A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in :

### Concept & Ratio of Time When multiple people have comparative times for completing work, define one person's time as a variable and express the others in terms of it. Formula for combined work: $$ \frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{T_{total}} $$ ### Step-by-Step Solution * **Defining Individual Times:** Let the time taken by A to finish the work be $x$ days. Given: A takes twice as much time as B $\Rightarrow B = \frac{x}{2}$ days. Given: A takes thrice as much time as C $\Rightarrow C = \frac{x}{3}$ days. * **Setting up the Equation:** Working together, they take $2$ days. Their 1-day work combined: $\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{2}$ Substitute the values: $\frac{1}{x} + \frac{1}{\frac{x}{2}} + \frac{1}{\frac{x}{3}} = \frac{1}{2}$ $\frac{1}{x} + \frac{2}{x} + \frac{3}{x} = \frac{1}{2}$ * **Solving for x:** $\frac{1 + 2 + 3}{x} = \frac{1}{2}$ $\frac{6}{x} = \frac{1}{2}$ $x = 12$ days (This is A's time). * **Finding B's Time:** Time taken by B = $\frac{x}{2} = \frac{12}{2} = 6$ days. ### Exam Strategy & Shortcut Assume the total work is the LCM of the comparative time ratios. Let A take $6t$ days. Then B takes $3t$ days and C takes $2t$ days. Efficiencies (work per day): A = $\frac{1}{6t}$, B = $\frac{1}{3t}$, C = $\frac{1}{2t}$. Sum of efficiencies = $\frac{1+2+3}{6t} = \frac{6}{6t} = \frac{1}{t}$. This combined efficiency equals $\frac{1}{2}$ work/day. Thus $t = 2$. B takes $3t = 3(2) = 6$ days. ### Common Pitfall A frequent error is assigning the ratios directly to efficiencies rather than time. "Takes twice as much time" means half the efficiency. ### Final Answer Therefore, the correct answer is **6 days**.