A girl walks 20 meters towards North. Then, turning to her left, she walks 50 meters. Then, turning to her right, she walks 40 metres. Again, she turns to her right and walks 50 metres. How far is she from her initial position?

The correct answer is 60 metres.

To find the total distance the girl is from her initial position, let us trace her path step-by-step starting from an initial point, which we will call $A$:
1. First movement: She walks 20 meters towards North. Let this bring her to point $B$. The distance $AB=20\text{metres}$ northwards.
2. Second movement: Facing North at point $B$, she turns to her left (which is West) and walks 50 meters to reach point $C$. The distance $BC=50\text{metres}$ westwards.
3. Third movement: Facing West at point $C$, she turns to her right (which is North) and walks 40 meters to reach point $D$. The distance $CD=40\text{metres}$ northwards.
4. Fourth movement: Facing North at point $D$, she turns to her right (which is East) and walks 50 meters to reach point $E$. The distance $DE=50\text{metres}$ eastwards.

Now, let us analyze her final position $E$ relative to her starting point $A$:
Since $BC$ represents a 50-meter walk West and $DE$ represents a 50-meter walk East, the horizontal (East-West) displacement cancels out because: $BC=DE=50\text{metres}$ Therefore, point $E$ lies directly north of point $B$, and the segment $BE$ is parallel and equal in length to segment $CD$.

Thus, we have: $BE=CD=40\text{metres}$

The total distance from her initial position $A$ to her final position $E$ is the straight-line distance along the North direction, which is: $AE=AB+BE$ Substituting the values: $AE=20\text{metres}+40\text{metres}=60\text{metres}$

She is exactly 60 metres away from her initial position.