Significant figures — rules for operations: Exercises

Before we begin, one word on what "decimal places" and "significant figures" mean physically, so the figures below make sense.

Figure — Significant figures — rules for operations

Look at the number line above. The red measurement is only trustworthy down to the tenths mark — everything finer is a guess. When we add it to a finer number, the whole sum inherits that coarse red resolution. When we multiply, what matters instead is the count of trustworthy digits. Keep this picture in mind for every problem.

Level 1 — Recognition

Goal: read a number and state how much information it carries.

Problem 1.1

How many significant figures are in each? (a) $0.00405$ (b) $2.300$ (c) $4500$ (d) $6.022 \times 1 0^{23}$

Recall Solution 1.1

What we do: apply the counting rules from the parent note, digit by digit.

(a) $0.00405$ — leading zeros ( $0.00$ ) are just place-holders → not significant. The digits $4$ , $0$ , $5$ remain; the middle $0$ sits between non-zeros so it counts. → 3 sig figs.
(b) $2.300$ — the $2$ and $3$ count; the two trailing zeros come after a decimal point, so they are deliberate claims of precision → they count. → 4 sig figs.
(c) $4500$ — trailing zeros with no decimal point are ambiguous. Could be 2, 3, or 4. → ambiguous (write $4.5 \times 1 0^{3}$ , $4.50 \times 1 0^{3}$ , etc. to fix it).
(d) $6.022 \times 1 0^{23}$ — in scientific notation only the mantissa $6.022$ counts. → 4 sig figs.

Problem 1.2

Which rule (sig figs or decimal places) applies to each operation, and why? (a) $3.2 \times 4.115$ (b) $3.2 + 4.115$

Recall Solution 1.2

Why we ask this first: choosing the rule is the single most important decision — get it wrong and every later step is wrong.

(a) Multiplication → propagates relative error → count significant figures. Keep $min (2, 4) = 2$ .
(b) Addition → propagates absolute error → count decimal places. Keep $min (1, 3) = 1$ . Same two numbers, different rule, because the type of error is different.

Level 2 — Application

Goal: run one rule cleanly to a rounded answer.

Problem 2.1

Compute $8.24 \times 0.61$ to the correct number of sig figs.

Recall Solution 2.1

Sig figs: $8.24 \to 3$ , $0.61 \to 2$ .
Why this step? Multiplication → keep $min (3, 2) = 2$ sig figs.
Raw product $= 5.0264$ .
Round to 2 sig figs → the third digit is $2 < 5$ , drop it → $5.0$ . Note we keep the trailing zero — dropping it to " $5$ " would falsely claim only 1 sig fig.

Problem 2.2

Compute $45.76 - 2.1$ to the correct precision.

Recall Solution 2.2

Decimal places: $45.76 \to 2$ , $2.1 \to 1$ .
Why this step? Subtraction → keep $min (2, 1) = 1$ decimal place.
Raw difference $= 43.66$ .
Round to 1 decimal: digit after cutoff is $6 > 5$ → round up → $43.7$ .

Problem 2.3

Compute $\frac{9.00}{4}$ where the $4$ is an exact counting number (4 identical objects).

Recall Solution 2.3

Why this matters: an exact number has infinite sig figs — it never limits the result. So only $9.00$ (3 sig figs) counts.
Raw quotient $= 2.25$ .
Keep 3 sig figs → $2.25$ .

Level 3 — Analysis

Goal: pick the right rule under pressure, and watch precision behave.

Problem 3.1

Compute $12.11 + 0.3 + 1.012$ and then multiply that sum by $2.5$ . Give each answer to correct precision, rounding only at the very end.

Recall Solution 3.1

Step 1 (addition): raw sum $= 13.422$ . Least decimal places among terms is $0.3$ with 1 decimal. Guard-digit trick: do not round to $13.4$ yet — carry the full $13.422$ into step 2. Step 2 (multiplication): $13.422 \times 2.5 = 33.555$ .

How many sig figs may the final answer have? The multiplication limits us to $min (sig figs)$ . Here $2.5$ has 2 sig figs, and $13.422$ (carried) represents a number whose precision-limiting factor was 1 decimal place ≈ 3 sig figs. So keep 2 sig figs.
Round $33.555$ to 2 sig figs → $34$ . If you had prematurely rounded to $13.4$ , you'd get $13.4 \times 2.5 = 33.5 \to 34$ — same here, but early rounding is not guaranteed to be safe, so never rely on it.

Problem 3.2

The subtraction of near-equal numbers: compute $5.004 - 4.999$ and comment on the surviving precision.

Recall Solution 3.2

Both have 3 decimal places, so keep 3 decimal places.
Raw $= 0.005$ .
Answer $= 0.005$ , which has only 1 sig fig!
What it looks like: two numbers each known to 4 sig figs, subtracted, leave a result with only 1 sig fig. This is catastrophic cancellation — the reliable leading digits cancel, exposing the uncertain tail. The decimal-place rule handled it automatically; sig-fig counting on the inputs would have misled you.

Level 4 — Synthesis

Goal: combine rules across a multi-step physical formula.

Problem 4.1

A rectangular plate has length $L = 12.5 cm$ and width $W = 3.20 cm$ . Compute the area and the perimeter, each to correct precision.

Recall Solution 4.1

Area (multiplication):

$A = L \times W = 12.5 \times 3.20 = 40.0$ (raw $= 40.00$ ).
Sig figs: $12.5 \to 3$ , $3.20 \to 3$ → keep 3 → $A = 40.0 cm^{2}$ .

Perimeter (addition, with an exact multiplier):

$P = 2 (L + W)$ . The $2$ is exact — infinite sig figs.
Inner sum $L + W = 12.5 + 3.20 = 15.70$ ; least decimals is $12.5$ (1 decimal) → precision-limited to 1 decimal → $15.7$ .
Multiply by exact $2$ : $P = 2 \times 15.7 = 31.4$ (the exact $2$ doesn't reduce precision).
$P = 31.4 cm$ . Notice: two different rules inside one problem — × for area, + for the perimeter's inner step.

Problem 4.2

Density: a sample of mass $m = 24.0 g$ occupies volume $V = 5.1 cm^{3}$ . Find the density $ρ = m / V$ .

Recall Solution 4.2

Division → sig figs. $24.0 \to 3$ , $5.1 \to 2$ → keep $min (3, 2) = 2$ .
Raw $ρ = 24.0/5.1 = 4.7058 \dots g/cm^{3}$ .
Round to 2 sig figs: third digit $0 < 5$ → drop → $ρ = 4.7 g/cm^{3}$ .

Level 5 — Mastery

Goal: full reasoning — choose rules, guard digits, exact numbers, and cancellation all at once.

Problem 5.1

A car covers $d = 1000.0 m$ (5 sig figs) in $t = 20.1 s$ . It then covers a second stretch, from marker $x_{1} = 850.42 m$ to $x_{2} = 851.00 m$ , in $t_{2} = 0.30 s$ . Find the average speed over each stretch to correct precision, and say which speed is less trustworthy and why.

Recall Solution 5.1

Stretch 1: $v_{1} = d / t = 1000.0/20.1$ .

Division → sig figs. $1000.0 \to 5$ , $20.1 \to 3$ → keep 3.
Raw $= 49.751 \dots$ → 3 sig figs → cutoff digit $5$ with $1$ after → round up → $v_{1} = 49.8 m/s$ .

Stretch 2 distance (subtraction first!): $Δ x = 851.00 - 850.42$ .

Subtraction → decimal places. Both have 2 decimals → keep 2 decimals.
Raw $= 0.58 m$ — but this has only 2 sig figs (catastrophic cancellation: 5-sig-fig inputs, 2-sig-fig output).

Stretch 2 speed: $v_{2} = Δ x / t_{2} = 0.58/0.30$ .

Division → sig figs. $0.58 \to 2$ , $0.30 \to 2$ → keep 2.
Raw $= 1.9333 \dots$ → 2 sig figs → $v_{2} = 1.9 m/s$ .

Which is less trustworthy? $v_{2}$ . Even though its input positions were each known to 5 sig figs, the subtraction destroyed most of that precision, leaving only 2 sig figs — so the derived speed is far coarser than $v_{1}$ 's. The lesson: precision is not conserved through subtraction of near-equal numbers.

Problem 5.2

You measure the period of a pendulum by timing $n = 10$ swings (an exact count) with a stopwatch reading $t = 18.46 s$ . The period is $T = t / n$ . Then the theory says $g = \frac{4 π ^{2} L}{T ^{2}}$ with $L = 0.9930 m$ . Find $T$ and then $g$ to correct precision. Treat $π$ as exact.

Recall Solution 5.2

Period: $T = t / n = 18.46/10$ .

$n = 10$ is exact (infinite sig figs) → only $18.46$ (4 sig figs) limits us.
$T = 1.846 s$ (keep 4 sig figs) — carry as guard value.

Compute $g = 4 π^{2} L / T^{2}$ (all multiplications/divisions; $4$ and $π$ are exact):

Sig-fig-limiting factors: $L = 0.9930$ (4 sig figs), $T = 1.846$ (4 sig figs) → final answer keeps 4 sig figs.
$T^{2} = 1.84 6^{2} = 3.40772$ (guard digits kept).
$4 π^{2} = 39.4784 \dots$ (exact, carry plenty of digits).
$g = 39.4784 \times 0.9930/3.40772 = 39.2001/3.40772 = 11.503 \dots m/s^{2}$ .
Round to 4 sig figs → $g = 11.50 m/s^{2}$ . (Physically high because a real pendulum has other errors — but the sig-fig bookkeeping is exactly this.)

Recall One-line summary of the whole page

Operation type first ( $\times \div \Rightarrow$ sig figs, $+ - \Rightarrow$ decimals) ::: carry guard digits ::: ignore exact numbers ::: round once at the end ::: beware subtraction wiping out precision.

Connections

Error propagation — relative vs absolute — the rigorous "why" behind both rules used here.
Measurement & uncertainty — where the input uncertainties come from.
Scientific notation — how we fixed the ambiguous $4500$ in Problem 1.1(c).
Orders of magnitude & estimation — when even 2 sig figs is more than you need.
Dimensional analysis — checks the units of $g$ ; sig figs check its precision.