Deepak starts walking straight towards the East from a point and covers 75 metres. He then turns to his left and walks 25 metres straight, turns left again and walks 40 metres straight, and once more turns to his left and walks another 25 metres. How far is Deepak now from his starting point?

Introduction / Context:
This question describes Deepak walking along four straight segments with three left turns. We are asked to find the straight-line distance from his final position back to his starting point. Because the path contains several right-angle turns, the resultant displacement can be found using basic coordinate geometry and Pythagoras theorem after simplifying the net horizontal and vertical movements.

Given Data / Assumptions:

• Deepak starts walking towards the East and covers 75 metres.
• He then turns left and walks 25 metres straight.
• Again, he turns left and walks 40 metres straight.
• Once more, he turns left and walks 25 metres.
• All turns are 90 degree left turns and all segments are straight.
• Standard direction convention is used: from East, left is North; from North, left is West; from West, left is South.

Concept / Approach:
We assign coordinates to each position by treating the starting point as the origin. East–West movements change the x-coordinate, and North–South movements change the y-coordinate. After tracking all four segments, we compute the straight-line distance between the starting and final coordinates using Pythagoras theorem: distance^2 = (Δx)^2 + (Δy)^2. Because some vertical movements cancel, the computation turns out to be straightforward.

Step-by-Step Solution:
Step 1: Place Deepak's starting point at (0, 0). Step 2: He walks 75 m East, reaching (75, 0). He is facing East. Step 3: From facing East, a left turn means facing North. Walking 25 m North takes him to (75, 25). Step 4: From facing North, a left turn means facing West. Walking 40 m West brings him to (35, 25). Step 5: From facing West, another left turn faces him South. Walking 25 m South takes him to (35, 0). Step 6: His final coordinates are (35, 0), while his starting point is (0, 0). Therefore, the straight-line distance between the two points is the horizontal difference: |35 − 0| = 35 metres.

Verification / Alternative check:
Instead of computing coordinates, we can consider net horizontal and vertical displacement. Horizontally, Deepak walks 75 m East and 40 m West, so net Eastward displacement is 75 − 40 = 35 m. Vertically, he walks 25 m North and 25 m South, which cancel each other out, leaving zero net vertical displacement. Thus, his final position is directly East of the starting point by 35 m. There is no need for Pythagoras here because the displacement is purely horizontal, confirming that 35 m is the correct distance.

Why Other Options Are Wrong:
50 m would arise if someone incorrectly subtracted or added the wrong distances or assumed a diagonal separation. 115 m and 140 m are simply the sums of some path segments and do not represent the straight-line distance from start to end. 25 m might be mistaken for the vertical leg alone, ignoring the larger Eastward net displacement. Since the net vertical displacement is zero, any answer that does not reflect a pure 35 m Eastward movement is incorrect.

Common Pitfalls:
Many students mistakenly add all distances to get the answer, confusing path length with displacement. Others might misinterpret left turns or neglect the fact that some movements cancel each other out. Carefully tracking each segment on a simple diagram, or writing down the cumulative x and y coordinates after each move, avoids such errors and helps you see that the final position is just 35 m East of the starting point.

Final Answer:
Deepak is now 35 m away from his starting point, directly to the East.