Time, speed and distance for RRB NTPC: methods, patterns and practice

Speed questions feel natural in a railway exam, and they are a steady part of NTPC Maths. They also feed two other chapters: problems on trains and boats and streams are this chapter with one extra idea each. Get the basics solid here and those two become easy.

The good news is that almost everything comes from one relationship. The bad news is that most wrong answers come from careless units and from averaging speeds the wrong way. Both are fixable.

The one relationship

• Distance = Speed × Time
• Speed = Distance ÷ Time
• Time = Distance ÷ Speed

Before you calculate anything, make the units agree. If speed is in km/h, time must be in hours and distance in km. If speed is in m/s, use seconds and metres.

Converting km/h and m/s

Why 5/18: 1 km/h means 1,000 m in 3,600 s, which is 1,000/3,600 = 5/18 m/s. Going the other way, you multiply by the reciprocal, 18/5.

Speeds that are multiples of 18 km/h convert to whole numbers: 18 → 5, 36 → 10, 54 → 15, 72 → 20, 90 → 25, 108 → 30 m/s. Learn this list and many train questions become mental arithmetic.

Proportion shortcuts

Many questions change only one quantity. Then you do not need the full formula.

Why: if d = s × t and d is fixed, raising s must lower t by the same factor. So if a person walks at 3/4 of the usual speed, the same trip takes 4/3 of the usual time. The extra 1/3 of the usual time is the delay, and that single fact solves a whole family of "late" questions.

Average speed

Average speed = total distance ÷ total time. It is not the average of the speeds.

• Equal distances at speeds a and b: average = 2ab ÷ (a + b).
• Equal times at speeds a and b: average = (a + b) ÷ 2.

Why they differ: on a round trip you spend more time at the slower speed, so the slow speed pulls the average down. For a distance d each way, total time is d/a + d/b and total distance is 2d. Dividing gives 2ab/(a + b). When the times are equal instead, each speed counts equally, so the plain average is correct.

For three equal distances at a, b and c, average = 3 ÷ (1/a + 1/b + 1/c).

Relative speed: meeting and chasing

Why: when two people approach each other, the gap shrinks by both their distances every hour. When one chases the other, the gap shrinks only by the difference.

If one starts later, first work out the gap at the moment the second one starts. Then apply relative speed.

Late and early

A person travels the same distance d at two speeds. At the slower speed they are late; at the faster speed they are early. The total time gap is (minutes late + minutes early), because both are measured against the same correct time.

d/s₁ − d/s₂ = total gap in hours, which gives d = s₁ × s₂ × gap ÷ (s₂ − s₁).

Stoppages

A train that averages S km/h without stops and s km/h with stops loses (S − s) km of running in each hour. At speed S, covering that takes (S − s)/S hours.

Minutes stopped per hour = (S − s) ÷ S × 60.

Worked questions at NTPC level

Q1. A bus covers 162 km in 3 hours. Find its speed in m/s.

Speed = 162 ÷ 3 = 54 km/h. 54 × 5/18 = 15 m/s.

Q2. A person travels to a town at 40 km/h and returns at 60 km/h. Find the average speed.

Equal distances: 2 × 40 × 60 ÷ (40 + 60) = 4,800 ÷ 100 = 48 km/h.
Check with a distance of 120 km: 3 h + 2 h = 5 h for 240 km, which is 48 km/h.

Q3. Walking at 3/4 of the usual speed, a person reaches the office 20 minutes late. What is the usual time?

New time = 4/3 of usual, so the extra 1/3 of usual time = 20 minutes.
Usual time = 60 minutes.

Q4. Walking at 4 km/h, a person reaches the station 10 minutes late. At 5 km/h, they are 5 minutes early. Find the distance.

Total gap = 15 minutes = 1/4 hour.
d = 4 × 5 × 1/4 ÷ (5 − 4) = 5 km.
Check: 75 minutes at 4 km/h and 60 minutes at 5 km/h, a gap of 15 minutes.

Q5. Stations P and Q are 300 km apart. A train leaves P at 8 a.m. towards Q at 60 km/h. Another leaves Q at 9 a.m. towards P at 40 km/h. When do they meet?

By 9 a.m. the first train has covered 60 km, so the gap is 240 km.
Relative speed = 100 km/h. Time = 240 ÷ 100 = 2.4 h = 2 h 24 min.
They meet at 11:24 a.m., 204 km from P.

Q6. Without stoppages a train runs at 60 km/h; with stoppages its average is 45 km/h. How many minutes per hour does it stop?

(60 − 45) ÷ 60 × 60 = 15 minutes.

Common mistakes

• Mixing units. A speed in km/h with a time in minutes gives nonsense. Convert first.
• Averaging speeds directly. 40 and 60 on a round trip give 48, not 50.
• Using only the late minutes. In late-and-early questions the gap is late + early.
• Forgetting the head start. If one person starts later, find the gap at that moment first.
• Adding speeds when chasing. Same direction means subtract.

Practice set

1. Convert 54 km/h into m/s.
2. Convert 25 m/s into km/h.
3. A person goes at 30 km/h and returns at 20 km/h over the same road. Find the average speed.
4. Two cars 400 km apart travel towards each other at 45 km/h and 55 km/h. When do they meet?
5. At 5/6 of the usual speed, a person is 12 minutes late. What is the usual time?
6. A police officer spots a thief 200 m ahead. The thief runs at 10 km/h and the officer chases at 12 km/h. How long does the chase last?
7. A person covers three equal stretches at 10, 20 and 60 km/h. Find the average speed.
8. A bus averages 54 km/h without stoppages and 45 km/h with them. For how many minutes per hour does it stop?

Answers:

1. 15 m/s. 54 × 5/18.
2. 90 km/h. 25 × 18/5.
3. 24 km/h. 2 × 30 × 20 ÷ 50.
4. 4 hours. 400 ÷ (45 + 55).
5. 60 minutes. Time becomes 6/5 of usual; the extra 1/5 is 12 minutes.
6. 6 minutes. Relative speed 2 km/h; 0.2 km ÷ 2 = 0.1 h.
7. 18 km/h. 3 ÷ (1/10 + 1/20 + 1/60) = 3 ÷ (10/60).
8. 10 minutes. (54 − 45) ÷ 54 × 60.

What to do next

• Learn the 18-multiples conversion list until 72 → 20 and 90 → 25 are instant.
• Solve 15 average-speed questions and label each one "equal distances" or "equal times" before calculating.
• Do 10 meeting and chasing questions, writing the starting gap each time.
• Move on to problems on trains, then boats and streams.