SSC CGL Exam > SSC CGL Notes > Famous Books Course for Competitive Exams > RS Aggarwal Summary: Time and Distance
RS Aggarwal Summary: Time and Distance
Introduction
Time and Distance is one of the most frequently tested topics in competitive examinations. Questions from this chapter appear consistently across various recruitment tests including banking, railways, SSC, campus placements, and management entrance exams. The chapter forms the foundation for understanding motion, speed calculations, and relative velocity concepts. Mastering these formulas and problem-solving techniques is crucial as they not only help in direct questions but also in solving problems related to trains, boats, races, and circular motion. The practical applications and logical approach required make this topic both scoring and essential for exam success.
Important Formulas and Concepts
1. Basic Speed-Distance-Time Relationship
The fundamental relationship between speed, distance, and time:
Formulas:
\[\text{Speed} = \frac{\text{Distance}}{\text{Time}}\] \[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\] \[\text{Distance} = \text{Speed} \times \text{Time}\]
\[\text{Speed} = \frac{\text{Distance}}{\text{Time}}\] \[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\] \[\text{Distance} = \text{Speed} \times \text{Time}\]
Example: A train travels 82.6 km/hr. How many metres will it travel in 15 minutes?
Solution:
Distance travelled in 1 min = \(\frac{82.6}{60}\) km
Distance travelled in 15 min = \(\frac{82.6 \times 15}{60}\) km = 20.65 km = 20,650 m
Distance travelled in 1 min = \(\frac{82.6}{60}\) km
Distance travelled in 15 min = \(\frac{82.6 \times 15}{60}\) km = 20.65 km = 20,650 m
2. Unit Conversion
Conversion between km/hr and m/sec:
km/hr to m/sec: Multiply by \(\frac{5}{18}\)
\[x \text{ km/hr} = x \times \frac{5}{18} \text{ m/sec}\]
m/sec to km/hr: Multiply by \(\frac{18}{5}\)
\[x \text{ m/sec} = x \times \frac{18}{5} \text{ km/hr}\]
\[x \text{ km/hr} = x \times \frac{5}{18} \text{ m/sec}\]
m/sec to km/hr: Multiply by \(\frac{18}{5}\)
\[x \text{ m/sec} = x \times \frac{18}{5} \text{ km/hr}\]
Example: How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?
Solution:
Aditya's speed = 20 km/hr = \(20 \times \frac{5}{18} = \frac{50}{9}\) m/sec
Time taken = \(400 \div \frac{50}{9} = 400 \times \frac{9}{50} = 72\) sec = 1 min 12 sec
Aditya's speed = 20 km/hr = \(20 \times \frac{5}{18} = \frac{50}{9}\) m/sec
Time taken = \(400 \div \frac{50}{9} = 400 \times \frac{9}{50} = 72\) sec = 1 min 12 sec
3. Ratio of Speeds and Times
Formula: If the ratio of speeds of A and B is a : b, then the ratio of times taken by them to cover the same distance is \(\frac{1}{a} : \frac{1}{b}\) or b : a
Note: Speed and time are inversely proportional for the same distance
Example: A man walked at a speed of 4 km/hr from point A to B and came back from point B to A at the speed of 6 km/hr. What would be the ratio of the time taken by the man in walking from point A to B to that from point B to A?
Solution:
Ratio of speeds = 4 : 6 = 2 : 3
Ratio of times taken = \(\frac{1}{2} : \frac{1}{3}\) = 3 : 2
Ratio of speeds = 4 : 6 = 2 : 3
Ratio of times taken = \(\frac{1}{2} : \frac{1}{3}\) = 3 : 2
4. Average Speed for Equal Distances
Formula: When a man covers a certain distance at x km/hr and an equal distance at y km/hr, the average speed during the whole journey is:
\[\text{Average Speed} = \frac{2xy}{x + y} \text{ km/hr}\]
\[\text{Average Speed} = \frac{2xy}{x + y} \text{ km/hr}\]
Example: A man travelled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.
Solution:
Average speed = \(\frac{2 \times 25 \times 4}{25 + 4} = \frac{200}{29}\) km/hr
Distance in 5 hrs 48 min = \(\frac{200}{29} \times \frac{29}{5} = 40\) km
Distance of post-office from village = \(\frac{40}{2} = 20\) km
Average speed = \(\frac{2 \times 25 \times 4}{25 + 4} = \frac{200}{29}\) km/hr
Distance in 5 hrs 48 min = \(\frac{200}{29} \times \frac{29}{5} = 40\) km
Distance of post-office from village = \(\frac{40}{2} = 20\) km
5. Relative Speed
When moving in the same direction:
If two persons are moving in the same direction at u m/s and v m/s (where u > v), then:
\[\text{Relative Speed} = (u - v) \text{ m/s}\]
\[\text{Relative Speed} = (u - v) \text{ m/s}\]
When moving in opposite directions:
If two persons are moving in opposite directions at u m/s and v m/s, then:
\[\text{Relative Speed} = (u + v) \text{ m/s}\]
\[\text{Relative Speed} = (u + v) \text{ m/s}\]
Example: A carriage driving in a fog passed a man who was walking at the rate of 3 kmph in the same direction. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 m. What was the speed of the carriage?
Solution:
Let speed of carriage = x kmph
Relative speed = (x - 3) kmph
Distance in 4 min \(\left(\frac{1}{15}\right)\) hr = 100 m = \(\frac{1}{10}\) km
\((x - 3) \times \frac{1}{15} = \frac{1}{10}\)
\(x - 3 = \frac{3}{2}\)
\(x = 4.5\) kmph
Let speed of carriage = x kmph
Relative speed = (x - 3) kmph
Distance in 4 min \(\left(\frac{1}{15}\right)\) hr = 100 m = \(\frac{1}{10}\) km
\((x - 3) \times \frac{1}{15} = \frac{1}{10}\)
\(x - 3 = \frac{3}{2}\)
\(x = 4.5\) kmph
6. Meeting Point Problems
Formula: If two persons A and B start at the same time in opposite directions from two points and after passing each other they complete the journeys in 'a' and 'b' hours respectively, then:
\[\text{A's speed} : \text{B's speed} = \sqrt{b} : \sqrt{a}\]
\[\text{A's speed} : \text{B's speed} = \sqrt{b} : \sqrt{a}\]
Example: Two boys A and B start at the same time to ride from Delhi to Meerut, 60 km away. A travels 4 km an hour slower than B. B reaches Meerut and at once turns back meeting A 12 km from Meerut. Find A's speed.
Solution:
Let A's speed = x km/hr, B's speed = (x + 4) km/hr
B covers 72 km (60 + 12) while A covers 48 km (60 - 12)
Time is same: \(\frac{72}{x + 4} = \frac{48}{x}\)
\(72x = 48(x + 4)\)
\(72x = 48x + 192\)
\(24x = 192\)
\(x = 8\) km/hr
Let A's speed = x km/hr, B's speed = (x + 4) km/hr
B covers 72 km (60 + 12) while A covers 48 km (60 - 12)
Time is same: \(\frac{72}{x + 4} = \frac{48}{x}\)
\(72x = 48(x + 4)\)
\(72x = 48x + 192\)
\(24x = 192\)
\(x = 8\) km/hr
7. Speed Variation Problems
When speed changes, the time also changes proportionally:
If speed becomes \(\frac{m}{n}\) of original, time becomes \(\frac{n}{m}\) of original
Example: By walking at 3/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. Find the usual time taken by him to reach his office.
Solution:
New speed = \(\frac{3}{4}\) of usual speed
New time = \(\frac{4}{3}\) of usual time
\(\frac{4}{3} \times \text{usual time} - \text{usual time} = 20\) min
\(\frac{1}{3} \times \text{usual time} = 20\) min
Usual time = 60 min = 1 hr
New speed = \(\frac{3}{4}\) of usual speed
New time = \(\frac{4}{3}\) of usual time
\(\frac{4}{3} \times \text{usual time} - \text{usual time} = 20\) min
\(\frac{1}{3} \times \text{usual time} = 20\) min
Usual time = 60 min = 1 hr
8. Time Difference Problems
Example: A person reaches his destination 40 minutes late if his speed is 3 km/hr, and reaches 30 minutes before time if his speed is 4 km/hr. Find the distance of his destination from his starting point.
Solution:
Let distance = x km
Difference in times = 70 min = \(\frac{7}{6}\) hr
\(\frac{x}{3} - \frac{x}{4} = \frac{7}{6}\)
\(\frac{4x - 3x}{12} = \frac{7}{6}\)
\(\frac{x}{12} = \frac{7}{6}\)
\(x = 14\) km
Let distance = x km
Difference in times = 70 min = \(\frac{7}{6}\) hr
\(\frac{x}{3} - \frac{x}{4} = \frac{7}{6}\)
\(\frac{4x - 3x}{12} = \frac{7}{6}\)
\(\frac{x}{12} = \frac{7}{6}\)
\(x = 14\) km
9. Stoppages and Effective Speed
Example: Excluding the stoppages, the speed of a bus is 64 km/hr and including the stoppages, the speed of the bus is 48 km/hr. For how many minutes does the bus stop per hour?
Solution:
Due to stoppage, bus covers (64 - 48) = 16 km less per hour
Time to cover 16 km at 64 km/hr = \(\frac{16}{64} \times 60 = 15\) min
Stoppage time = 15 min per hour
Due to stoppage, bus covers (64 - 48) = 16 km less per hour
Time to cover 16 km at 64 km/hr = \(\frac{16}{64} \times 60 = 15\) min
Stoppage time = 15 min per hour
10. Accidents and Speed Change
Example: A train after travelling 150 km meets with an accident and then proceeds at 3/5 of its former speed and arrives at its destination 8 hours late. Had the accident occurred 360 km further, it would have reached the destination 4 hours late. What is the total distance travelled by the train?
Solution:
Let original speed = x km/hr
\(\frac{360}{\frac{3x}{5}} - \frac{360}{x} = 4\)
\(\frac{360 \times 5}{3x} - \frac{360}{x} = 4\)
\(\frac{600}{x} - \frac{360}{x} = 4\)
\(\frac{240}{x} = 4\)
\(x = 60\) km/hr
Let total distance = y km
\(\frac{150}{60} + \frac{y-150}{\frac{3 \times 60}{5}} = \frac{y}{60} + 8\)
\(2.5 + \frac{y-150}{36} = \frac{y}{60} + 8\)
Solving: \(y = 870\) km
Let original speed = x km/hr
\(\frac{360}{\frac{3x}{5}} - \frac{360}{x} = 4\)
\(\frac{360 \times 5}{3x} - \frac{360}{x} = 4\)
\(\frac{600}{x} - \frac{360}{x} = 4\)
\(\frac{240}{x} = 4\)
\(x = 60\) km/hr
Let total distance = y km
\(\frac{150}{60} + \frac{y-150}{\frac{3 \times 60}{5}} = \frac{y}{60} + 8\)
\(2.5 + \frac{y-150}{36} = \frac{y}{60} + 8\)
Solving: \(y = 870\) km
10 Complete Solved Examples
Example 1: Basic Speed Calculation
Problem: A cyclist covers a distance of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?
Solution:
Speed = \(\frac{750}{150}\) m/sec = 5 m/sec (since 2 min 30 sec = 150 sec)
Speed = \(5 \times \frac{18}{5} = 18\) km/hr
Speed = \(\frac{750}{150}\) m/sec = 5 m/sec (since 2 min 30 sec = 150 sec)
Speed = \(5 \times \frac{18}{5} = 18\) km/hr
Example 2: Comparing Speeds
Problem: A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.
Solution:
Let distance in 1 leap of dog = x, hare = y
\(3x = 4y \rightarrow x = \frac{4y}{3}\)
In same time: Dog covers 4x, Hare covers 5y
Ratio = \(4 \times \frac{4y}{3} : 5y = \frac{16y}{3} : 5y = 16 : 15\)
Let distance in 1 leap of dog = x, hare = y
\(3x = 4y \rightarrow x = \frac{4y}{3}\)
In same time: Dog covers 4x, Hare covers 5y
Ratio = \(4 \times \frac{4y}{3} : 5y = \frac{16y}{3} : 5y = 16 : 15\)
Example 3: Multi-Mode Journey
Problem: A trip to a destination is made as: 900 km by train at 60 km/hr, 3000 km by plane at 500 km/hr, 400 km by boat at 25 km/hr, 15 km by taxi at 45 km/hr. What is the average speed for the entire journey?
Solution:
Total distance = 4315 km
Total time = \(\frac{900}{60} + \frac{3000}{500} + \frac{400}{25} + \frac{15}{45} = 15 + 6 + 16 + \frac{1}{3} = \frac{112}{3}\) hr
Average speed = \(4315 \times \frac{3}{112} = 115.54\) km/hr ≈ 115 km/hr
Total distance = 4315 km
Total time = \(\frac{900}{60} + \frac{3000}{500} + \frac{400}{25} + \frac{15}{45} = 15 + 6 + 16 + \frac{1}{3} = \frac{112}{3}\) hr
Average speed = \(4315 \times \frac{3}{112} = 115.54\) km/hr ≈ 115 km/hr
Example 4: Fractional Distance Problems
Problem: One-third of a certain journey was covered at 20 km/hr, one-fourth at 30 km/hr and the rest at 50 km/hr. Find the average speed for the whole journey.
Solution:
Let total distance = x km
Distance at 20 km/hr = \(\frac{x}{3}\), at 30 km/hr = \(\frac{x}{4}\), at 50 km/hr = \(\frac{5x}{12}\)
Total time = \(\frac{x/3}{20} + \frac{x/4}{30} + \frac{5x/12}{50} = \frac{x}{60} + \frac{x}{120} + \frac{x}{120} = \frac{x}{30}\) hrs
Average speed = \(\frac{x}{x/30} = 30\) km/hr
Let total distance = x km
Distance at 20 km/hr = \(\frac{x}{3}\), at 30 km/hr = \(\frac{x}{4}\), at 50 km/hr = \(\frac{5x}{12}\)
Total time = \(\frac{x/3}{20} + \frac{x/4}{30} + \frac{5x/12}{50} = \frac{x}{60} + \frac{x}{120} + \frac{x}{120} = \frac{x}{30}\) hrs
Average speed = \(\frac{x}{x/30} = 30\) km/hr
Example 5: Square Path Problem
Problem: An aeroplane flies along the four sides of a square at speeds of 100, 200, 300 and 400 km/hr. Find the average speed around the field.
Solution:
Let each side = x km
Total distance = 4x km
Total time = \(\frac{x}{100} + \frac{x}{200} + \frac{x}{300} + \frac{x}{400} = \frac{25x}{1200}\) hrs
Average speed = \(\frac{4x}{25x/1200} = \frac{4 \times 1200}{25} = 192\) km/hr
Let each side = x km
Total distance = 4x km
Total time = \(\frac{x}{100} + \frac{x}{200} + \frac{x}{300} + \frac{x}{400} = \frac{25x}{1200}\) hrs
Average speed = \(\frac{4x}{25x/1200} = \frac{4 \times 1200}{25} = 192\) km/hr
Example 6: Train Speed Comparison
Problem: A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speeds of the two trains.
Solution:
Let speed of fast train = x km/hr
Speed of slow train = (x - 10) km/hr
\(\frac{600}{x-10} - \frac{600}{x} = 3\)
\(600x - 600(x-10) = 3x(x-10)\)
\(6000 = 3x^2 - 30x\)
\(x^2 - 10x - 2000 = 0\)
\((x - 50)(x + 40) = 0\)
\(x = 50\) km/hr (fast), 40 km/hr (slow)
Let speed of fast train = x km/hr
Speed of slow train = (x - 10) km/hr
\(\frac{600}{x-10} - \frac{600}{x} = 3\)
\(600x - 600(x-10) = 3x(x-10)\)
\(6000 = 3x^2 - 30x\)
\(x^2 - 10x - 2000 = 0\)
\((x - 50)(x + 40) = 0\)
\(x = 50\) km/hr (fast), 40 km/hr (slow)
Example 7: Late Start Problem
Problem: An aeroplane started 30 minutes later than scheduled from a place 1500 km away. To reach at scheduled time, pilot increased speed by 250 km/hr. What was the speed during the journey?
Solution:
Let original speed = x km/hr
\(\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}\)
\(1500(x+250) - 1500x = \frac{x(x+250)}{2}\)
\(375000 = x^2 + 250x\)
\(x^2 + 250x - 750000 = 0\)
\((x - 750)(x + 1000) = 0\)
\(x = 750\) km/hr (original)
Speed during journey = 1000 km/hr
Let original speed = x km/hr
\(\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}\)
\(1500(x+250) - 1500x = \frac{x(x+250)}{2}\)
\(375000 = x^2 + 250x\)
\(x^2 + 250x - 750000 = 0\)
\((x - 750)(x + 1000) = 0\)
\(x = 750\) km/hr (original)
Speed during journey = 1000 km/hr
Example 8: Two-Train Meeting
Problem: A and B are 390 km apart. A train starts from A at 10 a.m. at 65 kmph towards B. Another starts from B at 11 a.m. at 35 kmph towards A. At what time do they meet?
Solution:
Let they meet x hours after 10 a.m.
Distance by first train + Distance by second train = 390
\(65x + 35(x-1) = 390\)
\(100x = 425\)
\(x = 4.25\) hours = 4 hrs 15 min
Meeting time = 2:15 p.m.
Let they meet x hours after 10 a.m.
Distance by first train + Distance by second train = 390
\(65x + 35(x-1) = 390\)
\(100x = 425\)
\(x = 4.25\) hours = 4 hrs 15 min
Meeting time = 2:15 p.m.
Example 9: Opposite Direction Travel
Problem: Two places A and B are 80 km apart. A car starts from A and another from B at the same time. Moving in same direction they meet in 8 hours. Moving in opposite directions they meet in 1 hour 20 minutes. Determine the speeds.
Solution:
Let speeds be x and y kmph
Same direction: \(x - y = \frac{80}{8} = 10\) ... (i)
Opposite direction: \(x + y = \frac{80}{4/3} = 60\) ... (ii)
Adding: \(2x = 70 \rightarrow x = 35\) kmph
From (i): \(y = 25\) kmph
Let speeds be x and y kmph
Same direction: \(x - y = \frac{80}{8} = 10\) ... (i)
Opposite direction: \(x + y = \frac{80}{4/3} = 60\) ... (ii)
Adding: \(2x = 70 \rightarrow x = 35\) kmph
From (i): \(y = 25\) kmph
Example 10: Pickup Time Problem
Problem: Sneha is picked up by her father at 4 p.m. daily. One day college ended an hour early. She started walking. Her father meets her on the way and they reach home 15 minutes early. Find the ratio of father's driving speed to Sneha's walking speed.
Solution:
15 minutes saved means father drives meeting point to college and back in 15 min
One way = 7.5 min
Father reaches college at 4 p.m., so they meet at 3:52:30 p.m.
Sneha walked for 52.5 min covering distance father covers in 7.5 min
Speed ratio = 52.5 : 7.5 = 7 : 1
15 minutes saved means father drives meeting point to college and back in 15 min
One way = 7.5 min
Father reaches college at 4 p.m., so they meet at 3:52:30 p.m.
Sneha walked for 52.5 min covering distance father covers in 7.5 min
Speed ratio = 52.5 : 7.5 = 7 : 1
Key Points to Remember
- Always convert units properly before solving
- Speed and time are inversely proportional for constant distance
- For average speed with equal distances, use harmonic mean formula
- Relative speed concept is crucial for meeting/overtaking problems
- Draw diagrams for complex problems involving multiple people/vehicles
- Check if question asks for one-way or round-trip distance
- Be careful with time units (hours, minutes, seconds)
- Practice fractional speed problems thoroughly
- Understand the difference between speed excluding and including stoppages
- Master the relationship between speed change and time change
The document RS Aggarwal Summary: Time and Distance is a part of the SSC CGL Course Famous Books Course for Competitive Exams.
All you need of SSC CGL at this link: SSC CGL
FAQs on RS Aggarwal Summary: Time and Distance
| 1. What are the key concepts involved in the topic of time and distance? |  |
Ans. The key concepts in time and distance include the relationship between speed, distance, and time, often expressed through the formula: Distance = Speed × Time. Understanding concepts such as relative speed, average speed, and the effect of time on distance is also crucial for solving related problems.
| 2. Can you explain the important formulas related to time and distance? | |
Ans. Important formulas related to time and distance include: 1. Distance = Speed × Time 2. Speed = Distance / Time 3. Time = Distance / Speed Additionally, for relative speed, when two objects move in the same direction, Relative Speed = Speed of A - Speed of B; and for opposite directions, Relative Speed = Speed of A + Speed of B.
| 3. How do you calculate average speed in a journey with different speeds? | |
Ans. To calculate average speed for a journey with different speeds, use the formula: Average Speed = Total Distance / Total Time. If the distance for each segment is the same but speeds differ, the formula becomes Average Speed = 2 × (Speed₁ × Speed₂) / (Speed₁ + Speed₂) for two segments.
| 4. What are some common types of problems encountered in time and distance questions? | |
Ans. Common types of problems include calculating the time taken for a journey at a certain speed, determining the distance travelled given a speed and time, solving relative speed problems involving two moving objects, and problems involving trains or vehicles passing each other.
| 5. What are the key points to remember when solving time and distance problems in exams? | |
Ans. Key points to remember include: carefully noting the units of speed, distance, and time; understanding the direction of movement for relative speed calculations; applying the correct formulas based on the problem type; and practising various types of problems to become familiar with different scenarios.
About this Document
4.68/5 Rating
Apr 19, 2026 Last updated
Related Exams
Document Description: RS Aggarwal Summary: Time and Distance for SSC CGL 2026 is part of Famous Books Course for Competitive Exams preparation. The notes and questions for RS Aggarwal Summary: Time and Distance have been prepared according to the SSC CGL exam syllabus. Information about RS Aggarwal Summary: Time and Distance covers topics like and RS Aggarwal Summary: Time and Distance Example, for SSC CGL 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Summary: Time and Distance.
Introduction of RS Aggarwal Summary: Time and Distance in English is available as part of our Famous Books Course for Competitive Exams for SSC CGL & RS Aggarwal Summary: Time and Distance in Hindi for Famous Books Course for Competitive Exams course. Download more important topics related with notes, lectures and mock test series for SSC CGL Exam by signing up for free. SSC CGL: RS Aggarwal Summary: Time and Distance
Description
RS Aggarwal Summary: Time & Distance of Famous Books Course - covers all the important topics from the book, explained in easy language for better understanding. Free PDF available for download.
Information about RS Aggarwal Summary: Time and Distance
In this doc you can find the meaning of RS Aggarwal Summary: Time and Distance defined & explained in the simplest way possible. Besides explaining types of RS Aggarwal Summary: Time and Distance theory, EduRev gives you an ample number of questions to practice RS Aggarwal Summary: Time and Distance tests, examples and also practice SSC CGL tests
Related Searches
RS Aggarwal Summary: Time and Distance, shortcuts and tricks, RS Aggarwal Summary: Time and Distance, study material, Important questions, mock tests for examination, pdf , Previous Year Questions with Solutions, video lectures, practice quizzes, ppt, Free, Semester Notes, Extra Questions, past year papers, Objective type Questions, Sample Paper, Viva Questions, Exam, RS Aggarwal Summary: Time and Distance, MCQs, Summary;