SOLUTION: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate?

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate?      Log On


   

Question 190688: Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.12x+.08(7000-x)=764
.12x+560-.08x=764
.04x=764-560
.04x=204
x=204/.04
x=$5,100 is the amount invested @ 12%.
7,000-5,100=$1,900 is the amounr invested @ 8%.
Proof:
.12*5,100+.08*1,900=764
612+152=764
764=764