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Two plumbers received a job. At first, one of the plumbers worked alone for 1 hour, and then they worked together for the next 4 hours. After this 40% of the job was finished. How long would it take each plumber to do the whole job by himself if it is known that it would take the first plumber 5 more hours to finish the job than it would take the second plumber?

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3 answers

  1. If the fast plumber takes x hours, then the slow one takes x+5 hours.

    1/(x+5) + 4(1/x + 1/(x+5)) = 2/5
    x = 20

    So, the 1st plumber takes 25 hours.

    Check:
    In 1 hour, the 1st plumber finishes 1/25 of the job.

    During each of the next 4 hours, the two together finish 1/25 + 1/20 = 9/100 of the job.

    1/25 + 36/100 = 40/100 = 40%

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  2. 1st Plumber = 25 hrs, 2nd Plumber = 20 hrs.

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  3. 37

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