A car travels 30 miles in 20 minutes. What is its speed in miles per hour?

90 mph. 'Per hour' means in one hour, and 20 minutes is only ⅓ of an hour, so the car goes three times as far in a whole hour: 30 ÷ ⅓ = 30 × 3 = 90 mph. The common wrong answer 1.5 comes from 30 ÷ 20, treating the minutes as hours — but 1.5 mph is slower than a person walking, which is the tell that the time was never converted.

How do you find a speed from a distance–time graph?

Use the gradient: the change in distance divided by the change in time. If a segment runs from 10 km at 1 hour to 70 km at 3 hours, the speed is (70 − 10) ÷ (3 − 1) = 60 ÷ 2 = 30 km/h. Reading a single point, 70 ÷ 3 ≈ 23.3, is wrong because it ignores the distance and time already covered before that segment.

How many seconds are in a minute when I convert a time?

60. Time runs in sixties, not hundreds. So 1200 seconds is 1200 ÷ 60 = 20 minutes, not 1200 ÷ 100 = 12. Dividing by 100, as if a minute were 100 seconds, is a common slip when converting seconds to minutes.

GCSE Maths Foundation · AQA · Proportion, rates & compound measures

Speed, distance, time & rates: convert the time before you divide

Speed questions trip students who treat a rate as “distance ÷ the number you are given”. Told a car travels 30 miles in 20 minutes, they answer 1.5 mph — $30 \div 20$ — without noticing the time is in minutes, not hours. But “miles per hour” means miles in one hour, and 20 minutes is only a third of an hour, so the car goes three times as far: $30 \div \tfrac{1}{3} = 90$ mph.

The thirty-second fix: convert the time to the rate’s unit first. “Per hour” means in one hour, so change minutes to hours before you divide. And a speed off a distance–time graph is the gradient — the change in distance over the change in time, never a single point.

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How to spot it in your own work

You divided the distance by the minutes — writing $30 \div 20 = 1.5$ instead of $30 \div \tfrac{1}{3} = 90$ .
Your vehicle speed came out slower than walking (about 3 mph) — a sure sign the time was never converted.
You changed seconds to minutes by dividing by 100, as if a minute were 100 seconds, instead of dividing by 60.
You read a speed off a distance–time graph from one point — $70 \div 3$ — instead of the change, $(70 - 10) \div (3 - 1)$ .
You treated “5 minutes” as hours, writing $5 \times 4 = 20$ mph instead of $4 \times 12 = 48$ mph.

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Jun23 P1 Q16 / Nov24 P1 Q17 shape, non-calculator):

A car travels 30 miles in 20 minutes. Work out its average speed in miles per hour.

The misconception answer is $30 \div 20 = 1.5$ — dividing by the minutes. But 1.5 mph is slower than a person walking, which cannot be a car’s speed.

Convert the time first: 20 minutes is $\tfrac{1}{3}$ hour, so $30 \div \tfrac{1}{3} = 30 \times 3 = 90$ mph.

Why students fall for this

Division is the first move students reach for when they see a distance and a time and want a rate. But the “hour” in “miles per hour” sets the unit the time must be measured in. Twenty minutes is not 20 hours; it is $\tfrac{20}{60} = \tfrac{1}{3}$ hour. Dividing distance by the raw minutes answers a different question entirely, and the giveaway is the size: a car at 1.5 mph would be slower than walking.

The same belief — that the time scale factor is invisible — shows up when converting seconds. Students write “1 minute = 100 seconds” and divide by 100, so 1200 seconds becomes 12 minutes instead of the correct $1200 \div 60 = 20$ minutes. Time runs in sixties, not hundreds.

On a distance–time graph the belief reappears as reading a single point. A speed is a gradient: the change in distance divided by the change in time. AQA Foundation papers exploit every face of this — 18 miles in 20 minutes (Jun23 P1 Q16), 4 miles in 5 minutes (Nov24 P1 Q17), a speed off a distance–time graph (Jun24 P2 Q25), and a seconds-to-minutes conversion (Jun24 P3 Q24a).

The fix: Convert the time to the rate's unit first; a graph speed is the change in distance over the change in time

“Per hour” means in one hour, so convert minutes to hours. 30 miles in 20 minutes: 20 minutes is $\tfrac{1}{3}$ hour, so $30 \div \tfrac{1}{3} = 90$ mph. If a vehicle speed comes out slower than walking, the time was not converted.

Time runs in sixties, not hundreds. A minute is 60 seconds, so $1200 \div 60 = 20$ minutes — never divide by 100.

A speed off a graph is the gradient. Take the change in distance over the change in time: $(70 - 10) \div (3 - 1) = 30$ km/h, not the single point $70 \div 3$ .

Worked example

A cyclist rides 18 miles in 20 minutes. Work out the average speed in miles per hour.

Convert the time to the rate’s unit. 20 minutes is $\tfrac{20}{60} = \tfrac{1}{3}$ hour.
Divide distance by the converted time.
$\text{speed} = 18 \div \tfrac{1}{3} = 18 \times 3 = 54 \text{ mph}$
Sense-check the size. 54 mph is a believable cycling-then-coasting speed; the trap $18 \div 20 = 0.9$ mph is slower than walking, so it is wrong.

The same habit reads a graph speed as the change: $(70 - 10) \div (3 - 1) = 60 \div 2 = 30 \text{ km/h}$ — not $70 \div 3 \approx 23.3$ . Convert the time, take the change.

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Common questions

Why can’t I just divide the distance by the minutes?: Because “miles per hour” means miles in one hour, and minutes are not hours. 20 minutes is $\tfrac{1}{3}$ hour, so the car goes three times as far in a whole hour: $30 \div \tfrac{1}{3} = 90$ mph. Dividing by 20 gives 1.5 mph, slower than walking — impossible for a car.
How do I turn minutes into a fraction of an hour?: Put the minutes over 60. So 20 minutes is $\tfrac{20}{60} = \tfrac{1}{3}$ hour, 30 minutes is $\tfrac{1}{2}$ hour, and 12 minutes is $\tfrac{12}{60} = \tfrac{1}{5}$ hour. Then divide the distance by that fraction, which is the same as multiplying by its reciprocal.
How many seconds are in a minute?: 60, not 100. Time runs in sixties. So 1200 seconds is $1200 \div 60 = 20$ minutes. Dividing by 100 to get 12 is the “a minute is 100 seconds” slip.
How do I read a speed off a distance–time graph?: Use the gradient: the change in distance divided by the change in time. From 10 km at 1 hour to 70 km at 3 hours, that is $(70 - 10) \div (3 - 1) = 30$ km/h. Never read a single point like $70 \div 3$ — it ignores what was already covered before that segment.

Related misconceptions

Comparing rates & density: convert to a common unit before comparingThe neighbouring rates misconception — convert to a common unit before comparing, and remember that density multiplies.
Direct vs inverse proportion: why more can mean lessThe proportion companion — telling direct from inverse proportion, where more of one quantity can mean less of the other.

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