A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:

15 days
26 days
310 days
4 $10 \frac{1}{2}$ $10\frac{1}{2}$ days

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (B + C)'s 1 day's work \\ = (\frac{1}{9} + \frac{1}{12}) = \frac{7}{36} \\ Work done by B and C in 3 days \\ = (\frac{7}{36} \times 3) = \frac{7}{12} \\ Remaining work \\ = (1 - \frac{7}{12}) = \frac{5}{12} \\ Now, \frac{1}{24} work is done by A in 1 day \\ So, \frac{5}{12} work is done by A in \\ (24 \times \frac{5}{12}) = 10 days \end{aligned}

$\eqalign{ & \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {\frac{1}{9} + \frac{1}{{12}}} \right) = \frac{7}{{36}} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{B}}\,{\text{and}}\,{\text{C}}\,{\text{in}}\,{\text{3}}\,{\text{days}} \cr & = \left( {\frac{7}{{36}} \times 3} \right) = \frac{7}{{12}} \cr & {\text{Remaining}}\,{\text{work}} \cr & = \left( {1 - \frac{7}{{12}}} \right) = \frac{5}{{12}} \cr & {\text{Now}},\,\frac{1}{{24}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & {\text{So}},\,\frac{5}{{12}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & \left( {24 \times \frac{5}{{12}}} \right) = 10\,{\text{days}} \cr}$