What is the difference between estimation and rounding in Mathematics?

Estimation involves finding an approximate value that is close enough to the right answer, often used to simplify calculations. Rounding, on the other hand, is a specific process of adjusting numbers to a nearby value, typically to a certain degree of precision, such as the nearest ten or hundred. Both techniques are essential in the Cambridge IGCSE Mathematics curriculum for simplifying complex calculations and making quick decisions.

How do you round numbers to the nearest ten in the Cambridge IGCSE Mathematics curriculum?

To round a number to the nearest ten, look at the digit in the units place. If it is 5 or greater, increase the tens digit by one and change the units digit to zero. If it is less than 5, simply change the units digit to zero. For example, rounding 47 to the nearest ten results in 50, while rounding 43 results in 40.

Why is estimation important in solving real-world problems in Mathematics?

Estimation is crucial in real-world applications because it allows for quick decision-making and problem-solving without needing exact calculations. In the Cambridge IGCSE Mathematics curriculum, estimation helps students develop number sense and the ability to judge the reasonableness of answers, which is vital in fields like engineering, finance, and everyday life scenarios.

What are some common strategies for estimating sums and differences in Mathematics?

Common strategies for estimating sums and differences include rounding numbers to the nearest ten, hundred, or other place values before performing the operation. Another method is using compatible numbers, which are numbers that are easy to compute mentally. These strategies are emphasized in the Cambridge IGCSE Mathematics curriculum to help students efficiently manage calculations and verify the plausibility of their answers.

How can rounding numbers be useful in data analysis and interpretation?

Rounding numbers simplifies data, making it easier to analyze and interpret trends without getting bogged down by insignificant details. In the Cambridge IGCSE Mathematics curriculum, students learn how rounding can help in presenting data clearly and concisely, which is especially useful in statistics and when communicating findings to a non-technical audience.

Why is "Dropping trailing zeros after rounding" a common mistake in Estimation and Rounding Numbers?

Why it happens: Students think 13.00 "means the same as" 13, so they drop the zeros. How to avoid it: Trailing zeros after the decimal point indicate accuracy and must be kept when the question asks for a fixed number of d.p.

Why is "Counting leading zeros as significant" a common mistake in Estimation and Rounding Numbers?

Why it happens: Confusing the position of digits with their significance. How to avoid it: Significant figures start from the **first non-zero digit**. 0.00463 has 3 s.f., not 5.

Why is "Forgetting that rounding can cascade" a common mistake in Estimation and Rounding Numbers?

Why it happens: 9.99 rounded to 1 d.p. is 10.0, but students often write 9.9 or 10. How to avoid it: Treat the round-up like an addition: 9.99 + 0.005 = 9.995 10.0.

Why is "Estimating a calculation but keeping some numbers unrounded" a common mistake in Estimation and Rounding Numbers?

Why it happens: Students round only the "awkward" numbers and leave nice numbers alone. How to avoid it: Estimate means **every** number to 1 s.f. — even the easy ones. Source: 0580/22 May/Jun 2024 — examiner report Q3

Cambridge IGCSE Mathematics (0580)

Estimation and Rounding Numbers

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Short Notes - Estimation and Rounding Numbers

Detailed Study Notes

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

Estimation and Rounding — Cambridge IGCSE 0580 Maths Extended (2026)

Round numbers to a given number of decimal places or significant figures, then estimate calculator-paper answers using sensible 1-significant-figure approximations. Two skills, both worth easy marks if you stay disciplined.

At a glance

Decimal places (d.p.) count digits AFTER the decimal point.
Significant figures (s.f.) count from the first NON-ZERO digit.
Rounding rule: digit after the cut is $\ge 5$ ? Round up. Less than $5$ ? Round down.
$0.0342$ rounded to 2 s.f. is $0.034$ — leading zeros don't count as significant.
Estimate = round each number to 1 s.f., compute, then state the estimate.
An estimate uses ' $\approx$ ' (approximately) — never '=' for rounded answers.
Mark schemes credit the rounded values shown, not just the final answer.

What you’ll learn

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

E1.10 — Round values to a specified number of decimal places, significant figures, or to the nearest 10, 100, 1000, etc.
E1.11 — Estimate the results of calculations to check answers using 1 significant-figure approximations.

Rounding to a number of decimal places

Count digits AFTER the point. Look at the next digit; round the last one up if it's 5 or more.

Decimal places (d.p.) count digits after the decimal point.

To round to $n$ d.p.:

Locate the digit in the $n$ -th decimal place.
Look at the digit immediately after it (the "decider").
If decider $\ge 5$ , round the last kept digit UP. If decider $< 5$ , leave it.
Drop everything after the $n$ -th decimal place.

Worked walkthroughs.

Round $7.6843$ to 2 d.p. → second decimal is $8$ , decider is $4$ → round down → $7.68$ .
Round $0.0497$ to 3 d.p. → third decimal is $9$ , decider is $7$ → round UP. $9$ becomes $10$ , which carries: $0.0497 \to 0.050$ .
Round $5.999$ to 2 d.p. → decider is $9$ , round up. The $9 \to 10$ carries up the chain: $5.999 \to 6.00$ .

When rounding causes the digits to roll over (e.g. $9.99 \to 10.00$ ), keep the trailing zeros — they show the precision.

7.6843 is left of the halfway mark 7.685, so it rounds down to 7.68.

Look at the single digit after the cut — that decider tells you whether to round up or down.

Identify the digit in the $n$ -th decimal place.
Look at the next digit only (don't double-round).
$\ge 5$ → round up; $< 5$ → round down.
Keep trailing zeros: $5.999 \to 6.00$ (NOT $6$ ).

Rounding to significant figures

Count from the first non-zero digit. Trailing zeros AFTER a non-zero digit do count.

Significant figures (s.f.) count digits starting from the first non-zero digit.

Rules for what's significant:

ALL non-zero digits are significant.
Zeros BETWEEN non-zero digits are significant ( $1{,}007$ has 4 s.f.).
Leading zeros (before the first non-zero digit) are NOT significant ( $0.0034$ has 2 s.f.).
Trailing zeros in a decimal ARE significant ( $2.50$ has 3 s.f.).
Trailing zeros in a whole number are AMBIGUOUS without context ( $1{,}500$ could be 2, 3 or 4 s.f.).

Counting starts at the first non-zero digit; trailing zeros in a whole number (like 1500) stay ambiguous.

Significant figures begin at the first non-zero digit — leading zeros are only place-holders.

Method to round to $n$ s.f.:

Find the $n$ -th significant digit.
Look at the next digit (the decider).
Round up if decider $\ge 5$ ; round down if $< 5$ .
Replace any trailing whole-number digits with zeros to preserve the magnitude.

Worked walkthroughs.

$7{,}486$ to 1 s.f. → first significant is $7$ , decider is $4$ → round down → $7{,}000$ . (Place-fillers, not omitted!)
$0.04785$ to 2 s.f. → first two are $4, 7$ . Decider is $8$ → round up → $0.048$ .
$59.96$ to 3 s.f. → first three are $5, 9, 9$ . Decider is $6$ → round up. The $9$ s carry: $59.96 \to 60.0$ .

That last example is a classic Paper-2 trap. Trailing $.0$ matters — drop it and you've reported 2 s.f. (60), not 3.

Count from the first NON-ZERO digit.
Leading zeros never count.
Trailing zeros in a decimal DO count.
Use place-filler zeros for whole numbers ( $7{,}486 \to 7{,}000$ to 1 s.f.).

Estimation by 1 significant figure

On the calculator paper, an 'estimate' answer uses 1-s.f. inputs. Cambridge wants both the rounded inputs AND the rough computation visible.

Cambridge's "estimate" means: round every number to 1 significant figure, then compute by hand. The estimate doesn't need to match the exact answer — it just needs to be a sensible rough check.

Standard procedure.

Replace each number with its 1-s.f. round.
Compute (without a calculator).
Write the answer prefixed with " $\approx$ ".
Quote the estimate; if asked, compare with the calculator value.

Worked. Estimate $\dfrac{29.6 \times 5.21}{0.486}$ .

Round to 1 s.f.: $29.6 \to 30$ , $5.21 \to 5$ , $0.486 \to 0.5$ .
Compute: $\dfrac{30 \times 5}{0.5} = \dfrac{150}{0.5} = 300$ .
Estimate $\approx 300$ .

Rounded inputs give a quick hand calculation; the true value 317.3 is close, so no digit slipped.

The calculator value is $\dfrac{29.6 \times 5.21}{0.486} \approx 317.3$ . Our estimate of $300$ is close enough — it tells us the calculator answer is sensible (not, say, $30$ or $3{,}000$ from a typo).

When the estimate disagrees, redo the calculation. A calculator that returns $31.7$ for the example above shouldn't be trusted — the estimate of $300$ flags the misplacement of the decimal.

Round each number to 1 s.f.
Compute without the calculator.
Always write ' $\approx$ ' for an estimate (not '=').
Use the estimate to spot calculator/typo errors.

Rounding to the nearest 10, 100, 1000

Same idea as decimal places, just on the other side of the point. Watch the carry on $9$ s.

To round to the nearest $10, 100, 1000, \ldots$ :

Identify the digit in that place.
Look at the digit immediately to the right (the decider).
Apply the same $\ge 5$ rule.
Replace digits to the right of the rounded place with zeros.

Worked walkthroughs.

$1{,}476$ to nearest 10 → tens digit is $7$ , decider is $6$ → round up → $1{,}480$ .
$1{,}476$ to nearest 100 → hundreds digit is $4$ , decider is $7$ → round up → $1{,}500$ .
$9{,}962$ to nearest 100 → hundreds digit is $9$ , decider is $6$ → round up. $9 \to 10$ carries: $9{,}962 \to 10{,}000$ .

Same pitfall as before: don't double-round. Always go from the original number straight to the target precision.

Identify the place digit, look at the next digit, apply $\ge 5$ rule.
Trailing places become zeros.
Carrying $9 \to 10$ propagates leftward.
Don't go via an intermediate rounding.

Quick recap

$n$ d.p. = $n$ digits AFTER the point. $n$ s.f. = $n$ digits starting from the first non-zero.
Look at ONE digit beyond the cut to decide: $\ge 5$ rounds up.
Don't double-round; never truncate.
Trailing zeros in $5.00$ or $60.0$ are significant — keep them.
Estimate = round each number to 1 s.f., compute by hand, write ' $\approx$ '.
Use estimates to sanity-check calculator answers — wrong order of magnitude = wrong button pressed.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Decimal place — a digit position to the right of the decimal point.
Significant figure — a digit that contributes to the precision of a number, starting from the first non-zero digit.
Round — adjust a number to a stated level of precision using the standard $\ge 5$ rule.
Estimate — an approximate value, usually obtained by rounding each input to 1 significant figure.

How it’s examined

Rounding appears in 1-mark questions throughout the paper — "give your answer to 3 s.f." is the most common phrasing. Estimation appears as a 2-3 mark stand-alone question, usually on Paper 2, where Cambridge wants to see the 1-s.f. rounded values AND the rough calculation. Examiner reports highlight three recurring slips: truncating instead of rounding, dropping trailing zeros that show precision, and rounding $0.486$ to $0$ instead of $0.5$ .

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.10, E1.11); 0580/22 May/Jun 2024 — Q3 (rounding to s.f.); 0580/42 Oct/Nov 2024 — Q1 (estimation); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Estimation and Rounding Numbers

Step-by-step solutions to past-paper-style questions on estimation and rounding numbers, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Estimation and Rounding Numbers

Almost every estimation and rounding numbers 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 — round to … d.p./s.f., estimate, find the bounds, calculate the percentage error — on a given value or expression.

How to approach it

Identify the deciding digit (the one just past the required place), round it, and keep trailing zeros to show the stated accuracy. For estimation, round every number to $1$ s.f. before evaluating.

Common trap

Examiner reports flag dropped trailing zeros ( $13$ instead of $13.00$ ), leading zeros counted as significant, and dividing percentage error by the approximate rather than the exact value.

Word problem

Recognise it by

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

How to approach it

Find the $\pm\frac{u}{2}$ bound on each measurement, then pick the combination that maximises or minimises the result: a sum uses bounds the same way, but $\frac{A}{B}$ maximum needs $A_{\max}$ with $B_{\min}$ .

Common trap

Examiner reports flag mixing bounds — e.g. using $A_{\max}$ with $B_{\max}$ for a quotient — as the single biggest error in bounds questions.

Show that / prove

Recognise it by

The question asks you to decide whether a result is reasonable or to justify a check — the comparison, not just a number, is the answer.

How to approach it

Produce the estimate by rounding to $1$ s.f., then write an explicit sentence comparing it with the value being checked.

Common trap

Examiner reports flag answers that give only the estimate — a bare number scores the method mark but not the conclusion; state the comparison in words.

1Round to a given number of decimal places
CoreDirect calculation• decimal places
Question
Round (a) $4.5867$ to $2$ d.p., (b) $0.05649$ to $3$ d.p., (c) $12.9954$ to $2$ d.p.
Step-by-step solution
1. Step 1
  (a) $4.58|67$ . Look at the third digit ( $6$ ) — round up: $4.59$ .
2. Step 2
  (b) $0.056|49$ . Fourth digit ( $4$ ) — round down: $0.056$ .
3. Step 3
  (c) $12.99|54$ . Third digit $5$ → round up. $12.99 + 0.01 = 13.00$ .
Answer
(a) $4.59$ (b) $0.056$ (c) $13.00$
Examiner tip
Always write trailing zeros — $13.00$ , not $13$ — to show 2 d.p. accuracy. Mark schemes deduct for this.
2Round to significant figures
CoreDirect calculation• significant figures
Question
Round (a) $0.0046382$ to $3$ s.f., (b) $48{,}573$ to $2$ s.f., (c) $9.9999$ to $3$ s.f.
Step-by-step solution
1. Step 1
  (a) Significant figures start at the first non-zero digit ( $4$ ): $0.00463|82$ → fourth s.f. is $8$ , round up → $0.00464$ .
2. Step 2
  (b) $48|573$ → third digit $5$ → round up → $49{,}000$ .
3. Step 3
  (c) $9.99|99$ . Round up cascades: $10.0$ (which is $3$ s.f. in this context).
Answer
(a) $0.00464$ (b) $49{,}000$ (c) $10.0$
Examiner tip
Trailing zeros after a decimal point are significant — $10.0$ has $3$ s.f. but $10$ has only $1$ or $2$ . Be precise.
3Estimate a calculation by rounding to 1 s.f.
CoreDirect calculation• Adapted from 0580/22 May/Jun 2024 Q3• estimation
Question
Estimate the value of $\dfrac{29.7 \times 4.13}{0.488}$ by rounding each number to $1$ significant figure.
Step-by-step solution
1. Step 1
  Round: $29.7 \to 30$ , $4.13 \to 4$ , $0.488 \to 0.5$ .
2. Step 2
  Substitute and evaluate.
  $\dfrac{30 \times 4}{0.5} = \dfrac{120}{0.5} = 240$
Answer
Approximately $240$
Examiner tip
Estimation questions on 0580 require all numbers to be rounded to 1 s.f. — no exceptions. Don't keep $4.13$ as $4.13$ .
4Find upper and lower bounds for a rounded value
ExtendedDirect calculation• Adapted from 0580/42 Feb/Mar 2024 Q12• bounds, limits of accuracy
Question
A length is given as $14.6\,\text{cm}$ to $1$ d.p. State the lower and upper bounds.
Step-by-step solution
1. Step 1
  $1$ d.p. means the value is rounded to the nearest $0.1$ . The error is half of $0.1$ , i.e. $0.05$ .
2. Step 2
  Lower bound: $14.6 - 0.05 = 14.55\,\text{cm}$ .
3. Step 3
  Upper bound: $14.6 + 0.05 = 14.65\,\text{cm}$ .
Answer
Lower bound $14.55$ cm; upper bound $14.65$ cm.
5Round a quotient to a stated number of significant figures
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2023 Q4• significant figures, calculation
Question
Calculate $\dfrac{7.84^{2}}{0.39}$ and give your answer correct to (a) $4$ s.f., (b) $2$ d.p.
Step-by-step solution
1. Step 1
  Evaluate to enough precision to round safely. $7.84^{2} = 61.4656$ , then $61.4656 \div 0.39 = 157.6041025\dots$
  $\dfrac{7.84^{2}}{0.39} = 157.6041025\dots$
2. Step 2
  (a) Four s.f.: count the first four non-zero digits → $157.6|041\dots$ Next digit $0$ → round down. Result: $157.6$ .
3. Step 3
  (b) Two d.p.: $157.60|41\dots$ Next digit $4$ → round down. Result: $157.60$ .
Answer
(a) $157.6$ (b) $157.60$
Examiner tip
Mark schemes accept the rounded final value only if the intermediate calculation is shown to at least $1$ extra digit. The 2023 examiner report notes premature rounding as the leading cause of lost accuracy marks.
6Use estimation to check a calculator answer
ExtendedShow that / prove• Adapted from 0580/22 Oct/Nov 2024 Q6• estimation, checking
Question
A student types $\dfrac{4.92 \times 19.7}{0.51}$ into a calculator and reads off $190.06$ . Use an estimate (each number to $1$ s.f.) to decide whether the calculator answer is reasonable.
Step-by-step solution
1. Step 1
  Round each number to $1$ s.f.: $4.92 \to 5$ , $19.7 \to 20$ , $0.51 \to 0.5$ .
2. Step 2
  Substitute.
  $\dfrac{5 \times 20}{0.5} = \dfrac{100}{0.5} = 200$
3. Step 3
  The estimate is $200$ , and the calculator value $190.06$ is close to $200$ — the calculation is reasonable.
Answer
Estimate $= 200$ ; the calculator value $190.06$ is consistent with the estimate, so it is reasonable.
Examiner tip
Examiner reports flag that 'is the answer reasonable?' demands an explicit comparison sentence. State both the estimate and the comparison — a bare estimate scores only the method mark.
7Use bounds in a perimeter calculation
ExtendedWord problem• Adapted from 0580/42 May/Jun 2024 Q7• bounds, limits of accuracy, applied
Question
A rectangle has length $8.5\,\text{cm}$ and width $3.2\,\text{cm}$ , each given to $1$ d.p. Find the upper bound for the perimeter.
Step-by-step solution
1. Step 1
  Rounding to $1$ d.p. gives an error of $\pm 0.05$ . The upper bound for each side is value $+ 0.05$ .
2. Step 2
  Upper bound of length $= 8.5 + 0.05 = 8.55\,\text{cm}$ . Upper bound of width $= 3.2 + 0.05 = 3.25\,\text{cm}$ .
3. Step 3
  Perimeter $= 2(\text{length} + \text{width})$ . Upper bound combines both upper bounds.
  $\text{Perimeter}_{\max} = 2(8.55 + 3.25) = 2(11.80) = 23.6$
Answer
Upper bound of perimeter $= 23.6\,\text{cm}$ .
Examiner tip
The 2024 examiner report stresses that the upper bound of a sum uses upper bounds of both terms, while the lower bound of a difference uses upper of one minus lower of the other. Always check which combination the question demands.
8Find the lower bound of an area
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2023 Q11• bounds, area, applied
Question
A square tile has side $12\,\text{cm}$ measured to the nearest cm. Find (a) the lower bound for its side, (b) the lower bound for its area.
Step-by-step solution
1. Step 1
  Nearest cm means $\pm 0.5$ cm error.
2. Step 2
  (a) Lower bound of side: $12 - 0.5 = 11.5\,\text{cm}$ .
3. Step 3
  (b) Lower bound of area uses the lower bound of each factor: $\text{area}_{\min} = 11.5^{2}$ .
  $\text{area}_{\min} = 11.5 \times 11.5 = 132.25$
Answer
(a) $11.5\,\text{cm}$ (b) $132.25\,\text{cm}^{2}$ .
Examiner tip
The mark scheme awards the accuracy mark only when both bounds are correctly distinguished. A common slip is computing $12.5 \times 11.5$ — mixing bounds. Use the lower bound for both factors when finding the lower area.
9Percentage error from rounding
ChallengeDirect calculation• Adapted from 0580/42 May/Jun 2024 Q22• percentage error, rounding, stretch
Question
The exact value of a length is $\pi$ cm. A student rounds it to $3.14$ cm. Calculate the percentage error caused by this rounding, giving your answer to $3$ s.f.
Step-by-step solution
1. Step 1
  Percentage error $= \dfrac{|\text{approx} - \text{exact}|}{\text{exact}} \times 100\%$ .
2. Step 2
  Compute the absolute error: $|3.14 - \pi| = |3.14 - 3.14159265\dots| = 0.00159265\dots$
  $|3.14 - \pi| \approx 0.00159265$
3. Step 3
  Divide by the exact value and convert to a percentage.
  $\dfrac{0.00159265}{\pi} \times 100 = 0.0506766\dots \%$
4. Step 4
  Round to $3$ s.f.: $0.0507\%$ .
Answer
$\approx 0.0507\%$
Examiner tip
The examiner report flags candidates who divide by the approximate value instead of the exact one. Convention on 0580: percentage error is always relative to the exact (true) value.
10Combined upper and lower bounds in a speed calculation
ChallengeWord problem• Adapted from 0580/42 Feb/Mar 2024 Q19• bounds, speed, stretch
Question
A car travels a distance of $80\,\text{km}$ , measured to the nearest km, in a time of $1.2\,\text{hours}$ measured to $1$ d.p. Find the upper bound for the average speed in km/h, giving your answer to $3$ s.f.
Step-by-step solution
1. Step 1
  Bounds: distance to nearest km gives $\pm 0.5$ km, so upper bound $80.5$ km and lower bound $79.5$ km. Time to $1$ d.p. gives $\pm 0.05$ h, so upper bound $1.25$ h and lower bound $1.15$ h.
2. Step 2
  Speed $= \dfrac{\text{distance}}{\text{time}}$ . To maximise speed, use the largest distance and the smallest time.
  $\text{speed}_{\max} = \dfrac{80.5}{1.15}$
3. Step 3
  Evaluate.
  $\text{speed}_{\max} = \dfrac{80.5}{1.15} = 70.0\,\text{km/h}$
Answer
Upper bound of speed $= 70.0\,\text{km/h}$ (to $3$ s.f.).
Examiner tip
The 2024 examiner report notes that picking the wrong combination is the single biggest error here. Rule: for $\dfrac{A}{B}$ , the maximum uses $A_{\max}$ and $B_{\min}$ ; the minimum uses $A_{\min}$ and $B_{\max}$ .

Key Formulae — Estimation and Rounding Numbers

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

Bounds when rounded to nearest unit $u$
$\text{lower bound} = x - \dfrac{u}{2},\quad \text{upper bound} = x + \dfrac{u}{2}$
$x$
the rounded value
$u$
the rounding unit (e.g. $0.1$ for 1 d.p., $1$ for nearest integer)
When to use
Any question asking for upper or lower bounds.

Key Definitions and Keywords — Estimation and Rounding Numbers

Definitions to memorise and the exact keywords mark schemes credit for estimation and rounding numbers answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Decimal place (d.p.)
Examiner keyword
A digit position after the decimal point.
Significant figure (s.f.)
Examiner keyword
A digit that contributes to the precision of a number, starting from the first non-zero digit.
Estimation
Examiner keyword
Approximating a calculation, usually by rounding each value to $1$ s.f.
Upper bound
Examiner keyword
The largest possible value of a measurement before it would round to a higher value.
Lower bound
Examiner keyword
The smallest possible value of a measurement that still rounds to the given value.

Common Mistakes and Misconceptions — Estimation and Rounding Numbers

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

✕Dropping trailing zeros after rounding
Why it happens
Students think $13.00$ "means the same as" $13$ , so they drop the zeros.
How to avoid it
Trailing zeros after the decimal point indicate accuracy and must be kept when the question asks for a fixed number of d.p.
✕Counting leading zeros as significant
Why it happens
Confusing the position of digits with their significance.
How to avoid it
Significant figures start from the first non-zero digit. $0.00463$ has $3$ s.f., not $5$ .
✕Forgetting that rounding can cascade
Why it happens
$9.99$ rounded to 1 d.p. is $10.0$ , but students often write $9.9$ or $10$ .
How to avoid it
Treat the round-up like an addition: $9.99 + 0.005 = 9.995 \Rightarrow 10.0$ .
✕Estimating a calculation but keeping some numbers unrounded
0580/22 May/Jun 2024 — examiner report Q3
Why it happens
Students round only the "awkward" numbers and leave nice numbers alone.
How to avoid it
Estimate means every number to $1$ s.f. — even the easy ones.