Direction Sense — PPSC previous year questions

Track the facing direction step by step. > West + 45° clockwise → North-West > North-West + 180° clockwise → South-East > South-East + 270° anticlockwise (= 90° clockwise) → South-West Net rotation is 45° + 180° − 270° = −45°. That is 45° anticlockwise from West, i.e. South-West.

Take the first move as north. She walks 10 m north, then 10 m east. Turning left each time, she then walks 5 m north, 15 m west and 15 m south. So she ends 5 m west of the start, and level with it. > E–W: +10 − 15 = −5 m > N–S: +10 + 5 − 15 = 0 m ```puzzle:direction start: A A > N 10 > B B > E 10 > C C > N 5 > D D > W 15 > E E > S 15 > F ``` So she is 5 metres from her starting point.

South-West to North-West is a 90° clockwise turn. But their new names go from North to West, a 90° anticlockwise turn. So the compass has been flipped (mirrored), not merely rotated. Measure directions clockwise from North. The rule is new = 225° − old. So SW (225°) → 0° = N and NW (315°) → −90° = W, as given. East (90°) → 225° − 90° = 135°, which is South-East.

Trace the walk. He goes 3 km north. A left turn faces west (2 km). The next left faces south (3 km), cancelling the north leg. The last left faces east (3 km). > North: 3 − 3 = 0 km > East: 3 − 2 = 1 km ```puzzle:direction start: S S > N 3 > A A > W 2 > B B > S 3 > C C > E 3 > D ``` He ends 1 km east of the start.

Treat the three villages as a 3-4-5 right triangle: the 5 km leg to the uncle's village is the hypotenuse, the 4 km leg south is one side, and the distance from the father-in-law's village back to home is the third side. > 5² − 4² = 25 − 16 = 9 → 3 km The uncle's village is north-east of home and the father-in-law's village is due south of it, so the father-in-law's village lies to the east of home: 3 km East. ```puzzle:direction start: Home Home > NE 5 > Uncle Uncle > S 4 > Father-in-law ``` Taken exactly, 5 km north-east is about 3.5 km north and 3.5 km east, so the true position is about 3.6 km away, a little south of due east; the question uses the 3-4-5 triangle as an approximation, and 3 km East is the only option that fits.

Kunal ends up 10 − 6 = 4 km north of the start. He then moves 3 km east. So his displacement forms a right triangle with legs 4 km and 3 km. > Net north = 10 − 6 = 4 km > Distance = sqrt(4² + 3²) = sqrt(25) = 5 km ```puzzle:direction start: A A > N 10 > B B > S 6 > C C > E 3 > D ``` He is 5 km from the start, north and east of it. That is 5 km North-east.

Let the boy first ride x km north. Turning left he rides 1 km west, turning left again he rides 2 km south. He ends 1 km west of the start, so his north–south displacement must be zero: x − 2 = 0, giving x = 2 km. ```puzzle:direction start: S S > N 2 > A A > W 1 > B B > S 2 > C ``` So he rode 2 km northwards.

Take A as the origin. B is 3 km east of A. Turning left, he goes north 3 × 3 = 9 km to C. Turning left again, he goes west 5 × 3 = 15 km to D. Relative to A, D is 15 − 3 = 12 km west and 9 km north. > AD = sqrt(12² + 9²) = sqrt(144 + 81) = sqrt(225) = 15 km ```puzzle:direction start: A A > E 3 > B B > N 9 > C C > W 15 > D ``` So the shortest distance is 15 km.

Take X to Y as north. Each right turn turns him clockwise. So he walks north 80 m, east 50 m, south 70 m and west 50 m. The east and west legs cancel. That leaves 80 − 70 = 10 m north of X. ```puzzle:direction start: X X > N 80 > Y Y > E 50 > P P > S 70 > Q Q > W 50 > R ``` So Raj is 10 metres from the starting point.

Start facing east. A right turn faces south, 20 m. A left turn faces east, 20 m. A right turn faces south, 20 m. The next right turn faces west, 40 m. The final right turn faces north, 40 m. > Southward: 20 + 20 = 40 m, cancelled by 40 m north → net north–south displacement 0 > Eastward 20 m against westward 40 m → net 20 m west ```puzzle:direction start: O O > S 20 > A A > E 20 > B B > S 20 > C C > W 40 > D D > N 40 > E ``` The final position is 20 m due west of the start.