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Complex Map or Route Puzzle

Introduction

Complex Map or Route Puzzles involve multi-step journeys across a map - often with different legs, bearings, distances, and intermediate stops. These problems test your ability to combine displacement, direction sense, vector addition, and map-visualization to determine final location, bearing, or distance between places.

This pattern is important because many competitive exams (SSC, CAT, railway exams) include multi-leg route questions that require methodical tracing rather than intuition.

Pattern: Complex Map or Route Puzzle

Pattern

Key concept: Break the entire route into sequential vector components (Δx, Δy) using a consistent coordinate system, sum components, then compute displacement/bearing using signs and Pythagoras.

Practical rules

  • Choose a fixed origin and axis: East = +x, North = +y. Mark each leg as (Δx, Δy).
  • For bearings given as N60°E, convert to components: Δx = d·sin(θ), Δy = d·cos(θ) (θ measured from North toward East).
  • Sum all x-components and y-components separately to get total (ΣΔx, ΣΔy).
  • Compute straight-line distance = √(ΣΔx² + ΣΔy²). Determine bearing from signs: quadrant → use arctan(|ΣΔx/ΣΔy|) to find angle from cardinal axis.
  • If map contains returns or loops, simplify by cancelling opposite components before heavy computation.

Step-by-Step Example

Question

From City A a traveller goes 10 km North to B, then 6 km at bearing N60°E to C, then 8 km East to D, and finally 12 km at bearing S30°E to E. Find the straight-line distance and bearing of E from A.

Solution

  1. Step 1: How to begin:

    Set A at origin (0,0). Use axes: East = +x, North = +y. Convert each leg into (Δx,Δy).

  2. Step 2: Leg AB (10 km North)

    Δx₁ = 0, Δy₁ = +10 → Position B = (0, 10).

  3. Step 3: Leg BC (6 km, bearing N60°E)

    Bearing N60°E means 60° east of North. Components: Δx₂ = 6·sin(60°) = 6 × (√3/2) = 3√3 ≈ 5.196; Δy₂ = 6·cos(60°) = 6 × (1/2) = 3. Position C = B + (Δx₂, Δy₂) = (0 + 3√3, 10 + 3) ≈ (5.196, 13).

  4. Step 4: Leg CD (8 km East)

    Δx₃ = +8, Δy₃ = 0. Position D = C + (8,0) ≈ (5.196 + 8, 13) = (13.196, 13).

  5. Step 5: Leg DE (12 km, bearing S30°E)

    S30°E = 30° east of South. Components: moving South (-y) and East (+x). Δx₄ = 12·sin(30°) = 12 × 0.5 = 6. Δy₄ = -12·cos(30°) = -12 × (√3/2) = -6√3 ≈ -10.392. Position E = D + (6, -6√3) ≈ (13.196 + 6, 13 - 10.392) = (19.196, 2.608).

  6. Step 6: Step - compute net displacement from A

    ΣΔx = 19.196, ΣΔy = 2.608. Distance AE = √(19.196² + 2.608²) ≈ √(368.48 + 6.80) ≈ √375.28 ≈ 19.37 km.

  7. Step 7: Step - compute bearing from A to E

    Since Σx > 0 and Σy > 0 → quadrant = North-East. Angle from North toward East = arctan(Σx / Σy) = arctan(19.196 / 2.608) ≈ arctan(7.36) ≈ 82.3°. Bearing ≈ N82.3°E (or equivalently about 7.7° East of North).

  8. Final Answer:

    Distance ≈ 19.37 km, Bearing ≈ N82.3°E

  9. Quick Check:

    Most displacement is East (large x), small positive y → distance ≈ 19.4 and NE quadrant with a large angle from North - consistent. ✅

Quick Variations

1. Replace exact bearings with cardinal-turn steps (e.g., right 45°) - convert turns to bearings before component math.

2. Include loops where one leg retraces part of a previous leg - cancel components to simplify.

3. Time-parametric routes (who meets whom) - convert to parametric equations and solve simultaneous positions.

4. Map grid with blocks (city blocks) - use Manhattan distance when required, but use vector method for straight-line answers.

Trick to Always Use

  • Step 1: Convert every leg to (Δx, Δy) immediately - keep a running sum.
  • Step 2: Cancel opposite components early to reduce arithmetic.
  • Step 3: For bearing, compute arctan(|Σx/Σy|) and use the signs of Σx, Σy to fix quadrant (NE/NW/SE/SW).

Summary

Summary

  • Always fix an origin and axes (East = +x, North = +y) before tracing routes.
  • Convert bearings to components: use sin for east/west component and cos for north/south when angle measured from North (or vice versa when measured from East).
  • Sum components to get final coordinates; use Pythagoras for distance and arctan for bearing.
  • Cancel opposite displacements early and use squared distances to compare before taking square roots.

Example to remember:
Break each leg into Δx,Δy → sum → distance = √(ΣΔx² + ΣΔy²) → bearing from signs & arctan.

Practice

(1/5)
1. A person travels 8 km North, then 6 km East, and finally 10 km South. What is his final displacement from the starting point (approximate)?
easy
A. 6.3 km South-East
B. 4 km South-East
C. 5.4 km South-West
D. 2 km South-East

Solution

  1. Step 1: Represent each leg as coordinates

    North = +y, East = +x, South = -y.
    Total Δx = +6, Total Δy = (8 - 10) = -2.

  2. Step 2: Compute displacement

    √(6² + (-2)²) = √(36 + 4) = √40 ≈ 6.32 km.

  3. Step 3: Determine direction

    Positive x (East), Negative y (South) → South-East.

  4. Final Answer:

    6.3 km South-East → Option A

  5. Quick Check:

    East + South → SE; larger east component ✅
Hint: Subtract vertical components and use Pythagoras for final displacement.
Common Mistakes: Adding distances instead of vector subtraction for opposite directions.
2. A person goes 5 km East, then 12 km North, then 5 km West. How far is he from the starting point?
easy
A. 5 km
B. 10 km
C. 13 km
D. 12 km

Solution

  1. Step 1: Compute net movement

    East 5 - West 5 = 0 (cancelled); only 12 km North remains.

  2. Step 2: Displacement

    √(0² + 12²) = 12 km.

  3. Step 3: Check options

    12 km corresponds to the straight-line path.

  4. Final Answer:

    12 km → Option D

  5. Quick Check:

    Horizontal cancelled, only 12 km North ✅
Hint: Cancel exact opposite directions before calculating distance.
Common Mistakes: Forgetting that East-West cancel out completely.
3. A car moves 10 km North, 6 km East, 8 km South, and 6 km West. What is its final position relative to the start?
easy
A. 2 km North
B. 4 km North
C. 2 km South
D. At the same point

Solution

  1. Step 1: Combine vertical

    North 10 - South 8 = 2 km North.

  2. Step 2: Combine horizontal

    East 6 - West 6 = 0.

  3. Step 3: Final position

    2 km North, same longitude.

  4. Final Answer:

    2 km North → Option A

  5. Quick Check:

    Horizontal cancels completely ✅
Hint: Subtract opposite directions before computing displacement.
Common Mistakes: Forgetting to check both axes separately.
4. Ravi travels 6 km North, then 8 km at bearing N45°E. Find his final displacement from the start (approx).
medium
A. 13.0 km
B. 12.5 km
C. 10 km
D. 9.8 km

Solution

  1. Step 1: Break second leg into components

    For N45°E both components are equal: Δx₂ = 8·sin45° = 8/√2 ≈ 5.6569, Δy₂ = 8·cos45° = 8/√2 ≈ 5.6569.

  2. Step 2: Sum totals

    Σx = 0 + 5.6569 = 5.6569; Σy = 6 + 5.6569 = 11.6569.

  3. Step 3: Displacement

    Distance = √(Σx² + Σy²) = √(5.6569² + 11.6569²) ≈ √(32.00 + 135.96) ≈ √167.96 ≈ 12.96 ≈ 13.0 km.

  4. Final Answer:

    13.0 km → Option A

  5. Quick Check:

    Both legs lie in the NE quadrant; combined y (≈11.66) is about twice x (≈5.66), giving hypotenuse ≈13 km ✅
Hint: Resolve the bearing leg into equal x/y components for 45° bearings, then apply Pythagoras.
Common Mistakes: Rounding intermediate components too coarsely or treating the bearing leg as pure N+E without trigonometry.
5. A pilot flies 100 km North, 100 km East, and then 100 km at bearing S45°E. Approximate straight-line distance from start?
medium
A. 141 km
B. 173 km
C. 200 km
D. 224 km

Solution

  1. Step 1: Compute components of each leg

    Leg1: Δx₁=0, Δy₁=+100; Leg2: Δx₂=+100, Δy₂=0; Leg3 (S45°E): Δx₃=100·sin(45)=70.71, Δy₃=-100·cos(45)=-70.71.

  2. Step 2: Net totals

    Σx = 0 + 100 + 70.71 = 170.71; Σy = 100 - 70.71 = 29.29.

  3. Step 3: Displacement

    √(170.71² + 29.29²) ≈ √(29142 + 858) ≈ √30000 = 173.2 km.

  4. Final Answer:

    173 km → Option B

  5. Quick Check:

    Mostly eastward motion → distance ≈ 170+ ✅
Hint: For bearing legs, decompose using sin-cos before summation.
Common Mistakes: Ignoring negative sign for southward component.

Mock Test

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