25 thoughts on “Time and Work – 3

  1. Sir,
    Can you post some questions related to work done on alternative days and also wages related concepts?
    Regards,
    Adorini

  2. The sum of odd numbers is n^2 which will not give the result, We take into account that masons show up after every alternate day : so manhours look like 1+2+3+3+4+4+5 for 7 days making up 28 manhours. The rest of 120 manhours(5mason * 20days) = 120-28 = 92 are completed in 92/5 = 18.4 days.Total days required = 46.4.

  3. Hi !
    I solved it considering the availability of only 5 masons to do the work. Accordingly, I got (1+1+2+2+3+3+4+4+5) = 25 for the initial 9 days. And then, all 5 workers work to complete the remaining 75 man days work in 75/15 = 15 days, totaling 24 days.

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