Correct Answer: $8\frac{2}{7}$

Solution : A's 1 day's work = $\frac{1}{10}$
B's 1 day's work = $\frac{1}{20}$
C's 1 day's work = $\frac{1}{40}$
(A + B + C)'s 1 day's work = $\frac{1}{10} + \frac{1}{20}+\frac{1}{40}$ = $\frac{7}{40}$
Work done in three days will be the sum of A's two days' work and (A + B + C)'s 1 day's work.
A's two days work = $2 \times \frac{1}{10}$ = $\frac{1}{5}$
Therefore, the work is done in three days = $\frac{1}{5} + \frac{7}{40}$ = $\frac{15}{40}$ = $\frac{3}{8}$
$\frac{3}{8}$th part of the job is done in 3 days.
6 days work = $\frac{6}{8}$, thus remaining work = $\frac{2}{8}$ = $\frac{1}{4}$
7th and 8th day work = two days work of A = $\frac{1}{5}$
Thus remaining work = $\frac{1}{4} -\frac{1}{5} = \frac{1}{20}$
$\therefore \frac{1}{20}$ work done by A, B, and C together in = $\frac{\frac{1}{20}}{\frac{7}{40}}=\frac{2}{7}$ days
So, the entire job will be done by them in = $ 8\frac{2}{7}$ days
Hence, the correct answer is $ 8\frac{2}{7}$.