Aamir and Birju can cut $5000\;\text{g}$ of wood in $20\; \text{min} \implies A + B$ efficiency $ = \dfrac{5kg}{20\;\text{min}} = \dfrac{1}{4}\;\text{kg/min}$
Birju and Charles can cut $5000\;\text{g}$ of wood in $40\; \text{min} \implies B + C$ efficiency $ = \dfrac{5\;\text{kg}}{40\;\text{min}} = \dfrac{1}{8}\;\text{kg/min}$
Charles and Aamir cut $5\;\text{kg}$ of wood in $30\; \text{min} \implies C+A$ efficiency $ = \dfrac{5\;\text{kg}}{30\;\text{min}} = \dfrac{1}{6}\;\text{kg/min}$
Efficiency of $2(A+B+C) = \left(\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{6}\right)\;\text{kg/min} = \dfrac{12 + 6 + 8}{48}\;\text{kg/min}= \dfrac{26}{48}\;\text{kg/min} = \dfrac{13}{24}\;\text{kg/min}$
Efficiency of $A+B+C = \dfrac{13}{48}\;\text{kg/min}$
Now, the efficiency of $C = \dfrac{13}{48}\;\text{kg/min} – \dfrac{1}{4}\;\text{kg/min} = \dfrac{13-12}{48}\;\text{kg/min} = \dfrac{1}{48}\;\text{kg/min}$
The time Charles will take to cut $5\;\text{kg}$ wood alone $ = \dfrac{5kg}{\frac{1}{48}\;\text{kg/min}} = 240\;\text{min}.$
$\textbf{OR}$
Aamir and Birju can cut $5000\;\text{g}$ of wood in $20\; \text{min.}$ Birju and Charles can cut $5000\;\text{g}$ of wood in $40\; \text{min}.$ Charles and Aamir cut $5\;\text{kg} = 5000\;\text{g}$ of wood in $30\; \text{min}.$
Total work $ = $LCM of $(20,40,30) = 120\;\text{unit}$
- Efficiency of $A+B = 6\;\text{unit/min}$
- Efficiency of $B + C = 3\;\text{unit/min}$
- Efficiency of $C + A = 4\;\text{unit/min}$
Efficiency of $2(A+B + C) = 13\;\text{unit/min}$
$\implies$ Efficiency of $A+B+C = \dfrac{13}{2}\;\text{unit/min}$
$\implies$ Efficiency of $C = \dfrac{13}{2}\;\text{unit/min} – 6\;\text{unit/min} = \dfrac{13 – 12}{2}\;\text{unit/min} = \dfrac{1}{2}\;\text{unit/min}$
The time Charles will take to do $120$ unit work $ = \dfrac{120\;\text{unit}}{\frac{1}{2}\;\text{unit/min}} = 240\;\text{min}.$
So, the correct answer is $(C).$