📑 On this page
⚡ The short answer
Average speed = total distance ÷ total time. Never average the speeds blindly. Two equal distances at x and y: 2xy/(x + y). Two equal times: (x + y)/2. The trap: the slow leg lasts longer, so it drags the answer below the midpoint, towards the slower speed.
🎬 The 30-second story
The 50 km/h promise
1
“Fest is 120 km away. I'll do 60 going. Easy.”
🚗
Aarav
Aarav drives Zoya to a fest.
2
“And on the way back? The highway's jammed in the evening.”
🚦
Zoya
The return trip crawls at 40 km/h.
3
“60 and 40. Average 50. So 240 km in 4.8 hours. Tell your mum.”
🧮
Aarav
Zoya texts home: back by 9:48.
4
“It's 10 o'clock, Aarav. Where did 12 minutes go?”
⌚
Zoya
Five hours, not 4.8.
Tara's maths: Going: 120 ÷ 60 = 2 h. Coming back: 120 ÷ 40 = 3 h. Total 240 km in 5 h = 48 km/h. The car spent three of its five hours at 40, so 40 gets the bigger say.
What is average speed, and why isn't it the average of the speeds?
Average speed answers one honest question: what single steady speed would have covered the same distance in the same time? Aarav needed a steady 48 km/h to do 240 km in 5 hours. That is the average speed.
🟦 Picture the percent
Aarav's 5 hours, shrunk to 100 squares
5 h on the road
Same distance each way, but not the same time.
At 60 km/h (2 h)At 40 km/h (3 h)
Averaging 60 and 40 gives each speed 50 squares. The clock gave them 40 : 60.
Here is the whole secret. A speed gets counted once for every hour you spend at it. The slow leg lasts longer, so it gets counted more. That is why, over equal distances, the average speed always falls below the plain average of the speeds.
What is the average speed formula?
🧮 The formula
The only formula you truly need
🛣️
Total distance
add every leg
÷
⏱️
Total time
add every leg, stops included
=
🎯
Average speed
one steady speed for the whole trip
For each leg, time = distance ÷ speed. Add the times, never the speeds.
📌 Time is the weight
Average speed is a weighted average of the speeds, weighted by time: (s₁t₁ + s₂t₂) ÷ (t₁ + t₂). For Aarav, (60 × 2 + 40 × 3) ÷ 5 = 48. Speed × time is distance, so the top line is simply the total distance.
🎛️ Try it yourself — change any number
Aarav's fest trip: change the legs
Average speed
48
Average of the speeds (trap)
50
Total distance ÷ total time = 240 ÷ 5 = 48
Set both distances to 50: still 48, because equal distances cancel. Now drop the return speed to 10 and watch the average sink to about 17, nowhere near 35.
Where does the 2xy/(x + y) shortcut come from, and when does it work?
Call each leg d km, at x then y. Average speed = 2d ÷ (d/x + d/y). Divide through by d and it vanishes: 2xy/(x + y). So for equal distances you never need the distance. For 60 and 40: 4,800 ÷ 100 = 48.
⚔️ Side by side
Equal distances vs equal times
📏Equal distances
There and back; first half, second half
- •Slow leg lasts longer, so it dominates
- •Answer sits below the midpoint
⏲️Equal times
1 hour at x, 1 hour at y
- •Each speed gets the same weight
- •Plain average is correct
VS
| Test | Equal distances | Equal times |
|---|---|---|
| Formula | 2xy/(x + y) | (x + y)/2 |
| 60 and 40 | 48 km/h | 50 km/h |
Ask which thing is equal, distance or time. That picks the formula.
| Two speeds (equal distances) | Average of speeds (trap) | True average speed | Gap |
|---|---|---|---|
| 40 and 60 | 50 | 48 | 2 |
| 30 and 60 | 45 | 40 | 5 |
| 20 and 30 | 25 | 24 | 1 |
| 10 and 15 | 12.5 | 12 | 0.5 |
→ scroll to see more
The gap is (x − y)² ÷ 2(x + y): close speeds barely differ, far-apart speeds open a big gap. That is why CLAT loves 30 and 60: the trap 45 sits a full 5 above the truth.
How do you solve average speed questions step by step?
✍️ Worked example
Warm-up
⏱ 20 sec
Kabir's delivery van
Kabir's van drives for 30 minutes at 36 km/h through the market, then 30 minutes at 54 km/h on the bypass. What is its average speed?
- 1Distances: 36 × ½ = 18 km and 54 × ½ = 27 km. Total 45 km.
- 2Total time = 30 + 30 min = 1 h.
- 3Average speed = 45 ÷ 1 = 45 km/h.
Answer: 45 km/h. The plain average is right here.
⚡ The exam-hall shortcut
Equal times, so average the speeds: (36 + 54) ÷ 2 = 45.
✍️ Worked example
CLAT level
⏱ 60 sec
Zoya's three-leg commute
Zoya rides the metro 12 km at 48 km/h, takes an auto 3 km at 18 km/h, then walks 1 km at 4 km/h. What is her average speed for the whole commute?
- 1Time for each leg: 12 ÷ 48 = ¼ h = 15 min; 3 ÷ 18 = ⅙ h = 10 min; 1 ÷ 4 = ¼ h = 15 min.
- 2Total distance = 12 + 3 + 1 = 16 km. Total time = 40 min = ⅔ h.
- 3Average speed = 16 ÷ ⅔ = 24 km/h.
Answer: 24 km/h, not the plain average of about 23.3.
⚡ The exam-hall shortcut
16 km in 40 min is 0.4 km a minute; × 60 = 24. Unequal legs? Only totals will do.
✍️ Worked example
Tricky
⏱ 75 sec
How fast must Meher ride back?
Meher cycles 6 km to coaching at 12 km/h. How fast must she ride back to average 15 km/h for the round trip? Could she average 24 km/h?
- 1Round trip = 12 km. To average 15, total time must be 12 ÷ 15 = 0.8 h = 48 min.
- 2Going took 6 ÷ 12 = 30 min. So the return gets 48 − 30 = 18 min.
- 3Return speed = 6 ÷ (18/60) = 20 km/h. Check: 2 × 12 × 20 ÷ 32 = 15. ✓
- 4For 24 km/h, total time must be 12 ÷ 24 = 30 min, all of it already used going. Impossible.
Answer: 20 km/h. 24 km/h is out of reach at any speed.
⚡ The exam-hall shortcut
Think in time budgets. Over two equal halves you can never reach double the first speed.
Is there a faster way? The cross, the LCM trick and estimation
✖️ The alligation cross
Aarav's speeds on the alligation cross
Back (slow) : Going (fast) = 3 : 2
12 : 8 = 3 : 2 is the ratio of times (3 h at 40, 2 h at 60), not distances: time is the weight.
Want distances back? Multiply each time by its speed: 3 × 40 : 2 × 60 = 120 : 120. Equal, as it should be. And for three equal stretches with no distance given, pick one every speed divides (the LCM). At 3, 4 and 6 km/h, take 12 km each: 4 + 3 + 2 = 9 h for 36 km, so 4 km/h, not the 4.33 plain average.
💡 Kill options before you calculate
Equal distances at 30 and 60? The answer is below 45 and above 30. Strike 45 (the trap) and anything outside the range; often one option survives.
How does CLAT hide average speed inside a caselet?
CLAT won't say "find the average speed". It describes a rider, a bus route or a courier's shift, leg by leg, then asks about the whole journey. Build a tiny table of distance, speed and time, then divide total by total.
📊 The data
Rohan's evening delivery shift: speed on each leg
Leg 1 · 6 km, 15 min
24 km/h
Leg 2 · 4 km, 15 min
16 km/h
Leg 3 · 9 km, 15 min
36 km/h
Leg 4 · 5 km, 15 min
20 km/h
Riding only · 24 km in 60 min
24 km/h
With 20-min wait · 24 km in 80 min
18 km/h
Every leg takes 15 minutes, so the plain average of the speeds (96 ÷ 4 = 24) is right for riding alone. The wait drags it to 18.
✍️ Worked example
Caselet
⏱ 45 sec
The shift, wait included
Using the chart, Rohan also waited 20 minutes at a restaurant between legs 2 and 3. What was his average speed for the whole shift?
- 1Each leg takes ¼ h = 15 min (6 ÷ 24, 4 ÷ 16, 9 ÷ 36, 5 ÷ 20). Riding time = 60 min.
- 2Add the wait: 80 min = 4/3 h. Distance = 6 + 4 + 9 + 5 = 24 km.
- 3Average speed = 24 ÷ 4/3 = 18 km/h.
Answer: 18 km/h. Leave out the wait and you get the 24 km/h trap.
⚡ The exam-hall shortcut
24 km in 80 min is 0.3 km a minute = 18 km/h.
What traps does CLAT set in average speed questions?
🧭 Decision flow
Which average speed formula?
❓ Is the time spent on each leg equal?
Yes →
Plain average of the speeds works.
No ↓
❓ Are the distances equal (there and back, halves, equal stretches)?
Yes →
2xy/(x + y), or pick an LCM distance.
No ↓
❓ Do you know each leg's distance and speed (or time)?
Yes →
Total distance ÷ total time.
No →
Cannot be determined.
You only know a range between the speeds.
⚠️ Three silent mark-killers
Minutes: 40 min at 30 km/h is 30 × 40/60 = 20 km, not 1,200. Units: m/s × 18/5 = km/h; km/h × 5/18 = m/s. Halts: average speed counts every stop unless the question asks for the moving or running speed.
How did CLAT ask average speed questions?
📝 Asked in CLAT 2019
· Q54
Equal time slots: average the speeds
A car driver increases the average speed of his car by 3 km/hr every hour. The total distance travelled in 7 hours if the distance covered in first hour was 30 km, is
A266 km
B273 km
C280 km
D287 km
Reveal answer + reasoning
Answer: B
No official key exists for 2019; this is our solution. Hourly speeds: 30, 33, …, 48 km/h. Each hour is an equal time slot, so average the speeds: (30 + 48) ÷ 2 = 39 km/h, the middle (4th) hour's speed. Distance = 39 × 7 = 273 km. Time: about 30 seconds.
Why the others fail
- A — 7 × 38: the middle-hour speed is one too low.
- C — 7 × 40: the middle-hour speed is one too high.
- D — 7 × 41: the middle-hour speed is two too high, e.g. counting seven increases instead of six.
🧪 CLAT-style practice (not a past question)
The cannot-be-determined trap
In a 10 km charity run, Aarav ran the first part of the course at 12 km/h and the rest at 8 km/h. He did not stop at any point.
What was Aarav's average speed for the run?
A10 km/h
B9.6 km/h
C9.8 km/h
DCannot be determined
Reveal answer + reasoning
Answer: D
How long was the "first part"? Not given. Equal halves would give 2 × 12 × 8 ÷ 20 = 9.6; a 2 km first part gives about 8.57; an 8 km one about 10.91. Different splits, different answers: cannot be determined. Time: 15 seconds, once you ask "equal what?"
Why the others fail
- A — The plain average, which needs equal times; nothing says so.
- B — Correct only if the two parts were equal halves, which the passage never says.
- C — A guess between the two; it has no basis in the data.
Check yourself: myths and a five-question quiz
💥 Myth vs fact
Average speed myths
🙅 Myth
Average speed = (first speed + second speed) ÷ 2.
Tap to bust it →
✅ Fact
Only when the times are equal. For equal distances, use 2xy/(x + y).
🙅 Myth
For a round trip, you need the distance to find the average speed.
Tap to bust it →
✅ Fact
It cancels. 60 and 40 give 48 for a 5 km trip or a 500 km one.
🙅 Myth
Half the distance at 30? Just do the second half at 90 and you average 60.
Tap to bust it →
✅ Fact
Averaging 60 over equal halves leaves 0 time for the second half. Impossible.
🙅 Myth
Tea breaks don't count; you weren't moving.
Tap to bust it →
✅ Fact
They count. Average speed uses total time unless the question asks for moving speed.
🎯 Quick quiz
Five quick ones
0 / 5
A train goes from P to Q at 80 km/h and returns at 120 km/h along the same track.
Q1. What is its average speed for the round trip?
2 × 80 × 120 ÷ 200 = 19,200 ÷ 200 = 96. 100 is the plain-average trap.A cyclist rides for 2 hours at 15 km/h and then for 3 hours at 20 km/h.
Q2. What is the average speed?
Distance = 30 + 60 = 90 km in 5 h = 18. Times are unequal, so 17.5 is wrong.A car covers the first half of a journey at 30 km/h.
Q3. At what speed must it cover the second half to average 40 km/h overall?
Take 60 km each half. Allowed time = 120 ÷ 40 = 3 h; the first half used 2 h, so 60 km/h.A bus covers three equal stretches at 10, 20 and 60 km/h.
Q4. What is its average speed?
Take 60 km each (LCM): 6 + 3 + 1 = 10 h for 180 km = 18. 30 is the plain average.Meher walks 600 m in 10 minutes, then jogs 1.8 km in the next 10 minutes.
Q5. What is her average speed in km/h?
2.4 km in 20 min = 7.2 km/h. Equal times, so (3.6 + 10.8) ÷ 2 works too.🎯 Average speed in a nutshell
- Total distance ÷ total time. Add times, never speeds.
- Equal times: plain average. Equal distances: 2xy/(x + y).
- The slow leg lasts longer, so the answer leans towards the slower speed.
- No distance given for equal stretches? Pick the LCM of the speeds.
- Split unknown? Cannot be determined.
📎 Formula card
- Average speed = total distance ÷ total time
- Time-weighted form: (s₁t₁ + s₂t₂) ÷ (t₁ + t₂)
- Equal times: (x + y) ÷ 2
- Equal distances: 2xy ÷ (x + y)
- Three equal distances: 3xyz ÷ (xy + yz + zx)
- Gap below the plain average: (x − y)² ÷ 2(x + y)
- Return speed for target A: xA ÷ (2x − A), possible only if A < 2x
- Units: km/h × 5/18 = m/s; m/s × 18/5 = km/h
Frequently asked questions
What is the formula for average speed?
Average speed = total distance ÷ total time. Find the time for each leg as distance ÷ speed, add the times, and divide the whole distance by that total. For 120 km at 60 km/h and 120 km back at 40 km/h, it is 240 ÷ 5 = 48 km/h.
Why is average speed not the average of the two speeds?
Because you spend more time at the slower speed. Average speed weights each speed by the time spent at it, so the slow leg counts for more. The plain average of the speeds is correct only when you spend equal time at each speed, such as one hour at each.
When can I use the 2xy/(x + y) formula?
Only when the two legs cover equal distances: a round trip on the same road, or the first and second halves of a journey. The distance cancels out, so you don't need it. For equal times use (x + y)/2, and for anything else go back to total distance ÷ total time.
Does stopping time count in average speed?
Yes. Average speed uses the total time from start to finish, so halts, waits and breaks all count. Leave them out only if the question asks for the moving or running speed. A 20-minute wait can turn a 24 km/h shift into an 18 km/h one.
If I drive half the distance at 30 km/h, can I average 60 km/h?
No. To average 60 over the whole trip you need total time = distance ÷ 60, and at 30 km/h the first half alone already takes exactly that long. The second half would need zero time. Over equal halves, you can never reach double the first speed.
Keep going
📚
Part of the chapter
Averages, Mixtures & Alligation for CLAT Quantitative Techniques
See how this topic fits the whole of Averages, Mixtures & Alligation →
Connected topics in other chapters
Weighted Average
Average speed is a weighted average in disguise: speeds weighted by time.
Speed, Distance and Time
The distance = speed × time triangle every average-speed question stands on.
Alligation: the Cross Method
Run the cross on two speeds and it hands you the ratio of the times.
Trains, Races and Relative Speed
The next step up in speed questions: two things moving at once.
Every topic in Averages, Mixtures & Alligation
- 1.Average: Formula, Meaning and the Balance Idea
- 2.When Someone Joins or Leaves: How the Average Shifts
- 3.Weighted Average: Why You Can't Just Average the Averages
- 4.Alligation: the Cross Method
- 5.Mixtures, Concentration and Replacement
- 6.Average Speed: Why It Is Not the Average of Speeds (you're here)