Kendra invested twice as much money at an annual rate of 7% as she did at an annual rate of 5.5%. How much did she invest at the higher rate

Question

Kendra invested twice as much money at an annual rate of 7% as she did at an annual rate of 5.5%. How much did she invest at the higher rate if her income from these investments for one year totaled $175.50?

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Aaliyah 2 weeks 2021-10-03T22:29:55+00:00 1 Answer 0

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    2021-10-03T22:31:47+00:00

    Answer:

    $1,679.42

    Step-by-step explanation:

    Let x be the amount Kendra invested at 5.5%. Since the amount invested at 7% is twice the amount she invested at 5.5%, she invested 2x at 7%.

    Remember that I=Prt , where:

    I is the interest earned

    P is the principal or initial investment

    r is the interest rate in decimal form

    t is the time in years

    – For the 5.5% investment:

    P=x

    r=\frac{5.5}{100} =0.055

    t=1

    Let’s replace the values in our formula, I=x(0.055)(1), which simplified is I=0.055x

    – For the 7% investment:

    P=2x

    r=\frac{7.7}{100} =0.077

    t=1

    I=2x(0.077)(1)

    I=0.154x

    Now, we that the the amount of interest earned combining the two investment is $175.50, so:

    0.055x+0.154x=175.50

    0.209x=175.50

    x=\frac{175.50}{0.209}

    x=839.71

    She invested $839.71 at 5.5%, and since she invested twice that amount at 7%, she invested 2*($839.71 ) = $1,679.42 at 7% (the higher rate).

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