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 ->  Percentage-and-ratio-word-problems -> 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 187624: 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?
Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
"Mike invested $6000 for one year" ----> x+y=6000 ----> y=6000-x


"He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest" ----> 0.09x+0.11y=624 ----> 9x+11y=62400


9x+11y=62400

9x+11(6000-x)=62400

9x+66000-11x=62400

-2x+66000=62400

-2x=62400-66000

-2x=-3600

x=1800


He invested $1,800 at 9%

y=6000-x

y=6000-1800

y=4200


He invested $4,200 at 11%