Directions & Distance
Key Concepts & Formulas
Provide 5-7 essential concepts for Directions & Distance:
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | CARDINAL DIRECTIONS | North (N), South (S), East (E), West (W) form 90° angles. Remember: “Never Eat Soggy Wheat” clockwise order |
| 2 | CLOCKWISE ROTATION | Right turn = 90° clockwise. From North: Right→East, Left→West, U-turn→South |
| 3 | SHADOW FORMULA | Morning (6-12): Shadow points West. Evening (12-6): Shadow points East. At 12 noon: No shadow |
| 4 | PYTHAGOREAN DISTANCE | When moving at right angles: Distance = √(North-South movement² + East-West movement²) |
| 5 | DIRECTION CODES | N=0°, E=90°, S=180°, W=270°. Add/subtract turns from current bearing |
| 6 | RELATIVE POSITIONS | Face North: Behind=South, Left=West, Right=East. Always establish reference point first |
10 Practice MCQs
Generate 10 MCQs with increasing difficulty (Q1-3: Easy, Q4-7: Medium, Q8-10: Hard)
Q1. A train moves 15 km North from Station A to B, then 8 km East to Station C. What is the shortest distance between A and C? A) 17 km B) 23 km C) 7 km D) 13 km
Answer: A) 17 km
Solution: Using Pythagoras theorem: AC = √(AB² + BC²) = √(15² + 8²) = √(225 + 64) = √289 = 17 km
Shortcut: Remember 8-15-17 is a Pythagorean triplet
Concept: Directions & Distance - Right angle movement calculation
Q2. Ravi faces East. He turns 135° clockwise, then 225° anticlockwise. Which direction does he face now? A) North B) South C) West D) East
Answer: A) North
Solution: Starting East (90°)
- 135° clockwise: 90° + 135° = 225° (South-West)
- 225° anticlockwise: 225° - 225° = 0° (North)
Shortcut: Net rotation = 135° clockwise - 225° anticlockwise = 90° anticlockwise from East = North
Concept: Directions & Distance - Compound rotation
Q3. A railway employee walks 2 km towards sunrise in the morning, then turns left and walks 3 km. Which direction is he from the starting point? A) North-West B) South-East C) North-East D) South-West
Answer: C) North-East
Solution: Morning sunrise = East direction
- Walks 2 km East
- Turns left (North) and walks 3 km
- Final position: 2 km East, 3 km North = North-East from start
Shortcut: Morning + East walk + Left turn = North-East quadrant
Concept: Directions & Distance - Sunrise direction application
Q4. A goods train travels 40 km North, then 30 km South, then 120 km East. How far and in which direction is it from the starting point? A) 120 km East B) 50 km North-East C) 120 km North-East D) 50 km East
Answer: D) 50 km East
Solution:
- Net North-South: 40N - 30S = 10 km North
- East movement: 120 km East
- Distance = √(10² + 120²) = √(100 + 14400) = √14500 ≈ 120.4 km
- Direction: Mostly East with slight North (≈ East)
Shortcut: When one component » other, direction ≈ larger component
Concept: Directions & Distance - Net displacement calculation
Q5. At 3:30 PM, a 15-meter electric pole casts a 15√3 meter shadow. Which direction does the shadow point? A) East B) West C) North D) South
Answer: A) East
Solution: 3:30 PM = Afternoon (after 12 PM) Afternoon shadow points East Length calculation confirms: tan θ = 15/(15√3) = 1/√3 → θ = 30° sun angle
Shortcut: PM time = Afternoon = Shadow East
Concept: Directions & Distance - Shadow direction based on time
Q6. An express train goes from Delhi to Mumbai: 800 km South-West, then 600 km South-East. Find the direct distance and bearing from Delhi to Mumbai. A) 1000 km South B) 1400 km South C) 1000 km South-East D) 1000 km South-West
Answer: A) 1000 km South
Solution:
- South-West = 45° between South and West
- South-East = 45° between South and East
- Net East-West: 800cos45° West - 600cos45° East = 200cos45° West ≈ 141 km West
- Net North-South: 800sin45° + 600sin45° = 1400sin45° ≈ 990 km South
- Direct distance = √(141² + 990²) ≈ 1000 km
- Direction ≈ South (West component negligible)
Shortcut: SW + SE movements → Predominantly South when distances similar
Concept: Directions & Distance - Vector addition of diagonal movements
Q7. A railway track inspector walks 3 km North from point P, sees his shadow falling to his right at 4 PM. He turns 90° right and walks 4 km. Find his final position relative to P. A) 4 km East, 3 km North B) 4 km West, 3 km North C) 4 km East, 7 km North D) 4 km West, 1 km North
Answer: C) 4 km East, 7 km North
Solution:
- 4 PM: Shadow points East (afternoon)
- Shadow on right → He faces North (East is right of North)
- Walks 3 km North: Position = 0E, 3N
- Turns 90° right → Faces East
- Walks 4 km East: Position = 4E, 3N
- Wait! Correction: After 3 km North, at 4 PM, shadow confirms North direction
- Turns 90° right → Faces East, walks 4 km
- Final: 4 km East, 3 km North from P
Shortcut: 4 PM shadow right when facing North → Confirm North direction
Concept: Directions & Distance - Shadow confirmation with movement
Q8. Two trains start from the same junction. Train A goes 60 km/h North for 2 hours, Train B goes 80 km/h East for 1.5 hours. Find the distance between them and direction of B from A. A) 120 km North-East B) 120 km South-East C) 150 km South-East D) 150 km North-East
Answer: C) 150 km South-East
Solution:
- Train A: 60 × 2 = 120 km North
- Train B: 80 × 1.5 = 120 km East
- Distance between = √(120² + 120²) = 120√2 ≈ 170 km
- Direction of B from A: South-East (B is East and South of A)
Shortcut: Equal N-S and E-W distances → 45° diagonal = √2 × side
Concept: Directions & Distance - Relative positions with speed-time
Q9. A circular railway track (radius 7 km) has stations at N, S, E, W points. An employee travels from North to East station via the shorter arc, then South to West via longer arc. Find total distance traveled. A) 22 km B) 44 km C) 33 km D) 55 km
Answer: C) 33 km
Solution:
- Circumference = 2πr = 2 × (22/7) × 7 = 44 km
- North to East (shorter arc) = 1/4 circumference = 11 km
- South to West (longer arc) = 3/4 circumference = 33 km
- Total = 11 + 33 = 44 km
Shortcut: Quarter circle = πr/2, Three-quarter circle = 3πr/2
Concept: Directions & Distance - Circular movement calculation
Q10. At a railway crossing, a person sees a goods train's shadow (length 200m) covering exactly 100m on the ground when the sun is 30° above the horizon. If the train moves North and the shadow points West, find the train's height and sun's position. A) 100√3 m, East B) 115 m, West C) 200 m, East D) 173 m, West
Answer: D) 173 m, West
Solution:
- tan 30° = height/shadow = h/200 = 1/√3
- Height = 200/√3 = 200√3/3 ≈ 115.5 m
- Shadow West → Sun in East
- Morning time (sun in East)
Shortcut: Height = Shadow length × tan(sun angle)
Concept: Directions & Distance - Shadow length with sun position
5 Previous Year Questions
Generate PYQ-style questions with authentic exam references:
PYQ 1. A man walks 5 km East, then 4 km North, then 2 km West. How far is he from the starting point? [RRB NTPC 2021 CBT-1]
Answer: C) √29 km
Solution:
- Net East: 5 - 2 = 3 km East
- Net North: 4 km North
- Distance = √(3² + 4²) = √25 = 5 km
Exam Tip: Always calculate net movements first, then apply Pythagoras
PYQ 2. A railway employee faces South. He turns 135° anticlockwise, then 180° clockwise. Which direction does he face now? [RRB Group D 2022]
Answer: B) North-West
Solution:
- Start: South (180°)
- 135° anticlockwise: 180° - 135° = 45° (North-East)
- 180° clockwise: 45° + 180° = 225° (South-West)
Exam Tip: Net rotation = 180° - 135° = 45° clockwise from South = South-West
PYQ 3. At 9 AM, a 12-meter train coach casts a 12√3 meter shadow. What is the sun's elevation angle? [RRB ALP 2018]
Answer: B) 30°
Solution:
- tan θ = height/shadow = 12/(12√3) = 1/√3
- θ = 30°
Exam Tip: Morning shadow West confirms calculation validity
PYQ 4. Two stations are 100 km apart on a map. If the scale is 1 cm = 25 km, what is the map distance? [RRB JE 2019]
Answer: A) 4 cm
Solution:
- Map distance = Actual distance / Scale = 100 / 25 = 4 cm
Exam Tip: Scale questions often combine with direction problems
PYQ 5. A goods train moves 120 km North, then 160 km East. Find the direct distance and bearing from start. [RPF SI 2019]
Answer: C) 200 km North-East
Solution:
- Distance = √(120² + 160²) = √(14400 + 25600) = √40000 = 200 km
- tan θ = 160/120 = 4/3 → θ ≈ 53° East of North = North-East
Exam Tip: 120-160-200 is a 3-4-5 triangle multiple
Speed Tricks & Shortcuts
For Directions & Distance, provide exam-tested shortcuts:
| Situation | Shortcut | Example |
|---|---|---|
| Pythagorean Triplets | Remember common triplets: 3-4-5, 5-12-13, 8-15-17, 7-24-25 | If movements are 60N and 80E, distance = 100 km (3-4-5 × 20) |
| Clockwise Rotation | Right turn = +90°, Left = -90°, U-turn = ±180° | Face North → Right → East → Right → South |
| Shadow Time | AM: Shadow West, PM: Shadow East, 12 noon: No shadow | 8 AM shadow West, 4 PM shadow East |
| Net Movement | Calculate N-S and E-W separately, then combine | 30N + 20S = 10N net, 40E + 10W = 30E net |
| Circular Track | Quarter circle = πr/2, Half circle = πr | Radius 7 km track: Quarter arc = 11 km (using π = 22/7) |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Ignoring Net Movement | Adding all distances without considering direction | Always subtract opposite directions first |
| Wrong Shadow Direction | Confusing morning/evening shadow directions | Remember: AM-West, PM-East, Noon-No shadow |
| Rotation Error | Adding instead of setting new bearing | Use degree calculation: Current° ± Turn° = New° |
| Pythagoras Miscalculation | Forgetting to square both components | Distance = √(ΔN-S² + ΔE-W²), never add directly |
| Reference Point Confusion | Calculating “A from B” instead of “B from A” | Always establish “from where to where” first |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| Cardinal Directions Clockwise | N → E → S → W → N |
| 9 AM Shadow Direction | Points West |
| 3 PM Shadow Direction | Points East |
| Pythagorean Formula | Distance = √(N-S² + E-W²) |
| Right Turn from North | Faces East |
| Left Turn from North | Faces West |
| U-turn from any direction | Opposite direction |
| Sun at 30° elevation | Shadow = √3 × Height |
| Quarter circle distance | πr/2 |
| 3-4-5 triangle multiple | Most common in exams |
Topic Connections
How Directions & Distance connects to other RRB exam topics:
- Direct Link: Speed, Time & Distance - Many direction problems involve movement with speed
- Combined Questions: Trains & Boats - Often combined with relative speed concepts
- Foundation For: Navigation & Mapping - Advanced railway signaling systems use these principles
- Scale Connection: Mensuration - Map scale calculations frequently appear with direction problems
- Vector Applications: Physics - Force and velocity problems use similar vector addition principles
BETA
