Nico is saving money for his college education. He invests some money at 9%, and $1100 less than that amount at 6%. The investments produced a total of $249 interest in 1 yr. How much did he invest at each rate?
To find out how much Nico invested at each rate, we can set up two equations based on the information given.
Let's say Nico invested x dollars at 9% interest. According to the problem, he invested $1100 less than that amount at 6% interest. So, the amount Nico invested at 6% would be (x - $1100).
Now, we'll set up the equation for the total interest earned:
0.09x + 0.06(x - $1100) = $249
Simplifying the equation, we have:
0.09x + 0.06x - 0.066 = $249
Combining like terms, we get:
0.15x - 0.066 = $249
Adding 0.066 to both sides of the equation, we have:
0.15x = $249 + $0.066
0.15x = $249.066
Finally, dividing both sides of the equation by 0.15, we find:
x = $249.066 / 0.15
x ≈ $1660.44
So, Nico invested approximately $1660.44 at 9% interest.
To find the amount Nico invested at 6% interest, we'll substitute the value of x into the equation:
(x - $1100) ≈ $1660.44 - $1100
(x - $1100) ≈ $560.44
Therefore, Nico invested approximately $560.44 at 6% interest.
In conclusion, Nico invested approximately $1660.44 at 9% interest and $560.44 at 6% interest.