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28 November, 01:42

Two investments total $1000. If the combined annual interest is $68, how much was invested at 8% and how much at 5%?

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  1. 28 November, 02:30
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    600 was invested at 8% and $400 at 5%

    Step-by-step explanation:

    Let x represent the amount of $ invested at 8% and y the amount at 5%.

    Then x + y = $1000 is the total investment. Therefore, y = 10000 - x

    Interest earned is then:

    0.08x + 0.05 (y) = 68, or (after substitution)

    0.08x + 0.05 (1000) - 0.05x = 68

    0.08x + 50 - 0.05x = 68, or

    0.03x = 18. Dividing both sides by 0.03, we find that x = 18/0.03 = 600.

    If x = $600, then y = $1000 - $600 = $400.

    $600 was invested at 8% and $400 at 5%.
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