Asked by miranda
wyatt invests $6000 in two accounts, the first pays 10 percent simple interest, the second pays 11 percent simple interest. At the end of the year he has earned $640 in interest. How much is invested at each rate?
Answers
Answered by
oobleck
this is <u>pre</u>-algebra? It sure looks like algebra to me.
If he invests $x at 10%, then the rest (6000-x) is at 11%. So, add up the interest:
0.10x + 0.11(6000-x) = 640
Now just solve for x.
If he invests $x at 10%, then the rest (6000-x) is at 11%. So, add up the interest:
0.10x + 0.11(6000-x) = 640
Now just solve for x.
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