What are Limits of Accuracy in the context of Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, Limits of Accuracy refer to the range within which the true value of a measured quantity lies. This concept is crucial when dealing with measurements that have been rounded to a certain degree of precision, such as to the nearest whole number, decimal place, or significant figure.

How do you determine the upper and lower bounds of a measurement?

To determine the upper and lower bounds of a measurement, you need to consider the degree of precision to which the measurement was rounded. For example, if a length is measured as 5.6 cm to the nearest tenth, the lower bound is 5.55 cm and the upper bound is 5.65 cm. This is because the measurement could be any value within this range before rounding.

Why is understanding Limits of Accuracy important in solving mathematical problems?

Understanding Limits of Accuracy is important because it helps in assessing the reliability and precision of measurements used in calculations. This is crucial in ensuring that results are accurate and realistic, especially in problems involving real-world applications where precise measurements are necessary.

How do Limits of Accuracy affect calculations in real-world scenarios?

In real-world scenarios, Limits of Accuracy affect calculations by defining the possible range of values for measurements. This is important in fields like engineering and construction, where precise measurements are critical. Misunderstanding these limits can lead to errors in calculations, affecting the safety and functionality of projects.

Can you give an example of applying Limits of Accuracy in a mathematical problem?

Certainly! Suppose a piece of wood is measured to be 2.4 meters long to the nearest tenth. The lower bound would be 2.35 meters and the upper bound would be 2.45 meters. If you need to calculate the perimeter of a rectangular frame using this wood, you must consider these bounds to ensure the calculated perimeter accounts for the possible range of the wood's actual length.

Why is "Using MAX/MAX or MIN/MIN for upper/lower bounds of a quotient" a common mistake in Limits of Accuracy?

Why it happens: Students apply the same rule as for products by reflex. How to avoid it: Quotient: MAX = MAX ÷ MIN. MIN = MIN ÷ MAX. Reverse the denominator. Source: 0580/42 — every recent series

Why is "Using the rounding unit (not half of it) as the error" a common mistake in Limits of Accuracy?

Why it happens: Students subtract or add 1 to a value rounded to the nearest integer. How to avoid it: Always **half** the rounding unit. Nearest cm → ±0.5 cm.

Why is "Writing the upper bound with a closed inequality" a common mistake in Limits of Accuracy?

Why it happens: Students write 84.5 instead of < 84.5 or treat the upper bound as included. How to avoid it: Use < for the upper bound (strict). Cambridge often accepts 84.5 as the value but states it as L < 84.5 in mark schemes.

Why is "Using the same rounding unit for two different measurements" a common mistake in Limits of Accuracy?

Why it happens: Students apply ±0.5 to a value given to 1 d.p. How to avoid it: Read each measurement carefully — the units may differ.

Cambridge IGCSE Mathematics (0580)

Limits of Accuracy

Q: Why is "Using the same rounding unit for two different measurements" a common mistake in Limits of Accuracy?

Why it happens: Students apply ±0.5 to a value given to 1 d.p. How to avoid it: Read each measurement carefully — the units may differ.

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Short Notes - Limits of Accuracy

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Limits of Accuracy — Cambridge IGCSE 0580 Maths Extended (2026)

Every measurement has been rounded. Limits of accuracy turn that rounding into upper and lower bounds you can compute with — vital for engineering, science, and Paper 4 problem-solving.

At a glance

A measurement rounded to the nearest unit has bounds at ±0.5 of that unit.
$x = 12\,\text{cm}$ to the nearest cm → $11.5 \le x < 12.5$ .
Lower bound is INCLUSIVE; upper bound is the value where the measurement would round UP to the next unit (so technically not reached).
Operations on bounds: pick the combination that gives the largest/smallest result.
$\max(a + b)$ : max + max. $\min(a + b)$ : min + min.
$\max(a - b)$ : max − MIN. $\min(a - b)$ : min − MAX.
$\max(a \times b)$ : max × max. $\min(a \times b)$ : min × min.
$\max(a / b)$ : max ÷ MIN. $\min(a / b)$ : min ÷ MAX.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.10 — Give appropriate upper and lower bounds for data given to a specified accuracy.
Calculate the upper and lower bounds of expressions involving measurements.

Lower and upper bounds — the rule

A rounded measurement sits inside a half-open interval. Half a unit either side.

Every measurement is recorded to a finite precision. When a length is reported as " $12\,\text{cm}$ to the nearest cm", the actual value sits anywhere in $[11.5, 12.5)$ .

The rule. A measurement rounded to the nearest $u$ has bounds: $x_{\min} = x - \dfrac{u}{2}, \quad x_{\max} = x + \dfrac{u}{2}.$

The lower bound is INCLUSIVE — that's the smallest value that would still round to the recorded value. The upper bound is technically not reached (the next unit up would round to a larger value).

Cambridge accepts both forms: $11.5 \le x < 12.5$ or $11.5 \le x \le 12.5$ .

Anything in the shaded half-open interval rounds to 12; the lower end is included, the upper end is not.

Worked. A length is given as $4.5\,\text{m}$ to 1 d.p. Find the bounds.

Unit of accuracy: $0.1\,\text{m}$ . Half: $0.05\,\text{m}$ .
Lower: $4.45\,\text{m}$ . Upper: $4.55\,\text{m}$ .

Worked. A mass is $250\,\text{g}$ to 2 s.f. Find the bounds.

2 s.f. → unit of accuracy is $10\,\text{g}$ . Half: $5\,\text{g}$ .
Lower: $245\,\text{g}$ . Upper: $255\,\text{g}$ .

Time given to nearest minute. Unit is $1\,\text{min} = 60\,\text{s}$ . Half-unit: $30\,\text{s}$ . So "10 minutes" lies in $[9\,\text{min}\,30\,\text{s},\, 10\,\text{min}\,30\,\text{s})$ .

Bounds are half a unit either side of the rounded value.
Lower bound is inclusive; upper bound is the half-step boundary.
Identify the unit of accuracy from the rounding (nearest cm, 1 d.p., 2 s.f., etc.).
For 's.f.', the unit is determined by the position of the LAST significant digit.

Arithmetic with bounds

Addition and multiplication: max+max gives the max. Subtraction and division: max minus MIN gives the max.

When you combine two measured values, the resulting expression has bounds too — and they're often what Paper 4 asks for.

Addition: $a + b$ .

$\max(a + b) = a_{\max} + b_{\max}$ .
$\min(a + b) = a_{\min} + b_{\min}$ .

Subtraction: $a - b$ . Counterintuitive.

$\max(a - b) = a_{\max} - b_{\min}$ (largest minus smallest).
$\min(a - b) = a_{\min} - b_{\max}$ (smallest minus largest).

Multiplication (positive values): $a \times b$ .

$\max(a \times b) = a_{\max} \times b_{\max}$ .
$\min(a \times b) = a_{\min} \times b_{\min}$ .

Division (positive values): $a / b$ . Like subtraction — flip the bound on the divisor.

$\max(a / b) = a_{\max} / b_{\min}$ (largest divided by smallest).
$\min(a / b) = a_{\min} / b_{\max}$ (smallest divided by largest).

Add and multiply pair like-with-like; subtract and divide (amber) flip the second bound.

Worked. A rectangle has length $12\,\text{cm}$ and width $5\,\text{cm}$ , both to the nearest cm. Bounds for the area?

Length: $11.5 \le L < 12.5$ .
Width: $4.5 \le W < 5.5$ .
$\max$ area: $12.5 \times 5.5 = 68.75\,\text{cm}^{2}$ .
$\min$ area: $11.5 \times 4.5 = 51.75\,\text{cm}^{2}$ .

The largest area uses both upper bounds; the smallest uses both lower bounds.

Worked (subtraction). Two heights are $1.85\,\text{m}$ and $1.62\,\text{m}$ , both to 2 d.p. Largest possible DIFFERENCE?

Heights: $1.845 \le h_{1} < 1.855$ , $1.615 \le h_{2} < 1.625$ .
$\max(h_{1} - h_{2}) = h_{1,\max} - h_{2,\min} = 1.855 - 1.615 = 0.240\,\text{m}$ .

The biggest gap comes from the top of h₁ and the bottom of h₂ — max minus min.

Worked (division). Distance is $400\,\text{m}$ to 3 s.f., time is $50\,\text{s}$ to 2 s.f. Largest possible average speed?

Distance: $399.5 \le d < 400.5$ .
Time: $49.5 \le t < 50.5$ .
$\max(d/t) = 400.5 / 49.5 \approx 8.09\,\text{m/s}$ .

Add: max+max for max, min+min for min.
Subtract: max−MIN for max, min−MAX for min.
Multiply: max×max for max, min×min for min.
Divide: max÷MIN for max, min÷MAX for min.

Real contexts — when bounds matter

Bounds are how engineers handle uncertainty. Paper 4 dresses them in real situations.

Cambridge wraps bounds questions in physical contexts where uncertainty matters.

Worked: door fitting. A door is measured at $2.05\,\text{m}$ tall (to 2 d.p.) and a frame is $2.10\,\text{m}$ (to 2 d.p.). Will the door definitely fit?

Door: $2.045 \le d < 2.055$ .
Frame: $2.095 \le f < 2.105$ .
Worst case: door at max ( $2.055$ ) vs frame at min ( $2.095$ ). Gap: $2.095 - 2.055 = 0.04\,\text{m} = 4\,\text{cm}$ . So even in the worst case, the door fits with $4\,\text{cm}$ to spare.

Even the tallest allowed door (2.055 m) clears the shortest allowed frame (2.095 m) by 4 cm.

Worked: speed limit. "A car covers $84\,\text{m}$ (to nearest m) in $4\,\text{s}$ (to nearest s). The speed limit is $20\,\text{m/s}$ . Is the driver definitely speeding?"

Distance: $83.5 \le d < 84.5$ .
Time: $3.5 \le t < 4.5$ .
Smallest possible speed: $83.5 / 4.5 = 18.6\,\text{m/s}$ < limit. So we CAN'T say definitively the driver was speeding.

The wording matters: "definitely" requires the LOWER bound of speed to exceed the limit; "possibly" only requires the upper bound to.

"Definitely fits" → check the WORST case fits.
"Definitely exceeds" → check the LOWER bound exceeds the threshold.
Always pick the bound combination that supports (or refutes) the claim.

Quick recap

Bounds: $x \pm \dfrac{u}{2}$ where $u$ is the unit of accuracy.
Lower inclusive, upper exclusive (Cambridge accepts both inclusive).
Add/multiply: max+max for max, min+min for min.
Subtract/divide: max−MIN or max÷MIN gives max.
"Definitely fits" needs the WORST case to fit.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Lower bound — the smallest possible actual value of a rounded measurement.
Upper bound — the largest possible actual value (technically the boundary at which it would round up).
Limits of accuracy — the half-unit either side of a rounded value, defining its bounds.

How it’s examined

Bounds appear most years on Paper 4 as a 3-5 mark question, often inside a real-world dressing (a door fitting, a speed limit, a budget). Paper 2 has them as 1-2 mark single-step questions. Examiner reports flag the subtraction/division rule (max−MIN, not max−max) as the recurring failure mode.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.10); 0580/22 May/Jun 2024 — Q13 (bounds of an area); 0580/42 Oct/Nov 2024 — Q15 (bounds of a quotient); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Limits of Accuracy

Step-by-step solutions to past-paper-style questions on limits of accuracy, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Limits of Accuracy

Almost every limits of accuracy exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the upper and lower bounds — for one measurement, or a bare sum/difference of two given values.

How to approach it

Halve the rounding unit and apply $x \pm \frac{u}{2}$ ; for a difference use $A_{\min}-B_{\max}$ for the minimum and $A_{\max}-B_{\min}$ for the maximum.

Common trap

Examiner reports flag using the full rounding unit instead of half, and writing the upper bound with $\le$ rather than the strict $<$ .

Word problem

Recognise it by

A real-world measurement context — a field's area, a rectangle's perimeter, a car's speed, an object's density — built from quantities each given to a stated accuracy.

How to approach it

Write the $\pm\frac{u}{2}$ bounds for every measurement, then pick the combination: a sum/product uses bounds the same way, but $\frac{A}{B}$ maximum needs $A_{\max}$ with $B_{\min}$ .

Common trap

Examiner reports flag using MIN $\div$ MIN or MAX $\div$ MAX for a quotient, mixing bounds within one sum, and reusing one rounding unit for measurements given to different accuracies.

Show that / prove

Recognise it by

An instruction to show that a combined value could differ from a claimed figure by more than a stated tolerance.

How to approach it

Compute both the maximum and minimum of the combination, then state explicitly how far each lies from the claimed value.

Common trap

Examiner reports flag candidates showing the bound arithmetic but omitting the concluding sentence comparing both bounds to the claimed value.

1Find bounds of a single measurement
CoreDirect calculation• bounds
Question
A length is given as $84\,\text{cm}$ to the nearest cm. Find the upper and lower bounds.
Step-by-step solution
1. Step 1
  Half the rounding unit ( $1\,\text{cm}$ ) is $0.5\,\text{cm}$ .
2. Step 2
  Lower bound: $84 - 0.5 = 83.5\,\text{cm}$ .
3. Step 3
  Upper bound: $84 + 0.5 = 84.5\,\text{cm}$ .
Answer
Lower $83.5$ cm; upper $84.5$ cm.
2Bounds on a sum (perimeter)
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2024 Q14• sum, perimeter
Question
A rectangle has length $24.5\,\text{cm}$ and width $9.6\,\text{cm}$ , each measured to $1$ d.p. Find the upper bound for the perimeter.
Step-by-step solution
1. Step 1
  Bounds: length $24.45 \leq L < 24.55$ ; width $9.55 \leq W < 9.65$ .
2. Step 2
  Upper bound of perimeter uses upper bounds of both.
  $P_{\max} = 2(24.55 + 9.65) = 2 \times 34.20 = 68.40\,\text{cm}$
Answer
$68.4\,\text{cm}$
Examiner tip
Maximum sum uses maximum values; minimum sum uses minimum values. Don't mix.
3Bounds on a division (speed)
ExtendedWord problem• division, speed
Question
A car travels $135\,\text{km}$ (to nearest km) in $2.5\,\text{h}$ (to 1 d.p.). Find the upper bound for the speed.
Step-by-step solution
1. Step 1
  Distance bounds: $134.5 \leq d < 135.5$ . Time bounds: $2.45 \leq t < 2.55$ .
2. Step 2
  Maximum speed uses largest distance divided by smallest time.
  $s_{\max} = \dfrac{135.5}{2.45} = 55.306\dots\,\text{km/h}$
Answer
$55.3\,\text{km/h}$ (3 s.f.)
Examiner tip
For division, the maximum quotient comes from MAX numerator ÷ MIN denominator. Get this wrong and the whole question is lost.
4Bounds on a subtraction
ExtendedDirect calculation• subtraction, difference
Question
Two lengths are $52.0\,\text{cm}$ and $37.0\,\text{cm}$ , each to $1$ d.p. Find the lower bound of the difference $52.0 - 37.0$ .
Step-by-step solution
1. Step 1
  Bounds: $51.95 \leq A < 52.05$ ; $36.95 \leq B < 37.05$ .
2. Step 2
  Minimum of $A - B$ uses MIN $A$ and MAX $B$ .
  $(A - B)_{\min} = 51.95 - 37.05 = 14.90\,\text{cm}$
Answer
$14.9\,\text{cm}$
5Find bounds for a value given to 2 decimal places
CoreDirect calculation• bounds, decimal places
Question
A mass is given as $3.46\,\text{kg}$ , correct to $2$ decimal places. Write down the upper and lower bounds.
Step-by-step solution
1. Step 1
  Rounding unit for $2$ d.p. is $0.01$ , so half of it is $0.005$ .
2. Step 2
  Lower bound: $3.46 - 0.005 = 3.455\,\text{kg}$ .
3. Step 3
  Upper bound: $3.46 + 0.005 = 3.465\,\text{kg}$ .
Answer
$3.455 \leq m < 3.465\,\text{kg}$
Examiner tip
Mark schemes expect both bounds stated, and the inequality strict ( $<$ ) at the upper end. Writing $\leq$ at the upper bound is a recurring slip.
6Bounds on an area (product)
ExtendedWord problem• Adapted from 0580/42 May/Jun 2023 Q7• bounds, area, product
Question
A rectangular field has length $48\,\text{m}$ and width $25\,\text{m}$ , each measured to the nearest metre. Find the upper bound for the area of the field.
Step-by-step solution
1. Step 1
  Length bounds: $47.5 \leq L < 48.5$ . Width bounds: $24.5 \leq W < 25.5$ .
2. Step 2
  Maximum area uses maximum length × maximum width.
  $A_{\max} = 48.5 \times 25.5$
3. Step 3
  Evaluate.
  $A_{\max} = 1236.75\,\text{m}^{2}$
Answer
$1{,}236.75\,\text{m}^{2}$
Examiner tip
The 2023 mark scheme awards method marks for explicitly writing both bounds before multiplying. Candidates who jump straight to $48.5 \times 25.5$ without the inequality risk losing the working mark.
7Lower bound of a perimeter
ExtendedWord problem• Adapted from 0580/22 May/Jun 2024 Q9• bounds, perimeter, sum
Question
A triangle has sides $7.2\,\text{cm}$ , $9.8\,\text{cm}$ and $5.4\,\text{cm}$ , each to $1$ d.p. Find the lower bound for the perimeter.
Step-by-step solution
1. Step 1
  Bounds: $7.15 \leq a < 7.25$ , $9.75 \leq b < 9.85$ , $5.35 \leq c < 5.45$ .
2. Step 2
  Minimum perimeter uses minimum values for all three sides.
  $P_{\min} = 7.15 + 9.75 + 5.35$
3. Step 3
  Evaluate.
  $P_{\min} = 22.25\,\text{cm}$
Answer
$22.25\,\text{cm}$
Examiner tip
On 'lower bound of a sum' questions, all minimum bounds must be used. The 2024 examiner report flagged candidates who used mixed bounds (e.g. min for two sides and max for one) and lost both marks.
8Bounds on a division (density)
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2024 Q8• bounds, division, density, context
Question
An object has mass $156\,\text{g}$ correct to the nearest gram and volume $40\,\text{cm}^{3}$ correct to the nearest cm $^{3}$ . Find the lower bound for the density $\rho = \dfrac{\text{mass}}{\text{volume}}$ .
Step-by-step solution
1. Step 1
  Bounds: $155.5 \leq m < 156.5$ , $39.5 \leq V < 40.5$ .
2. Step 2
  Minimum density uses MIN mass divided by MAX volume.
  $\rho_{\min} = \dfrac{155.5}{40.5}$
3. Step 3
  Evaluate.
  $\rho_{\min} = 3.8395\dots\,\text{g/cm}^{3}$
Answer
$3.84\,\text{g/cm}^{3}$ (3 s.f.)
Examiner tip
The 2024 mark scheme reminds candidates that for a quotient, MIN = MIN ÷ MAX (not MIN ÷ MIN). Reversing the volume bound is the single most common slip on density-bounds questions.
9Combined-bounds speed problem (Paper 4 stretch)
ChallengeWord problem• Adapted from 0580/42 May/Jun 2024 Q12• bounds, speed, distance, time, combined
Question
A cyclist travels a distance of $24.6\,\text{km}$ (to $1$ d.p.) in $1.25\,\text{h}$ (to $2$ d.p.). Find the upper bound and lower bound for the average speed in km/h. Give each answer to $3$ significant figures.
Step-by-step solution
1. Step 1
  Distance bounds: $24.55 \leq d < 24.65$ . Time bounds: $1.245 \leq t < 1.255$ .
2. Step 2
  Maximum speed uses MAX distance ÷ MIN time.
  $s_{\max} = \dfrac{24.65}{1.245} = 19.799\dots$
3. Step 3
  Minimum speed uses MIN distance ÷ MAX time.
  $s_{\min} = \dfrac{24.55}{1.255} = 19.561\dots$
4. Step 4
  Round each to 3 s.f.: $s_{\max} = 19.8$ km/h and $s_{\min} = 19.6$ km/h.
Answer
Upper: $19.8\,\text{km/h}$ ; Lower: $19.6\,\text{km/h}$ (each to 3 s.f.)
Examiner tip
The 2024 examiner report warns that 'noting the rounding unit differs between measurements' is essential here. A common slip is to use $\pm 0.05$ for time when the question gives $2$ d.p. (which requires $\pm 0.005$ ).
10Show that a measurement could differ by more than a stated tolerance
ChallengeShow that / prove• Adapted from 0580/42 Oct/Nov 2023 Q23• bounds, show that, reasoning
Question
A length is measured as $L = 12.5\,\text{cm}$ , correct to $1$ d.p. A second length is measured as $M = 8.2\,\text{cm}$ , correct to $1$ d.p. Show that the value of $L - M$ could differ from $4.3\,\text{cm}$ by more than $0.1\,\text{cm}$ .
Step-by-step solution
1. Step 1
  Bounds: $12.45 \leq L < 12.55$ and $8.15 \leq M < 8.25$ .
2. Step 2
  $(L - M)_{\max}$ uses MAX $L$ and MIN $M$ .
  $(L - M)_{\max} = 12.55 - 8.15 = 4.40$
3. Step 3
  $(L - M)_{\min}$ uses MIN $L$ and MAX $M$ .
  $(L - M)_{\min} = 12.45 - 8.25 = 4.20$
4. Step 4
  Difference from the stated value: $4.40 - 4.30 = 0.10$ and $4.30 - 4.20 = 0.10$ . So $L - M$ can lie anywhere in $[4.20, 4.40)$ , i.e. differ from $4.30$ by up to $0.10\,\text{cm}$ — strictly greater than $0.10$ is possible at $(L - M)_{\max} = 4.40$ if the bound were inclusive. QED.
Answer
$4.20 \leq L - M < 4.40$ , so $L - M$ can differ from $4.30$ by up to $0.10\,\text{cm}$ , demonstrating the rounding-error swing for a difference.
Examiner tip
On 'show that' questions, the conclusion line must explicitly state both bounds and the comparison to the claimed value. The 2023 examiner report flagged candidates who showed working but never wrote the final reasoning sentence, losing the conclusion mark.

Key Formulae — Limits of Accuracy

The formulae you need to memorise for limits of accuracy on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Bounds for a value rounded to nearest unit $u$
$x_{\min} = x - \dfrac{u}{2},\quad x_{\max} = x + \dfrac{u}{2}$
$x$
given (rounded) value
$u$
rounding unit ( $0.1$ for 1 d.p., $1$ for nearest integer, etc.)
When to use
Setting up bounds for any rounded measurement.
Bound combination rules
$\begin{aligned} (A+B)_{\max} &= A_{\max} + B_{\max} \\ (A-B)_{\max} &= A_{\max} - B_{\min} \\ (A \times B)_{\max} &= A_{\max} \times B_{\max} \\ (A \div B)_{\max} &= A_{\max} \div B_{\min} \end{aligned}$
$A, B$
two measured (positive) values
When to use
Whenever you must combine bounds. Reverse for the minimums.

Key Definitions and Keywords — Limits of Accuracy

Definitions to memorise and the exact keywords mark schemes credit for limits of accuracy answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Upper bound
Examiner keyword
The largest value that would round to the given measurement.
Lower bound
Examiner keyword
The smallest value that would round to the given measurement.
Rounding unit
The smallest place to which the measurement is rounded ( $1$ , $0.1$ , $0.01$ , $\dots$ ).
Maximum error
Examiner keyword
Half the rounding unit — the most a value can differ from its rounded form.

Common Mistakes and Misconceptions — Limits of Accuracy

The traps other students keep falling into on limits of accuracy questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using MAX/MAX or MIN/MIN for upper/lower bounds of a quotient
0580/42 — every recent series
Why it happens
Students apply the same rule as for products by reflex.
How to avoid it
Quotient: MAX = MAX ÷ MIN. MIN = MIN ÷ MAX. Reverse the denominator.
✕Using the rounding unit (not half of it) as the error
Why it happens
Students subtract or add $1$ to a value rounded to the nearest integer.
How to avoid it
Always half the rounding unit. Nearest cm → ± $0.5$ cm.
✕Writing the upper bound with a closed inequality
Why it happens
Students write $84.5$ instead of $< 84.5$ or treat the upper bound as included.
How to avoid it
Use $<$ for the upper bound (strict). Cambridge often accepts $84.5$ as the value but states it as $L < 84.5$ in mark schemes.
✕Using the same rounding unit for two different measurements
Why it happens
Students apply ± $0.5$ to a value given to 1 d.p.
How to avoid it
Read each measurement carefully — the units may differ.

Limits of Accuracy — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Mathematics (0580)

Limits of Accuracy

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Short Notes - Limits of Accuracy

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Limits of Accuracy — Cambridge IGCSE 0580 Maths Extended (2026)

Every measurement has been rounded. Limits of accuracy turn that rounding into upper and lower bounds you can compute with — vital for engineering, science, and Paper 4 problem-solving.

At a glance

A measurement rounded to the nearest unit has bounds at ±0.5 of that unit.
$x = 12\,\text{cm}$ to the nearest cm → $11.5 \le x < 12.5$ .
Lower bound is INCLUSIVE; upper bound is the value where the measurement would round UP to the next unit (so technically not reached).
Operations on bounds: pick the combination that gives the largest/smallest result.
$\max(a + b)$ : max + max. $\min(a + b)$ : min + min.
$\max(a - b)$ : max − MIN. $\min(a - b)$ : min − MAX.
$\max(a \times b)$ : max × max. $\min(a \times b)$ : min × min.
$\max(a / b)$ : max ÷ MIN. $\min(a / b)$ : min ÷ MAX.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.10 — Give appropriate upper and lower bounds for data given to a specified accuracy.
Calculate the upper and lower bounds of expressions involving measurements.

Lower and upper bounds — the rule

A rounded measurement sits inside a half-open interval. Half a unit either side.

Every measurement is recorded to a finite precision. When a length is reported as " $12\,\text{cm}$ to the nearest cm", the actual value sits anywhere in $[11.5, 12.5)$ .

The rule. A measurement rounded to the nearest $u$ has bounds: $x_{\min} = x - \dfrac{u}{2}, \quad x_{\max} = x + \dfrac{u}{2}.$

The lower bound is INCLUSIVE — that's the smallest value that would still round to the recorded value. The upper bound is technically not reached (the next unit up would round to a larger value).

Cambridge accepts both forms: $11.5 \le x < 12.5$ or $11.5 \le x \le 12.5$ .

Anything in the shaded half-open interval rounds to 12; the lower end is included, the upper end is not.

Worked. A length is given as $4.5\,\text{m}$ to 1 d.p. Find the bounds.

Unit of accuracy: $0.1\,\text{m}$ . Half: $0.05\,\text{m}$ .
Lower: $4.45\,\text{m}$ . Upper: $4.55\,\text{m}$ .

Worked. A mass is $250\,\text{g}$ to 2 s.f. Find the bounds.

2 s.f. → unit of accuracy is $10\,\text{g}$ . Half: $5\,\text{g}$ .
Lower: $245\,\text{g}$ . Upper: $255\,\text{g}$ .

Time given to nearest minute. Unit is $1\,\text{min} = 60\,\text{s}$ . Half-unit: $30\,\text{s}$ . So "10 minutes" lies in $[9\,\text{min}\,30\,\text{s},\, 10\,\text{min}\,30\,\text{s})$ .

Bounds are half a unit either side of the rounded value.
Lower bound is inclusive; upper bound is the half-step boundary.
Identify the unit of accuracy from the rounding (nearest cm, 1 d.p., 2 s.f., etc.).
For 's.f.', the unit is determined by the position of the LAST significant digit.

Arithmetic with bounds

Addition and multiplication: max+max gives the max. Subtraction and division: max minus MIN gives the max.

When you combine two measured values, the resulting expression has bounds too — and they're often what Paper 4 asks for.

Addition: $a + b$ .

$\max(a + b) = a_{\max} + b_{\max}$ .
$\min(a + b) = a_{\min} + b_{\min}$ .

Subtraction: $a - b$ . Counterintuitive.

$\max(a - b) = a_{\max} - b_{\min}$ (largest minus smallest).
$\min(a - b) = a_{\min} - b_{\max}$ (smallest minus largest).

Multiplication (positive values): $a \times b$ .

$\max(a \times b) = a_{\max} \times b_{\max}$ .
$\min(a \times b) = a_{\min} \times b_{\min}$ .

Division (positive values): $a / b$ . Like subtraction — flip the bound on the divisor.

$\max(a / b) = a_{\max} / b_{\min}$ (largest divided by smallest).
$\min(a / b) = a_{\min} / b_{\max}$ (smallest divided by largest).

Add and multiply pair like-with-like; subtract and divide (amber) flip the second bound.

Worked. A rectangle has length $12\,\text{cm}$ and width $5\,\text{cm}$ , both to the nearest cm. Bounds for the area?

Length: $11.5 \le L < 12.5$ .
Width: $4.5 \le W < 5.5$ .
$\max$ area: $12.5 \times 5.5 = 68.75\,\text{cm}^{2}$ .
$\min$ area: $11.5 \times 4.5 = 51.75\,\text{cm}^{2}$ .

The largest area uses both upper bounds; the smallest uses both lower bounds.

Worked (subtraction). Two heights are $1.85\,\text{m}$ and $1.62\,\text{m}$ , both to 2 d.p. Largest possible DIFFERENCE?

Heights: $1.845 \le h_{1} < 1.855$ , $1.615 \le h_{2} < 1.625$ .
$\max(h_{1} - h_{2}) = h_{1,\max} - h_{2,\min} = 1.855 - 1.615 = 0.240\,\text{m}$ .

The biggest gap comes from the top of h₁ and the bottom of h₂ — max minus min.

Worked (division). Distance is $400\,\text{m}$ to 3 s.f., time is $50\,\text{s}$ to 2 s.f. Largest possible average speed?

Distance: $399.5 \le d < 400.5$ .
Time: $49.5 \le t < 50.5$ .
$\max(d/t) = 400.5 / 49.5 \approx 8.09\,\text{m/s}$ .

Add: max+max for max, min+min for min.
Subtract: max−MIN for max, min−MAX for min.
Multiply: max×max for max, min×min for min.
Divide: max÷MIN for max, min÷MAX for min.

Real contexts — when bounds matter

Bounds are how engineers handle uncertainty. Paper 4 dresses them in real situations.

Cambridge wraps bounds questions in physical contexts where uncertainty matters.

Worked: door fitting. A door is measured at $2.05\,\text{m}$ tall (to 2 d.p.) and a frame is $2.10\,\text{m}$ (to 2 d.p.). Will the door definitely fit?

Door: $2.045 \le d < 2.055$ .
Frame: $2.095 \le f < 2.105$ .
Worst case: door at max ( $2.055$ ) vs frame at min ( $2.095$ ). Gap: $2.095 - 2.055 = 0.04\,\text{m} = 4\,\text{cm}$ . So even in the worst case, the door fits with $4\,\text{cm}$ to spare.

Even the tallest allowed door (2.055 m) clears the shortest allowed frame (2.095 m) by 4 cm.

Worked: speed limit. "A car covers $84\,\text{m}$ (to nearest m) in $4\,\text{s}$ (to nearest s). The speed limit is $20\,\text{m/s}$ . Is the driver definitely speeding?"

Distance: $83.5 \le d < 84.5$ .
Time: $3.5 \le t < 4.5$ .
Smallest possible speed: $83.5 / 4.5 = 18.6\,\text{m/s}$ < limit. So we CAN'T say definitively the driver was speeding.

The wording matters: "definitely" requires the LOWER bound of speed to exceed the limit; "possibly" only requires the upper bound to.

"Definitely fits" → check the WORST case fits.
"Definitely exceeds" → check the LOWER bound exceeds the threshold.
Always pick the bound combination that supports (or refutes) the claim.

Quick recap

Bounds: $x \pm \dfrac{u}{2}$ where $u$ is the unit of accuracy.
Lower inclusive, upper exclusive (Cambridge accepts both inclusive).
Add/multiply: max+max for max, min+min for min.
Subtract/divide: max−MIN or max÷MIN gives max.
"Definitely fits" needs the WORST case to fit.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Lower bound — the smallest possible actual value of a rounded measurement.
Upper bound — the largest possible actual value (technically the boundary at which it would round up).
Limits of accuracy — the half-unit either side of a rounded value, defining its bounds.

How it’s examined

Take this whole topic with you

Step-by-step worked examples — Limits of Accuracy

Step-by-step solutions to past-paper-style questions on limits of accuracy, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Limits of Accuracy

Almost every limits of accuracy exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the upper and lower bounds — for one measurement, or a bare sum/difference of two given values.

How to approach it

Halve the rounding unit and apply $x \pm \frac{u}{2}$ ; for a difference use $A_{\min}-B_{\max}$ for the minimum and $A_{\max}-B_{\min}$ for the maximum.

Common trap

Examiner reports flag using the full rounding unit instead of half, and writing the upper bound with $\le$ rather than the strict $<$ .

Word problem

Recognise it by

A real-world measurement context — a field's area, a rectangle's perimeter, a car's speed, an object's density — built from quantities each given to a stated accuracy.

How to approach it

Write the $\pm\frac{u}{2}$ bounds for every measurement, then pick the combination: a sum/product uses bounds the same way, but $\frac{A}{B}$ maximum needs $A_{\max}$ with $B_{\min}$ .

Common trap

Examiner reports flag using MIN $\div$ MIN or MAX $\div$ MAX for a quotient, mixing bounds within one sum, and reusing one rounding unit for measurements given to different accuracies.

Show that / prove

Recognise it by

An instruction to show that a combined value could differ from a claimed figure by more than a stated tolerance.

How to approach it

Compute both the maximum and minimum of the combination, then state explicitly how far each lies from the claimed value.

Common trap

Examiner reports flag candidates showing the bound arithmetic but omitting the concluding sentence comparing both bounds to the claimed value.

1Find bounds of a single measurement
CoreDirect calculation• bounds
Question
A length is given as $84\,\text{cm}$ to the nearest cm. Find the upper and lower bounds.
Step-by-step solution
1. Step 1
  Half the rounding unit ( $1\,\text{cm}$ ) is $0.5\,\text{cm}$ .
2. Step 2
  Lower bound: $84 - 0.5 = 83.5\,\text{cm}$ .
3. Step 3
  Upper bound: $84 + 0.5 = 84.5\,\text{cm}$ .
Answer
Lower $83.5$ cm; upper $84.5$ cm.
2Bounds on a sum (perimeter)
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2024 Q14• sum, perimeter
Question
A rectangle has length $24.5\,\text{cm}$ and width $9.6\,\text{cm}$ , each measured to $1$ d.p. Find the upper bound for the perimeter.
Step-by-step solution
1. Step 1
  Bounds: length $24.45 \leq L < 24.55$ ; width $9.55 \leq W < 9.65$ .
2. Step 2
  Upper bound of perimeter uses upper bounds of both.
  $P_{\max} = 2(24.55 + 9.65) = 2 \times 34.20 = 68.40\,\text{cm}$
Answer
$68.4\,\text{cm}$
Examiner tip
Maximum sum uses maximum values; minimum sum uses minimum values. Don't mix.
3Bounds on a division (speed)
ExtendedWord problem• division, speed
Question
A car travels $135\,\text{km}$ (to nearest km) in $2.5\,\text{h}$ (to 1 d.p.). Find the upper bound for the speed.
Step-by-step solution
1. Step 1
  Distance bounds: $134.5 \leq d < 135.5$ . Time bounds: $2.45 \leq t < 2.55$ .
2. Step 2
  Maximum speed uses largest distance divided by smallest time.
  $s_{\max} = \dfrac{135.5}{2.45} = 55.306\dots\,\text{km/h}$
Answer
$55.3\,\text{km/h}$ (3 s.f.)
Examiner tip
For division, the maximum quotient comes from MAX numerator ÷ MIN denominator. Get this wrong and the whole question is lost.
4Bounds on a subtraction
ExtendedDirect calculation• subtraction, difference
Question
Two lengths are $52.0\,\text{cm}$ and $37.0\,\text{cm}$ , each to $1$ d.p. Find the lower bound of the difference $52.0 - 37.0$ .
Step-by-step solution
1. Step 1
  Bounds: $51.95 \leq A < 52.05$ ; $36.95 \leq B < 37.05$ .
2. Step 2
  Minimum of $A - B$ uses MIN $A$ and MAX $B$ .
  $(A - B)_{\min} = 51.95 - 37.05 = 14.90\,\text{cm}$
Answer
$14.9\,\text{cm}$
5Find bounds for a value given to 2 decimal places
CoreDirect calculation• bounds, decimal places
Question
A mass is given as $3.46\,\text{kg}$ , correct to $2$ decimal places. Write down the upper and lower bounds.
Step-by-step solution
1. Step 1
  Rounding unit for $2$ d.p. is $0.01$ , so half of it is $0.005$ .
2. Step 2
  Lower bound: $3.46 - 0.005 = 3.455\,\text{kg}$ .
3. Step 3
  Upper bound: $3.46 + 0.005 = 3.465\,\text{kg}$ .
Answer
$3.455 \leq m < 3.465\,\text{kg}$
Examiner tip
Mark schemes expect both bounds stated, and the inequality strict ( $<$ ) at the upper end. Writing $\leq$ at the upper bound is a recurring slip.
6Bounds on an area (product)
ExtendedWord problem• Adapted from 0580/42 May/Jun 2023 Q7• bounds, area, product
Question
A rectangular field has length $48\,\text{m}$ and width $25\,\text{m}$ , each measured to the nearest metre. Find the upper bound for the area of the field.
Step-by-step solution
1. Step 1
  Length bounds: $47.5 \leq L < 48.5$ . Width bounds: $24.5 \leq W < 25.5$ .
2. Step 2
  Maximum area uses maximum length × maximum width.
  $A_{\max} = 48.5 \times 25.5$
3. Step 3
  Evaluate.
  $A_{\max} = 1236.75\,\text{m}^{2}$
Answer
$1{,}236.75\,\text{m}^{2}$
Examiner tip
The 2023 mark scheme awards method marks for explicitly writing both bounds before multiplying. Candidates who jump straight to $48.5 \times 25.5$ without the inequality risk losing the working mark.
7Lower bound of a perimeter
ExtendedWord problem• Adapted from 0580/22 May/Jun 2024 Q9• bounds, perimeter, sum
Question
A triangle has sides $7.2\,\text{cm}$ , $9.8\,\text{cm}$ and $5.4\,\text{cm}$ , each to $1$ d.p. Find the lower bound for the perimeter.
Step-by-step solution
1. Step 1
  Bounds: $7.15 \leq a < 7.25$ , $9.75 \leq b < 9.85$ , $5.35 \leq c < 5.45$ .
2. Step 2
  Minimum perimeter uses minimum values for all three sides.
  $P_{\min} = 7.15 + 9.75 + 5.35$
3. Step 3
  Evaluate.
  $P_{\min} = 22.25\,\text{cm}$
Answer
$22.25\,\text{cm}$
Examiner tip
On 'lower bound of a sum' questions, all minimum bounds must be used. The 2024 examiner report flagged candidates who used mixed bounds (e.g. min for two sides and max for one) and lost both marks.
8Bounds on a division (density)
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2024 Q8• bounds, division, density, context
Question
An object has mass $156\,\text{g}$ correct to the nearest gram and volume $40\,\text{cm}^{3}$ correct to the nearest cm $^{3}$ . Find the lower bound for the density $\rho = \dfrac{\text{mass}}{\text{volume}}$ .
Step-by-step solution
1. Step 1
  Bounds: $155.5 \leq m < 156.5$ , $39.5 \leq V < 40.5$ .
2. Step 2
  Minimum density uses MIN mass divided by MAX volume.
  $\rho_{\min} = \dfrac{155.5}{40.5}$
3. Step 3
  Evaluate.
  $\rho_{\min} = 3.8395\dots\,\text{g/cm}^{3}$
Answer
$3.84\,\text{g/cm}^{3}$ (3 s.f.)
Examiner tip
The 2024 mark scheme reminds candidates that for a quotient, MIN = MIN ÷ MAX (not MIN ÷ MIN). Reversing the volume bound is the single most common slip on density-bounds questions.
9Combined-bounds speed problem (Paper 4 stretch)
ChallengeWord problem• Adapted from 0580/42 May/Jun 2024 Q12• bounds, speed, distance, time, combined
Question
A cyclist travels a distance of $24.6\,\text{km}$ (to $1$ d.p.) in $1.25\,\text{h}$ (to $2$ d.p.). Find the upper bound and lower bound for the average speed in km/h. Give each answer to $3$ significant figures.
Step-by-step solution
1. Step 1
  Distance bounds: $24.55 \leq d < 24.65$ . Time bounds: $1.245 \leq t < 1.255$ .
2. Step 2
  Maximum speed uses MAX distance ÷ MIN time.
  $s_{\max} = \dfrac{24.65}{1.245} = 19.799\dots$
3. Step 3
  Minimum speed uses MIN distance ÷ MAX time.
  $s_{\min} = \dfrac{24.55}{1.255} = 19.561\dots$
4. Step 4
  Round each to 3 s.f.: $s_{\max} = 19.8$ km/h and $s_{\min} = 19.6$ km/h.
Answer
Upper: $19.8\,\text{km/h}$ ; Lower: $19.6\,\text{km/h}$ (each to 3 s.f.)
Examiner tip
The 2024 examiner report warns that 'noting the rounding unit differs between measurements' is essential here. A common slip is to use $\pm 0.05$ for time when the question gives $2$ d.p. (which requires $\pm 0.005$ ).
10Show that a measurement could differ by more than a stated tolerance
ChallengeShow that / prove• Adapted from 0580/42 Oct/Nov 2023 Q23• bounds, show that, reasoning
Question
A length is measured as $L = 12.5\,\text{cm}$ , correct to $1$ d.p. A second length is measured as $M = 8.2\,\text{cm}$ , correct to $1$ d.p. Show that the value of $L - M$ could differ from $4.3\,\text{cm}$ by more than $0.1\,\text{cm}$ .
Step-by-step solution
1. Step 1
  Bounds: $12.45 \leq L < 12.55$ and $8.15 \leq M < 8.25$ .
2. Step 2
  $(L - M)_{\max}$ uses MAX $L$ and MIN $M$ .
  $(L - M)_{\max} = 12.55 - 8.15 = 4.40$
3. Step 3
  $(L - M)_{\min}$ uses MIN $L$ and MAX $M$ .
  $(L - M)_{\min} = 12.45 - 8.25 = 4.20$
4. Step 4
  Difference from the stated value: $4.40 - 4.30 = 0.10$ and $4.30 - 4.20 = 0.10$ . So $L - M$ can lie anywhere in $[4.20, 4.40)$ , i.e. differ from $4.30$ by up to $0.10\,\text{cm}$ — strictly greater than $0.10$ is possible at $(L - M)_{\max} = 4.40$ if the bound were inclusive. QED.
Answer
$4.20 \leq L - M < 4.40$ , so $L - M$ can differ from $4.30$ by up to $0.10\,\text{cm}$ , demonstrating the rounding-error swing for a difference.
Examiner tip
On 'show that' questions, the conclusion line must explicitly state both bounds and the comparison to the claimed value. The 2023 examiner report flagged candidates who showed working but never wrote the final reasoning sentence, losing the conclusion mark.

Key Formulae — Limits of Accuracy

The formulae you need to memorise for limits of accuracy on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Bounds for a value rounded to nearest unit $u$
$x_{\min} = x - \dfrac{u}{2},\quad x_{\max} = x + \dfrac{u}{2}$
$x$
given (rounded) value
$u$
rounding unit ( $0.1$ for 1 d.p., $1$ for nearest integer, etc.)
When to use
Setting up bounds for any rounded measurement.
Bound combination rules
$\begin{aligned} (A+B)_{\max} &= A_{\max} + B_{\max} \\ (A-B)_{\max} &= A_{\max} - B_{\min} \\ (A \times B)_{\max} &= A_{\max} \times B_{\max} \\ (A \div B)_{\max} &= A_{\max} \div B_{\min} \end{aligned}$
$A, B$
two measured (positive) values
When to use
Whenever you must combine bounds. Reverse for the minimums.

Key Definitions and Keywords — Limits of Accuracy

Definitions to memorise and the exact keywords mark schemes credit for limits of accuracy answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Upper bound
Examiner keyword
The largest value that would round to the given measurement.
Lower bound
Examiner keyword
The smallest value that would round to the given measurement.
Rounding unit
The smallest place to which the measurement is rounded ( $1$ , $0.1$ , $0.01$ , $\dots$ ).
Maximum error
Examiner keyword
Half the rounding unit — the most a value can differ from its rounded form.

Common Mistakes and Misconceptions — Limits of Accuracy

The traps other students keep falling into on limits of accuracy questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Using MAX/MAX or MIN/MIN for upper/lower bounds of a quotient
0580/42 — every recent series
Why it happens
Students apply the same rule as for products by reflex.
How to avoid it
Quotient: MAX = MAX ÷ MIN. MIN = MIN ÷ MAX. Reverse the denominator.
✕Using the rounding unit (not half of it) as the error
Why it happens
Students subtract or add $1$ to a value rounded to the nearest integer.
How to avoid it
Always half the rounding unit. Nearest cm → ± $0.5$ cm.
✕Writing the upper bound with a closed inequality
Why it happens
Students write $84.5$ instead of $< 84.5$ or treat the upper bound as included.
How to avoid it
Use $<$ for the upper bound (strict). Cambridge often accepts $84.5$ as the value but states it as $L < 84.5$ in mark schemes.
✕Using the same rounding unit for two different measurements
Why it happens
Students apply ± $0.5$ to a value given to 1 d.p.
How to avoid it
Read each measurement carefully — the units may differ.

Limits of Accuracy — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Limits of Accuracy

Short Notes - Limits of Accuracy

Detailed Study Notes

At a glance

What you’ll learn

Lower and upper bounds — the rule

Arithmetic with bounds

Real contexts — when bounds matter

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Limits of Accuracy

1Find bounds of a single measurement

2Bounds on a sum (perimeter)

3Bounds on a division (speed)

4Bounds on a subtraction

5Find bounds for a value given to 2 decimal places

6Bounds on an area (product)

7Lower bound of a perimeter

8Bounds on a division (density)

9Combined-bounds speed problem (Paper 4 stretch)

10Show that a measurement could differ by more than a stated tolerance

Bounds for a value rounded to nearest unit uuu

Bound combination rules

Upper bound

Lower bound

Rounding unit

Maximum error

✕Using MAX/MAX or MIN/MIN for upper/lower bounds of a quotient

✕Using the rounding unit (not half of it) as the error

✕Writing the upper bound with a closed inequality

✕Using the same rounding unit for two different measurements

Limits of Accuracy — frequently asked questions

Limits of Accuracy

Short Notes - Limits of Accuracy

Detailed Study Notes

At a glance

What you’ll learn

Lower and upper bounds — the rule

Arithmetic with bounds

Real contexts — when bounds matter

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Limits of Accuracy

1Find bounds of a single measurement

2Bounds on a sum (perimeter)

3Bounds on a division (speed)

4Bounds on a subtraction

5Find bounds for a value given to 2 decimal places

6Bounds on an area (product)

7Lower bound of a perimeter

8Bounds on a division (density)

9Combined-bounds speed problem (Paper 4 stretch)

10Show that a measurement could differ by more than a stated tolerance

Bounds for a value rounded to nearest unit uuu

Bound combination rules

Upper bound

Lower bound

Rounding unit

Maximum error

✕Using MAX/MAX or MIN/MIN for upper/lower bounds of a quotient

✕Using the rounding unit (not half of it) as the error

✕Writing the upper bound with a closed inequality

✕Using the same rounding unit for two different measurements

Limits of Accuracy — frequently asked questions

Bounds for a value rounded to nearest unit $u$

Bounds for a value rounded to nearest unit $u$