Average speed is total distance divided by total time — a definition, not an average of speeds. The slow half of a trip eats more of the clock, and the answer leans its way.
The problem
You drive to a city at 60 km/h and return over the same road at 40 km/h. What is your average speed for the whole trip?
Total over total: with 120 km each way, the trip takes 2 h + 3 h = 5 h for 240 km — 48 km/h. A is the famous trap: 60 + 402 pretends both speeds got equal time, but the slow leg took 3 of the 5 hours, dragging the average below 50.
average speed = total distancetotal time
always — no exceptions, no cases
Whatever the trip looks like — two legs, five legs, a stopover — compute the whole distance, compute the whole time, and divide once. Every wrong method on this topic is a shortcut that skips one of the totals.
The distance cancels in the end, so pick a convenient one — 120 km divides nicely by both speeds.
going: 120 ÷ 60 = 2 h
the fast leg
returning: 120 ÷ 40 = 3 h
the slow leg takes longer
2405 = 48 km/h
total distance over total time
average = 2aba + b = 2 · 60 · 40100 = 48
same answer, one line
That formula is the harmonic mean of the two speeds — the same 2aba + b from the harmonic sequences lesson. It applies only when the two legs cover equal distances, which round trips always do.
| The two legs have equal… | Then the average speed is… |
|---|---|
| distances (a round trip) | 2aba + b — below the midpoint |
| times (one hour each, say) | a + b2 — the plain average, legitimately |
Equal distances give the slow speed more TIME, so it drags the average down below the midpoint — the true answer always sits closer to the slower speed. That fact alone eliminates choices on sight.
Unequal times: find each distance first, as speed × time. 1 hour at 30 km/h and 3 hours at 50 km/h cover 30 + 150 = 180 km in 4 hours, so the average is 180 ÷ 4 = 45 km/h — not 40, because the faster leg ran longer.
Pick a distance, then find the time left for the return.
pick 60 km each way
divides by 30 and 40
going: 60 ÷ 30 = 2 h
whole trip: 120 ÷ 40 = 3 h
at the target average
return: 3 − 2 = 1 h
the time left
60 ÷ 1 = 60 km/h
check: 2 · 30 · 60 ÷ 90 = 40 ✓
30 + 502 = 40 only works for equal times. With equal distances the slow leg uses so much time that the return must be much faster. An average of 60 here would leave zero time to come back — impossible.
Total distance ÷ total time, every time; only the bookkeeping changes.
Minutes and a rest: a cyclist rides 18 km in 45 minutes, rests 15 minutes, then rides 12 km in 30 minutes.
distance: 18 + 12 = 30 km
the rest adds no distance
time: 45 + 15 + 30 = 90 min = 1.5 h
the rest does add time
30 ÷ 1.5 = 20 km/h
divide once
Leaving out the rest gives 30 ÷ 1.25 = 24 km/h, the moving average. The question decides which is meant; “for the whole trip” includes stops.
Unequal shares: one third of a trip at 20 km/h, the rest at 60 km/h.
pick 60 km: 20 km and 40 km
a total that splits into thirds
20 ÷ 20 = 1 h; 40 ÷ 60 = 23 h
time for each leg
60 ÷ (53) = 36 km/h
total over total
2aba + b is only for equal distances; here it would give 30. Picking a distance works for any split.
(13)(20) + (23)(60) weights the speeds by distance. Speeds average by time, and the slow third takes more than half the time.
The distance is the unknown: a driver goes to a town at 60 km/h and returns at 40 km/h. The round trip takes 5 hours. How far is the town?
d60 + d40 = 5
the two times add to 5
2d + 3d120 = 5
LCD 120
5d = 600, so d = 120 km
check: 2 h + 3 h = 5 h ✓
The unknown sits in the numerators, so it is time that adds, not speed.
Write each time as distance ÷ speed, then subtract.
d45 − d60 = 13
20 min = 13 h
4d − 3d180 = 13
LCD 180
d = 60 km
check: 80 min − 60 min = 20 min ✓
Speeds are per hour, so the 20 minutes must enter as 13, not 20.
Average-speed items hide the plain-average trap: always total distance ÷ total time, stops included. For a round trip use 2aba + b or pick a convenient distance, and check that the answer lies nearer the slower speed.
The problem
A cyclist rides out at 30 km/h and returns along the same route at 60 km/h. What is the average speed for the round trip?
2aba + b = 2(30)6090 = 3,60090 = 40 km/h — or with 60 km each way: 2 h out, 1 h back, 120 km in 3 h = 40. A is the midpoint trap again; the answer must sit closer to 30 than to 60, because the slow leg owns two thirds of the clock.
