(A + B)'s 1 day's work = \(\frac{1}{10}\)

C's 1 day's work = \(\frac{1}{50}\)

(A + B + C)'s 1 day's work = \(\left(\frac{1}{10}+\frac{1}{50}\right)=\frac{6}{50}= \frac{3}{25}. ......(i)\)

A's 1 day's work = (B + C)'s 1 day's work .... (ii)

From (i) and (ii), we get: 2 x (A's 1 day's work) = \(\frac{3}{25}\)

A's 1 day's work =  \(\frac{3}{50}\)

So, B's 1 day's work \(\left(\frac{1}{10}-\frac{3}{50}\right)=\frac{2}{50} = \frac{1}{25}.\)

So, B alone could do the work in 25 days.