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the annual interest on an 18,000 investment exceeds the interest earned on a 9000 investment by 648. the 18,000 is invested at a 0.6% higher rate of interest than the 9000. what is the interest rate of
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the annual interest on an 18,000 investment exceeds the interest earned on a 9000 investment by 648. the 18,000 is invested at a 0.6% higher rate of interest than the 9000. what is the interest rate of each investment

User RemoteCTO
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Let x=annual interest rate of 18000 investment.
Then
18000*x - 9000(x-0.006)=648
Solve for x
18000x-9000x+9000*0.006=648
9000x=648-54=594
x=594/9000=0.066
Answer:
the annual interest rate of the 18000 investment is 6.6%
the annual interest rate of the 9000 investment is 6.0%

User Stoni
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