SOLUTION: A total of $50,000 is invested in two funds paying 8% and 8.5% simple interest. The annual interest is $4120. How much is invested in each fund?

Question 179787: A total of $50,000 is invested in two funds paying 8% and 8.5% simple interest. The annual interest is $4120. How much is invested in each fund?
Found 3 solutions by Mathtut, checkley77, solver91311:
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
let x and y be the amount invested at 8% and 8.5% respectively
:
x+y=50000........eq 1
.08x+.085y=4120.eq 2
:
rewrite eq 1 to x=50000-y and plug that value into eq 2
:
.08(50000-y)+.085y=4120
:
4000-.08y+.085y=4120
:
.005y=120
:
amount invested at 8.5%
:
amount invested at 8%
Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
.085x+.08(50,000-x)=4120
.085x+4,000-.08x=4120
.005x=4,120-4,000
.005x=120
x=120/.005
x=120/.005
x=24,000 invested @ 8.5%
50,000-24,000=26,000 invested @ 8%
Proof:
.085*24,000+.08*26,000=4,120
2040+2080=4,120
4,120=4,120
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Let x be the amount invested at 8.5%. Then the amount invested at 8.0% must be $50,000 - x.

The amount of interest on the 8.5% investment must then be and the amount of interest on the 8.0% investment must be . The sum of these two amounts is the total interest, so:

Solve for x to get one of the investment amounts. Subtract that amount from 4120 to get the other amount.

John

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