SOLUTION: The annual interest on a $16,000 investment exceeds the interest earned on a $6000 investment by $830. The $16,000 is invested at a 0.5% higher rate of interest than the $6000. W
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Question 1049541: The annual interest on a $16,000 investment exceeds the interest earned on a $6000 investment by $830. The $16,000 is invested at a 0.5% higher rate of interest than the $6000. What is the interest rate of each investment? Answer by jorel555(1290) (Show Source):
You can put this solution on YOUR website! Let r be the interest rate on the $6000 investment. Then the interest rate on the $16000 investment is r+.005. So:
16000(r+.005)=6000(r)+830
16000r+80=6000r+830
10000r=750
r=.075, or 7.5% on the $6000 investment; and 8% on the $16000 investment. ☺☺☺☺
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