Significant figures and rounding rules: Worked examples

The scenario matrix

Every sig-fig question is really one of these cases. Read the table as: "what makes THIS case tricky?"

Cell	Case class	What makes it tricky	Hit by example
A	Counting sig figs — leading + sandwiched + trailing zeros all in one number	Three different zero rules collide	Ex 1
B	Trailing-zero ambiguity in a whole number	$1500$ could be 2, 3, or 4 sf	Ex 2
C	Multiplication / division — fewest sig figs wins	Relative uncertainty; ignore decimal places	Ex 3
D	Addition / subtraction — fewest decimal places wins	Absolute uncertainty; ignore sig figs	Ex 4
E	Exact number in the mix (counted/defined)	Must NOT limit the answer	Ex 5
F	Banker's rounding — the "half" edge case, both parities	Round-half-to-even, up vs down	Ex 6
G	Degenerate / limiting input — subtracting near-equal numbers	"Catastrophic" loss of sig figs	Ex 7
H	Real-world word problem (density, multi-step)	Round ONCE, choose the right rule per step	Ex 8
I	Exam twist — mixed $+$ then $\times$ in one expression	Rule switches mid-calculation	Ex 9

We now walk every cell. Prerequisites you may want open: Scientific notation and orders of magnitude, Accuracy vs precision, and Uncertainty and error propagation.

Figure — Significant figures and rounding rules

Figure s01 (the zero-rule map). Read it as a colour code we will reuse on EVERY worked example: red bars = leading zeros (placeholders, NOT significant), blue bar = a sandwiched zero (significant), green bars = non-zero measured digits (always significant), yellow bar = a trailing zero after the decimal (significant). Every time the text below says "the red/blue/green/yellow bar," look back here — and each example carries its own colour-coded overlay so you never have to hold the map in your head.

Worked example Ex 1 — Cell A: all three zero rules in one number

How many significant figures are in $0.02090$ ?

Forecast: Guess a number 1–5 before reading on. Which zeros are "real"?

Split the number into zones. Write it as 0.0 2 0 9 0. Why this step? Each zone obeys a different zero rule, so we must isolate them before judging.
Leading zeros ( $0.0$ before the $2$ ) → not significant. Why this step? They only tell you where the decimal sits; they are placeholders, not measured digits. In figure s02 these are the red bars.
The non-zero digits $2$ and $9$ → always significant. Why this step? Non-zero digits are real measured readings by definition — the green bars in s02, the backbone every count starts from.
The middle $0$ (between $2$ and $9$ ) → significant (sandwiched). Why this step? You cannot measure the $9$ without passing through that $0$ — it is real information (the blue bar in s02).
The trailing $0$ (after the $9$ ) → significant. Why this step? It sits after a decimal point; nobody writes it unless they measured it (the yellow bar in s02).
Count the significant digits: green $2$ , blue $0$ , green $9$ , yellow $0$ = 4 sig figs.

Verify: Rewrite in scientific notation — every digit in the coefficient must be significant: $2.090 \times 1 0^{- 2}$ . The coefficient $2.090$ has 4 digits. ✓ Placeholders vanished, real digits survived.

Worked example Ex 2 — Cell B: the ambiguous whole number

A balance reads " $4500$ g". State its sig figs, then rewrite it to unambiguously mean 3 sig figs.

Forecast: Is $4500$ 2, 3, or 4 sig figs? (Trick: you cannot tell from the plain number.)

Recognise the ambiguity. Trailing zeros in a whole number with no decimal point are undecidable — $4500$ could be 2, 3, or 4 sf. In figure s03, the two trailing zeros are drawn grey/striped — "could be red placeholder OR yellow measured, we can't tell." Why this step? The trailing-zero (no decimal) rule says: with no decimal point written, those final zeros might be placeholders or measured — the plain notation cannot tell you which, so it is genuinely ambiguous.
Fix it with scientific notation. For exactly 3 sig figs write $4.50 \times 1 0^{3}$ g — in s03's fixed version the green $4, 5$ and a yellow $0$ are now unmistakable. Why this step? In $N \times 1 0^{n}$ , every digit of $N$ counts — so $4.50$ pins it at 3, no guessing.
Note the alternatives so you can spot them: $4.5 \times 1 0^{3}$ (2 sf), $4.500 \times 1 0^{3}$ (4 sf).

Verify: $4.50 \times 1 0^{3} = 4500$ . ✓ Same value, but now the precision promise is explicit.

Worked example Ex 3 — Cell C: multiplication, fewest sig figs wins

Compute $3.14 \times 0.0026$ to the correct precision.

Forecast: Which factor is the "weak link"? Guess the final digit count.

Raw product first. $3.14 \times 0.0026 = 0.008164$ . Why this step? Carry full digits; decide precision after, never before.
Count each factor's sig figs. $3.14$ → 3 sf (three green bars in s04). $0.0026$ → 2 sf (two red leading zeros, then green $2, 6$ ). Why this step? For $\times / \div$ we track relative uncertainty, measured by sig figs.
Take the minimum: $min (3, 2) = 2$ sig figs. Why this step? The answer can't be more precise than its least-precise ingredient — mud in the water.
Round $0.008164$ to 2 sf. Significant digits are $8, 1, \dots$ ; decider (next digit) is $6 > 5$ → round the $1$ up to $2$ : $0.0082$ . Why this step? We must trim to the 2-sf cap from step 3; the first dropped digit ( $6 > 5$ ) pushes the last kept digit up.

Verify: $0.0082 = 8.2 \times 1 0^{- 3}$ , coefficient has 2 digits. ✓ Matches the required 2 sf.

Worked example Ex 4 — Cell D: subtraction, fewest decimal places wins

Compute $18.9 - 0.036$ .

Forecast: Sig-fig rule or decimal-place rule? (It's $-$ , so...)

Raw difference. $18.9 - 0.036 = 18.864$ . Why this step? Get the exact arithmetic before trimming.
Count decimal places, not sig figs. $18.9$ has 1 decimal place; $0.036$ has 3. Figure s05 stacks the two numbers by decimal column, so the tenths/hundredths/thousandths align. Why this step? Addition/subtraction stacks absolute uncertainty (same units), so we match the coarsest decimal position — see Uncertainty and error propagation.
Keep 1 decimal place. Fewest = 1. Decider is the hundredths digit $6 > 5$ → round the tenths $8$ up to $9$ : $18.9$ . Why this step? The answer's precision is set by the coarsest term ( $18.9$ , 1 dp), so we trim to 1 dp; the $6$ in the hundredths place is the first dropped digit, and $6 > 5$ forces a round-up.

Verify: $18.9$ has 1 decimal place ✓. Sanity: $0.036$ is far below the precision of $18.9$ , so it barely moves the tenths digit — exactly why we can't claim more places.

Worked example Ex 5 — Cell E: an exact number must NOT limit

A single tablet has mass $0.2506$ g. What is the mass of exactly $8$ tablets?

Forecast: Does the " $8$ " cap the answer at 1 sig fig? (No — and here's why.)

Multiply. $0.2506 \times 8 = 2.0048$ g. Why this step? Straightforward arithmetic first.
Classify the $8$ . It is a counted quantity → an exact number with infinite sig figs. Why this step? Counting 8 objects is not an estimate; there is zero uncertainty, so it can't be the weak link.
Apply min rule using only measured values. Only $0.2506$ (4 sf) counts. So the answer must be reported to 4 sig figs. Why this step? Exact numbers are invisible to the sig-fig cap.
Round $2.0048$ to 4 sig figs. Its digits in order are $2, 0, 0, 4, 8$ — that is five significant digits, so it must be trimmed. The four we keep are $2, 0, 0, 4$ ; the first dropped digit is $8$ , and $8 > 5$ → round the $4$ up to $5$ : $2.005$ g. Why this step? We keep exactly 4 sig figs (set by the measured mass), and the standard rounding rule — look at the first dropped digit — decides whether the last kept digit moves up.

Verify: $0.2506 \times 8 = 2.0048$ ; rounded to 4 sf = $2.005$ . ✓ If we had wrongly used "1 sf" we'd have gotten $2$ g — throwing away three real digits.

Worked example Ex 6 — Cell F: banker's rounding, both directions

Round $6.25$ and $6.35$ each to 2 sig figs (dropping the trailing $5$ with nothing after it).

Forecast: Both end in $\dots 5$ . Do they both go up? (No — that's the whole point.)

Identify the tie case. In each, the dropped digit is exactly $5$ with nothing after → use round-half-to-even. Why this step? Always-round-up would bias averages upward over many calculations (see Accuracy vs precision).
$6.25$ : kept digit is the $2$ (even). Round to keep it even → stays $2$ → $6.2$ . Why this step? $6.2$ keeps an even last digit; $6.3$ would make it odd. In s06 the $5$ sits exactly halfway (the yellow marker on the number line) so parity, not magnitude, decides.
$6.35$ : kept digit is $3$ (odd). Round up to reach even $4$ → $6.4$ . Why this step? $6.4$ has an even last digit; we move toward even.
Observe: one rounded down, one rounded up — the anti-bias in action (green arrow down, red arrow up in s06).

Verify: $6.25 \to 6.2$ (last digit $2$ , even ✓); $6.35 \to 6.4$ (last digit $4$ , even ✓).

Worked example Ex 7 — Cell G: degenerate input, catastrophic subtraction

Two lengths are measured: $5.437$ cm and $5.432$ cm, each good to 4 sig figs. Find their difference and its sig figs.

Forecast: The inputs have 4 sig figs each. Guess the sig figs of the difference.

Subtract. $5.437 - 5.432 = 0.005$ cm. Why this step? Use the decimal-place rule: both have 3 decimal places, so keep 3 → $0.005$ .
Now count sig figs of the result. $0.005$ has only 1 sig fig (the green $5$ ; the leading zeros are red placeholders). In s07 you see the shared green digits $5.43$ cancel, leaving one lonely surviving digit. Why this step? Subtracting two nearly equal numbers cancels the leading digits, leaving only the uncertain tail. This is loss of significance — a degenerate/limiting case.
Lesson: decimal-place rule was obeyed, yet precision collapsed from 4 sf to 1 sf. Why this step? The rule preserves decimal places, not sig figs — subtraction can silently destroy relative precision. See Uncertainty and error propagation.

Verify: $5.437 - 5.432 = 0.005$ ; $0.005 = 5 \times 1 0^{- 3}$ → 1 sig fig. ✓ A dramatic drop, and totally correct.

Worked example Ex 8 — Cell H: real-world word problem (density, multi-step)

A metal cube has edge $2.10$ cm and mass $61.8$ g. Find its density to the correct precision. Round only at the end.

Forecast: Volume needs a cube; edge has 3 sf, mass has 3 sf. Guess the final sig-fig count.

Volume = edge $^{3}$ . $V = 2.1 0^{3} = 9.261$ cm³. Why this step? Cubing is repeated multiplication → sig-fig rule. Edge $2.10$ has 3 sf, so $V$ will carry 3 sf — but carry the extra digit $9.261$ for now.
Density = mass / volume. $ρ = \frac{61.8}{9.261} = 6.6731 \dots$ g/cm³. Why this step? Division → sig-fig rule again.
Apply the min rule once, at the end. Mass $61.8$ → 3 sf; edge (hence $V$ ) → 3 sf. $min (3, 3) = 3$ . Why this step? Rounding only now avoids error snowballing through the cube and the divide.
Round $6.6731$ to 3 sf. Digits $6, 6, 7$ ; the first dropped digit is $3$ , and $3 < 5$ → keep the $7$ : $6.67$ g/cm³. Why this step? Step 3 fixed the cap at 3 sig figs, so we must trim the fourth digit; the standard rounding rule — look at the first dropped digit ( $3 < 5$ ) — tells us to leave the last kept digit unchanged.

Verify: $2.1 0^{3} = 9.261$ ; $61.8/9.261 = 6.673 \dots$ ; to 3 sf $= 6.67$ g/cm³. Units: g / cm³ = g/cm³ ✓ (a valid density unit; check with Units and dimensional analysis).

Worked example Ex 9 — Cell I: exam twist, the rule switches mid-expression

Evaluate $(12.11 + 0.3) \times 2.0$ to correct precision.

Forecast: Which rule for the bracket, which for the multiply? They are different.

Do the addition inside the bracket first. $12.11 + 0.3 = 12.41$ . Why this step? Brackets before multiply, and $+$ obeys the decimal-place rule: $12.11$ (2 dp), $0.3$ (1 dp) → keep 1 dp.
Round the intermediate for its sig-fig count only. $12.41 \to 12.4$ (1 dp). Now $12.4$ has 3 sig figs. Why this step? The addition rule fixed the decimal place; that in turn fixes how many sig figs the bracket can pass forward (3).
Now multiply — switch to the sig-fig rule. $12.4 \times 2.0 = 24.8$ . Sig figs: $12.4$ (3 sf), $2.0$ (2 sf) → $min = 2$ . Why this step? Multiplication tracks relative uncertainty, so the rule changes from decimal places to sig figs.
Round $24.8$ to 2 sf. Digits $2, 4$ ; the first dropped digit is $8$ , and $8 > 5$ → round the $4$ up to $5$ : $25$ (or $2.5 \times 1 0^{1}$ ). Why this step? The multiplication rule caps the answer at 2 sig figs, so we trim the third digit; standard rounding on that first dropped digit ( $8 > 5$ ) pushes the kept digit up.

Verify: bracket $= 12.41 \to 12.4$ ; $12.4 \times 2.0 = 24.8 \to$ 2 sf $= 25$ . ✓ The precision was throttled first by decimals ( $0.3$ ), then by sig figs ( $2.0$ ).

Recall Quick self-test on the matrix

Which rule governs subtraction, sig figs or decimal places? ::: Decimal places (absolute uncertainty). Does an exact counted number ever limit the sig figs of an answer? ::: No — it has infinite sig figs. Round $8.45$ to 2 sf using banker's rounding. ::: $8.4$ (keep the even $4$ ; the $8.45$ 's kept digit $4$ is already even). Why can $5.437 - 5.432$ collapse to 1 sig fig? ::: Subtracting near-equal numbers cancels the certain leading digits (loss of significance).