Mohit starts from point Y and drives 6 km towards south.

Solution

Place Y at the origin (0,0)(0,0) ( 0 , 0 ) . Take north as +y+y + y , east as +x+x + x .

Follow Mohit’s moves step by step:

1. 6 km south → (0,−6)(0,-6) ( 0 , − 6 )

2. Left (facing south → left = east) 23 km → (23,−6)(23,-6) ( 23 , − 6 )

3. Right (facing east → right = south) 27 km → (23,−33)(23,-33) ( 23 , − 33 )

4. Right (facing south → right = west) 11 km → (12,−33)(12,-33) ( 12 , − 33 )

5. Right (facing west → right = north) 13 km → (12,−20)(12,-20) ( 12 , − 20 )

6. Right (facing north → right = east) 19 km → (31,−20)(31,-20) ( 31 , − 20 )

7. Left (facing east → left = north) 20 km → (31,0)(31,0) ( 31 , 0 ) = point Z

So Z is at (31,0)(31,0) ( 31 , 0 ) and Y is at (0,0)(0,0) ( 0 , 0 ) . The displacement from Z to Y is (−31,0)(-31,0) ( − 31 , 0 ) , i.e. 31 km due west.

Answer: He must drive 31 km towards the west to return to point Y.