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    Topics || Problems

    Divide 80 into two parts such that the ratio of one part, decreased by 4, to the other part, decreased by 8, is 1:3.

    Solution:

    Let \(x\) be the first part and \(y\) be the second part.

    \(x+y = 80\)

    \(\frac{x-4}{y-8}=\frac{1}{3}\)

    \(3(x-4) = y-8\)

    \(y = 80-x\)

    \(3(x-4) =80-x-8\)

    \(3x+x = 72 +12\)

    \(x = 21\)

    \(y = 80-21 = 59\)

    Thus, the parts are 59 and 21