SOLUTION: The student council invested $4000, part at 8.5% per annum and the remainder at 9.5%. After one year the interest from the 8.5% was $16 more than the interest from the 9.5% investm

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The student council invested $4000, part at 8.5% per annum and the remainder at 9.5%. After one year the interest from the 8.5% was $16 more than the interest from the 9.5% investm      Log On


   



Question 947464: The student council invested $4000, part at 8.5% per annum and the remainder at 9.5%. After one year the interest from the 8.5% was $16 more than the interest from the 9.5% investment. How much was invested at 9.5%?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The student council invested $4000, part at 8.5% per annum and the remainder at 9.5%.
After one year the interest from the 8.5% was $16 more than the interest from the 9.5% investment.
How much was invested at 9.5%?
:
let x = amt invested at 9.5%
the total is $4000, therefore
(4000-x) = amt invested at 8.5%
:
8.5 int - 9.5 int = $16
.085(4000-x) - .095x = 16
340 - .085x - .095x = 16
-.18x = 16 - 340
-.18x = -324
x = -324/-.18
x = $1800 invested at 9.5%
:
:
See if that checks out, find the actual interest in each
.085(2200) = 187
.095(1800) = 171
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difference: 16