SOLUTION: last year, John had $1,200 to invest. He invested some of the money at 3% interest and the rest at 1.5% interest. At the end of the year, his investments had earned $28.50. How muc

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: last year, John had $1,200 to invest. He invested some of the money at 3% interest and the rest at 1.5% interest. At the end of the year, his investments had earned $28.50. How muc      Log On

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Question 1009002: last year, John had $1,200 to invest. He invested some of the money at 3% interest and the rest at 1.5% interest. At the end of the year, his investments had earned $28.50. How much did he invest at each rate?
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Call what he invested at 3%, x.
Then the amount invested at 1.5% would be 1200-x.
The setup is like this
.03x + .015(1200-x) = 28.50
.03x + 18 - .015x = 28.50
.015x = 10.50
x = $700 at 3%
1200-x = $500 at 1.5%