Step 1: Set up the same two equations, then solve by elimination instead of by plotting points.
Let the price of one pencil be Rs $x$ and the price of one chocolate be Rs $y$. From the given information:
\[ 2x + 3y = 11 \quad \text{...(1)} \]
\[ x + 2y = 7 \quad \text{...(2)} \]
Step 2: Make the $x$-coefficients match so one variable can be cancelled by subtraction.
Multiply equation (2) by 2, so its $x$-coefficient becomes 2, matching equation (1):
\[ 2x + 4y = 14 \quad \text{...(3)} \]
Step 3: Subtract equation (1) from equation (3) to eliminate $x$.
\[ (2x + 4y) - (2x + 3y) = 14 - 11 \]
\[ y = 3 \]
Step 4: Substitute this value of $y$ back into equation (2) to find $x$.
\[ x + 2(3) = 7 \]
\[ x + 6 = 7 \]
\[ x = 1 \]
Step 5: Interpret the result, and check that the two lines from these equations do meet exactly at this point, matching the graphical picture.
Since $(x, y) = (1, 3)$ satisfies both original equations at once, this is exactly the point where the two lines representing them would cross if plotted, confirming the same intersection point the graphical method is meant to find.
Step 6: Final Answer.
The price of one pencil is Rs 1 and the price of one chocolate is Rs 3, so option (A) is correct.
\[ \boxed{\text{Pencil} = \text{Rs } 1,\ \text{Chocolate} = \text{Rs } 3} \]