Why is the average speed for a there-and-back journey at 30 mph and 60 mph equal to 40 mph and not 45 mph?

Because the outward leg at 30 mph takes twice as long as the return at 60 mph. Average speed is total distance divided by total time. For distance d each way, total time equals d divided by 30 plus d divided by 60, which equals d over 20. Average equals 2d divided by d over 20, which equals 40 mph. The arithmetic mean 45 ignores that more time is spent at 30 mph.

When is the arithmetic mean of two speeds the correct average speed?

Only when both legs take the same amount of time. If you drive at 30 mph for 2 hours and then at 60 mph for 2 hours, the average is (30 plus 60) divided by 2 equals 45 mph, which is correct because each speed contributes for equal time. For equal-distance legs the arithmetic mean overstates the average.

What is the formula for average speed when two equal distances are covered at speeds a and b?

The formula is 2ab divided by the sum a plus b, known as the harmonic mean. It is always less than or equal to the arithmetic mean, with equality only when both speeds are the same. For a equals 30 and b equals 60: 2 times 30 times 60 divided by 90 equals 40 mph.

GCSE Maths Higher · AQA · Compound measures and rates

Average speed is not the mean of the speeds

For a journey with legs at different speeds, the instinct is to average the speeds: $(30 + 60) / 2 = 45$ mph. That answer is wrong whenever the legs do not take the same time. The correct rule is always average speed = total distance ÷ total time, and the two legs must be timed individually before the division.

For a there-and-back journey at 30 mph and 60 mph, the outward leg takes twice as long as the return. The slower speed dominates the total time and pulls the average well below the midpoint: the answer is 40 mph, not 45. This is one of the most reliable grade 7 to 9 mark-losers on AQA Higher compound-measures papers.

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How to spot it in your own work

You added the two speeds and divided by 2 to find the average speed for a there-and-back or two-stage journey.
You wrote $(30 + 60) / 2 = 45$ mph when the correct answer is $\text{total distance} / \text{total time} = 40$ mph.
You did not calculate the time for each individual leg before dividing total distance by total time.
You assumed averaging the speeds was valid without checking whether both legs took the same amount of time.

An exam question that triggers it

Here is the structure of a typical AQA Higher compound-measures question:

A driver travels from town A to town B at 30 mph.

She returns from town B to town A at 60 mph.

The distance from A to B is the same as the distance from B to A.

Work out the average speed for the whole journey.

The misconception writes $(30 + 60) / 2 = 45$ mph. The fix uses total distance / total time: for distance $d$ each way, total time $= d/30 + d/60 = d/20$ , so average speed $= 2d / (d/20) = 40$ mph.

Why students fall for this

Students learn the arithmetic mean early and use it as the default “average” operation. When a question mentions two speeds, the instinct is to add them and halve the result, treating speed like any other set of data values.

The error is a mismatch of structure: the arithmetic mean weights each value equally, but a journey weights each speed by the time spent at it, not by the number of legs. The slower speed demands more time for the same distance, so it carries more weight in the total, and the true average is pulled towards it.

The misconception is reinforced by the fact that averaging the speeds gives a plausible-looking number in the right ballpark. A student who writes 45 mph often has no sense that anything is wrong, which is what makes this a reliable mark-loser at grades 7 to 9.

The fix: Total distance divided by total time, leg by leg

Step 1: find the time for each leg separately. Use $\text{time} = \text{distance} / \text{speed}$ for every leg. If you do not have a numeric distance, call it $d$ and it will cancel later.

Step 2: sum the times. Add all leg times to get the total journey time.

Step 3: divide total distance by total time. Average speed $= \text{total distance} / \text{total time}$ . Never divide the total of the speeds by the number of legs.

Step 4: sense-check the direction. For equal-distance legs at different speeds the answer must be below the arithmetic mean of the speeds and closer to the slower one. If your answer equals or exceeds the arithmetic mean, recheck the time calculation.

Worked example

There-and-back: 30 mph out, 60 mph back, equal distances.

Times. Let distance each way be $d$ . Time out $= d/30$ ; time back $= d/60$ ; total time $= d/30 + d/60 = 2d/60 + d/60 = 3d/60 = d/20$ .
Divide.
$\text{average speed} = \frac{2d}{d/20} = 2d \times \frac{20}{d} = 40 \text{ mph}$
Trap: $(30 + 60)/2 = 45$ , which is too high.

Symbolic result for equal-distance legs at speeds $a$ and $b$ :

Total time. $d/a + d/b = d(a+b)/(ab)$ .
Average.
$\text{average} = \frac{2d}{d(a+b)/(ab)} = \frac{2ab}{a+b}$
Check: $2 \times 30 \times 60 / 90 = 40$ mph. ✓

Equal-time legs (arithmetic mean IS correct): 30 mph for 2 h and 60 mph for 2 h. Distances: 60 miles and 120 miles. Total 180 miles in 4 h. Average = $180/4 = 45$ mph, which equals $(30+60)/2$ only because both legs share the same 2-hour block.

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Common questions

Why is the average speed for 30 mph out and 60 mph back equal to 40 mph, not 45?: The outward leg at 30 mph takes twice as long as the return at 60 mph. Average speed = total distance / total time. For distance $d$ each way, total time $= d/30 + d/60 = d/20$ , so average $= 2d / (d/20) = 40$ mph.
When can you use the arithmetic mean of two speeds?: Only when both legs take exactly the same time. If you drive at 30 mph for 2 h then at 60 mph for 2 h, the average is $(30 + 60)/2 = 45$ mph, which is correct. For equal-distance legs the arithmetic mean overstates the average, because the slower leg uses more time.
What is the formula for equal-distance two-speed journeys?: $2ab/(a+b)$ , the harmonic mean of the two speeds. It is always at most the arithmetic mean $(a+b)/2$ , with equality only when $a = b$ . For 30 and 60 mph: $3600/90 = 40$ mph.

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