College exam

Question

A person turns 6 times left, 7 times right and again turns 2 times left. If he finds himself in south direction, then find initial direction

Answer 1

Inital position + 4 times rotate in one direction = initial position.

Therefore 6 times left = 6-4 = 2 times left and 7 times right = 3 times right

Total shift from initial position = 2 left + 3 right + 2 left = 4 left + 3 right = 1 left from initial

if final = south, then initial = south - left = south + right = West

Comments

Let’s break down the problem step by step:

1. Initial Direction:

   • Let’s assume the person starts facing north initially.

2. Left Turns:

   • The person turns 6 times left.
   • Each left turn corresponds to a 90-degree anticlockwise rotation.
   • So, the total angle turned left is:

     6 \times 90^\circ = 540^\circ6×90∘=540∘

     .

3. Right Turns:

   • The person then turns 7 times right.
   • Each right turn corresponds to a 90-degree clockwise rotation.
   • So, the total angle turned right is:

     7 \times 90^\circ = 630^\circ7×90∘=630∘

     .

4. Final Left Turn:

   • Finally, the person turns 2 times left.
   • The total angle turned left is:

     2 \times 90^\circ = 180^\circ2×90∘=180∘

     .

5. Net Angle Turned:

   • The net angle turned is the difference between the total left turns and the total right turns:

     \text{Net Angle} = 540^\circ - 630^\circ + 180^\circ = 90^\circNet Angle=540∘−630∘+180∘=90∘

so the intial direction was west