, let's calculate the annual interest for each investment:
The interest earned from the investment at 5% is calculated as: 5% of x = 0.05x.
The interest earned from the investment at 7% is calculated as: 7% of (x + $200) = 0.07(x + $200).
The interest earned from the investment at 9% is calculated as: 9% of (x + $200) = 0.09(x + $200).
According to the problem, the total annual interest from the three investments is $2,600, so we can write the equation:
0.05x + 0.07(x + $200) + 0.09(x + $200) = $2,600.
Now, let's solve this equation to find the value of x, which represents the amount invested at 5%:
0.05x + 0.07x + 0.07($200) + 0.09x + 0.09($200) = $2,600.
0.05x + 0.07x + $14 + 0.09x + $18 = $2,600.
Combining like terms: 0.21x + $32 = $2,600.
Subtracting $32 from both sides: 0.21x = $2,600 - $32.
Simplifying: 0.21x = $2,568.
Dividing both sides by 0.21: x = $2,568 / 0.21.
Calculating: x ≈ $12,228.57.
So, approximately $12,228.57 was invested at 5%.
The amount invested at 7% = $12,228.57 + $200 = $12,428.57.
The amount invested at 9% = $12,228.57 + $200 = $12,428.57.
Therefore, approximately $12,228.57 was invested at 5%, approximately $12,428.57 was invested at 7%, and approximately $12,428.57 was invested at 9%.