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A certain amount of money is invested at 6% per year. A second amount is P5000 larger than the first is invested at 8% per year. The interest from the investment at the higher rate exceeds the income from the lower investment by P500. Find the investment at each rate.

  • investment -

    $X @ 6%.
    $(X+5000) @ 8%.

    (x+5000)0.08 = 0.06x + 500.
    0.08x + 400 = 0.06x+500
    0.02x = 500-400 = 100
    X = $5000 @ 6%.
    X+5000 = 5000+5000 = $10,000 @ 8%

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