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?
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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
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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