Part $30000 is invested at 6% another part at 7%
and the remainder at 8% yearly interest the total yearly income from the 3 investments is $2200
To solve this problem, we can use the concept of percentages and algebraic equations. Let's break down the information given:
Let's assume the part invested at 6% is X dollars.
The part invested at 7% will be (30000 - X) dollars.
The remainder invested at 8% will be (30000 - X) dollars.
Now we can set up the equation to find the total yearly income:
0.06X + 0.07(30000 - X) + 0.08(30000 - X) = 2200
We use the decimal equivalents of the percentages (0.06, 0.07, and 0.08) and multiply them by their respective amounts to find the income earned from each investment.
Let's solve the equation step-by-step:
0.06X + 0.07(30000 - X) + 0.08(30000 - X) = 2200
0.06X + 2100 - 0.07X + 2400 - 0.08X = 2200
-0.09X + 4500 = 2200
-0.09X = 2200 - 4500
-0.09X = -2300
X = (-2300) / (-0.09)
X ≈ 25555.56
Therefore, approximately $25,555.56 was invested at 6%, and the remaining amount ($30,000 - $25,555.56) ≈ $4,444.44 was invested at 7% and 8% respectively.