A can complete one-third of a work in 5 days and B can do $$\frac{2}{5}$$th of the same work in 10 days. They work together for 6 days. The remaining work is completed by C in 18 days. C alone will do the same work in:

Given : 

A complete one-third of work in 5 days.

i.e; $$\frac{1}{3}\longrightarrow\ 5$$

So, Total work finish by A in  : $$3\times\ \ 5=15\ days$$

Similarly, 

B complete $$\frac{2}{5}^{ }th$$ of the work in 10 days

So, Total work finish by B in : 25 days 

Let the total work = 75 unit (LCM of 15 & 25)

$$\therefore\ total\ work=efficiency\times\ time$$

Efficiency of A = $$\frac{75}{15}=5\ units$$

Similarly, Efficiency of B = $$\frac{75}{25}=3\ units$$

A and B together work for 6 days (given)

Work done by them in 6 days = $$\left(5+3\right)\times\ 6=48\ units$$

Remaining work = 75 - 48 = 27 units

Remaining work is finished by c in 18 days (given)

Efficiency of C = $$\frac{27}{18}$$

$$\therefore\ $$ Time taken by C to finish total work alone : 

= $$\frac{75}{\left(\frac{27}{18}\right)}=50\ days$$