SOLUTION: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

Algebra ->  Sequences-and-series -> SOLUTION: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?       Log On


   

Question 142441This question is from textbook
: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?
This question is from textbook

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.11x+.09(6000-x)=624
.11x+540-.09x=624
.02x=624-540
.02x=84
x=84/.02
x=4200 invested @ 11%
6000-4200=1800 amount invested @ 9%
Proof:
.11*4200+.09*1800=624
462+162=624
624=624