A and B can do a work in 40 days and 60 days respectively. They work together. After 6 days B leaves from the work and A continues. After 5 more days C joins with A and both completes the remaining in 5 days. In how many days C alone can do it.

1 day work of $A=\frac{1}{40}$

1 day work of $B=\frac{1}{60}$

$A+B=\frac{1}{40}+\frac{1}{60}\Rightarrow \frac{3+2}{120}\Rightarrow \frac{5}{120}=\frac{1}{24}$

6 days work of A and B$=6\times \frac{1}{24}=\frac{1}{4}$

5 days work of$A=5\times \frac{1}{40}=\frac{1}{8}$

Total completed work $=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$

Remaining work $=1-\frac{3}{8}=\frac{5}{8}$

Both A and C completes the remaining work is 5 days

$(\frac{1}{40}+\frac{1}{x})5=\frac{5}{8}$

$\frac{1}{x}=\frac{1}{8}-\frac{1}{40}$$\to \frac{1}{x}=\frac{4}{40}$

x=40 days