Direction Sense Questions Placement
Direction Sense Questions for Placement Exams - Complete Question Bank
Last Updated: March 2026
Introduction to Direction Sense
Direction Sense is a fundamental topic in logical reasoning that tests your ability to understand and navigate spatial relationships. It evaluates how well you can track movements, determine positions, and calculate distances based on directional information. This topic is crucial for placement exams as it demonstrates your spatial reasoning capabilities—a skill valued in many corporate roles.
Why is Direction Sense Important?
- Universal Application: Questions appear in almost every placement and competitive exam
- Quick Solving: Most questions can be solved in 30-60 seconds with practice
- No Complex Formulas: Based on basic geometry and logical thinking
- High Accuracy: Once understood, accuracy rates are typically high
- Foundation for Advanced Topics: Essential for maps, seating arrangements, and route optimization
Types of Direction Sense Questions
- Simple Movement Tracking: Following a sequence of movements
- Distance and Direction: Finding final position relative to starting point
- Shadow-Based Problems: Using shadow direction to determine time/sun position
- Clock-Based Directions: Using clock hands to determine directions
- Distance Minimization: Finding shortest path
- Relative Positioning: Determining position of one object relative to another
Important Concepts and Formulas
Cardinal and Ordinal Directions
North (N)
|
|
West (W) -------+------- East (E)
|
|
South (S)
Ordinal Directions:
- North-East (NE): Between N and E
- North-West (NW): Between N and W
- South-East (SE): Between S and E
- South-West (SW): Between S and W
Direction Relationships
| Facing Direction | Left Turn | Right Turn | About Turn |
|---|---|---|---|
| North | West | East | South |
| South | East | West | North |
| East | North | South | West |
| West | South | North | East |
Distance Calculation Formulas
When movements are perpendicular:
Final Distance = √(x² + y²) (Pythagorean theorem)
Where x = total horizontal displacement
y = total vertical displacement
Shortcut for Common Distances:
- 3-4-5 triangle: If legs are 3 and 4, hypotenuse = 5
- 5-12-13 triangle: If legs are 5 and 12, hypotenuse = 13
- 8-15-17 triangle: If legs are 8 and 15, hypotenuse = 17
- 7-24-25 triangle: If legs are 7 and 24, hypotenuse = 25
Shadow Rules
Morning (Sunrise to Noon):
- Sun rises in the EAST
- Shadow falls towards WEST
- Shadow shortens as sun rises higher
Evening (Noon to Sunset):
- Sun sets in the WEST
- Shadow falls towards EAST
- Shadow lengthens as sun sets
At Noon:
- Shadow is shortest
- Shadow falls towards NORTH (in Northern Hemisphere)
Quick Reference: Turn Combinations
- Two left turns = About turn (180°)
- Two right turns = About turn (180°)
- Left + Right = Back to original direction (0°)
- Three left turns = One right turn (90° right)
- Three right turns = One left turn (90° left)
Practice Questions
Set 1: Basic Direction and Distance
Question 1 (Easy): A man walks 5 km towards North, turns left and walks 3 km, then turns left again and walks 5 km. How far is he from the starting point?
Solution:
- Start at origin (0, 0)
- Walk 5 km North → (0, 5)
- Turn left (now facing West), walk 3 km → (-3, 5)
- Turn left (now facing South), walk 5 km → (-3, 0)
Final position: (-3, 0) Starting point: (0, 0)
Distance = √((-3-0)² + (0-0)²) = √9 = 3 km
Shortcut: He walked North 5, then South 5 (cancels out). Only the 3 km West remains.
Question 2 (Easy): If South-East becomes North, then what will North become?
Solution: South-East to North is a 135° anti-clockwise rotation (or 225° clockwise).
Applying same rotation to North: North → 135° anti-clockwise → South-West
Question 3 (Medium): A person walks 10 m towards South, turns right and walks 8 m, then turns right and walks 10 m, then turns right and walks 8 m. Where is he now with respect to the starting point?
Solution:
- Start: (0, 0)
- South 10 m → (0, -10)
- Turn right (now West), walk 8 m → (-8, -10)
- Turn right (now North), walk 10 m → (-8, 0)
- Turn right (now East), walk 8 m → (0, 0)
Question 4 (Medium): Raju faces North. He turns 135° clockwise, then 180° anti-clockwise, then 90° clockwise. Which direction is he facing now?
Solution: Clockwise = negative, Anti-clockwise = positive (or vice versa, just be consistent)
Net rotation = -135° + 180° - 90° = -45°
Starting from North (0° or 360°): -45° clockwise = North-East direction (45° towards East from North, but clockwise means towards West... wait)
Actually: North = 0°/360° Clockwise from North: East = 90°, South = 180°, West = 270°
-135° clockwise: 0° - 135° = -135° = 225° (which is South-West) Wait let me recalculate.
Starting North (0°):
- 135° clockwise: facing 225° direction = South-West
- 180° anti-clockwise from SW: 225° - 180° = 45° = North-East
- 90° clockwise from NE: 45° + 90° = 135° = South-East? No, clockwise adds in my convention.
Let me use standard: clockwise from North goes through East. North (0°) → East (90°) → South (180°) → West (270°) → North (360°)
- 135° clockwise: 0 + 135 = 135° = South-East? No wait... Actually: 90° is East, so 135° is between East and South = South-East.
Wait, I think I confused myself. Let me use a clearer approach.
Facing North. 135° clockwise: This turns him towards SE direction (45° past East towards South). Actually: 90° = East, 135° = South-East? No, 135° from North clockwise:
- 0° = North
- 45° = North-East
- 90° = East
- 135° = South-East
- 180° = South
So 135° clockwise faces South-East.
180° anti-clockwise from South-East (135°): 135° - 180° = -45° = 315° = North-West
90° clockwise from North-West (315°): 315° + 90° = 405° = 45° = North-East
Question 5 (Hard): A man walks 6 km North, turns right and walks 4 km, turns left and walks 5 km, turns left and walks 4 km. How far and in which direction is he from the starting point?
Solution:
- Start: (0, 0)
- North 6 km → (0, 6)
- Turn right (East), walk 4 km → (4, 6)
- Turn left (North), walk 5 km → (4, 11)
- Turn left (West), walk 4 km → (0, 11)
Final position: (0, 11) Starting point: (0, 0)
Distance = √(0² + 11²) = 11 km Direction: North (same x-coordinate, higher y-coordinate)
Set 2: Shadow-Based Questions
Question 6 (Easy): A man is facing West in the evening. His shadow will be cast towards which direction?
Solution: Evening: Sun is in the West, setting. Shadow falls opposite to sun: East.
Facing West doesn't change where shadow falls—it falls towards East.
Question 7 (Medium): At sunrise, Rohan is facing the rising sun. He turns 90° clockwise, then 135° anti-clockwise. His shadow is now cast towards his left. What time of day is it?
Solution: First, determine Rohan's final facing direction:
- Sunrise: facing East (sun rises in East)
- 90° clockwise: facing South
- 135° anti-clockwise from South: South (180°) - 135° = 45° = North-East
Rohan is facing North-East. Shadow is cast towards his left. Facing NE, left side is towards NW.
Shadow towards NW means sun is in opposite direction: SE. Sun in South-East indicates morning time (before noon).
Question 8 (Medium): In the morning, two people A and B are standing such that A's shadow falls on B. If A is facing North, which direction is B with respect to A?
Solution: Morning: Sun in East, shadows towards West. A's shadow falls on B → B is to the West of A. A is facing North. West of A means B is to A's left.
Question 9 (Hard): At 4 PM, a 6 ft tall man casts a shadow of 9 ft. If he walks 20 ft towards the shadow, turns right and walks 15 ft, how far is he from the tip of his original shadow?
Solution: At 4 PM: Sun is in West, shadow towards East. Shadow is 9 ft long towards East from the man.
- Man starts at origin (0, 0), shadow tip is at (9, 0) assuming East is +x
- Walks 20 ft towards shadow (East) → (20, 0)
- Turns right (now facing South), walks 15 ft → (20, -15)
Tip of original shadow: (9, 0) Current position: (20, -15)
Distance = √((20-9)² + (-15-0)²) = √(11² + 15²) = √(121 + 225) = √346 ≈ 18.6 ft
Question 10 (Hard): Two people start from the same point. Person A walks 10 m in the direction of his shadow at noon. Person B walks 10 m in the opposite direction of A. At what distance and direction is B from A?
Solution: At noon: Shadow towards North (in Northern Hemisphere). A walks in direction of shadow → North 10 m. B walks opposite → South 10 m.
Distance between A and B = 10 + 10 = 20 m Direction of B from A: South
Set 3: Clock-Based Directions
Question 11 (Easy): A clock shows 3:30. If the hour hand points East, where does the minute hand point?
Solution: At 3:30:
- Hour hand is halfway between 3 and 4
- Minute hand points at 6
If hour hand (at 3.5) points East, then each hour = 360°/12 = 30°. 3.5 hours × 30° = 105° from 12 o'clock position. If 105° is East (90°), slight discrepancy... Let's say 3 points East.
If 3 points East: 6 is 3 hours away = 90° from 3. From East, 90° clockwise = South.
Question 12 (Medium): At 12:00 noon, the hour hand points North. At what time will the hour hand first point to the West?
Solution: 12:00: Hour hand at North (0° or 360°) West is at 270° or -90° from North (counter-clockwise) or 270° clockwise.
Hour hand moves 30° per hour (360°/12). To reach West (270° from North going clockwise): 270/30 = 9 hours.
12:00 + 9 hours = 9:00 PM
Question 13 (Medium): A watch shows 6:00. The minute hand points South. If a person faces the direction of the hour hand, turns 45° clockwise, then walks 10√2 m, how far South is he from the starting point?
Solution: At 6:00:
- Minute hand at 12, hour hand at 6
- Minute hand points South, so 12 is South, 6 is North
Hour hand at 6 points North. Person faces North, turns 45° clockwise → faces North-East. Walks 10√2 m NE.
NE direction: equal North and East components. North component = 10√2 × cos(45°) = 10√2 × (1/√2) = 10 m East component = 10 m
From starting point, he is 10 m North and 10 m East. Distance South from starting point = -10 m (or 10 m North)
If question asks how far in South direction: 10 m (but in North direction actually)
Re-reading: "how far South" → He is North, so technically 0 m South, or need to interpret as position.
Actually, position is 10 m North of start. Distance in South direction from start = he is at -10 m South (i.e., 10 m North).
Question 14 (Hard): A clock's hour hand points to a direction. After 4 hours and 30 minutes, it has turned 135°. What was the initial direction and what is the current direction?
Solution: Hour hand moves 30° per hour. In 4.5 hours, it moves 4.5 × 30 = 135°.
The question says it turned 135°, which matches. But we need to find directions.
Actually, this seems to confirm the movement. Initial direction could be anything; after 135° turn: If started at North, now at South-East (135° clockwise = SE).
Question 15 (Hard): Two clocks show different times. Clock A: hour hand points NE, minute hand at 3. Clock B: hour hand points NW. What's the angle between the hour hands of both clocks?
Solution: NE = 45° from North NW = 315° or -45° from North
Angle = 45° - (-45°) = 90°
Set 4: Complex Movement Patterns
Question 16 (Medium): A person starts from point A, walks 8 m North to B, turns right and walks 6 m to C, turns right and walks 8 m to D, turns right and walks 6 m to E. Where is E with respect to A?
Solution:
- A(0, 0) → B(0, 8)
- Turn right (East), 6 m → C(6, 8)
- Turn right (South), 8 m → D(6, 0)
- Turn right (West), 6 m → E(0, 0)
E is at (0, 0) which is A.
Question 17 (Medium): From point P, a man walks 12 m East to Q, turns left and walks 8 m to R, turns left and walks 5 m to S, turns left and walks 8 m to T. How far is T from P?
Solution:
- P(0, 0) → Q(12, 0)
- Turn left (North), 8 m → R(12, 8)
- Turn left (West), 5 m → S(7, 8)
- Turn left (South), 8 m → T(7, 0)
T is at (7, 0), P is at (0, 0). Distance = 7 m (East direction)
Question 18 (Hard): A man walks 15 m South, turns left and walks 20 m, turns left and walks 15 m, turns right and walks 10 m, turns left and walks 5 m. How far is he from the starting point and in which direction?
Solution:
- Start: (0, 0)
- South 15 m → (0, -15)
- Turn left (East), 20 m → (20, -15)
- Turn left (North), 15 m → (20, 0)
- Turn right (East), 10 m → (30, 0)
- Turn left (North), 5 m → (30, 5)
Final: (30, 5) Start: (0, 0)
Distance = √(30² + 5²) = √(900 + 25) = √925 = √(25 × 37) = 5√37 ≈ 30.41 m
Direction: tan⁻¹(5/30) = tan⁻¹(1/6) ≈ 9.46° North of East
Question 19 (Hard): Starting from point X, a person walks 10 m North-East, then 10 m South-East, then 10 m South-West, then 10 m North-West. Where does he end up?
Solution: Break down each movement:
NE 10 m: Δx = 10/√2 = 5√2, Δy = 5√2 Position: (5√2, 5√2)
SE 10 m: Δx = 5√2, Δy = -5√2 Position: (10√2, 0)
SW 10 m: Δx = -5√2, Δy = -5√2 Position: (5√2, -5√2)
NW 10 m: Δx = -5√2, Δy = 5√2 Position: (0, 0)
Shortcut: Walking equal distances in all four diagonal directions forms a square and returns to start.
Question 20 (Hard): A man walks 3 km towards North, turns right and walks 4 km, turns left and walks 5 km. Another man starts from the same point, walks 4 km East, turns right and walks 3 km, turns right and walks 5 km. What's the distance between them?
Solution:
Man 1:
- (0, 0) → (0, 3)
- Turn right (East), 4 km → (4, 3)
- Turn left (North), 5 km → (4, 8)
Man 2:
- (0, 0) → (4, 0)
- Turn right (South), 3 km → (4, -3)
- Turn right (West), 5 km → (-1, -3)
Man 1 at (4, 8), Man 2 at (-1, -3) Distance = √((4-(-1))² + (8-(-3))²) = √(5² + 11²) = √(25 + 121) = √146 ≈ 12.08 km
Set 5: Distance Minimization
Question 21 (Easy): A person walks 6 km East and 8 km North. What's the shortest distance back to the starting point?
Solution: Using Pythagorean theorem: Distance = √(6² + 8²) = √(36 + 64) = √100 = 10 km
Shortcut: Recognize 6-8-10 triangle (multiple of 3-4-5).
Question 22 (Medium): From point A to point B is 9 km East, then from B to C is 12 km North. What's the direct distance from A to C?
Solution: AC = √(9² + 12²) = √(81 + 144) = √225 = 15 km
Shortcut: 9-12-15 is a multiple of 3-4-5 triangle (×3).
Question 23 (Medium): A rectangular park is 40 m long (East-West) and 30 m wide (North-South). What's the shortest distance from the South-West corner to the North-East corner?
Solution: Diagonal = √(40² + 30²) = √(1600 + 900) = √2500 = 50 m
Shortcut: 30-40-50 is a multiple of 3-4-5 (×10).
Question 24 (Hard): A person needs to go from point P to point R via point Q. P to Q is 7 km at 30° North of East. Q to R is 24 km at 60° North of East. What's the direct distance from P to R?
Solution: Break into components:
PQ (7 km, 30° North of East):
- East component: 7 × cos(30°) = 7 × 0.866 = 6.06 km
- North component: 7 × sin(30°) = 7 × 0.5 = 3.5 km
QR (24 km, 60° North of East):
- East component: 24 × cos(60°) = 24 × 0.5 = 12 km
- North component: 24 × sin(60°) = 24 × 0.866 = 20.78 km
Total displacement from P:
- East: 6.06 + 12 = 18.06 km
- North: 3.5 + 20.78 = 24.28 km
Direct distance PR = √(18.06² + 24.28²) = √(326.16 + 589.52) = √915.68 ≈ 30.26 km
Question 25 (Hard): Three points A, B, C form a triangle. AB = 13 km, BC = 14 km, CA = 15 km. A person walks from A to B to C. What's the shortest distance saved by going directly from A to C instead?
Solution: Path A→B→C = 13 + 14 = 27 km Direct A→C = 15 km
Distance saved = 27 - 15 = 12 km
Set 6: Relative Positioning
Question 26 (Easy): A is 10 m North of B. C is 10 m East of B. Where is C with respect to A?
Solution: B at origin: B(0, 0) A is North 10 m: A(0, 10) C is East 10 m of B: C(10, 0)
C is South-East of A. Distance = √(10² + 10²) = √200 = 10√2 m
Question 27 (Medium): P is to the North of Q. R is to the East of Q. S is to the South of R. T is to the West of S and South of Q. What is the position of T with respect to P?
Solution: Place Q at origin: Q(0, 0) P is North of Q: P(0, p) where p > 0, say P(0, 10) R is East of Q: R(r, 0) where r > 0, say R(10, 0) S is South of R: S(10, -s) where s > 0 T is West of S: T has x < 10 T is South of Q: T has y < 0
From the relationships: T could be at (0, -s) directly South of Q, or anywhere West of S with y < 0.
If T is directly South of Q at (0, -5): P is at (0, 10), T is at (0, -5) T is directly South of P.
Distance = 15 units.
Question 28 (Medium): A is 5 m North of B. C is 5 m East of A. D is 5 m South of C. Where is D with respect to B?
Solution: B(0, 0) → A(0, 5) → C(5, 5) → D(5, 0)
D is at (5, 0), B is at (0, 0). D is 5 m East of B
This forms a square!
Question 29 (Hard): Building A is 20 m North of Building B. Building C is 15 m East of Building B. Building D is 25 m South of Building C. Building E is equidistant from A and D, and also equidistant from B and C. Where is E located?
Solution: B at origin: B(0, 0) A: (0, 20) C: (15, 0) D: (15, -25)
E equidistant from B(0,0) and C(15,0): Lies on perpendicular bisector: x = 7.5
E equidistant from A(0,20) and D(15,-25): Let E = (7.5, y) Distance EA = Distance ED
√((7.5-0)² + (y-20)²) = √((7.5-15)² + (y+25)²) √(56.25 + (y-20)²) = √((-7.5)² + (y+25)²) 56.25 + y² - 40y + 400 = 56.25 + y² + 50y + 625 -40y + 400 = 50y + 625 -90y = 225 y = -2.5
E is at (7.5, -2.5), which is 7.5 m East and 2.5 m South of B
Question 30 (Hard): Four friends A, B, C, D are standing at the corners of a square field. A is at the North-East corner. B is 50 m South of A. C is 50 m West of B. D is 50 m North of C. Where is D with respect to A?
Solution: A at NE corner of square. Square has sides of 50 m (given by the movements).
Let's place A at origin for calculation, then adjust: A(0, 0) B is 50 m South: B(0, -50) C is 50 m West of B: C(-50, -50) D is 50 m North of C: D(-50, 0)
D(-50, 0), A(0, 0) D is 50 m West of A
Actually, this means A and D are at opposite corners (NW and NE), so they are 50 m apart horizontally.
But wait - if A is NE corner and B is 50 m South, B is at SE corner. C is 50 m West (SW corner). D is 50 m North (NW corner). So D is NW corner, A is NE corner. D is 50 m West of A.
Companies and Exams That Frequently Ask Direction Sense
Campus Placement Exams
- TCS NQT: 2-3 direction sense questions
- Infosys: 2-4 questions (shadow-based, movement)
- Wipro: 2-3 questions (basic directions)
- Cognizant: 3-4 questions (mixed difficulty)
- HCL: 2-3 questions
- Accenture: 3-5 questions
- IBM: 2-4 questions (distance calculations)
- Capgemini: 2-3 questions
Government Exams
- IBPS PO/Clerk: 3-5 questions
- SBI PO/Clerk: 3-5 questions
- SSC CGL: 2-4 questions
- SSC CHSL: 2-3 questions
- Railway Exams: 4-6 questions
- Defense Exams (CDS, AFCAT): 5-8 questions
Preparation Tips for Direction Sense
1. Master the Basics
- Memorize the 8 directions and their relationships
- Practice left/right turns from each direction
- Understand shadow patterns thoroughly
2. Always Draw a Diagram
- Visual representation prevents confusion
- Use coordinate system for complex problems
- Mark North direction clearly
3. Use Coordinate System
- North = +y, South = -y
- East = +x, West = -x
- This makes calculations systematic
4. Memorize Pythagorean Triplets
- 3-4-5, 5-12-13, 8-15-17, 7-24-25
- Saves calculation time in exams
- Recognize multiples (6-8-10, etc.)
5. Practice Shadow Problems
- Morning: Sun East, Shadow West
- Evening: Sun West, Shadow East
- Noon: Shadow shortest, points North (Northern Hemisphere)
6. Time Your Practice
- Easy questions: 30-45 seconds
- Medium questions: 60-90 seconds
- Hard questions: 2 minutes maximum
7. Double-Check Directions
- Left/Right from which facing direction
- Clockwise vs anti-clockwise
- About turn = 180° reversal
Frequently Asked Questions (FAQ)
Q1: How do I remember left and right turns?
A: Use your hands! If facing North, left is West and right is East. Practice this physically until it becomes automatic. Create a mental compass with directions fixed.
Q2: What if multiple turns are given?
A: Track the current facing direction after each turn. Write it down if needed. For example: Start North → Left → West → Right → North → About turn → South.
Q3: How do I handle shadow problems at different times?
A: Remember:
- Sunrise: Sun East, Shadow West
- Noon: Sun overhead (South in Northern Hemisphere), Shadow North
- Sunset: Sun West, Shadow East
- Shadow is always opposite to sun direction
Q4: Are calculator approximations allowed?
A: Most exams expect exact values or simple approximations. Memorize √2 ≈ 1.414, √3 ≈ 1.732. For 3-4-5 triangles, exact values are expected.
Q5: How do I minimize calculation errors?
A: Use the coordinate method consistently. Check signs (+/-) for each direction. Verify your final answer makes sense (should be positive distance, reasonable direction).
Quick Reference: Common Pythagorean Triplets
| Leg 1 | Leg 2 | Hypotenuse |
|---|---|---|
| 3 | 4 | 5 |
| 5 | 12 | 13 |
| 8 | 15 | 17 |
| 7 | 24 | 25 |
| 9 | 40 | 41 |
| 20 | 21 | 29 |
| 12 | 35 | 37 |
| 11 | 60 | 61 |
Multiples: 6-8-10, 9-12-15, 12-16-20, etc.
Master direction sense through visualization and consistent practice!