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John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone $?$

Let the time taken by Jack to complete the work be $x$ days. Then time taken by John is $2x.$

$\begin{array} {lccc} & \text{John} & \text{Jack} & \text{Jack + Jim} \\ \text{Time}: & 2x & x & \frac {2x}{3} \end{array}$

Let the time taken by Jim be $y$ days to finish the work.

Now, $\frac{2x}{3} = \frac{xy}{x+y} \quad \left[\because \text{Time} \propto \frac{1}{\text{Efficiency}} \right]$

$\Rightarrow \frac{2}{3} = \frac{y}{x+y}$

$\Rightarrow 2(x+y) = 3y$

$\Rightarrow 2x + 2y = 3y$

$\Rightarrow \boxed {y = 2x}$

John takes three days more than, time is taken by all of them working together.

So, one day work by all of them $= \frac{1}{2x} + \frac{1}{x} + \frac{1}{y} \quad [\because \text{Efficiency for all of them}]$

$\qquad \qquad = \frac{1}{2x} + \frac{1}{x} + \frac{1}{2x} \quad [ \because y=2x]$

$\qquad \qquad = \frac{1+2+1}{2x} = \frac{4}{2x} = \frac{2}{x}$

Time taken by all of them $= \frac{x}{2}$

Now, $2x = \frac{x}{2} + 3$

$\Rightarrow 2x – \frac{x}{2} = 3$

$\Rightarrow \frac{4x-x}{2} = 3$

$\Rightarrow \frac{3x}{2} = 3$

$\Rightarrow \boxed {x = 2\;\text{days}}$

$\therefore$ Time taken by Jim alone to finish the work $= y = 2x = 2 (2)= 4 \; \text{days}.$

Correct Answer $: 4$

$\textbf{PS:}$ If one person takes $x$ days, and the second person takes $y$ days, then total time is taken by both of them working together $= \underbrace{\frac{1}{x} + \frac{1}{y}}_{\text{Efficiency}} = \frac{x + y}{xy} = \underbrace{\frac{xy}{x+y}}_{\text{Time}}. \quad [\because \text{Assume total work = 1 unit}]$
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