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Question

A girl starts from her school gate and walks 10 m East, then 4 m South, then 8 m West. She then walks 2 m North, turns right and walks 6 m. After that, she turns left and walks 2 m, and finally turns left again and walks 6 m. Where is she from the school gate?

The correct answer is
2 m East

Step-by-Step Solution: Girl's Walk from School

This problem requires us to determine the final position of a girl relative to her starting point (the school gate) after a series of movements. We can track her position using a coordinate system, where the school gate is the origin $(0, 0)$. We'll define East as the positive x-direction, West as the negative x-direction, North as the positive y-direction, and South as the negative y-direction.

Tracking the Girl's Movements

Let's follow each step of her journey:

  • Starting Point: School Gate at coordinates $(0, 0)$.
  • 1. Walks 10 m East:
    • Movement: $+10$ m along the x-axis.
    • New Position: $(0 + 10, 0) = (10, 0)$.
  • 2. Walks 4 m South:
    • Movement: $-4$ m along the y-axis.
    • New Position: $(10, 0 - 4) = (10, -4)$.
  • 3. Walks 8 m West:
    • Movement: $-8$ m along the x-axis.
    • New Position: $(10 - 8, -4) = (2, -4)$.
  • 4. Walks 2 m North:
    • Movement: $+2$ m along the y-axis.
    • New Position: $(2, -4 + 2) = (2, -2)$.
  • 5. Turns right and walks 6 m:
    • At this point, she is facing North (from the previous step). Turning right means she now faces East.
    • Movement: $+6$ m along the x-axis.
    • New Position: $(2 + 6, -2) = (8, -2)$.
  • 6. Turns left and walks 2 m:
    • She is currently facing East. Turning left means she now faces North.
    • Movement: $+2$ m along the y-axis.
    • New Position: $(8, -2 + 2) = (8, 0)$.
  • 7. Turns left again and walks 6 m:
    • She is currently facing North. Turning left means she now faces West.
    • Movement: $-6$ m along the x-axis.
    • Final Position: $(8 - 6, 0) = (2, 0)$.

Calculating Net Displacement

Alternatively, we can sum the displacements along each axis:

  • East-West (x-axis) Displacement:
    • Movements: $+10$ m (East), $-8$ m (West), $+6$ m (East), $-6$ m (West).
    • Net x-displacement = $10 - 8 + 6 - 6 = 2$ m.
  • North-South (y-axis) Displacement:
    • Movements: $-4$ m (South), $+2$ m (North), $+2$ m (North).
    • Net y-displacement = $-4 + 2 + 2 = 0$ m.

The net displacement is $2$ m in the East direction and $0$ m in the North-South direction. The final coordinates are $(2, 0)$.

Final Position Relative to School Gate

The final position $(2, 0)$ means the girl is 2 meters from the origin along the positive x-axis. Since the positive x-axis represents East, she is 2 meters East of the school gate.

Therefore, the girl is located 2 m East from the school gate.

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Important Questions from Direction and Distance

  1. Vijendra walks a certain distance, say X metres, from his home towards the west. Then he turns left and walks 23 metres. After that, he turns left and walks 36 metres. Then he turns left again to walk 23 metres. He finally turns left and walks 18 metres to reach his home. Find the value of X.
  2. Gaurav exits from the backdoor of his north-facing house and walks 25 m straight, then he takes a left turn and walks 36 m, then he turns left and walks 47 m. He turns left again and walks 36 m. How far and in which direction is he from his house now?

  3. Reeta is standing facing south-east. First, she turns 135° clockwise. After that, she turns 90 °  anticlockwise. Then she turns 45 °  clockwise, followed by a 135 ° anticlockwise  turn. In which direction is she facing now?

  4. Sahasra started running from her house towards the north. After 40 meters she turned left and ran for 75 meters. She then turned right and ran for 30 meters, and again turned right to run 75 meters. In which direction was she running finally?

  5. At the time of sunset, Lopa and Kritika are sitting facing each other. If the shadow of Lopa falls to the right of Kritika, in which direction is Kritika facing?

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