SOLUTION: A sum of money of $10000 is invested, partly at 4% and partly at 6%. If the yearly income from the two investments was $530. Find the amount invested at each rate.

Question 66157: A sum of money of $10000 is invested, partly at 4% and partly at 6%. If the yearly income from the two investments was $530. Find the amount invested at each rate.
Found 2 solutions by josmiceli, ptaylor:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
x is amount invested @ 4%
10,000 - x is amount invested @ 6%

amount invested @ 4%
amount invested @ 6%
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
A sum of money of $10000 is invested, partly at 4% and partly at 6%. If the yearly income from the two investments was $530. Find the amount invested at each rate.
Let x= amount invested at 4%
Then 10,000-x= amount invested at 6%

Now we are told that the interest accrued on the amount invested at 4% (.04)(x) plus the interest accrued on the amount invested at 6% (.06(10,000-x)) equals $530. So our equation to solve is:
.04x+.06(10000-x)=530
.04x+600-.06x=530 collect the x's together and subtract 600 from both sides:
-.02x=-600+530
-.02x=-70
x=$3500 amount invested at 4%
10,000-x=10,000-3500=$6500 amount invested at 6%
ck
.04(3500)+.06(6500)=530
140+390=530
$530=$530
Hope this helps-----ptaylor

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