If something is 8400 to the nearest 100, what are the bounds?

Half of 100 is 50, so the bounds sit 50 either side: from 8350 up to 8450. The lower bound 8350 is included (it rounds up to 8400). The upper edge 8450 is not included — it rounds up to 8500 — so for a whole-number count like attendance the maximum is 8449. The common wrong answers are 8300 and 8500 (using a whole 100) or 8499 (treating 'nearest 100' like 'nearest 1000').

Why isn't £12 to the nearest pound exactly £12?

Because £12 is rounded, so it stands for a range of true amounts. Any amount from £11.50 up to (but not reaching) £12.50 rounds to £12. The least it could be is the lower bound £11.50; the most is just under £12.50. Treating £12 as the exact amount is the core error examiners flag in 'show she has enough' questions.

How do I show someone 'definitely' has enough money when the amount is rounded?

Use the guaranteed minimum — the lower bound. If Rosie has £12 to the nearest pound, the least she could truly have is £11.50. Six drinks at £1.89 cost £11.34. Because £11.34 is less than £11.50, even with the least she could have she can afford them, so she definitely has enough. Comparing against the £12 face value ('66p left') is unsafe — it ignores that £12 is rounded.

GCSE Maths Foundation · AQA · Rounding & bounds

Error intervals and bounds: why £12 isn't exactly £12

A rounded number is the most ordinary thing on a Foundation paper — an attendance to the nearest hundred, a price to the nearest pound, a length to the nearest centimetre. The trap is reading it as if it were exact. AQA examiners flag the same errors every series: using a rounded figure at face value, going a whole unit either side instead of half, and writing "...99" for the top of the range. On 1F JUN24 Q20 the examiner reported that "common, incorrect answers were 8300 and 8500 or 8499".

The thirty-second fix is to treat the rounded number as an interval. It reaches half the rounding unit either side, with the lower edge included and the upper edge excluded. Once you read every rounded figure as a range, the bound questions and the "show she definitely has enough" questions collapse into one move.

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How to spot it in your own work

You wrote that 8400 to the nearest 100 has a maximum of $8499$ (or a minimum of $8300$ ).
You took "£12 to the nearest pound" as exactly $\pounds 12.00$ and reasoned "so there's 66p left".
You gave the bounds of a value as a whole unit either side, rather than half a unit.
You weren't sure whether the lower edge is included in the range or not.

An exam question that triggers it

Here is AQA 1F JUN24 Q20, the question the examiners singled out:

The attendance for a rugby match is 8400 people, to the nearest 100.

(a) Write down the minimum possible attendance.

(b) Write down the maximum possible attendance.

The common wrong answers are $8300$ and $8500$ (a whole 100 either side), or $8499$ for the maximum (reading "nearest 100" as if it were "nearest 1000").

The correct answers are $8350$ and $8449$ . Half of 100 is 50, so the range runs from $8400-50=8350$ up to $8400+50=8450$ . But 8450 itself rounds up to 8500, and you can't have part of a person, so the greatest whole-number attendance is $8449$ .

Why students fall for this

For most of school, a number simply is its value. The idea that a stated figure hides a range of true values is genuinely new, and it arrives late — this is a grade 4–5 idea sitting on top of place value. So students treat the rounding instruction ("to the nearest 100") as a label on an exact number rather than as a statement about a range. "8400" means 8400.

The second error is using the wrong gap. Students correctly sense that there is some room either side, but reach for the whole rounding unit (100) instead of half of it (50) — or, on a calculator paper, take the rounded amount at face value in a decision. On 3F NOV24 Q22 the examiner noted that students "stated 'Rosie has enough with 66p left' rather than dealing with the bounds of accuracy aspect": the £12 was rounded, but they spent it as if it were exact.

The fix: Treat the rounded number as an interval — half a unit either side

A rounded number stands for an interval of true values, reaching half the rounding unit either side of the stated figure. Find the rounding unit, halve it, then go that far down for the lower bound and that far up for the upper edge. The lower edge is included (it rounds up to the figure); the upper edge is not (it rounds up to the next figure), so it is a limit the true value approaches but never reaches.

This single rule fixes both failure modes. For a least/greatest question, subtract and add half the unit (then step back to the last whole item if the upper edge rounds up, as with whole people). For a "show she definitely has enough" decision, compare the cost against the guaranteed minimum — the lower bound — never the rounded figure itself: £12 to the nearest pound guarantees only $\pounds 11.50$ , so $\pounds 11.34$ in drinks is safe because $11.34 < 11.50$ .

Worked example

Show that Rosie definitely has enough. To the nearest pound, Rosie has £12. She wants 6 drinks at £1.89 each. (AQA 3F NOV24 Q22.)

The £12 is rounded, so find the guaranteed minimum. The unit is £1, half a unit is £0.50, so the least Rosie could truly have is the lower bound $12 - 0.50 = \pounds 11.50$ .
Work out the cost.
$6 \times \pounds 1.89 = \pounds 11.34$
Compare the cost against the guaranteed minimum. $\pounds 11.34 < \pounds 11.50$ , so even with the least she could have, Rosie can afford the drinks. Therefore she definitely has enough.
Why "66p left" is wrong. That answer takes £12 at face value ( $12 - 11.34 = 0.66$ ). If Rosie actually held only $\pounds 11.50$ — still "£12 to the nearest pound" — there would be just 16p left, not 66p. Only the lower bound proves the "definitely".

The same half-a-unit move handles the bound questions: 8400 to the nearest 100 runs from $8350$ to an upper edge of $8450$ (max whole person $8449$ ); 13 cm to the nearest cm runs from $12.5$ cm to $13.5$ cm.

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Common questions

What is the maximum possible attendance for 8400 to the nearest 100?: $8449$ . The upper edge is $8400 + 50 = 8450$ , but 8450 rounds up to 8500, so it is not in the range. The greatest whole-number attendance below it is $8449$ . The answer $8499$ is the common trap — it reads "nearest 100" as if it were "nearest 1000".
Is the lower bound included in the interval?: Yes. The lower bound is the smallest value that still rounds up to the figure, so it is included. For 8400 to the nearest 100, 8350 rounds to 8400, so 8350 is in. The upper edge (8450) rounds up to the next figure, so it is not included — it is the limit the true value can approach but not reach. This is why bounds are often written $8350 \le x < 8450$ .
Why is it half the unit and not the whole unit?: Because rounding sends each value to the nearest figure. The boundary between rounding down to 8400 and up to 8500 sits exactly halfway, at $8450$ — half of 100 above 8400. Likewise the boundary below is halfway down, at $8350$ . So the interval is half a unit either side; $8300$ and $8500$ (a whole unit) are the classic wrong answers.
How do I answer a "show she definitely has enough" question?: Use the lower bound — the guaranteed minimum — not the rounded amount. If Rosie has £12 to the nearest pound, the least she could have is $\pounds 11.50$ . Compare the cost ( $6 \times \pounds 1.89 = \pounds 11.34$ ) against £11.50: since $11.34 < 11.50$ , she can afford it even at the minimum, so she definitely has enough. "£12, so 66p left" uses the figure at face value and earns no marks for the bounds reasoning.

Related misconceptions

Decimal place valueBounds rest on place value: half of £1 is £0.50, half of 100 is 50.
Fraction additiveThe other place careless half-unit reasoning bites — adding parts without a common whole.
Fraction of as a labelReading a figure as a fixed label rather than an operation is the same face-value habit.

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