Measurement bounds and the limits of accuracy
Represent a rounded measurement as an interval and use its limits to judge possible calculated values.
Core: bounds of an individual rounded value. Extended also includes bounds of calculated results.
Before you begin
- Round to named place values and decimal places.
- Interpret strict and inclusive inequality signs.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How far from a positive recorded value can a number lie when rounded to the nearest 0.1?
What you will learn
- Find lower and upper bounds for rounded measurements.
- Express the corresponding interval accurately.
- Calculate bounds for positive quantities in Extended problems.
A rounded value represents many possible values
If a length is 8 cm to the nearest centimetre, the actual length may lie from 7.5 cm up to, but not including, 8.5 cm under the usual positive-value rounding convention. The lower bound is included; the upper bound would round to the next recorded centimetre. Write 7.5 ≤ length < 8.5.
The rounding unit determines the half-width. A mass of 4.2 kg to the nearest tenth has bounds 4.15 and 4.25 kg. A count of 240 rounded to the nearest ten has bounds 235 and 245. Read how the measurement was rounded; the written value alone may not establish its accuracy.
Significant figures still identify a rounding unit
For 3.6 recorded to two significant figures, the last significant digit is the tenths place, so use ±0.05. For 3600 recorded to two significant figures, the last significant digit is the hundreds place, so use ±50. The same number of significant figures does not mean the same absolute uncertainty.
Bounds describe permitted numerical values under the stated rounding model. They do not prove that the measuring instrument was calibrated or that all sources of experimental uncertainty have been included. An exact count of 12 objects is different from a measurement rounded to 12; do not invent a half-unit uncertainty for an exact count.
Extended: choose the extreme in the correct direction
For positive lengths a and b, the smallest possible product uses both lower bounds, while the upper limiting product uses both upper bounds. For a sum, add matching lower or upper bounds. For a difference a − b, its lower limit uses lower a minus upper b, and its upper limit uses upper a minus lower b.
For a positive quotient a/b, the lower limiting value uses lower a divided by upper b; the upper limit uses upper a divided by lower b. The denominator must stay positive. These shortcuts cannot be applied blindly to intervals crossing zero or containing negative values; consider how the expression changes in that domain.
Report limits without promising an attained maximum
When an upper endpoint is excluded, the corresponding upper bound may be approached without being reached. It is still called the upper bound. State the requested bound to a sensible precision; if you round a limit for a guarantee, round outward so the claimed interval still includes every permitted value.
To decide whether a claim is guaranteed, test it against the interval rather than the central rounded value. If a reported height of 180 cm is rounded to the nearest centimetre, the actual height can be below 180. A claim of at least 180 cm therefore needs more accurate evidence.
Pause and explain
Extended: how do you find the upper bound of positive distance/time?
Worked example
A length is recorded as 12.4 cm to the nearest 0.1 cm. State its bounds and interval.
Show the worked solution
- The rounding unit is 0.1 cm, so the half-width is 0.05 cm.
- Subtract and add 0.05: the limits are 12.35 and 12.45 cm.
- Include the lower bound and exclude the upper bound under the stated rounding convention.
Answer 12.35 ≤ length < 12.45 cm
From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A mass is 6.8 kg to the nearest 0.1 kg. Give the possible interval.
Give me a hint
Use half of 0.1 kg, not half of 1 kg.
Compare my reasoning
- The half-width is 0.05 kg.
- The lower limit is 6.75 kg and the upper limit is 6.85 kg.
- State 6.75 ≤ mass < 6.85 kg; the upper endpoint rounds to 6.9.
6.75 ≤ mass < 6.85 kg
Look for these in your work
- Used the stated measurement accuracy.
- Distinguished an included lower endpoint from an excluded upper one.
Extended: a rectangle measures 8 cm by 5 cm, each to the nearest centimetre. Find its lower and upper area bounds.
Give me a hint
Both lengths stay positive, so multiply matching endpoints.
Compare my reasoning
- The side intervals are [7.5, 8.5) and [4.5, 5.5) cm.
- Lower area = 7.5 × 4.5 = 33.75 cm².
- Upper area bound = 8.5 × 5.5 = 46.75 cm², approached but excluded.
33.75 cm² and 46.75 cm²
Look for these in your work
- Bounded both dimensions before multiplying.
- Used square units for area.
Extended: a journey is 120 km to the nearest kilometre and lasts 2.0 hours to the nearest 0.1 hour. Can you guarantee the average speed was below 61 km/h?
Give me a hint
Use the upper distance and lower time for the limiting speed.
Compare my reasoning
- Distance lies in [119.5,120.5) km; time lies in [1.95,2.05) hours.
- Upper speed bound is 120.5/1.95 ≈ 61.795 km/h.
- Permitted values can exceed 61 km/h, so the guarantee cannot be made from these measurements.
No; the upper speed bound is about 61.795 km/h.
Look for these in your work
- Used the smallest permitted positive denominator.
- Judged the guarantee from the interval, not the central value 60.
Common mistakes
- Adding and subtracting a whole rounding unit.
- Using two upper bounds for an upper quotient.
- Confusing a limiting upper bound with an attained maximum.
What can you explain now?
A positive value is 450 to the nearest ten. State the interval.
Compare with the explanation
445 ≤ value < 455
The rounding unit is ten, so each limit is five from the recorded value.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain the interval for 0.37 to two decimal places. Extended learners: derive bounds for a positive rectangle perimeter using both side intervals.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.
Original StudyForA teaching content · AI-assisted checks · Human teacher review pending.
Number contains 19 sequenced lessons; Algebra and graphs contains 21; Coordinate geometry contains 10 teaching lessons and a mixed checkpoint. Nine earlier overviews support selected topics across the syllabus. The other six syllabus areas, full cumulative assessment and human teacher review are not yet complete. Written lesson practice is self-checked, not automatically graded. The checkpoint samples skills and does not save an exam grade or certify mastery.
Check the official syllabus 2025–2027 Open related practice and resources