A can do a piece of work in $10$ days, B in $20$ days and C in $30$ days. If A is assisted by B and C turn by turn in alternate days, in how many days the work might have been completed?
My Attempt:
In $1$ day, A can do $\frac {1}{10}$ work.
In $1$ day, B can do $\frac {1}{20}$ work.
In $1$ day, C can do $\frac {1}{30}$ work.
Now, how should I complete?
I know the question already has an answer, but I could not understand that. Can anyone give me a clear idea?
On the first day, $\frac{1}{10} + \frac{1}{20} = \frac{3}{20} = \frac{9}{60}$ work gets done. On the second day, $\frac{1}{10} + \frac{1}{30} = \frac{4}{30} = \frac{8}{60}$ more work gets done, making $\frac{17}{60}$ in total. Total work continues to increase alternatingly by $\frac{9}{60}$ and $\frac{8}{60}$, and on the third and following days we have $\frac{26}{60},~ \frac{34}{60},~ \frac{43}{60},~ \frac{51}{60},$ and $ \frac{60}{60}$ work done in total, respectively. The work is therefore completed after $7$ days.