A man invests $5500 dollars in three accounts that pay 5%, 8%, and 9% in annual interest respectively. He has two times as much money invested at 9% as he does at 8%. If the total interest earned for the year is $449, how much is invested at 5%?
To solve this problem, we can set up a system of equations:
Let x be the amount invested at 5%.
Since the man has two times as much money invested at 9% as he does at 8%, the amount invested at 8% is x/2.
Therefore, the amount invested at 9% is 2 * (x/2) = x.
Now we can calculate the interest earned from each account:
Interest at 5% = (x * 5%) = 0.05x
Interest at 8% = ((x/2) * 8%) = 0.08(x/2) = 0.04x
Interest at 9% = (x * 9%) = 0.09x
The total interest earned is $449, so we can write the equation:
0.05x + 0.04x + 0.09x = 449
Combining like terms, we have:
0.18x = 449
To solve for x, we divide both sides of the equation by 0.18:
x = 449 / 0.18
x ≈ 2494.44
Therefore, approximately $2494.44 is invested at 5%.