Login
OR
Don't worry. We'll fix this for you!
Direction Distance Test is a vital part of verbal reasoning questions in competitive exams. It tests your ability to understand and analyze movements in different directions and calculate distances traveled or the shortest path between two points. These problems often involve directions like north, south, east, west, and their combinations, along with distances measured in metric units such as meters and kilometers.
Understanding these concepts is essential not only for exams but also for real-life situations like navigation, travel planning, and even delivery services in cities across India and worldwide. This chapter will guide you through the fundamental ideas, problem-solving strategies, and practical examples to master direction distance problems effectively.
Directions help us describe the position or movement of an object relative to a reference point. The most common directions are:
These directions can be represented on a plane using a compass rose, where each direction corresponds to a specific angle measured clockwise from North.
Why is this important? Knowing these directions and their relative angles helps you visualize and solve problems involving movement and position on a plane.
Before solving direction distance problems, it is crucial to understand two key terms:
Both distance and displacement are measured using the metric system, commonly in meters (m) or kilometers (km). Remember, 1 km = 1000 m.
In the diagram above, the path traveled is 60 m east, then 60 m north, totaling 120 m distance. However, the displacement is the straight line from start to end, which is shorter.
Direction distance problems can become complex with multiple turns and routes. To solve them efficiently, follow these steps:
graph TD A[Read the problem carefully] --> B[Draw a rough diagram] B --> C[Mark all directions and distances] C --> D[Use coordinate axes or vector components] D --> E[Calculate final position] E --> F[Find required distance or direction]
Breaking down problems into smaller steps helps avoid confusion and errors. Using diagrams and coordinate systems simplifies calculations, especially when directions involve angles.
Step 1: Represent the movements on a coordinate plane. Starting at origin (0,0), move 5 km north (along y-axis) to (0,5).
Step 2: From (0,5), move 12 km east (along x-axis) to (12,5).
Step 3: Calculate displacement using Pythagoras theorem:
\[ d = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ km} \]
Answer: The shortest distance from the starting point is 13 km.
Step 1: Start at origin (0,0).
Step 2: Move 3 km east: new position (3,0).
Step 3: Move 4 km north: new position (3,4).
Step 4: Move 5 km west: new position (3 - 5, 4) = (-2, 4).
Step 5: Calculate displacement from (0,0) to (-2,4):
\[ d = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 4.47 \text{ km (approx)} \]
Answer: The shortest distance from the starting point is approximately 4.47 km.
Step 1: Break each movement into x (east-west) and y (north-south) components using trigonometry.
For 10 km at 30° north of east:
\( x_1 = 10 \cos 30^\circ = 10 \times 0.866 = 8.66 \text{ km} \)
\( y_1 = 10 \sin 30^\circ = 10 \times 0.5 = 5 \text{ km} \)
For 8 km at 60° south of east:
\( x_2 = 8 \cos 60^\circ = 8 \times 0.5 = 4 \text{ km} \)
\( y_2 = -8 \sin 60^\circ = -8 \times 0.866 = -6.93 \text{ km} \) (negative because south)
Step 2: Add components:
\( x = x_1 + x_2 = 8.66 + 4 = 12.66 \text{ km} \)
\( y = y_1 + y_2 = 5 - 6.93 = -1.93 \text{ km} \)
Step 3: Calculate resultant displacement:
\[ d = \sqrt{12.66^2 + (-1.93)^2} = \sqrt{160.3 + 3.72} = \sqrt{164.02} = 12.81 \text{ km} \]
Step 4: Find direction angle \( \alpha \) from east:
\( \tan \alpha = \frac{|y|}{x} = \frac{1.93}{12.66} = 0.152 \Rightarrow \alpha = \tan^{-1}(0.152) = 8.7^\circ \)
Since y is negative, direction is 8.7° south of east.
Answer: Resultant displacement is approximately 12.81 km at 8.7° south of east.
Step 1: Represent the movements on a coordinate plane. Starting at origin (0,0), move 7 km north to (0,7).
Step 2: From (0,7), move 24 km east to (24,7).
Step 3: Calculate displacement:
\[ d = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ km} \]
Answer: The delivery person is 25 km away from the starting point.
Step 1: Break each movement into x (east-west) and y (north-south) components.
NE (6 km at 45°):
\( x_1 = 6 \cos 45^\circ = 6 \times 0.707 = 4.24 \text{ km} \)
\( y_1 = 6 \sin 45^\circ = 4.24 \text{ km} \)
SW (8 km at 225° or 45° south of west):
\( x_2 = -8 \cos 45^\circ = -8 \times 0.707 = -5.66 \text{ km} \)
\( y_2 = -8 \sin 45^\circ = -5.66 \text{ km} \)
SE (10 km at 135° or 45° south of east):
\( x_3 = 10 \cos 45^\circ = 7.07 \text{ km} \)
\( y_3 = -10 \sin 45^\circ = -7.07 \text{ km} \)
Step 2: Add components:
\( x = 4.24 - 5.66 + 7.07 = 5.65 \text{ km} \)
\( y = 4.24 - 5.66 - 7.07 = -8.49 \text{ km} \)
Step 3: Calculate resultant displacement:
\[ d = \sqrt{5.65^2 + (-8.49)^2} = \sqrt{31.92 + 72.05} = \sqrt{103.97} = 10.2 \text{ km} \]
Step 4: Find direction angle \( \alpha \) from east:
\( \tan \alpha = \frac{8.49}{5.65} = 1.5 \Rightarrow \alpha = \tan^{-1}(1.5) = 56.3^\circ \)
Since y is negative, direction is 56.3° south of east.
Answer: Final position is approximately 10.2 km at 56.3° south of east from the start.
When to use: At the start of every direction distance problem to visualize movements clearly.
When to use: When multiple directions and turns are involved to break down movements into x and y components.
When to use: When distances are given in mixed units to avoid calculation errors.
When to use: When movements are strictly north-south and east-west for quick displacement calculation.
When to use: When directions are given as angles rather than cardinal points for accurate results.
Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
Go to practice →
Your Score
Your Rank
| Rank | Name | Score |
|---|
