NSWMaths Standard 2Syllabus dot point

How accurate is a measurement, and how do you state its error and the range the true value must lie in?

Describe the limit of reading of an instrument, calculate the absolute error and percentage error of a measurement, and find the greatest possible error and the upper and lower bounds

A focused answer to the HSC Maths Standard 2 dot point on measurement error. The limit of reading of an instrument, absolute error as half the smallest division, percentage error, and the greatest possible error giving the upper and lower bounds, with worked Australian examples on rulers, scales and measuring jugs.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Every measurement has an error set by the limit of reading, the smallest division on the instrument. The absolute error (greatest possible error) is half of it: $\tfrac{1}{2} \times \text{limit of reading}$ . The true value lies between the lower bound $= \text{measurement} - \text{absolute error}$ and the upper bound $= \text{measurement} + \text{absolute error}$ . The percentage error $= \pm \dfrac{\text{absolute error}}{\text{measurement}} \times 100\%$ shows how good the reading is for its size; a smaller percentage error means a more accurate measurement. To find the smallest or largest possible perimeter or area, combine all the lower or all the upper bounds. This builds on units and unit conversion.

What this dot point is asking

NESA wants you to treat every measurement as an approximation that carries an error. You then have to describe that error precisely. Three quantities do all the work: the limit of reading (the smallest division marked on the instrument), the absolute error (half that smallest division), and the percentage error (the absolute error as a fraction of the measurement). From the absolute error you also state the greatest possible error and the upper and lower bounds, the range the true value must lie in. The arithmetic is light. The marks are won by reading the smallest division correctly and choosing the right one of these quantities for the question.

The answer

No instrument is perfect. A ruler can only be read to the nearest marked line, scales only to the nearest marked mass, a jug only to the nearest marked level. The smallest gap between marks is the limit of reading (also called the precision), and it sets a hard floor on how well you can know the true value.

The key idea is that when you read a measurement to the nearest division, the true value can be up to half a division away in either direction before the reading would round to the next mark. That half a division is the absolute error (the greatest possible error), and it puts the true value inside a band one half-division wide on each side of the reading.

The limit of reading

The limit of reading (or precision) is the smallest unit marked on the instrument. A $30$ cm ruler with millimetre marks has a limit of reading of $1$ mm. Kitchen scales marked every $2$ g have a limit of reading of $2$ g. A measuring jug marked every $25$ mL has a limit of reading of $25$ mL. Reading the smallest division correctly is the single most important step, because every later quantity is built from it. A slip here makes the absolute error, the bounds and the percentage error all wrong at once.

A common point of confusion is a digital instrument, which has no visible scale. The limit of reading is the smallest place value it displays: digital scales showing $0.1$ kg have a limit of reading of $0.1$ kg, and a stopwatch showing $0.01$ s has a limit of reading of $0.01$ s.

Absolute error: half the smallest division

The absolute error is the largest amount the measurement could be wrong by. Because a reading is taken to the nearest division, the true value can sit up to half a division away before it would round to the next mark, so

\text{absolute error} = \tfrac{1}{2} \times \text{limit of reading}.

This is also called the greatest possible error. For the millimetre ruler the absolute error is $\tfrac{1}{2} \times 1 = 0.5$ mm; for the $2$ g scales it is $\tfrac{1}{2} \times 2 = 1$ g; for the $25$ mL jug it is $\tfrac{1}{2} \times 25 = 12.5$ mL. The absolute error always carries the same unit as the measurement.

Upper and lower bounds

The true value is unknown, but it cannot be more than one absolute error away from the reading. That gives the range it must lie in:

\text{lower bound} = \text{measurement} - \text{absolute error},

\text{upper bound} = \text{measurement} + \text{absolute error}.

If kitchen scales marked every $2$ g read $250$ g, the absolute error is $1$ g, so the true mass is somewhere between $250 - 1 = 249$ g and $250 + 1 = 251$ g. The number line below shows the measured value in the centre with the bounds an equal step on each side; the true value is guaranteed to be inside the shaded interval.

Percentage error

The absolute error on its own does not tell you whether a measurement is good. An error of $0.5$ mm is trivial on a door frame but ruinous on a microchip. The percentage error fixes this by comparing the absolute error to the size of the thing being measured:

\text{percentage error} = \pm \frac{\text{absolute error}}{\text{measurement}} \times 100\%.

The $\pm$ sign is kept because the true value could be either above or below the reading. A ruler reading $86$ mm has a percentage error of $\pm \dfrac{0.5}{86} \times 100\% \approx \pm 0.6\%$ . The same $0.5$ mm absolute error on a $5$ mm reading would be a percentage error of $\pm 10\%$ . The larger the measurement relative to a fixed absolute error, the smaller the percentage error. This is why measuring a long distance in one go is more accurate than adding many short measurements that each carry their own error.

How exam questions ask about measurement error

The wording changes but each phrasing points to one of the four quantities:

"State the limit of reading / precision" asks for the smallest division marked on the instrument (or the smallest displayed place value on a digital device).
"Find the absolute error / greatest possible error / largest possible error" means halve the smallest division.
"Find the upper and lower bounds" or "between what values does the true ... lie" means take the measurement plus and minus the absolute error.
"Find the percentage error" means divide the absolute error by the measurement and multiply by $100$ ; watch the requested rounding ("correct to one decimal place").
"Which measurement is more accurate" compares percentage errors: the smaller the percentage error, the more accurate the measurement.
"Find the smallest / largest possible perimeter (or area)" means combine the bounds: use all lower bounds for the smallest, all upper bounds for the largest.

Measurement error across four Australian contexts

Four readings, each built the same way: read the smallest division, halve it for the absolute error, then form the bounds or the percentage error the question asks for. The contexts are a ruler, kitchen scales, a measuring jug and a measured garden bed.

State the absolute error and bounds of a ruler reading

A student reads the length of a pencil case as $86$ mm on a ruler marked in millimetres. State the limit of reading, the absolute error, and the upper and lower bounds, and find the percentage error correct to one decimal place.

Read the limit of reading. The smallest division on the ruler is $1$ mm, so the limit of reading is $1$ mm.

Halve it for the absolute error.

\text{absolute error} = \tfrac{1}{2} \times 1 = 0.5 \text{ mm}.

Form the bounds. The true length is the reading plus or minus the absolute error:

\text{lower bound} = 86 - 0.5 = 85.5 \text{ mm}, \qquad \text{upper bound} = 86 + 0.5 = 86.5 \text{ mm}.

Find the percentage error.

\text{percentage error} = \pm \frac{0.5}{86} \times 100\% \approx \pm 0.6\%

(the unrounded value is $0.5813\ldots\%$ , which rounds to $0.6\%$ ). So the pencil case is $86 \pm 0.5$ mm, with a percentage error of about $\pm 0.6\%$ .

Find the percentage error of a mass on kitchen scales

A cook reads $250$ g of flour on kitchen scales whose smallest division is $2$ g. Find the absolute error, the bounds of the true mass, and the percentage error correct to one decimal place.

Read the limit of reading. The marks are $2$ g apart, so the limit of reading is $2$ g.

Halve it.

\text{absolute error} = \tfrac{1}{2} \times 2 = 1 \text{ g}.

Form the bounds.

\text{lower bound} = 250 - 1 = 249 \text{ g}, \qquad \text{upper bound} = 250 + 1 = 251 \text{ g}.

Find the percentage error.

\text{percentage error} = \pm \frac{1}{250} \times 100\% = \pm 0.4\%.

So the flour is $250 \pm 1$ g, with a percentage error of $\pm 0.4\%$ , a precise reading for a kitchen task.

Find the greatest possible error of a measuring jug

Juice is poured to the $500$ mL line on a measuring jug whose markings are $25$ mL apart. Find the greatest possible error, the upper and lower bounds, and the percentage error correct to one decimal place.

Read the limit of reading. The markings are $25$ mL apart, so the limit of reading is $25$ mL. A coarse scale like this is where error really shows.

Halve it for the greatest possible error.

\text{greatest possible error} = \tfrac{1}{2} \times 25 = 12.5 \text{ mL}.

Form the bounds.

\text{lower bound} = 500 - 12.5 = 487.5 \text{ mL}, \qquad \text{upper bound} = 500 + 12.5 = 512.5 \text{ mL}.

Find the percentage error.

\text{percentage error} = \pm \frac{12.5}{500} \times 100\% = \pm 2.5\%.

So the true volume is between $487.5$ mL and $512.5$ mL, a percentage error of $\pm 2.5\%$ , much larger than the ruler or scales because the jug's divisions are so wide.

Find the smallest and largest possible perimeter of a garden bed

A rectangular garden bed is measured with a tape marked every $0.1$ m as $2.4$ m long and $1.8$ m wide. Find the absolute error in each side, then the smallest and largest possible perimeter.

Read the limit of reading and halve it. The smallest division is $0.1$ m, so each side has

\text{absolute error} = \tfrac{1}{2} \times 0.1 = 0.05 \text{ m}.

Form the bounds of each side.

\text{length: } 2.35 \text{ m to } 2.45 \text{ m}, \qquad \text{width: } 1.75 \text{ m to } 1.85 \text{ m}.

Combine the bounds for the perimeter. The perimeter is $P = 2(\text{length} + \text{width})$ . The smallest perimeter uses both lower bounds and the largest uses both upper bounds:

P_{\min} = 2(2.35 + 1.75) = 2 \times 4.1 = 8.2 \text{ m},

P_{\max} = 2(2.45 + 1.85) = 2 \times 4.3 = 8.6 \text{ m}.

So the perimeter is between $8.2$ m and $8.6$ m. (Check: the nominal perimeter is $2(2.4 + 1.8) = 8.4$ m, and four sides each $\pm 0.05$ m give a spread of $\pm 0.2$ m, which agrees.)

Final answers: the pencil case is $86 \pm 0.5$ mm ( $\pm 0.6\%$ ); the flour is $249$ g to $251$ g ( $\pm 0.4\%$ ); the juice is $487.5$ mL to $512.5$ mL ( $\pm 2.5\%$ ); and the garden bed's perimeter lies between $8.2$ m and $8.6$ m.

Common traps

Misreading the smallest division: The absolute error is half the smallest division, not half the gap between the big numbered marks. On a ruler numbered every centimetre but ruled in millimetres, the limit of reading is $1$ mm, so the absolute error is $0.5$ mm, not $0.5$ cm.
Forgetting to halve: The absolute error is half the limit of reading, not the whole division. A jug marked every $25$ mL has an absolute error of $12.5$ mL, not $25$ mL.
Putting the wrong number on the bottom of the percentage error: Divide the absolute error by the measurement, not by the absolute error or the limit of reading. The denominator is the size of the quantity you measured.
Dropping the units or the $\pm$: The absolute error and the bounds carry the same unit as the measurement, and the percentage error keeps a $\pm$ because the true value can be above or below the reading.
Mixing bounds when combining: For the smallest possible perimeter or area use all the lower bounds; for the largest use all the upper bounds. Pairing a lower bound with an upper bound gives neither extreme.
Rounding to the wrong precision: Percentage-error questions usually specify the rounding ("correct to one decimal place"); round only the final percentage, and keep full accuracy until then.

Exam technique

Write the limit of reading down explicitly first ("smallest division $= 2$ g"); that single line is where the method mark sits even if the arithmetic slips later. Then state the absolute error as half of it, and only then form the bounds or the percentage error. Keep the unit on every line for the absolute error and the bounds, and keep the $\pm$ on the percentage error. For percentage error, the formula mark is for $\dfrac{\text{absolute error}}{\text{measurement}} \times 100\%$ , so show that fraction before evaluating, and round only at the end to the requested precision. When a question asks for the smallest or largest possible perimeter or area, line up all the lower bounds for the smallest and all the upper bounds for the largest, and finish with a quick check that the nominal value sits midway between them.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC-style3 marksA surveyor measures the frontage of a block of land as

47.3

m using a trundle wheel marked every

0.1

m. (a) State the limit of reading and find the absolute error. (b) Write the lower and upper bounds of the true frontage. (c) Find the percentage error, correct to two decimal places.

Show worked answer →

Part (a) - limit of reading and absolute error. The smallest division on the trundle wheel is $0.1$ m, so that is the limit of reading. Halve it for the absolute error:

\text{absolute error} = \tfrac{1}{2} \times 0.1 = 0.05 \text{ m}

One mark for stating the limit of reading as $0.1$ m and halving it to $0.05$ m.

Part (b) - bounds. The true frontage is the reading plus or minus the absolute error:

\text{lower bound} = 47.3 - 0.05 = 47.25 \text{ m}, \qquad \text{upper bound} = 47.3 + 0.05 = 47.35 \text{ m}

One mark for both bounds, with the unit.

Part (c) - percentage error.

\text{percentage error} = \pm \frac{0.05}{47.3} \times 100\% \approx \pm 0.11\%

The unrounded value is $0.1057\ldots\%$ , which rounds to $0.11\%$ . One mark for the fraction $\dfrac{0.05}{47.3} \times 100\%$ and the rounded answer. Markers reward showing the fraction before evaluating, since that line carries the method mark even if the rounding slips.

2024 HSC-style4 marksA length of timber is measured as

3.6

m using a tape marked every

0.05

m. (a) Find the absolute error. (b) State the limits of accuracy (the lower and upper bounds) of the true length. (c) Find the percentage error, correct to one decimal place. (d) A second tape is marked every

0.01

m. State, with a reason, whether it would give a smaller percentage error for the same length.

Show worked answer →

Part (a) - absolute error. The smallest division is $0.05$ m, so

\text{absolute error} = \tfrac{1}{2} \times 0.05 = 0.025 \text{ m}

One mark.

Part (b) - limits of accuracy. The true length is the reading plus or minus the absolute error:

\text{lower bound} = 3.6 - 0.025 = 3.575 \text{ m}, \qquad \text{upper bound} = 3.6 + 0.025 = 3.625 \text{ m}

One mark for both bounds. A common slip is to write $3.55$ m and $3.65$ m by halving the wrong division.

Part (c) - percentage error.

\text{percentage error} = \pm \frac{0.025}{3.6} \times 100\% \approx \pm 0.7\%

The unrounded value is $0.6944\ldots\%$ , which rounds to $0.7\%$ . One mark for the fraction and the rounded answer.

Part (d) - the reason. Yes. A smaller smallest division ( $0.01$ m instead of $0.05$ m) gives a smaller absolute error, and dividing a smaller absolute error by the same measurement gives a smaller percentage error. One mark for the correct conclusion with this reason. Markers want the comparison stated in terms of the absolute error shrinking, not just "it is more precise".

2023 HSC-style4 marksTwo sets of scales are used to weigh the same parcel, whose mass is close to

200

g. Scales A are marked every

5

g and read

200

g. Scales B are marked every

1

g and read

200

g. (a) Find the percentage error of each reading, correct to two decimal places. (b) State, with a reason, which scales give the more accurate reading.

Show worked answer →

Part (a) - percentage error of each. For each, take half the smallest division for the absolute error, then divide by the measurement.

Scales A (smallest division $5$ g).

\text{absolute error} = \tfrac{1}{2} \times 5 = 2.5 \text{ g}, \qquad \text{percentage error} = \pm \frac{2.5}{200} \times 100\% = \pm 1.25\%

Scales B (smallest division $1$ g).

\text{absolute error} = \tfrac{1}{2} \times 1 = 0.5 \text{ g}, \qquad \text{percentage error} = \pm \frac{0.5}{200} \times 100\% = \pm 0.25\%

One mark for each correct percentage error (two marks total).

Part (b) - compare. Scales B give the more accurate reading because their percentage error of $0.25\%$ is smaller than Scales A's $1.25\%$ . One mark for naming Scales B and one mark for the reason that the smaller percentage error is the more accurate measurement. Markers will not award the reason mark for "B has finer marks" alone; the comparison must be made in terms of the percentage error.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksState the absolute error for each instrument. (a) A ruler whose smallest division is

1

mm. (b) Kitchen scales whose smallest division is

100

Show worked solution →

Use the rule: absolute error is half the smallest division. The absolute error is always $\tfrac{1}{2}$ of the limit of reading (the smallest unit the instrument shows).

Part (a) - ruler, smallest division $1$ mm.

\text{absolute error} = \tfrac{1}{2} \times 1 = 0.5 \text{ mm}

Part (b) - scales, smallest division $100$ g.

\text{absolute error} = \tfrac{1}{2} \times 100 = 50 \text{ g}

So a reading on the ruler is good to $\pm 0.5$ mm, and a reading on these scales only to $\pm 50$ g, which is why coarse scales are a poor choice for small masses.

foundation3 marksA length is measured as

156

cm using a tape marked in whole centimetres. (a) State the limit of reading. (b) Find the absolute error. (c) Write the lower and upper bounds of the true length.

Show worked solution →

Part (a) - limit of reading. This is the smallest division on the tape, which is $1$ cm.

Part (b) - absolute error. Half the smallest division:

\text{absolute error} = \tfrac{1}{2} \times 1 = 0.5 \text{ cm}

Part (c) - the bounds. The true length is the measurement plus or minus the absolute error:

\text{lower bound} = 156 - 0.5 = 155.5 \text{ cm}

\text{upper bound} = 156 + 0.5 = 156.5 \text{ cm}

So the true length lies between $155.5$ cm and $156.5$ cm, written $156 \pm 0.5$ cm.

core3 marksA person's mass is read as

72.5

kg on digital scales that display to the nearest

0.1

kg. (a) Find the absolute error. (b) Find the percentage error, correct to two decimal places. (c) State the lower and upper bounds.

Show worked solution →

Part (a) - absolute error. The limit of reading is $0.1$ kg, so the absolute error is half of that:

\text{absolute error} = \tfrac{1}{2} \times 0.1 = 0.05 \text{ kg}

Part (b) - percentage error. Divide the absolute error by the measurement and multiply by $100$ :

\text{percentage error} = \pm \frac{0.05}{72.5} \times 100\% \approx \pm 0.07\%

(The unrounded value is $0.0690\ldots\%$ , which rounds to $0.07\%$ to two decimal places.)

Part (c) - the bounds.

\text{lower bound} = 72.5 - 0.05 = 72.45 \text{ kg}, \qquad \text{upper bound} = 72.5 + 0.05 = 72.55 \text{ kg}

The tiny percentage error shows the reading is very precise relative to the size of the quantity.

core3 marksJuice is poured into a measuring jug marked every

10

mL, and the level reads

350

mL. (a) Find the absolute error. (b) Find the percentage error, correct to one decimal place. (c) State the bounds of the true volume.

Show worked solution →

Part (a) - absolute error. Half the smallest division, which is $10$ mL:

\text{absolute error} = \tfrac{1}{2} \times 10 = 5 \text{ mL}

Part (b) - percentage error.

\text{percentage error} = \pm \frac{5}{350} \times 100\% \approx \pm 1.4\%

(The unrounded value is $1.4285\ldots\%$ , rounding to $1.4\%$ to one decimal place.)

Part (c) - the bounds.

\text{lower bound} = 350 - 5 = 345 \text{ mL}, \qquad \text{upper bound} = 350 + 5 = 355 \text{ mL}

So the true volume is between $345$ mL and $355$ mL.

exam5 marksA rectangular tile is measured with a ruler marked in millimetres as

240

mm long and

160

mm wide. (a) State the absolute error in each measurement. (b) Write the bounds of the length and of the width. (c) Find the smallest and largest possible perimeter of the tile.

Show worked solution →

Part (a) - absolute error. The smallest division is $1$ mm, so each side has

\text{absolute error} = \tfrac{1}{2} \times 1 = 0.5 \text{ mm}

Part (b) - bounds of each side.

\text{length: } 240 - 0.5 = 239.5 \text{ mm to } 240 + 0.5 = 240.5 \text{ mm}

\text{width: } 160 - 0.5 = 159.5 \text{ mm to } 160 + 0.5 = 160.5 \text{ mm}

Part (c) - combine the bounds. The perimeter is $P = 2(\text{length} + \text{width})$ . The smallest perimeter uses both lower bounds, the largest uses both upper bounds.

P_{\min} = 2(239.5 + 159.5) = 2 \times 399 = 798 \text{ mm}

P_{\max} = 2(240.5 + 160.5) = 2 \times 401 = 802 \text{ mm}

So the perimeter lies between $798$ mm and $802$ mm. (Check: the nominal perimeter is $2(240 + 160) = 800$ mm, and combining four sides each $\pm 0.5$ mm gives a spread of $\pm 2$ mm, which matches.)

exam4 marksTwo people measure the same steel rod, whose length is close to

2

m. Aisha uses a tape marked every

0.1

m and records

2.0

m. Ben uses a tape marked every

0.01

m and records

2.00

m. (a) Find the percentage error of each measurement. (b) State, with a reason, whose measurement is more accurate.

Show worked solution →

Part (a) - percentage error of each. First find each absolute error as half the smallest division, then divide by the measurement.

Aisha (smallest division $0.1$ m).

\text{absolute error} = \tfrac{1}{2} \times 0.1 = 0.05 \text{ m}, \qquad \text{percentage error} = \pm \frac{0.05}{2.0} \times 100\% = \pm 2.5\%

Ben (smallest division $0.01$ m).

\text{absolute error} = \tfrac{1}{2} \times 0.01 = 0.005 \text{ m}, \qquad \text{percentage error} = \pm \frac{0.005}{2.0} \times 100\% = \pm 0.25\%

Part (b) - compare. Ben's percentage error of $0.25\%$ is one tenth of Aisha's $2.5\%$ , so Ben's measurement is more accurate. The finer the limit of reading, the smaller the absolute error and the smaller the percentage error, which is the whole reason a more finely divided instrument is better.

What this dot point is asking

The answer

The limit of reading

Absolute error: half the smallest division

Upper and lower bounds

Percentage error

How exam questions ask about measurement error

Exam-style practice questions

Practice questions

Related dot points