Question

A,B,C can do piece of work in 36,54,72 days respectively.they started the work but A left 8days before the completion of work while B left 12 days before the completion of work.find the no of days for which C workd?
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Answer 1

Solution: A does (1/36), B does (1/54) and C does (1/72) part of the work in a day.

Let the work be completed in x days, during which A for (x-8) days, B worked for (x-12) days and C work for all the x days.

So the equation is (x-8)/36 + (x-12)/54 + x/72 = 1 …(1)

Factors of

36 = 2x2x3x3

54 = 2x3x3x3

72 = 2x2x2x3x3

LCM = 2^3*3^3 = 216

So (1) becomes 6(x-8) + 4(x-12) +3x = 216

6x-48+4x-48+3x = 216

13x = 216+96 = 312

x = 312/13 = 24

So C worked for all 24 days, A for 16 days and B for 12 days. Answer.

Check: (16/36) + (12/54) + (24/72) = 1. Correct.