Master Speed, Time, and Distance for SSC CGL & CHSL: Formulas, Short Tricks & Solved Examples

Introduction to Speed, Time, and Distance

Speed, Time, and Distance (STD) is one of the most vital chapters in the Quantitative Aptitude section of Staff Selection Commission (SSC) examinations, including SSC CGL (Tier 1 & Tier 2) and SSC CHSL. Every year, 3 to 5 questions are directly or indirectly based on this topic. Mastering STD not only boosts your score in basic distance calculation problems but also lays the foundation for solving complex questions on Trains, Boats and Streams, and Races.

This comprehensive guide is designed to cover everything from fundamental definitions to high-level shortcut techniques. By the end of this post, you will be equipped to handle any Speed, Time, and Distance question with high speed and absolute accuracy.

Core Concepts and Fundamental Formulas

The entire topic of Speed, Time, and Distance rests upon a single fundamental relationship between three variables:

• Speed (\(S\)): The rate at which distance is covered per unit of time.
• Time (\(T\)): The duration taken to cover a specific distance.
• Distance (\(D\)): The total spatial length covered by a moving object.

1. Standard Mathematical Formulas

The core governing formulas are:

• \( \text{Distance } (D) = \text{Speed } (S) \times \text{Time } (T) \)
• \( \text{Speed } (S) = \frac{ \text{Distance } (D)}{ \text{Time } (T)} \)
• \( \text{Time } (T) = \frac{ \text{Distance } (D)}{ \text{Speed } (S)} \)

2. Unit Conversions

SSC exams frequently test your attention to detail by offering speed in kilometer per hour (km/h) and asking for time in seconds or distance in meters. Conversions must be instantaneous:

• To convert speed from km/h to m/s, multiply by \( \frac{5}{18} \).
  Formula: \( S_{ \text{m/s}} = S_{ \text{km/h}} \times \frac{5}{18} \)
• To convert speed from m/s to km/h, multiply by \( \frac{18}{5} \).
  Formula: \( S_{ \text{km/h}} = S_{ \text{m/s}} \times \frac{18}{5} \)

Proportion Techniques and Shortcuts

Using ratio and proportion methods allows you to solve STD problems mentally without complex algebraic calculations.

Case 1: Distance is Constant

When distance is fixed, Speed is inversely proportional to Time:

\( S eq 0 \text{ and } D = \text{Constant} ag{1} \)

\( S \text{ is proportional to } \frac{1}{T} ag{2} \)

This means if the ratio of speeds of two objects is \( a : b \), then the ratio of time taken by them to cover the same distance will be \( b : a \).

Case 2: Time is Constant

When time is fixed, Distance is directly proportional to Speed:

\( D \text{ is proportional to } S ag{3} \)

If the ratio of speeds of two objects is \( a : b \), the ratio of distance covered in equal time will also be \( a : b \).

Case 3: Speed is Constant

When speed is kept constant, Distance is directly proportional to Time:

\( D \text{ is proportional to } T ag{4} \)

Key Sub-Topics Explained

1. Average Speed

Average Speed is not the arithmetic mean of different speeds; it is defined as total distance divided by total time.

\( \text{Average Speed} = \frac{ \text{Total Distance Covered}}{ \text{Total Time Taken}} \)

• Special Case A: Equal Distances
  If a person travels a distance at speed \( x \) km/h and returns the same distance at speed \( y \) km/h, the average speed for the entire journey is:
  \( \text{Average Speed} = \frac{2xy}{x + y} \)
• Special Case B: Three Equal Distances
  If a body travels three equal distances at speeds \( x \), \( y \), and \( z \) respectively, then:
  \( \text{Average Speed} = \frac{3xyz}{xy + yz + zx} \)

2. Relative Speed

Relative speed is the effective speed of one body with respect to another moving body.

• Objects moving in Opposite Directions: Add their individual speeds.
  \( \text{Relative Speed} = S_1 + S_2 \)
• Objects moving in the Same Direction: Subtract the smaller speed from the larger speed.
  \( \text{Relative Speed} = S_1 - S_2 \) (where \( S_1 > S_2 \))

3. Problems on Trains

Train questions are an \textension of relative speed and length considerations:

• Train crossing a stationary point object (pole, man, tree):
  Total Distance = Length of the Train (\( L_T \)).
• Train crossing a platform, bridge, or tunnel (length \( L_P \)):
  Total Distance = \( L_T + L_P \).
• Two trains of lengths \( L_1 \) and \( L_2 \) crossing each other:
  Total Distance = \( L_1 + L_2 \). Speed used is the Relative Speed.

4. Boats and Streams

Let the speed of the boat in still water be \( u \) km/h and speed of the stream/river current be \( v \) km/h.

• Downstream Speed (\( D_s \)): Moving along the flow of current.
  \( D_s = u + v \)
• Upstream Speed (\( U_s \)): Moving against the flow of current.
  \( U_s = u - v \)
• Speed of Boat in Still Water (\( u \)):
  \( u = \frac{D_s + U_s}{2} \)
• Speed of Current (\( v \)):
  \( v = \frac{D_s - U_s}{2} \)

Solved Examples for Practice

Example 1: Basic Speed Conversion & Time Calculation

Question: A train 250 meters long is running at a speed of 72 km/h. How much time will it take to pass a telegraph post?

Solution:

Step 1: Convert speed from km/h to m/s.
\( \text{Speed} = 72 \times \frac{5}{18} = 20 \text{ m/s} \)

Step 2: Calculate time to cross the post.
Distance to cover = Length of the train = 250 m.
\( \text{Time} = \frac{ \text{Distance}}{ \text{Speed}} = \frac{250}{20} = 12.5 \text{ seconds} \)

Answer: 12.5 seconds

Example 2: Average Speed Shortcut

Question: A car travels from City A to City B at a speed of 40 km/h and returns to City A at a speed of 60 km/h. Find the average speed for the whole journey.

Solution:

Since the distance between A and B is constant, apply the harmonic mean formula:
\( \text{Average Speed} = \frac{2xy}{x + y} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h} \)

Answer: 48 km/h

Example 3: Ratio Method for Early/Late Problems

Question: Walking at \( \frac{4}{5} \) of his usual speed, a man reaches his office 15 minutes late. Find his usual time to reach the office.

Solution:

Step 1: Determine speed ratio.
\( \text{Speed Ratio (Usual : New)} = 5 : 4 \)

Step 2: Since distance is constant, Time ratio is inverse of Speed ratio.
\( \text{Time Ratio (Usual : New)} = 4 : 5 \)

Step 3: Calculate difference in time units.
Difference = \( 5 - 4 = 1 \text{ unit} \).
Given 1 unit = 15 minutes.
Usual Time = 4 units = \( 4 \times 15 = 60 \text{ minutes} = 1 \text{ hour} \).

Answer: 60 minutes (1 hour)

Example 4: Relative Speed & Train Crossing

Question: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. Find the ratio of their speeds.

Solution:

Let speed of Train 1 be \( S_1 \) and speed of Train 2 be \( S_2 \).
Length of Train 1 = \( 27 S_1 \)
Length of Train 2 = \( 17 S_2 \)

When crossing each other in opposite directions:
Total Distance = \( 27 S_1 + 17 S_2 \)
Relative Speed = \( S_1 + S_2 \)
Time taken = \( \frac{27 S_1 + 17 S_2}{S_1 + S_2} = 23 \)

Cross multiply:
\( 27 S_1 + 17 S_2 = 23 S_1 + 23 S_2 \)
\( 4 S_1 = 6 S_2 ightarrow \frac{S_1}{S_2} = \frac{6}{4} = \frac{3}{2} \)

Answer: 3 : 2

Example 5: Boats and Streams

Question: A boat takes 4 hours to travel 24 km upstream and 36 km downstream. It takes 5 hours to travel 36 km upstream and 24 km downstream. Find the speed of the stream.

Solution:

Let upstream speed be \( U \) and downstream speed be \( D \).

Equation 1: \( \frac{24}{U} + \frac{36}{D} = 4 \)
Equation 2: \( \frac{36}{U} + \frac{24}{D} = 5 \)

Let \( \frac{1}{U} = x \) and \( \frac{1}{D} = y \):
\( 24x + 36y = 4 ightarrow 6x + 9y = 1 \)
\( 36x + 24y = 5 \)

Solving these equations simultaneously yields:
\( x = \frac{1}{12} ightarrow U = 12 \text{ km/h} \)
\( y = \frac{1}{18} ightarrow D = 18 \text{ km/h} \)

Speed of stream \( v = \frac{D - U}{2} = \frac{18 - 12}{2} = 3 \text{ km/h} \).

Answer: 3 km/h

Practice Questions with Solutions

Q1. A thief is spotted by a policeman from a distance of 200 meters. The thief starts running and the policeman chases him. The thief and policeman run at the rate of 10 km/h and 11 km/h respectively. What is the distance between them after 6 minutes?

Solution:
Relative speed in same direction = \( 11 - 10 = 1 \text{ km/h} \).
Convert relative speed to m/min:
\( 1 \text{ km/h} = \frac{1000 \text{ meters}}{60 \text{ minutes}} = \frac{50}{3} \text{ m/min} \).
Distance covered in 6 minutes = \( \text{Speed} \times \text{Time} = \frac{50}{3} \times 6 = 100 \text{ meters} \).
Initial distance = 200 meters.
Remaining distance after 6 minutes = \( 200 - 100 = 100 \text{ meters} \).

Q2. A man travels 600 km by train at 80 km/h, 800 km by ship at 40 km/h, 500 km by aeroplane at 400 km/h and 100 km by car at 50 km/h. What is the average speed for the entire distance?

Solution:
Total Distance = \( 600 + 800 + 500 + 100 = 2000 \text{ km} \).
Total Time = \( \frac{600}{80} + \frac{800}{40} + \frac{500}{400} + \frac{100}{50} \)
Total Time = \( 7.5 + 20 + 1.25 + 2 = 30.75 \text{ hours} = \frac{123}{4} \text{ hours} \).
Average Speed = \( \frac{2000}{\frac{123}{4}} = \frac{8000}{123} \text{ km/h} \text{ (approx. } 65.04 \text{ km/h)} \).

Q3. A train overtakes two persons walking at 2 km/h and 4 km/h in the same direction in 9 seconds and 10 seconds respectively. Find the length of the train.

Solution:
Let length of train be \( L \) meters and speed be \( S \) km/h.
Speed 1 in m/s = \( (S - 2) \times \frac{5}{18} \)
Speed 2 in m/s = \( (S - 4) \times \frac{5}{18} \)
Distance (Length of train) is same:
\( L = (S - 2) \times \frac{5}{18} \times 9 \)
\( L = (S - 4) \times \frac{5}{18} \times 10 \)
Equating both equations:
\( 9(S - 2) = 10(S - 4) \)
\( 9S - 18 = 10S - 40 ightarrow S = 22 \text{ km/h} \).
Length \( L = (22 - 2) \times \frac{5}{18} \times 9 = 20 \times \frac{5}{2} = 50 \text{ meters} \).

Tips to Tackle STD Questions in SSC Exams

• Always check units first: Ensure distance is in meters if speed is in m/s and time in seconds. Convert before applying formulas.
• Master Ratio Methods: Avoid using variables like \( x \) for standard speed changes. Use inverse proportional ratios for time and speed.
• Memorize Pythogorean Triplets & Metric Multiples: For rapid calculation, know multiples of 18 (for km/h) and 5 (for m/s).
• Mock Practice: Practice 15-20 questions daily from Tier-1 and Tier-2 previous year question (PYQ) papers.