Ajit starts from a fixed point and walks 10 m towards the North.\nHe then turns to his right and walks another 10 m.\nAfter that, he successively turns to his left and walks 5 m, then turns left again and walks 10 m, and finally turns left once more and walks another 10 m.\nHow far is Ajit now from his starting point?

Introduction / Context:
Ajit's motion in this problem involves several turns and different distances, forming a somewhat zig-zag path on the plane. Our goal is to find his straight-line distance from the starting point after all movements. This is another classic direction sense and displacement question where we track each segment on a coordinate grid and then apply basic geometry to compute the final separation between the starting and ending positions.

Given Data / Assumptions:

• Ajit starts from a fixed point.
• He walks 10 m towards the North.
• He then turns right and walks 10 m.
• He successively turns left and walks 5 m, then left again and walks 10 m, and then left once more and walks another 10 m.
• All right and left turns are 90 degree turns.
• We use the standard convention for directions: from North, right is East; from East, left is North; from North, left is West; from West, left is South.

Concept / Approach:
It is convenient to represent Ajit's movement on a coordinate plane with the starting point as the origin. Each segment then corresponds to a change in x or y coordinates. By following the direction of each segment and updating the coordinates, we can determine the final point. The straight-line distance from the origin to this final point is then calculated using Pythagoras theorem if needed. In this specific problem, we will see that the final displacement is purely vertical, making the distance calculation straightforward.

Step-by-Step Solution:
Step 1: Place Ajit's starting point at (0, 0). Step 2: He walks 10 m North to reach (0, 10); he is facing North. Step 3: From facing North, a right turn makes him face East. Walking 10 m East takes him to (10, 10). Step 4: From facing East, a left turn faces him North. Walking 5 m North moves him to (10, 15). Step 5: From facing North, another left turn faces him West. Walking 10 m West takes him to (0, 15). Step 6: From facing West, another left turn faces him South. Walking 10 m South moves him to (0, 5). Step 7: His final coordinates are (0, 5). The starting point is (0, 0). The straight-line distance between these points is |5 − 0| = 5 m.

Verification / Alternative check:
Compute net horizontal and vertical movement. Horizontally, Ajit moves 10 m East and later 10 m West, which cancel each other, so his net East–West displacement is 0. Vertically, he moves 10 m North, then 5 m North, and finally 10 m South. Net vertical displacement = 10 + 5 − 10 = 5 m North. Therefore, he ends up 5 m directly North of the starting point, confirming that the distance from the start is 5 m and that the displacement is purely vertical.

Why Other Options Are Wrong:
15 m and 10 m are larger distances that might be obtained by incorrectly adding path segments without considering direction or cancellations. 3 m is not supported by any combination of right-angle geometry in this problem. 0 m would imply that Ajit returned exactly to his starting point, which is not the case here; his final coordinates are clearly different from the origin. Because only the 5 m displacement matches both the coordinate tracking and the net component calculation, the other options are incorrect.

Common Pitfalls:
Some learners confuse the sequence of left turns after the first right turn, accidentally swapping North and South or East and West at some step. Others might attempt to calculate distance by adding all the given segments, which gives the path length but not the displacement from the starting point. Maintaining a clear step-by-step record of coordinates or drawing the path on a simple grid is an effective strategy to avoid these mistakes.

Final Answer:
Ajit is now 5 m away from his starting point.