$A$ can do a piece of work in $10$ days, $B$ can do in $20$ days and $C$ can do in $30$ days. If $A$ is assisted by $B$ and $C$ turn by turn in alternate days respectively, in how many days the work might have been finished?
My attempt; In $10$ days, $A$ can do $1$ work
In $1$ day, $A$ can do $\frac{1}{10}$ work
Also, In $1$ day, $B$ can do $\frac{1}{20}$ work And, In $1$ day, $C$ can do $\frac{1}{30}$ work. Now, Whay should I do after this?
You're definitely on track to the correct solution. The first day, $\frac1{10} + \frac{1}{20} = \frac{3}{20}$ work is done. The next day, $\frac1{10} + \frac{1}{30} = \frac{4}{30}$ work is done. After two days, therefore, $\frac{3}{20} + \frac{4}{30} = \frac{17}{60}$ work is done in total. Can you see how to finish from here?