A and B can do a piece of work in 8 days, B and C can do it in 24 days, while C and A can do it in $$8 \frac{4}{7}$$ days. In how many days can C do it alone?

Let total work = L.C.M. (8,24,$$\frac{60}{7}$$) = 120 units

Let efficiencies of A, B and C are $$a,b$$ and $$c$$ respectively.

A and B can do the piece of work in 8 days = $$a+b=\frac{120}{8}=15$$ units/day -------------(i)

Similarly, $$b+c=\frac{120}{24}=5$$ units/day --------------(ii)

And $$c+a=\frac{120}{\frac{60}{7}}=14$$ units/day --------------(iii)

Adding the three equations, we get :

=> $$2(a+b+c)=15+5+14$$

=> $$(a+b+c)=\frac{34}{2}=17$$

Substituting value of $$a+b$$ from equation (i) in above equation, => $$15+c=17$$

=> $$c=17-15=2$$ units/day

$$\therefore$$ Time taken by C alone to finish the work = $$\frac{120}{2}=60$$ days

=> Ans - (A)