Direction Sense: previous-year questions

Answer: (D) 1 km West

Opposite moves cancel out: 500 m North-East and 500 m South-West cancel each other, and 1 km South and 1 km North cancel each other. The only move left is the 1 km walked after 'turning right' while facing South, and a right turn from South faces West. So he ends 1 km West of the start (D). With coordinates: (354, 354), then (354, -646), then (-646, -646), then (-1000, -1000), and finally (-1000, 0). Tip: in direction-sense problems, pair off equal and opposite legs before doing any calculation, and remember that turning right means clockwise (N to E to S to W).

Answer: (B) South-East

A navigation map in heading-up mode always shows the direction of travel at the top of the screen. If the North arrow points to the bottom-right corner, North lies $135^\circ$ clockwise from the car's heading, so the car is heading $135^\circ$ anticlockwise from North, i.e. South-West. A $90^\circ$ left (anticlockwise) turn from South-West gives South-East (B). Quick check: when you face South-West, your left hand points South-East and your right hand North-West, so option D is what a right turn would give, the common trap. In compass bearings: South-West is $225^\circ$, and $225^\circ - 90^\circ = 135^\circ$, which is South-East.

Answer: (B) 10 km North-East

Net movement is 15 − 9 = 6 km North and 8 km East. The straight-line distance is $\sqrt{6^2 + 8^2} = 10$ km, and he is north and east of the start, i.e. North-East. South-West reverses the direction.

Answer: (A) South-East

Take X as the origin. Naseer walks 5 km west, turns left (now facing south) and walks 3 km, then turns left again (now facing east) and walks 7 km. Net east-west displacement $=7-5=2$ km east; net displacement southwards $=3$ km. So he is 2 km east and 3 km south of X, i.e., to the South-East, at a straight-line distance of $\sqrt{2^2+3^2}=\sqrt{13}\approx3.6$ km. Remember: facing west, a left turn points south; facing south, a left turn points east.

Answer: (C) North West

Plotting the path on a compass grid shows Radhika ends up south-east of her starting house, so walking back "to her house" means heading north-west.

Answer: (B) South

Before sunset the sun is in the west, so shadows fall towards the east. Amal's shadow is to his right, so his right hand points east, which means he is facing north. Ajoy stands face to face with him and therefore faces south. Remember the morning case too: the sun is in the east and shadows fall to the west.

Answer: (D) 3 km

Plotting the path: starting at $(0,0)$, moving 8 km east to $(8,0)$, then turning left (now facing north) 3 km to $(8,3)$, then turning left again (now facing west) 8 km back to $(0,3)$. The distance from the starting point $(0,0)$ to $(0,3)$ is 3 km.

Answer: (C) 10 km

Using coordinate geometry: Start at (0,0). South 4km → (0,−4). East 6km → (6,−4). North 12km → (6,8). Distance from origin = $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ km.

Answer: (B) West

Start facing North; walk 3 km North. Turn left (now facing West); walk 2 km West. Turn left again (now facing South); walk 3 km South. Turn right (from South, turning right faces West). He is now walking West.

Answer: (D) South

After sunrise, shadows fall towards the West. If Bikram's shadow falls to his left, then Bikram is facing North (because standing North, left is West). Since Bikram and Shailesh have their backs to each other, Shailesh faces the opposite direction — South.