Episode 8 — Aptitude and Reasoning / 8.20 — Direction Sense
8.20.c Solved Examples -- Direction Sense
Example 1: Basic Two-Turn Problem
Problem: Ravi walks 5 km towards North, then turns right and walks 3 km. In which direction is he from the starting point?
Solution:
Step 1: Draw the path.
B --------3 km-------> C
|
5 km (North)
|
A (Start)
Step 2: From A, point C is towards the upper-right = North-East.
Answer: North-East
Example 2: Three Turns with Distance
Problem: A man walks 4 km East, turns left and walks 3 km, then turns left again and walks 4 km. How far is he from the starting point?
Solution:
Step 1: Plot each movement.
D (End) C
| |
4 km (West) 3 km (North)
| |
A -------4 km (East)---> B
Wait -- let me re-trace:
- Start at A, walk 4 km East to B
- Turn left (from East, left = North), walk 3 km North to C
- Turn left (from North, left = West), walk 4 km West to D
D <------4 km (West)---- C
|
3 km (North)
|
A -------4 km (East)---> B
Step 2: Net displacement:
- East-West: 4 km East - 4 km West = 0
- North-South: 3 km North
Step 3: Distance from A to D = 3 km (directly North).
Answer: 3 km
Example 3: Finding Direction After Multiple Turns
Problem: Starting from point A, Meena walks 2 km North to B, turns right and walks 3 km to C, turns right and walks 5 km to D, turns right and walks 3 km to E. What is the direction of E from A?
Solution:
Step 1: Trace the path.
- A to B: 2 km North
- B to C: Turn right (East), 3 km East
- C to D: Turn right (South), 5 km South
- D to E: Turn right (West), 3 km West
B --------3 km-------> C
| |
2 km (N) 5 km (S)
| |
A E <---3 km--- D
Step 2: Coordinates:
- A = (0, 0)
- B = (0, 2)
- C = (3, 2)
- D = (3, -3)
- E = (0, -3)
Step 3: E is at (0, -3) relative to A at (0, 0).
- E is directly South of A.
- Distance = 3 km.
Answer: South
Example 4: Shadow-Based Direction (Morning)
Problem: One morning, Suresh was walking. His shadow fell to his left. Which direction was he facing?
Solution:
Step 1: Morning -> Sun is in the East -> Shadow falls to the West.
Step 2: Shadow is to his left.
Step 3: If West is to your left, you must be facing North.
N (Facing)
^
|
W (Shadow/Left) ---- Person ---- E (Sun/Right)
|
S
Answer: North
Example 5: Shadow-Based Direction (Evening)
Problem: In the evening, Ravi was standing such that his shadow fell in front of him. Which direction was he facing?
Solution:
Step 1: Evening -> Sun is in the West -> Shadow falls to the East.
Step 2: Shadow is in front of him -> He faces the direction of the shadow = East.
Answer: East
Example 6: Pythagoras Application
Problem: Aman walks 6 km South, then turns left and walks 8 km. What is the shortest distance from the starting point?
Solution:
A (Start)
|
6 km (South)
|
B --------8 km (East)-------> C (End)
Turning left from South = East.
Shortest distance (A to C):
AC = sqrt(6^2 + 8^2)
= sqrt(36 + 64)
= sqrt(100)
= 10 km
Answer: 10 km
Example 7: Complex Path (Z-Shape)
Problem: Rahul walks 10 km North, turns right and walks 6 km, turns right and walks 3 km, then turns left and walks 4 km. How far is he from the starting point?
Solution:
Step 1: Trace the path.
- A to B: 10 km North
- B to C: 6 km East (right from North)
- C to D: 3 km South (right from East)
- D to E: 4 km East (left from South)
B ------6 km------> C
| |
10 km 3 km
| |
A D ---4 km---> E
Step 2: Coordinates:
- A = (0, 0)
- B = (0, 10)
- C = (6, 10)
- D = (6, 7)
- E = (10, 7)
Step 3: Displacement from A(0,0) to E(10, 7):
Distance = sqrt(10^2 + 7^2)
= sqrt(100 + 49)
= sqrt(149)
~ 12.2 km
Answer: sqrt(149) km (approximately 12.2 km)
Example 8: Finding Direction of X from Y
Problem: Point A is 5 km North of point B. Point C is 5 km East of point A. Point D is 5 km South of point C. What is the direction of D from B?
Solution:
A ------5 km (E)----> C
| |
5 km (N) 5 km (S)
| |
B D
Step 1: Coordinates:
- B = (0, 0)
- A = (0, 5)
- C = (5, 5)
- D = (5, 0)
Step 2: D is at (5, 0) relative to B at (0, 0).
- D is directly East of B.
Answer: East
Example 9: Starting Direction Unknown
Problem: A man walks 1 km towards East, turns left, walks 1 km, turns left, walks 1 km, turns left, walks 2 km, turns left, walks 1 km. How far is he from the starting point?
Solution:
Step 1: Trace all movements.
- A to B: 1 km East
- B to C: 1 km North (left from East)
- C to D: 1 km West (left from North)
- D to E: 2 km South (left from West)
- E to F: 1 km East (left from South)
D <----1 km---- C
| |
| 1 km (N)
| |
| A ---1 km--> B
| |
2 km |
| |
E ---1 km (E)--> F
Step 2: Coordinates:
- A = (0, 0)
- B = (1, 0)
- C = (1, 1)
- D = (0, 1)
- E = (0, -1)
- F = (1, -1)
Step 3: Displacement from A(0,0) to F(1, -1):
Distance = sqrt(1^2 + (-1)^2) = sqrt(2) ~ 1.41 km
Answer: sqrt(2) km (approximately 1.41 km)
Example 10: Relative Direction
Problem: Town B is to the West of town A. Town C is to the South of town B. Town D is to the East of town C. Town D is in which direction relative to town A?
Solution:
A
| (B is West of A)
|
B ---+
| (C is South of B)
|
C ---------D (D is East of C)
Let's assign coordinates:
- A = (0, 0)
- B = (-x, 0) ... West of A
- C = (-x, -y) ... South of B
- D = (-x + z, -y) ... East of C
Without exact distances, D could be anywhere South-West to South-East of A. But typically in such problems, the distances are equal:
If all distances are equal (say d):
- A = (0, 0)
- B = (-d, 0)
- C = (-d, -d)
- D = (0, -d)
D is directly South of A.
Answer: South (assuming equal distances)
Example 11: About Turn Problem
Problem: Facing North, Anil turns 90 degrees clockwise, then takes an about turn, then turns 90 degrees anti-clockwise. Which direction is he facing now?
Solution:
Step 1: Start facing North.
Step 2: 90 degrees clockwise -> East.
Step 3: About turn (180 degrees) -> West.
Step 4: 90 degrees anti-clockwise -> from West, ACW 90 = South.
Turn sequence:
North --90 CW--> East --180--> West --90 ACW--> South
Answer: South
Example 12: Sunrise Shadow with Distance
Problem: One morning Rekha was walking towards the sun. She turned left and walked 2 km, then turned right and walked 3 km. In which direction is she now from the starting point?
Solution:
Step 1: Morning -> Walking towards the sun = Walking East.
Step 2: Turns left (from East, left = North) -> walks 2 km North.
Step 3: Turns right (from North, right = East) -> walks 3 km East.
C ----3 km (E)----> D (End)
|
2 km (N)
|
A ---(towards sun/East)---> (turned at some point, but
let's assume she turned immediately)
Actually, re-reading: she was walking towards sun, then turned left, then right.
B ----2 km (N)----> C ----3 km (E)----> D
|
(Walking East initially, turned left at B)
|
A (Start)
Hmm, she doesn't walk East any specified distance. Let's assume she turns immediately:
- A = (0, 0)
- Turn left (North), walk 2 km: B = (0, 2)
- Turn right (East), walk 3 km: C = (3, 2)
Direction of C from A: C is at (3, 2) -- North-East.
Answer: North-East
Example 13: Multiple Right Turns
Problem: Mohan walks 3 km North, turns right, walks 4 km, turns right, walks 7 km, turns right, walks 4 km. How far is he from the starting point and in which direction?
Solution:
Step 1: Trace:
- A to B: 3 km North
- B to C: 4 km East (right from North)
- C to D: 7 km South (right from East)
- D to E: 4 km West (right from South)
B ------4 km (E)-----> C
| |
3 km (N) 7 km (S)
| |
A |
. |
. |
E <-----4 km (W)------ D
Step 2: Coordinates:
- A = (0, 0)
- B = (0, 3)
- C = (4, 3)
- D = (4, -4)
- E = (0, -4)
Step 3: E is at (0, -4) from A(0, 0) -> directly South, 4 km.
Answer: 4 km, South
Example 14: Intercardinal Direction
Problem: Priya walks 5 km towards North-East and then 5 km towards South-East. How far is she from the starting point?
Solution:
Step 1: NE movement (5 km):
- X change = 5/sqrt(2), Y change = 5/sqrt(2)
Step 2: SE movement (5 km):
- X change = 5/sqrt(2), Y change = -5/sqrt(2)
Step 3: Total:
- X = 5/sqrt(2) + 5/sqrt(2) = 10/sqrt(2) = 5*sqrt(2)
- Y = 5/sqrt(2) - 5/sqrt(2) = 0
B (after NE)
/ \
5 km 5 km
(NE) (SE)
/ \
A C
A and C are on the same horizontal line.
AC = 5*sqrt(2) ~ 7.07 km (due East)
Answer: 5*sqrt(2) km (approximately 7.07 km), due East
Example 15: Who is to whose direction?
Problem: A is to the South of B. C is to the East of B. D is to the North of C. If all distances are equal, what is the direction of D from A?
Solution:
Placing B at origin:
B = (0, 0)
A = (0, -d) ... South of B
C = (d, 0) ... East of B
D = (d, d) ... North of C
D (d, d)
|
| d
|
B ---+-------> C (d, 0)
(0,0) d
|
| d
|
A (0, -d)
Direction of D from A:
- D is at (d, d), A is at (0, -d)
- Relative: D is (d, 2d) from A
- That is East and North -> North-East
More precisely: tan(theta) = d/(2d) = 1/2, so theta ~ 26.6 degrees from North toward East.
Answer: North-East
Example 16: Tricky "Left of" Problem
Problem: A walks 3 km North, then 4 km to the left, then 3 km to the right. In which direction and how far is he from the starting point?
Solution:
Step 1: Walking North, left = West, then from West, right = North.
- A to B: 3 km North -> B = (0, 3)
- B to C: 4 km West -> C = (-4, 3)
- C to D: 3 km North -> D = (-4, 6)
D (End)
|
3 km (N)
|
C <---4 km(W)-- B
|
3 km (N)
|
A (Start)
Step 2: D is at (-4, 6) from A(0, 0).
- Distance = sqrt(16 + 36) = sqrt(52) = 2*sqrt(13) ~ 7.21 km
- Direction: North-West
Answer: 2*sqrt(13) km (~7.21 km), North-West direction
Example 17: Angle-Based Turning
Problem: Raj faces North. He turns 135 degrees clockwise. Which direction is he facing?
Solution:
Starting: North = 0 degrees After turning: 0 + 135 = 135 degrees clockwise from North.
0 = N, 45 = NE, 90 = E, 135 = SE
Answer: South-East
Example 18: Mirror Image / Opposite Direction
Problem: Renu is facing South-West. She turns 90 degrees anti-clockwise, then makes an about turn. Which direction is she facing?
Solution:
Step 1: Facing SW. Step 2: 90 degrees ACW from SW -> SE (counter-clockwise: SW -> S -> SE at 90 degrees... wait, let me recalculate).
Using angles: SW = 225 degrees ACW 90 degrees: 225 - 90 = 135 degrees = SE
Step 3: About turn (180 degrees): 135 + 180 = 315 degrees = NW
Answer: North-West
Example 19: Complex Multi-Leg Journey
Problem: A person starts from point P, walks 4 km East to Q, turns left and walks 3 km to R, turns left and walks 8 km to S, turns left and walks 6 km to T, turns left and walks 4 km to U. Find the distance between U and P.
Solution:
Step 1: Trace:
- P to Q: 4 km East -> Q = (4, 0)
- Q to R: 3 km North (left from East) -> R = (4, 3)
- R to S: 8 km West (left from North) -> S = (-4, 3)
- S to T: 6 km South (left from West) -> T = (-4, -3)
- T to U: 4 km East (left from South) -> U = (0, -3)
S <--------8 km (W)--------- R
| |
| 3 km (N)
| |
6 km (S) P ---4 km (E)---> Q
| |
| |
T ---4 km (E)-> U
Step 2: P = (0, 0), U = (0, -3)
- U is directly South of P, 3 km away.
Answer: 3 km
Example 20: Sunset Shadow Problem
Problem: One evening, two friends A and B are standing facing each other. A's shadow falls to his right. Which direction is B facing?
Solution:
Step 1: Evening -> Sun in West -> Shadow falls to East.
Step 2: A's shadow falls to his right -> East is to A's right -> A faces North.
Step 3: A and B face each other -> B faces South.
A (faces North)
^
|
| (facing each other)
v
B (faces South)
Answer: B is facing South
Example 21: Displacement = 0
Problem: Anil walks 5 km East, 4 km North, 5 km West, and 4 km South. How far is he from the starting point?
Solution:
D <----5 km (W)---- C
| |
4 km (S) 4 km (N)
| |
A ----5 km (E)----> B
Net E-W: 5 - 5 = 0 Net N-S: 4 - 4 = 0
He is back at the starting point.
Answer: 0 km (he returned to the start)
Example 22: Three-Dimensional Thinking (Distance Between Two People)
Problem: A and B start from the same point. A walks 6 km North and then 8 km East. B walks 8 km South and then 6 km West. What is the distance between A and B?
Solution:
Step 1: A's position:
- Start (0, 0) -> 6 km N -> (0, 6) -> 8 km E -> (8, 6)
Step 2: B's position:
- Start (0, 0) -> 8 km S -> (0, -8) -> 6 km W -> (-6, -8)
Step 3: Distance between A(8, 6) and B(-6, -8):
Distance = sqrt((8-(-6))^2 + (6-(-8))^2)
= sqrt(14^2 + 14^2)
= sqrt(196 + 196)
= sqrt(392)
= 14*sqrt(2)
~ 19.8 km
Answer: 14*sqrt(2) km (approximately 19.8 km)
Example 23: "Left" and "Right" Relative to Facing
Problem: Sunil faces West. He turns right, walks 5 m, turns left, walks 10 m, turns left, walks 5 m, turns right and walks 5 m. In which direction is he from the starting point?
Solution:
- Start facing West.
- Turns right (from West, right = North), walks 5 m -> (0, 5)
- Turns left (from North, left = West), walks 10 m -> (-10, 5)
- Turns left (from West, left = South), walks 5 m -> (-10, 0)
- Turns right (from South, right = West), walks 5 m -> (-15, 0)
Starting at origin, facing West:
C <---10 m (W)---- B Legend:
| | A = Start (0,0)
5m (S) 5m (N) E = End (-15, 0)
| |
D ----5 m (W)----> A
|
5m (W)...
Wait, let me redo with coordinates:
Let me use a grid:
- A = (0, 0)
- After 5m North: B = (0, 5)
- After 10m West: C = (-10, 5)
- After 5m South: D = (-10, 0)
- After 5m West: E = (-15, 0)
E is at (-15, 0) from A -> directly West, 15 m.
Answer: West, 15 m from the starting point
Example 24: Finding the Starting Direction
Problem: A person walks 4 km, turns right, walks 3 km, and reaches a point that is 5 km from the starting point. What was his initial direction?
Solution:
Step 1: This forms a right triangle with sides 4, 3, and hypotenuse 5.
Step 2: Since 3-4-5 is a valid Pythagorean triplet, the path forms a right angle at the turning point. This is consistent since a right turn is 90 degrees.
Step 3: The question doesn't specify the initial direction, so any cardinal direction works. But the answer is: Any direction (the problem only confirms the distances are consistent).
If additional information says "he ended up to the North-East of start," we can determine the initial direction. Without that, the initial direction could be N, S, E, or W.
Answer: The initial direction cannot be determined uniquely from the given information (any cardinal direction is possible).
Example 25: Clock-Position Direction
Problem: If North is at the 12 o'clock position of a clock, in which direction does the 5 o'clock position point?
Solution:
Step 1: Each hour = 360/12 = 30 degrees.
Step 2: 5 o'clock = 5 x 30 = 150 degrees from North (clockwise).
Step 3: 150 degrees is between SE (135) and S (180), closer to SE.
12 = N (0)
1 = 30 (NNE)
2 = 60 (ENE)
3 = 90 (E)
4 = 120 (ESE)
5 = 150 (SSE)
6 = 180 (S)
7 = 210 (SSW)
8 = 240 (WSW)
9 = 270 (W)
10 = 300 (WNW)
11 = 330 (NNW)
Answer: South-South-East (SSE), or approximately South-East
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