divide 116 among four part such that 5 added to first,4 subtracted from the second,third multiplied by 3 and f?

divide 116 among four part such that 5 added to first,4 subtracted from the second,third multiplied by 3 and fourth divided by 2 give the same result.

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  • Anonymous
    8 years ago
    Favorite Answer

    A + B + C + D = 116

    A + 5 = B - 4 = 3C = D/2

    B = A + 9

    C = (A + 5)/3

    D = 2(A + 5)

    Using the first equation: A + (A + 9) + (A + 5)/3 + 2(A + 5) = 116

    4A + A/3 = 116 - 9 - 10 - 5/3

    13A = 286

    A = 22

    You take it from there.

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  • 8 years ago

    a+5 = b -4 = 3*c = d/2

    a=3c-5

    b=3c+4

    d=6c

    a+b+c+d=116

    3c-5 + 3c +4 +c +6c =116

    13c = 117

    c=9

    a=22

    b=31

    d=54

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  • 8 years ago

    5+a=b-4=c*3=d/2=x

    a=x-5

    b=x+4

    c=x/3

    d=x*2

    add all these

    so 4x+x/3-1=116

    13x/3=117

    x=27

    numbers are 22,31,9,54

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