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)