To prove an area is at least a given value, which bounds do you test?

All lower bounds, the smallest the area could be. An 'at least' claim is hardest to satisfy at the smallest case, so if even that clears the bar, every case does. For a floor 2.4 m by 2.9 m to 2 significant figures, the smallest area is 2.35 x 2.85 = 6.6975, which is at least 6.51, so the claim is proved. The nominal 2.4 x 2.9 = 6.96 reaches the same verdict but proves nothing.

To prove a total stays safely under a limit, which bounds do you test?

All upper bounds, the heaviest the total could be. A 'stays under a limit' claim is hardest at the largest case. For a van loaded with 1200 kg to the nearest 50 kg plus a table of 140 kg to the nearest 10 kg, the heaviest total is 1225 + 145 = 1370, which is under the 1375 kg limit, so it is safe. The stated 1200 + 140 = 1340 proves nothing.

Why does the nominal answer earn no marks even when the verdict is correct?

Because it tests only one possibility, the rounded values, not the worst case. The claim is about every possible measurement, so a proof must show even the hardest case clears the bar. The nominal answer can reach the right verdict by luck and still earn no bounds marks, because it has not tested the case that could fail.

GCSE Maths Higher · AQA · Bounds & error intervals

Worst-case bound in a decision: test the hardest case, not the rounded values

When a question says show that the area is at least X, she definitely has enough, or it can be safely added, it is a worst-case question. The instinct is to test the rounded values you were given and check the verdict. But those values are just one possibility, and the claim is about every possible measurement.

To prove it, you test the single hardest case. An at least claim is hardest when the quantities are smallest, so you test all lower bounds. A stays under a limit claim is hardest when the quantities are largest, so you test all upper bounds. The nominal answer often reaches the right verdict but proves nothing, so it earns no bounds marks: a reliable grade 7 to 9 separator.

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How to spot it in your own work

You proved an at least claim by multiplying or adding the measured values, instead of the smallest case.
You proved a stays under a limit claim from the stated values, not the heaviest case.
You used all lower bounds for a "stays under" claim, or all upper bounds for an "at least" claim, the wrong extreme.
You reached the right yes or no, but did not test the case that could have failed, so the proof is missing.

An exam question that triggers it

Here is the structure of JUN22 Paper 2 Question 27:

A floor measures $2.4$ m by $2.9$ m, each to 2 significant figures.

To be rented, the area must be at least $6.51$ m squared. Show that the floor is large enough.

The misconception tests the nominal values, $2.4 \times 2.9 = 6.96$ , and stops. The fix reads "at least" as a worst-case claim and tests the smallest area, both sides at their lower bound: $2.35 \times 2.85 = 6.6975$ . Since $6.6975 \ge 6.51$ , it is proved for every possible measurement.

Why students fall for this

"Show that the area is at least 15" reads like "work out the area". The student computes the area from the figures given, checks it clears the bar, and feels finished, never noticing that the figures were rounded and the real value could be different.

The worst-case idea is counter-intuitive. To prove something always holds, you deliberately pick the case most likely to break it. Students expect a proof to use the "normal" values, not the extreme ones.

Even students who know to use bounds often pick the wrong extreme. They push everything up out of habit, when an "at least" claim needs everything pushed down, so the same reflex that helps one claim hurts the other.

The fix: Read the claim direction, test the single worst case

Step 1: read the claim direction. Is it "at least / definitely enough", or "safely / stays under a limit"?

Step 2: for "at least", test all lower bounds. The claim is hardest when the quantities are smallest, so compute the smallest possible result.

Step 3: for "stays under a limit", test all upper bounds. The claim is hardest when the quantities are largest, so compute the largest possible result.

Step 4: conclude only if the worst case clears the bar. If even the hardest case satisfies the claim, it is proved for every possible measurement. The nominal answer is not the proof.

Worked example

JUN22·2·27 structure (an "at least" claim, all lower bounds): a floor $2.4$ m by $2.9$ m to 2 s.f. Prove the area is at least $6.51$ m squared.

Intervals. $2.4 \in [2.35, 2.45)$ and $2.9 \in [2.85, 2.95)$ .
Test the smallest area. "At least" is hardest at the smallest, so both sides at their lower bound.
$2.35 \times 2.85 = 6.6975 \ge 6.51$
Proved. Trap: the nominal $2.4 \times 2.9 = 6.96$ proves nothing.

NOV23·2·21 structure (a "stays under" claim, all upper bounds): a van is safe up to $1375$ kg, loaded with $1200$ kg (nearest 50) plus a table of $140$ kg (nearest 10). Can it be safely added?

Intervals. $1200 \in [1175, 1225)$ and $140 \in [135, 145)$ .
Test the heaviest total. "Stays under" is hardest at the heaviest, so both at their upper bound.
$1225 + 145 = 1370 < 1375$
Safe. Trap: the stated $1200 + 140 = 1340$ proves nothing, and the wrong bounds $1250 + 150 = 1400$ would give a false no.

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Common questions

For an "at least" claim, which bounds do you test?: All lower bounds, the smallest the result could be. If even the smallest case clears the bar, every case does. For the floor, $2.35 \times 2.85 = 6.6975 \ge 6.51$ , not the nominal $2.4 \times 2.9 = 6.96$ .
For a "stays safely under a limit" claim, which bounds do you test?: All upper bounds, the largest the result could be. For the van, $1225 + 145 = 1370 < 1375$ , not the stated $1200 + 140 = 1340$ .
Why does the nominal answer earn no marks if the verdict is right?: Because it only tests one possibility, not the worst case. The claim is about every possible measurement, so a proof must show even the hardest case clears the bar. A correct verdict from the nominal values is luck, not proof.

Related misconceptions

Bounds of a calculationA subtracted or divided-by quantity takes the opposite bound: A minus B is biggest when B is smallest, and A over B is biggest when B is smallest, so the all-upper habit is wrong.
Error intervals and boundsWriting the half-unit error interval for a rounded value, the foundation skill that this worst-case decision work builds on.

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