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 48802This question is from textbook survey of math with applications
: 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 survey of math with applications

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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?
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9% DATA:
amount= x dollars ; interest=0.09x dollars
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11% Data
amount = 6000-x dollars ; interest = 0.11(6000-x)=660-0.11x
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EQUATION:
int + int = 624 dollars
0.09x+660-0.11x=624
0.02x =660-624
0.02x=36
x=$1800 (amount invested at 9%)
6000-x=$4200 (amount invested at 11%)
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Cheers,
Stan H.

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