Question:

help needed in math question?

by Guest6320  |  11 years ago

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William opened two investment accounts for his grandson’s college fund. The first year, these investments, which totaled $18,000, yielded $831 in simple interest. Part of the money was invested at 5.5% and the rest at 4%. How much was invested at each rate?

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4 ANSWERS

  1. Guest7654
    x+y=18000

    .04x + .055y = 831

    .04x + .055(18000-x) = 831

    .04x + 990 - .055x = 831

    .015x = 159

    x = 10600

    y = 18000 - 10600 = 7400

  2. Guest3973
    $18,000 was invested, so split that into the two parts. $x amount was invested at 5.5% and ($18,000-x) was invested at 4%

    .055x + (18,000 - x).04 = 831

    .055x + 720 - .04x = 831

    .015x = 111

    x = 7,400

    $7,400 was invested at 5.5%

    $10,600 was invested at 4.0%

    Piece of cake.
  3. Guest2404
    interest = p * 18000 * 0.055 + (1-p) * 18000 * 0.04

    = 831

    => 270 p + 720 = 831

    => 270 p = 111

    => p = 0.4111111....

    So 41.111 % of the money was invested at 5.5 %

    or 7400 $. The rest (10600 $) was invested at 4 %.
  4. Guest6863
    let x and y be the two parts

    x + y = 18000

    x(0.055) + y(0.04) = 831

    x = 18000 - y

    (18000 - y)(0.055) + y(0.04) = 831

    990 - 0.055y + 0.04y = 831

    0.015y = 159

    y = 10600

    x = 7400

    Check

    7400(0.055) + 10600(0.04) = 831

    407 + 424 = 831

    .

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