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One day, Rahul started a work at $9 \; \text{AM}$ and Gautam joined him two hours later. They then worked together and completed the work at $5 \; \text{PM}$ the same day. If both had started at $9 \; \text{AM}$ and worked together, the work would have been completed $30 \; \text{minutes}$ earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

1. $10$
2. $12$
3. $12.5$
4. $11.5$

Let’s draw the table for better understanding.

$\begin{array}{lll} & \textbf{Rahul} & &\textbf{Gautam} \\ \text{Time:} & \underbrace{8\;\text{hours}}_{{\color{Teal} {\text{9 AM to 5 PM}}}} & & \underbrace{6\;\text{hours}}_{{\color{Blue}{\text{11 AM to 5 PM}}}}\\ & \underbrace{7.5\;\text{hours}}_{{\color{Green} {\text{9 AM to 4:30 PM}}}} & & \underbrace{7.5\;\text{hours}}_{{\color{Green} {\text{9 AM to 4:30 PM}}}} \\ \text{Efficiency:} & x\;\text{units/hour} & & y\;\text{units/hour} \\ \text{Total work:} & 8x+6y & = & 7.5x+7.5y \end{array}$

Now, the total work $= 8x+6y = 7.5x + 7.5y$

$\Rightarrow 0.5x = 1.5y$

$\Rightarrow \boxed{x=3y}$

$\Rightarrow \boxed{y=\frac{x}{3}}$

So, the total work $= 8x+6\left(\frac{x}{3}\right) = 8x + 2x = 10x\;\text{units}$

$\therefore$ The time taken by Rahul alone to complete the work $= \frac{10x\;\text{units}}{x\;\text{units/hour}} = 10 \; \text{hours}.$

Correct Answer $:\text{A}$
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