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Aamir and Birju can cut $5000\;\text{g}$ of wood in $20$ min. Birju and Charles can cut $5000\;\text{g}$ of wood in $40$ min. Charles and Aamir cut $5\;\text{kg}$ of wood in $30$ min. How much time Charles will take to cut $5\;\text{kg}$ wood alone?

  1. $120$ min
  2. $48$ min
  3. $240$ min
  4. $120/7$ min
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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 $(C).$

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