Speed, distance, and time is probably the most common aptitude topic. Almost every exam has at least 2-3 questions from this chapter. The formula is dead simple — Distance = Speed × Time — but the word problems can get creative. Let’s master the patterns.
Unit conversion trick: To go from km/h to m/s, we’re converting bigger units (km, hours) to smaller ones (meters, seconds). Just remember: 1 km/h = 5/18 m/s. So 36 km/h = 36 × 5/18 = 10 m/s. For reverse, multiply by 18/5.
Average Speed — The Classic Trap
This is where exams love to trick people. Average speed is NOT the average of two speeds. It depends on whether we travel equal distances or equal times.
Equal distances at speeds a and b: Average speed = 2ab/(a + b) (harmonic mean)
Equal times at speeds a and b: Average speed = (a + b)/2 (arithmetic mean)
In exams, 90% of the time it’s equal distances, so 2ab/(a+b) is the formula we need.
Example 1: Average speed trap
A person goes from A to B at 40 km/h and returns at 60 km/h. Find the average speed for the round trip.
Since the distances are equal (A to B = B to A):
Average speed = 2 × 40 × 60 / (40 + 60) = 4800/100 = 48 km/h
NOT 50 km/h! The person spends more time at the slower speed, which pulls the average down.
Relative Speed
When two objects are moving, what matters is how fast one is moving relative to the other.
Same direction: Relative speed = S₁ - S₂ (the faster one “gains” on the slower one)
Opposite direction: Relative speed = S₁ + S₂ (they approach each other faster)
Example 2: Chase problem
A thief starts running at 10 km/h. A cop starts from the same point 30 minutes later at 12 km/h. When does the cop catch the thief?
In 30 minutes, the thief covers: 10 × 0.5 = 5 km (head start).
Relative speed (same direction) = 12 - 10 = 2 km/h.
Time for cop to close 5 km gap = 5/2 = 2.5 hours after the cop starts.
Example 3: Meeting problem
Two people start from points 100 km apart and walk toward each other at 6 km/h and 4 km/h. When and where do they meet?
Relative speed (opposite direction) = 6 + 4 = 10 km/h.
Time to meet = 100/10 = 10 hours.
Meeting point from person 1: 6 × 10 = 60 km from person 1 (and 40 km from person 2).
The Proportion Shortcut
Speed and time are inversely proportional (when distance is constant). So if speed increases by 25% (becomes 5/4 of original), time decreases to 4/5 of original.
This is incredibly useful for problems like:
Example 4: Proportion shortcut
If a person walks at 5 km/h, they’re 30 minutes late. If they walk at 6 km/h, they’re 10 minutes early. Find the distance.
Time difference = 30 + 10 = 40 minutes = 2/3 hours.
At 5 km/h, time = D/5. At 6 km/h, time = D/6.
D/5 - D/6 = 2/3 (6D - 5D)/30 = 2/3 D/30 = 2/3 D = 20 km
This pattern — “late at one speed, early at another” — always works the same way. Set up the time difference equation and solve.
Example 5: Speed ratio to time ratio
A car covers a distance in 4 hours at 60 km/h. How long would it take at 80 km/h?
Speed ratio = 60:80 = 3:4. Time ratio = 4:3 (inverse of speed). New time = 4 × (3/4) = 3 hours.
Or just: Distance = 60 × 4 = 240 km. Time = 240/80 = 3 hours. Both work!
Round Trip Problems
For a round trip (going and coming back to the starting point), the total distance is 2D and we can use the average speed formula.
Example 6: Round trip with rest
A person cycles from A to B (30 km) at 15 km/h, rests for 1 hour, then returns at 10 km/h. What is the average speed for the entire journey?
Time going = 30/15 = 2 hours. Time returning = 30/10 = 3 hours. Total time including rest = 2 + 1 + 3 = 6 hours. Total distance = 30 + 30 = 60 km.
Average speed = 60/6 = 10 km/h.
Note: If the problem asks “average speed,” rest time is included. If it asks “average speed while moving,” rest time is excluded (60/5 = 12 km/h).
Multiple Meeting Problems
When two people walk toward each other repeatedly between two points:
First meeting: They together cover 1D (where D is the distance between points). Second meeting: They together cover 3D. Third meeting: They together cover 5D.
So the ratio of distances covered at each successive meeting follows 1:3:5:7… This means the first meeting point can tell us where subsequent meetings happen.
Shortcut for “Reaches X Minutes Late / Early”
When the same distance is covered at two different speeds and we know the time difference:
Distance = (S₁ × S₂ × Time difference) / (S₁ - S₂)
where S₁ < S₂ (slower and faster speeds), and time difference is in the same unit as distance/speed.
Using Example 4: D = (5 × 6 × 2/3) / (6 - 5) = 20/1 = 20 km. Quick!
Common Exam Variations
- Average speed for round trips or multi-leg journeys.
- Relative speed: chase problems, meeting problems.
- “Late at one speed, early at another” — find the distance.
- Speed increased by x%, time reduced by y% — find distance.
- Two people meet and continue — when do they reach each other’s starting points?
- Journey with stops or changing speeds midway.
Practice Problems
Problem 1: A car covers 300 km at 60 km/h and returns at 40 km/h. Find the average speed for the entire trip.
Problem 2: Two trains start from stations 400 km apart at the same time, heading toward each other at 70 km/h and 50 km/h. After how many hours will they meet?
Problem 3: Walking at 4 km/h, a man reaches his office 10 minutes late. Walking at 5 km/h, he reaches 5 minutes early. What is the distance to his office?
Answers
Problem 1: Average speed = 2 × 60 × 40 / (60 + 40) = 4800/100 = 48 km/h.
Problem 2: Relative speed = 70 + 50 = 120 km/h. Time = 400/120 = 10/3 hours = 3 hours 20 minutes.
Problem 3: Time difference = 10 + 5 = 15 minutes = 1/4 hour. D = (4 × 5 × 1/4) / (5 - 4) = 5/1 = 5 km.