Bluee 發問於 科學及數學數學 · 1 十年前

algebra

Three numbers are such that the second is the difference of three times the first and 6 while the third is the sum of 2 and 2/3 the second. The sum of the three numbers is 172. Find the largest number.

The yearly changes in the population of a city for three consecutive years are, respectively, 20% increase, 30% increase, and 20% decrease. What is the total percent change from the beginning to the end of the third year?

2 個解答

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  • 1 十年前
    最愛解答

    Let the first number be x,

    then the second number = 3x-6

    and the thrid number = 2+ 2/3 * (3x-6)

    As the sum of the three numbers is 172,

    so, x + (3x-6) + [2+ 2/3 * (3x-6)] = 172

    4x-6 + (2 + 2x - 4) = 172

    6x - 8 = 172

    x = 60

    so the first, second and third number are respectively = 60, 174, and 118

    so, the largest number = 174

    Let population in the beginning = y

    then population at the end of the third year

    = y * (1+20%) * (1+30%) * (1-20%)

    = 1.2 * 1.3 * 0.8y

    = 1.248 y

    so, the total percent change from the beginning to the end of the third year

    = (1.248y - y)/y * 100%

    = 0.248 * 100%

    = 24.8% (increase)

  • 1 十年前

    1)Let the 3 number as x ,y ,z

    y = 3x -6

    z = 2 + 2/3(y) = 2 + 2/3(3x-6) = 2x -2

    x + y + z = 172

    x + 3x - 6 + 2x - 2 = 172

    6x - 8 = 172

    6x = 180

    x = 30

    therefore: x = 30, y = 84, z = 58 ( Proof: 30 + 84 + 58 = 172)

    y is the largest...84

    2) let the original population as P

    Population at the end of 3rd year:

    (1 + 20%) * (1 + 30%) * (1 - 20%) * P

    = 1.2 * 1.3 * 0.8 * P

    = 1.248P

    the percent change in 3 consecutive years:

    (1.248P - P) / P

    = (1.248 - 1)P / P

    = 0.248

    = 24.8 %

    therefore, the city has a 24.8% population growth from the beginning to the end of the third year.

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