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5 January, 10:03

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

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  1. 5 January, 10:28
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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%
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