SOLUTION: A broker invested $10000, part earning interest at 8% per annum, and the remainder at 10% per annum. If the investments earned $872 in one year, how much is invested at 10%. (Use a
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-> SOLUTION: A broker invested $10000, part earning interest at 8% per annum, and the remainder at 10% per annum. If the investments earned $872 in one year, how much is invested at 10%. (Use a
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Question 1100317: A broker invested $10000, part earning interest at 8% per annum, and the remainder at 10% per annum. If the investments earned $872 in one year, how much is invested at 10%. (Use algebra) Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x=8% earned
10000-x=10% earned
.08x+.10(10000-x)=872
.08x+1000-.10x=872
-.02x=-128
multiply both sides by -50
x=6400 @8%=$512
10000-x=3600@10%=$360
