Correct Answer: $9 \frac{3}{8}$ days

Solution : Given: A, B, and C can do a piece of work in 30, 20, and 10 days, respectively.
Total work = time × efficiency
The total work of A, B, and C is LCM of (30, 20, 10) = 60 units
Efficiency of A is $\frac{\text{Total work}}{\text{Time}}=\frac{60}{30}$ = 2
Efficiency of B is $\frac{\text{Total work}}{\text{Time}}=\frac{60}{20}$ = 3
Efficiency of C is $\frac{\text{Total work}}{\text{Time}}=\frac{60}{10}$ = 6
A is assisted by B on one day and by C on the next day.
Work done in 2 days is = (Efficiency of A + Efficiency of B) + (Efficiency of A + Efficiency of C) = (2 + 3) + (2 + 6) = 13 units
Work done in 8 days is 52 units.
Work left is 60 – 52 = 8 units
Again (A + B) starts work and does 5 units of work on the 9th day.
Work done in 9 days is 57 units.
The work left is 3 units.
Now, (A + C) start the work and do 8 units work in 1 day then 3 units work in $\frac{3}{8}$ days.
So, the total time taken is 9 days + $\frac{3}{8}=9\frac{3}{8}$ days
Hence, the correct answer is $9\frac{3}{8}$ days.