Theory & Concepts
SSC CGL Time, Speed and Distance Questions, Formulas & Tricks
Prepare Time, Speed and Distance for SSC CGL with formulas, short tricks, solved examples, practice questions, PYQs, and free PDF notes for faster exam preparation.
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28 min readDifficulty: Intermediate
Speed, Distance, and Time (TSD) is a high-weightage topic in the Ssc cgl exam. Questions range from simple unit conversions to complex relative speed scenarios involving two moving trains or a police-thief chase.
This guide covers the core equations, the crucial rules for average speed, and the logic of relative speed—which is the key to solving crossing and overtaking problems without confusion.
Learning path
- Basic TSD relation
- KM/H to M/S conversion
- Average speed formulas
- Relative speed logic
- 10 standard solved problems
1. Fundamental formulas
The master equation
KM/H to M/S:
M/S to KM/H:
2. Average speed
If a person travels equal distances at speed and , the average speed for the whole journey is:
3. Relative speed logic
Relative speed is the speed of an object with respect to another moving object.
Opposite direction
Objects move towards or away from each other.
Same direction
One object chases or follows another.
4. Solved examples
Question 01Standard pattern
A person crosses a 600m long street in 5 minutes. What is his speed in km/hr?
7.2 km/hr
8.4 km/hr
3.6 km/hr
10 km/hr
Correct answer: a) 7.2 km/hr
Solution
Distance = \( 600 \text{ m} \).
Time = \( 5 \times 60 = 300 \text{ seconds} \).
Speed in m/s = \( 600 / 300 = 2 \text{ m/sec} \).
Speed in km/hr = \( 2 \times \frac{18}{5} = \frac{36}{5} = 7.2 \text{ km/hr} \).
Question 02Standard pattern
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
150 m
180 m
120 m
90 m
Correct answer: a) 150 m
Solution
Speed in m/s = \( 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/sec} \).
Time = \( 9 \text{ seconds} \).
Length (Distance) = \( \text{Speed} \times \text{Time} = \frac{50}{3} \times 9 \).
Result = \( 50 \times 3 = 150 \text{ metres} \).
Question 03Standard pattern
A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?
100 m
150 m
110 m
190 m
Correct answer: a) 100 m
Solution
Relative speed (Same direction) = \( 11 - 10 = 1 \text{ km/hr} \).
In m/min = \( \frac{1000 \text{ m}}{60 \text{ min}} = \frac{50}{3} \text{ m/min} \).
Distance covered in 6 min = \( \frac{50}{3} \times 6 = 100 \text{ m} \).
Remaining distance = \( 200 - 100 = 100 \text{ m} \).
Question 04Standard pattern
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
50 km/hr
54 km/hr
45 km/hr
60 km/hr
Correct answer: a) 50 km/hr
Solution
Let speed of train be \( x \) km/hr.
Relative speed = \( (x - 5) \) km/hr.
Relative speed in m/s = \( (x - 5) \times \frac{5}{18} \).
Equation: \( \frac{125}{10} = (x - 5) \times \frac{5}{18} \).
\( 12.5 = (x - 5) \times \frac{5}{18} \implies (x - 5) = \frac{12.5 \times 18}{5} = 45 \).
Result: \( x = 45 + 5 = 50 \text{ km/hr} \).
Question 05Standard pattern
An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1(2/3) hours, it must travel at a speed of:
720 kmph
600 kmph
360 kmph
480 kmph
Correct answer: a) 720 kmph
Solution
Distance = \( 240 \times 5 = 1200 \text{ km} \).
Target Time = \( 5/3 \text{ hours} \).
Required Speed = \( \frac{1200}{5/3} = \frac{3600}{5} = 720 \text{ km/hr} \).
Question 06Standard pattern
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
50 km
70 km
80 km
60 km
Correct answer: a) 50 km
Solution
Let actual time be \( T \) hours.
Distance diff = \( 14T - 10T = 20 \text{ km} \).
\( 4T = 20 \implies T = 5 \text{ hours} \).
Actual distance (at 10 km/hr) = \( 10 \times 5 = 50 \text{ km} \).
Question 07Standard pattern
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
120 kmph
100 kmph
110 kmph
130 kmph
Correct answer: a) 120 kmph
Solution
Let speed of car be \( v \) kmph. Speed of train = \( 1.5v \) kmph.
Time diff = \( \frac{75}{v} - \frac{75}{1.5v} = 12.5 \text{ min} = \frac{12.5}{60} \text{ hr} \).
\( \frac{75}{v} - \frac{50}{v} = \frac{25}{v} = \frac{1}{4.8} \).
Result: \( v = 25 \times 4.8 = 120 \text{ km/hr} \).
Question 08Standard pattern
Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?
11 : 9
11 : 10
10 : 9
5 : 4
Correct answer: a) 11 : 9
Solution
Distance covered by Train 1 = \( 110 \text{ km} \).
Distance covered by Train 2 = \( 200 - 110 = 90 \text{ km} \).
Since time is same, Ratio of speeds = Ratio of distances.
Ratio = \( 110 : 90 = 11 : 9 \).
Question 09Standard pattern
The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?
11 a.m.
10:30 a.m.
11:30 a.m.
12 noon
Correct answer: a) 11 a.m.
Solution
Distance covered by Train A in 1 hr (8 to 9 am) = \( 60 \text{ km} \).
Remaining distance at 9 am = \( 330 - 60 = 270 \text{ km} \).
Relative speed (Opposite direction) = \( 60 + 75 = 135 \text{ km/hr} \).
Meeting time = \( 270 / 135 = 2 \text{ hours} \).
Actual time = \( 9 \text{ a.m.} + 2 \text{ hours} = 11 \text{ a.m.} \).
Question 010Standard pattern
A train 240 m long passes a platform of equal length in 27 seconds. What is the speed of the train in km/hr?
64 km/hr
72 km/hr
60 km/hr
54 km/hr
Correct answer: a) 64 km/hr
Solution
Total distance to cover = \( 240 + 240 = 480 \text{ metres} \).
Time = \( 27 \text{ seconds} \).
Speed (m/s) = \( 480 / 27 = 160 / 9 \text{ m/sec} \).
Speed (km/hr) = \( \frac{160}{9} \times \frac{18}{5} = \frac{160 \times 2}{5} = 32 \times 2 = 64 \text{ km/hr} \).
5. Strategy errors to avoid
Error 01Unit neglect: Mixing km/hr and metres/second in the same equation without conversion.
Error 02Average speed trap: Summing speeds and dividing by 2 (e.g. (40+60)/2 = 50) when distances are equal (it should be 48).
Error 03Relative speed reversal: Subtracting speeds for opposite direction or adding them for same direction.
Error 04Length omission: Not adding the platform or bridge length to the train's length when calculating total distance to cross.
