Direction Sense
Direction Sense is a frequently tested area in CUET General Test. Work through these free NTA-style sample questions with full answers and explanations, then attempt all 35 in a timed practice test to build exam-day speed.
Snapshot
- Direction Sense describes a person walking and turning, then asks the final direction they face or their distance/direction from the start. Drawing the path on a compass turns it into simple geometry.
- The two ideas you need: the compass (N at top, E right, S down, W left, with the four diagonals) and the rule that a right turn is clockwise, a left turn anticlockwise — relative to the direction you are currently facing.
- This guide covers the compass, turning, shortest-distance (Pythagoras), shadows, and the "turn relative to facing" trap — with worked examples.
- Exam reality: +5 / −1. Sketch the path; don't track it in your head.
Part 1 — The compass and a worked path
Always orient your page with North up. Mark the start, then draw each leg with an arrow and its length. A left/right turn is taken from the direction you are facing now, not from North — this is the single biggest trap.
Part 2 — Shortest distance (the Pythagoras step)
If a person walks some distance North/South and some East/West, the straight-line distance from the start is the hypotenuse: distance = √(horizontal² + vertical²). In the path above (10 N, 10 E, 5 S) the net movement is 10 E and 5 N, so the distance from start = √(10² + 5²) = √125 ≈ 11.2, in the North-East direction. Watch for the Pythagorean triples (3-4-5, 6-8-10) the examiner loves.
Part 3 — Turning, shadows & special cases
- Turn relative to facing: facing South and turning left sends you East (not West) — rotate from the current heading.
- Clockwise / anticlockwise: a right turn is clockwise; two right turns reverse your direction; four bring you back.
- Shadows: at sunrise the sun is in the East, so shadows fall West; at sunset the sun is West and shadows fall East. At noon shadows are negligible. This is a recurring direction-sense question.
Part 4 — Speed techniques
- Always sketch with North up and arrows for each leg.
- Sum the N/S legs and the E/W legs separately — net vertical and net horizontal give both the distance (Pythagoras) and the final direction.
- For "facing" questions, rotate from the current heading, not from North.
- Memorise the triples to skip the square root.
- For shadows, fix sunrise = East light ⇒ West shadow.
Part 5 — Worked examples
1. A man walks 4 km North, then 3 km East. Distance from start? √(4²+3²) = 5 km (North-East).
2. Facing East, he turns right. Now facing? Right of East is South.
3. Walks 5 N, 5 E, 5 S. Net: 5 E, 0 N/S ⇒ he is 5 km East of start.
4. Facing North, turns left, walks 3, turns left, walks 3. He faces South and is at 3 W, 3 S ⇒ South-West of start, √18 ≈ 4.24 km away.
5. At sunrise a pole's shadow falls towards a boy. Which way does he face? Sunrise shadow points West, so he faces West.
6. Walks 10 E, 10 N, 10 E. Net 20 E, 10 N ⇒ √(400+100) = √500 ≈ 22.4 km, North-East.
7. Two right turns from facing West ⇒ now facing East (reversed).
8. Facing South, turns left (⇒ East), walks 6, turns right (⇒ South), walks 8. Distance from start = √(6²+8²) = 10.
Part 6 — Common traps
- Turning from North instead of the current heading — the #1 error.
- Forgetting Pythagoras for the shortest distance.
- Shadow direction — sunrise light East ⇒ shadow West, and vice-versa.
- Direction vs distance — read which the question wants.
- Net vs total distance — "distance from start" is the straight line, not the path length walked.
Part 7 — How to use this page
Sketch the compass once, re-solve the eight examples on paper with North up, then attempt the practice set and the timed test.
One-line revision: draw with North up, turn relative to your current facing, add N/S and E/W legs separately, use √(h²+v²) for distance, and remember sunrise light is East so shadows fall West.
Practice questions
Now test yourself. 8 free sample questions with explanations. 27 more in the timed practice test.
Q1. Rohan goes $5$ km North, $5$ km East, $5$ km South, $5$ km West. Where is he with respect to the start?
▸ Show answer & explanation
Answer: A
North–South cancel ($5-5=0$) and East–West cancel ($5-5=0$). He returns exactly to the starting point.
Q2. Sita walks $7$ km East, $7$ km North, $7$ km West and $7$ km North. How far is she from the start?
▸ Show answer & explanation
Answer: B
Net East–West: $7-7=0$. Net North: $7+7=14$ km. She ends $14$ km due North of start, so the distance is $14$ km.
Q3. A man walks $5$ km towards North, then turns right and walks $3$ km, then turns right again and walks $5$ km. In which direction is he now facing?
▸ Show answer & explanation
Answer: B
Start facing North. First right turn makes him face East; second right turn makes him face South. After the last walk he is still facing South.
Q4. A man drives $5$ km North, then $12$ km East, then $9$ km North, then $12$ km West. How far is he from the starting point?
▸ Show answer & explanation
Answer: C
Net East–West: $12-12=0$. Net North: $5+9=14$ km. He ends $14$ km due North; distance $=14$ km.
Q5. A person walks $12$ km East, then turns right and walks $5$ km. What is his straight-line distance from the starting point?
▸ Show answer & explanation
Answer: A
Facing East, a right turn gives South. The legs are perpendicular: $\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$ km.
Q6. A man starts walking towards South. After walking $40$ m he turns to his left and walks $30$ m. Then he turns right and walks $20$ m. In which direction is he from the starting point?
▸ Show answer & explanation
Answer: C
South $40$; left→East $30$; right→South $20$. Net South $=40+20=60$ m, net East $=30$ m. Being South and East of start, the direction is South-East.
Q7. A man goes $10$ m North, $6$ m East, $2$ m South and $6$ m West. How far is he from the start?
▸ Show answer & explanation
Answer: A
Net North–South: $10-2=8$ m North. Net East–West: $6-6=0$ m. So he is exactly $8$ m North of start, a straight-line distance of $8$ m.
Q8. A man walks $8$ km East, turns left and walks $6$ km, turns left and walks $16$ km. What is the shortest distance from his starting point?
▸ Show answer & explanation
Answer: C
East $8$; left→North $6$; left→West $16$. Net E–W: $16-8=8$ km West. Net North: $6$ km. Distance $=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$ km.
🔒 27 more questions
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