Railway Exams Arithmetic Maths Speed, Time & Distance

Speed, Time and Distance is one of the most important topics in the Arithmetic section of Railway exams and other competitive examinations. Questions from this chapter are frequently asked and test a candidate's ability to apply basic mathematical concepts in real-life scenarios involving travel, motion, and time calculations. A strong understanding of formulas, unit conversions, average speed concepts, and multi-stage journey problems can help candidates solve questions quickly and accurately.

This topic covers a wide range of concepts, including the relationship among speed, time, and distance; average speed calculations; late- and early-arrival problems; variable-speed journeys; and time-saving shortcuts. Regular practice of these concepts enables aspirants to improve their calculation speed, accuracy, and overall performance in Railway recruitment examinations.

Speed, Time, and Distance

The Speed, Time, and Distance chapter is a very important chapter for all competitive exams like SSC, Railway, and others, with questions consistently appearing. This section provides a clear understanding of core concepts and practical problem-solving methods through various examples, essential for mastering arithmetic for competitive examinations.

Problem 1: Square Field Circuit

Question: If a boy runs at a speed of 6 km/h and completes one round of a square field with a side of 38 meters, how many seconds will he take?

Solution Steps:

1. Given: Boy's speed = 6 km/h; Side of square field = 38 meters.

2. Calculate Distance (Perimeter): To complete one round, the boy travels the perimeter. Perimeter = 4 × side = 4 × 38 meters = 152 meters.

3. Convert Speed to m/s: Convert 6 km/h to m/s using the factor 5/18. Speed = 6 × (5/18) m/s = 5/3 m/s.

4. Calculate Time: Time = Distance / Speed = 152 meters / (5/3 m/s) = (152 × 3) / 5 seconds = 456 / 5 seconds = 91.2 seconds.

Problem 2: Average Speed (Constant Distance)

Question: Harit Teja walked to his school at a speed of 4 km/h and returned on a scooter at 20 km/h. What is his average speed during the two-way journey? 

Concept: When the distance travelled is the same in both directions, specific formulas or methods apply for average speed.

Method 1: Formulaic Approach

1. Formula for Average Speed (constant distance): Average Speed = (2 × S1 × S2) / (S1 + S2), where S1 = speed in one direction, S2 = speed in the other.

2. Given: S1 = 4 km/h, S2 = 20 km/h.

3. Calculation: Average Speed = (2 × 4 × 20) / (4 + 20) = (8 × 20) / 24 = 160 / 24 = 20/3 km/h.

Method 2: LCM / Work-and-Time Approach

1. Given Speeds: 4 km/h and 20 km/h.

2. Assume Total Distance (LCM): Find the Least Common Multiple (LCM) of the speeds. LCM(4, 20) = 20 km (one-way distance).

3. Calculate Time for Each Leg: Time 1 = 20 km / 4 km/h = 5 hours; Time 2 = 20 km / 20 km/h = 1 hour.

4. Total Distance and Total Time: Total Distance = 2 × 20 km = 40 km; Total Time = 5 hours + 1 hour = 6 hours.

5. Calculate Average Speed: Average Speed = Total Distance / Total Time = 40 km / 6 hours = 20/3 km/h.

Problem 3: Average Speed (Constant Time)

Example: A person travels at 45 km/h for a certain time and then at 40 km/h for the same amount of time. What is the average speed?

 Concept: When the time taken for each segment of the journey is the same, the average speed is the arithmetic mean of the speeds.

Solution:

1. Given: Speed 1 (S1) = 45 km/h, Speed 2 (S2) = 40 km/h. Time taken for each speed is Same Time.

2. Formula for Average Speed (constant time): Average Speed = (S1 + S2) / 2.

3. Calculation: Average Speed = (45 + 40) / 2 = 85 / 2 = 42.5 km/h.

Problem 4: Average Speed Application

Question: A person goes from Kanpur to Lucknow at a speed of 72 km/h and returns by the same route at a speed of 90 km/h. What is the average speed for the entire journey?

Concept: This scenario involves constant distance (Kanpur to Lucknow and back)

Solution Steps (using LCM method):

1. Given Speeds: S1 = 72 km/h, S2 = 90 km/h.

2. Assume Total Distance (LCM): LCM(72, 90) = 360 km (one-way distance).

3. Calculate Time for Each Leg: Time 1 = 360 km / 72 km/h = 5 hours; Time 2 = 360 km / 90 km/h = 4 hours.

4. Total Distance and Total Time: Total Distance = 2 × 360 km = 720 km; Total Time = 5 hours + 4 hours = 9 hours.

5. Calculate Average Speed: Average Speed = Total Distance / Total Time = 720 km / 9 hours = 80 km/h.

Problem 5: Multi-Stage Journey (Calculating Remaining Speed)

Question: Arjun has to reach Jaipur, 602 km away, in 22 hours. His starting speed for the first 7 hours was 29 km/h, and for the next 189 km, his speed was 21 km/h. At what speed must he travel now to reach Jaipur within 22 hours?

Solution Steps:

1. Total Journey: Total Distance = 602 km, Total Time = 22 hours.

2. First Stage: Speed = 29 km/h, Time = 7 hours. Distance covered = 29 × 7 = 203 km.

3. Second Stage: Distance = 189 km, Speed = 21 km/h. Time taken = 189 / 21 = 9 hours.

4. Remaining Journey:

• Distance covered = 203 + 189 = 392 km.

• Time elapsed = 7 + 9 = 16 hours.

• Remaining Distance = 602 - 392 = 210 km.

• Remaining Time = 22 - 16 = 6 hours.

5. Required Speed: To cover 210 km in 6 hours, Arjun must travel at 210 / 6 = 35 km/h.

Problem 6: Multi-Stage Journey (Similar to Problem 5)

Question: Aryan has to reach Ahmedabad, 889 km away, in 17 hours. His starting speed for the first 5 hours was 29 km/h. For the next 96 km, his speed was 24 km/h. At what speed must he travel now to reach Ahmedabad within the decided time?

Solution Steps:

1. Total Journey: Total Distance = 889 km, Total Time = 17 hours.

2. First Stage: Speed = 29 km/h, Time = 5 hours. Distance covered = 29 × 5 = 145 km.

3. Second Stage: Distance = 96 km, Speed = 24 km/h. Time taken = 96 / 24 = 4 hours.

4. Remaining Journey:

• Distance covered = 145 + 96 = 241 km.

• Time elapsed = 5 + 4 = 9 hours.

• Remaining Distance = 889 - 241 = 648 km.

• Remaining Time = 17 - 9 = 8 hours.

5. Required Speed: To cover 648 km in 8 hours, Aryan must travel at 648 / 8 = 81 km/h.

Problem 7: Speed Adjustment and New Time Calculation

Question: A car covers a distance of 275 km in 5 hours. If its speed is reduced by 5 km/h, how much time will it take to cover 250 km?

Solution Steps:

1. Calculate Original Speed: Original Speed = 275 km / 5 hours = 55 km/h.

2. Calculate New Speed: New Speed = 55 km/h - 5 km/h = 50 km/h.

3. Calculate Time for New Journey: Time = 250 km / 50 km/h = 5 hours.

Problem 8: Average Speed (Repeat Application)

Question: A truck goes from city P to Q at a speed of 60 km/h and returns by the same route at a speed of 100 km/h. What is the average speed for the entire journey?

Concept: This is another instance where the distance is the same. 

Solution Steps (using LCM method):

1. Given Speeds: S1 = 60 km/h, S2 = 100 km/h.

2. Assume Total Distance (LCM): LCM(60, 100) = 300 km (one-way distance).

3. Calculate Time for Each Leg: Time 1 = 300 km / 60 km/h = 5 hours; Time 2 = 300 km / 100 km/h = 3 hours.

4. Total Distance and Total Time: Total Distance = 2 × 300 km = 600 km; Total Time = 5 hours + 3 hours = 8 hours.

5. Calculate Average Speed: Average Speed = Total Distance / Total Time = 600 km / 8 hours = 75 km/h.

Problem 9: Variable Speed Over Time

Question: A motor car starts with a speed of 60 km/h and increases its speed by 15 km/h after every 2 hours. How much time will it take to cover a distance of 360 km? 

Solution Steps:

1. Stage 1: Speed = 60 km/h, Time = 2 hours. Distance = 60 × 2 = 120 km.

2. Stage 2: Speed increases to 60 + 15 = 75 km/h, Time = 2 hours. Distance = 75 × 2 = 150 km.

• Cumulative Distance after Stage 2 = 120 + 150 = 270 km.

• Remaining Distance = 360 - 270 = 90 km.

3. Stage 3: Speed increases to 75 + 15 = 90 km/h.

• Time to cover remaining 90 km = 90 km / 90 km/h = 1 hour.

4. Total Time: Total Time = 2 hours (Stage 1) + 2 hours (Stage 2) + 1 hour (Stage 3) = 5 hours.

Problem 10: Finding Missing Distance with Average Speed

Question: A person travels 80 km in 3 hours. He travels for an additional 2 hours. If the average speed for the entire journey is 30 km/h, find the distance covered in the final 2 hours.

Solution Steps:

1. Given: First Distance (D1) = 80 km, First Time (T1) = 3 hours, Second Time (T2) = 2 hours, Average Speed = 30 km/h. Let D2 be the unknown distance.

2. Total Journey: Total Distance = D1 + D2 = 80 + D2; Total Time = T1 + T2 = 3 + 2 = 5 hours.

3. Apply Average Speed Formula: Average Speed = Total Distance / Total Time.

• 30 km/h = (80 + D2) / 5 hours.

4. Solve for D2: 30 × 5 = 80 + D2 => 150 = 80 + D2 => D2 = 150 - 80 = 70 km.

Problem 11: Late/Early Arrival (Calculating Usual Time, Distance, and Speed)

Question: A bike, running at 50 km/h, reaches its destination 10 minutes late. If it runs at 60 km/h, it is late by only 5 minutes. How many minutes should the bike take to travel the same route to reach on time (usual time)?

 Concept: The distance covered is constant. We equate (Speed × Time) for both cases to find the usual time. Let the usual time be 't' minutes.

Solution Steps:

1. Set Up Equations: Convert time to hours (divide by 60).

• Case 1: Distance = 50 km/h × (t + 10)/60 hours.

• Case 2: Distance = 60 km/h × (t + 5)/60 hours.

2. Equate Distances: 50(t + 10)/60 = 60(t + 5)/60.

• 50(t + 10) = 60(t + 5)

• 5(t + 10) = 6(t + 5)

• 5t + 50 = 6t + 30

• t = 20 minutes.

• The usual time is 20 minutes.

Follow-up Calculations:

• Actual Distance: Using t = 20 mins in Case 1: Distance = 50 × (20 + 10)/60 = 50 × 30/60 = 25 km.

• Actual Speed: Speed = Distance / Usual Time = 25 km / (20/60 hours) = 25 / (1/3) = 75 km/h.

Problem 12: Comparing Travel Times

Question: Ishaan and Meghna have to travel from Delhi to Kanpur. Ishaan is driving at 77 km/h, while Meghna is driving at 33 km/h. Find the time taken by Meghna to reach Kanpur if Ishaan takes 12 hours.

Concept: The distance is the same for both travelers (Delhi to Kanpur).

Solution Steps:

1. Calculate Distance to Kanpur: Ishaan's Speed = 77 km/h, Ishaan's Time = 12 hours.

• Distance = 77 × 12 = 924 km.

2. Calculate Meghna's Time: Meghna's Speed = 33 km/h, Distance = 924 km.

• Meghna's Time = 924 km / 33 km/h = 28 hours.

Problem 13: Two Scenarios with Two Travelers

Question: In covering a distance of 182 km, Salaj takes 4 hours more than Kevin. If Salaj doubles his speed, he takes 9 hours less than Kevin. What is Salaj's speed?

 Solution Steps (Algebraic Method):

1. Define Variables: Distance (D) = 182 km. Let Salaj's original speed be S_Salaj.

2. Formulate Equations (Time = Distance / Speed):

• Scenario 1: Salaj takes 4 hours more than Kevin.

• (182 / S_Salaj) - (182 / S_Kevin) = 4 (Equation 1)

• Scenario 2: Salaj doubles speed (2 * S_Salaj), takes 9 hours less than Kevin.

• (182 / S_Kevin) - (182 / (2 * S_Salaj)) = 9 (Equation 2)

3. Solve the System:

• From (1), 182 / S_Kevin = (182 / S_Salaj) - 4. Substitute into (2):

• ((182 / S_Salaj) - 4) - (182 / (2 * S_Salaj)) = 9

• (182 / S_Salaj) - (182 / (2 * S_Salaj)) = 13

• (364 - 182) / (2 * S_Salaj) = 13

• 182 / (2 * S_Salaj) = 13

• 182 = 26 * S_Salaj

• S_Salaj = 182 / 26 = 7 km/h.

Problem 14: Direct Distance, Time, Speed Calculation

Question: A bus completes a journey at a speed of 35 km/h in 20 hours. If the same distance needs to be covered in 7 hours, what should be its speed?

Solution Steps:

1. Calculate Total Distance: Distance = Speed × Time = 35 km/h × 20 hours = 700 km.

2. Calculate New Speed: New Required Time = 7 hours.

• New Speed = Total Distance / New Required Time = 700 km / 7 hours = 100 km/h.