Time and Distance questions with answers and concepts

Learn time and distance with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

142 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

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Time, Speed and Distance is a fundamental topic in quantitative aptitude that deals with the relationship between three quantities: time taken, speed maintained, and distance covered. These problems appear frequently in competitive exams and require understanding of basic formulas, unit conversions, and various scenarios like relative speed, average speed, and meeting points.

Key formulas and rules

10 rules

Basic relationship: $Speed = \frac{Distance}{Time}$ , $Distance = Speed \times Time$ , $Time = \frac{Distance}{Speed}$

Unit conversion: $1 km/h = \frac{5}{18} m/s$ and $1 m/s = \frac{18}{5} km/h$

Average Speed: $Average Speed = \frac{Total Distance}{Total Time}$

Harmonic mean for equal distances: If equal distances are covered at speeds $x$ and $y$ , $Avg Speed = \frac{2 x y}{x + y}$

Three equal distances at $x, y, z$ : $Avg Speed = \frac{3 x y z}{x y + y z + z x}$

Equal time intervals: If equal times are spent at speeds $x$ and $y$ , $Avg Speed = \frac{x + y}{2}$

Relative speed (same direction): $S_{rel} = ∣ S_{1} - S_{2} ∣$

Relative speed (opposite direction): $S_{rel} = S_{1} + S_{2}$

Proportionality: When distance is constant, $Speed \propto \frac{1}{Time} ⟹ \frac{S _{1}}{S _{2}} = \frac{T _{2}}{T _{1}}$

Proportionality: When time is constant, $Distance \propto Speed ⟹ \frac{D _{1}}{D _{2}} = \frac{S _{1}}{S _{2}}$

Key concepts

Basic Relationships

The three fundamental quantities are related by Speed = Distance/Time. If any two are known, the third can be calculated. Always ensure units are consistent—convert km/hr to m/s when working with meters and seconds, or vice versa using the conversion factors 5/18 and 18/5.

Average Speed

Average speed is always Total Distance divided by Total Time, NOT the arithmetic mean of speeds. For equal distances traveled at different speeds, use the harmonic mean formula: 2xy/(x+y). For different distances, calculate total distance and total time separately.

Relative Speed

When two bodies move in the same direction, their relative speed is the difference of their speeds. When moving in opposite directions, their relative speed is the sum of their speeds. This concept is crucial for problems involving trains, races, and meeting points.

Trains and Platforms

A train crossing a stationary object (pole/person) covers distance equal to its own length. Crossing a platform or bridge requires covering the sum of train length and platform length. When two trains cross each other, the distance covered is the sum of both train lengths.

Boats and Streams

Upstream speed (against current) = Boat speed - Stream speed. Downstream speed (with current) = Boat speed + Stream speed. Still water speed = (Upstream + Downstream)/2. Stream speed = (Downstream - Upstream)/2.

Circular Tracks and Races

On a circular track, when two runners start together and run in the same direction, the faster gains one complete round every time the relative distance covered equals the track length. For opposite directions, they meet when the sum of distances equals the track length.

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Worked examples

Example 1

Convert a speed of 72 km/h into m/s, and convert 25 m/s into km/h.

Solution

Step 1: Convert km/h to m/s by multiplying by $\frac{5}{18}$ .

72 km/h = 72 \times \frac{5}{18} m/s = 4 \times 5 = 20 m/s

Step 2: Convert m/s to km/h by multiplying by $\frac{18}{5}$ .

25 m/s = 25 \times \frac{18}{5} km/h = 5 \times 18 = 90 km/h

Answer: 20 m/s and 90 km/h.

Example 2

A commuter drives from town A to town B at 60 km/h and returns along the same route at 40 km/h. What is the average speed for the entire round trip?

Solution

Step 1: Apply the harmonic mean average speed formula for equal distances.

v_{avg} = \frac{2 x y}{x + y}

Step 2: Substitute $x = 60$ and $y = 40$ .

v_{avg} = \frac{2 \times 60 \times 40}{60 + 40} = \frac{4 , 800}{100} = 48 km/h

Answer: 48 km/h.

Example 3

Walking at 5 km/h, a student reaches school 10 minutes late. Walking at 6 km/h, she reaches 15 minutes early. What is the distance to school?

Solution

Step 1: Calculate total time difference $Δ t$ .

Δ t = 10 min (late) - (- 15 min (early)) = 25 minutes = \frac{25}{60} = \frac{5}{12} hours

Step 2: Apply the dual-speed distance formula.

d = \frac{s _{1} \times s _{2}}{s _{2} - s _{1}} \times Δ t

d = \frac{5 \times 6}{6 - 5} \times \frac{5}{12} = 30 \times \frac{5}{12} = \frac{150}{12} = 12.5 km

Answer: 12.5 km.

Example 4

The speeds of A and B are in the ratio $3 : 4$ . A takes 30 minutes more than B to cover a given journey. What is the total travel time taken by A?

Solution

Step 1: Time is inversely proportional to speed for equal distance.

\frac{t _{A}}{t _{B}} = \frac{s _{B}}{s _{A}} = \frac{4}{3}

Step 2: Set up difference in time units.

t_{A} = 4 x, t_{B} = 3 x

t_{A} - t_{B} = 4 x - 3 x = x = 30 minutes

Step 3: Compute travel time for A.

t_{A} = 4 \times 30 minutes = 120 minutes = 2 hours

Answer: 2 hours (or 120 minutes).

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A thief escapes in a car at 80 km/h. A police car starts chasing after 30 minutes at 100 km/h. How long will it take the police to catch the thief?

1 hour
1.5 hours
2 hours
2.5 hours

Answer: C. 2 hours

Explanation

Step 1: Distance covered by thief in 30 $min = 80$ x 0.5 = 40 km
Step 2: Relative speed = 100 - 80 = 20 km/h (same direction, subtract speeds)
Step 3: Time to catch = Lead distance / Relative speed = 40 km / 20 $k m / h = 2$ hours
Answer: 2 hours

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A person covers a distance of 300 km. He travels the first 150 km at 50 km/h and the remaining distance at 75 km/h. Find his average speed.

60 km/h
62.5 km/h
65 km/h
58 km/h

Answer: A. 60 km/h

Explanation

Step 1: Time for first 150 $k m = 150/50 = 3$ hours
Step 2: Time for remaining 150 $k m = 150/75 = 2$ hours
Step 3: Total distance = 300 km, Total time = 5 hours
Step 4: Average speed = 300/5 = 60 km/h
Answer: 60 km/h

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A bus covers 4/5 of its journey at 40 km/h and the remaining at 20 km/h. If the total journey is 200 km, find the average speed.

33.33 km/h
35 km/h
30 km/h
36 km/h

Answer: A. 33.33 km/h

Explanation

Step 1: First part distance = 4/5 x $200 = 160$ km, Time = 160/40 = 4 hours
Step 2: Second part distance = 40 km, Time = 40/20 = 2 hours
Step 3: Total distance = 200 km, Total time = 6 hours
Step 4: Average speed = 200/6 = 33.33 km/h
Answer: 33.33 km/h

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always verify unit consistency before applying formulas—convert km/hr to m/s by multiplying by 5/18 when working with meter distances.
For average speed problems, remember it is total distance divided by total time, not the arithmetic mean of individual speeds.
In train problems, carefully identify what the train is crossing—a pole (just train length), a platform (train + platform), or another train (sum of both lengths).
When two objects move towards each other, add their speeds for relative speed. When moving in the same direction, subtract the slower from the faster.
For problems involving stoppages, calculate the running time and stopped time separately, then use total time for average speed calculations.
Practice converting between different time units (hours to minutes to seconds) and distance units (km to m) to avoid calculation errors under exam pressure.

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