A and B can finish a job in 10 days while B and C can do it in 18 days. A started the job, worked for 5 days, then B worked for 10 days and the remaining job was finished by C in 15 days. In how many days could C alone have finished the whole job?
Let total work to be done{ L.C.M. (10,18)} = 90 units
Let efficiency of A, B and C be $$x,y,z$$ respectively.
A and B can finish a job in 10 days, => $$x+y=\frac{90}{10}=9$$ --------------(i)
Similarly, $$y+z=5$$ --------------(ii)
Adding equations (i) and (ii), => $$x+2y+z=14$$ ----------(iii)
According to ques, A worked for 5 days, B for 10 days and C for 15 days
=> $$5x+10y+15z=90$$
=> $$x+2y+3z=18$$ ---------(iv)
Subtracting equation (iii) from (iv), we get : $$2z=4$$
=> $$z=2$$
$$\therefore$$ Time taken by C alone to finish the work = $$\frac{90}{2}=45$$ days
=> Ans - (C)
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