Question
tracy invested $6000 for one year, part at 10% annual and the balance at 13% annual interest. her total interest for the year was $712.5. how much money did she invested at each rate?
I tried using a table for this I just don't know how to set the problem up.
I tried using a table for this I just don't know how to set the problem up.
Answers
Let $x is what she invested at 10%
then $(6000-x) is what she invested at 13%.
We know that
0.10x + 0.13(6000-x) = 712.5
Solve for x to get the answers.
then $(6000-x) is what she invested at 13%.
We know that
0.10x + 0.13(6000-x) = 712.5
Solve for x to get the answers.
Related Questions
Rosa invested $8000 for one year, part ast 8% annual interest and the balance at 10% annual interest...
A person has invested $6000 . Part of the money is invested at 3% and the remaining amout at 4% . T...
A total of 6000 is invested part 8% and the remainder at 13% how much is invested at each rate if th...