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Exam Topic CUET General Test · 501 35 practice MCQs

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.

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Snapshot

Part 1 — The compass and a worked path

Compass & a sample pathNSEWNESESWNWWalk N 10, then E 10, then S 5start10 (N)10 (E)5 (S)net displacement

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

Part 4 — Speed techniques

  1. Always sketch with North up and arrows for each leg.
  2. 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.
  3. For "facing" questions, rotate from the current heading, not from North.
  4. Memorise the triples to skip the square root.
  5. 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

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.

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