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?

Question 177072This 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 checkley75(3666) (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=4,200 invested @ 11%
6,000-4,200=1,800 invested @ 9%
Proof:
.11*4,200+.09*1,800=624
462+162=624
624=624
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