Units and Measurements · Lesson 3 of 7

Calculations with Measured Values

“Carry measured precision through arithmetic and report uncertainty honestly.”

Learning Objectives

• Apply the correct arithmetic rule for measured data. • Round ties consistently and avoid early rounding. • Interpret absolute and percentage uncertainty. • Explain uncertainty in the area of a measured rectangle.

A calculator cannot improve the original measurements

Dividing a measured mass by a measured volume can produce many digits on a calculator. Those digits are arithmetic output, not new experimental information. The final result must reflect the precision of the input. Multiplication and division use significant figures, while addition and subtraction use decimal places after expressing all values in the same unit.

Operation	Reporting rule	Reason
Multiplication or division	Use the fewest significant figures among measured inputs	The least precise relative measurement limits the product or ratio
Addition or subtraction	Use the least precise decimal place	Absolute place-value uncertainty controls the result
Exact factors or counts	Do not restrict significant figures	They introduce no measurement uncertainty

Density from measured quantities

Problem
A sample has mass 5.74 g and volume 1.2 cm³. Find its density.

1.Use $ρ = m / V = 5.74/1.2 = 4.7833 \dots g c m^{- 3}$ .
2.Mass has three significant figures and volume has two. Report the result with two.
3.The density is $4.8 g c m^{- 3}$ .

Surface area and volume of a cube

Problem
Each side is 7.203 m. Report its surface area and volume.

1.The side has four significant figures. The exact geometric factor 6 does not reduce precision.
2. $A = 6 a^{2} = 311.299254 m^{2}$ , so report $311.3 m^{2}$ .
3. $V = a^{3} = 373.714754427 m^{3}$ , so report $373.7 m^{3}$ .

Addition and subtraction depend on place value

Combining masses

Problem
Add 436.32 g, 227.2 g and 0.301 g.

1.The unrounded sum is $663.821 g$ .
2.The least precise input ends at the tenths place. Round the sum to that place.
3.Report $663.8 g$ , even though it has four significant figures and one input has only three.

Subtraction can leave a small difference with fewer significant figures than either input. Subtracting 7.06 g from 12.9 g gives an unrounded 5.84 g, but the tenths place is the limit, so report 5.8 g. Similarly, 0.307 m − 0.304 m is 0.003 m; adding zeroes to write 0.00300 m would imply unsupported precision.

Rounding once, at the end

If the first discarded digit is below 5, leave the last retained digit unchanged; if above 5, raise it by one. For an exact halfway case, with discarded 5 followed only by zeroes, use the textbook’s tie-to-even rule: leave an even retained digit unchanged and raise an odd one. Thus 2.745 and 2.735 both become 2.74 to three significant figures. If nonzero digits follow the 5, the value is above halfway and rounds upward.

Keep at least one guard digit during intermediate work, or retain the calculator’s full internal precision. Rounding 1/9.58 immediately to 0.104 and then taking its reciprocal gives 9.62 to three significant figures. Keeping 0.1044 instead recovers 9.58. Round the final reported result, not every step.

Expressing and combining uncertainty

Relative and percentage uncertaintyLaTeX

relative uncertainty = \frac{Δ x}{∣ x ∣}, percentage uncertainty = \frac{Δ x}{∣ x ∣} \times 100%

Δx is an absolute uncertainty with the same unit as x. The ratio is dimensionless.

Equal significant-figure counts do not guarantee equal percentage uncertainty. An uncertainty of 0.01 g is about 1% of 1.02 g but only about 0.1% of 9.89 g. Interpret both the digits and the size of the value.

Area and its uncertainty

Problem
A rectangle has length (16.2 ± 0.1) cm and breadth (10.1 ± 0.1) cm. Estimate its area uncertainty.

1.The central area is $A = 16.2 \times 10.1 = 163.62 c m^{2}$ .
2.For small bounded uncertainties in a product, the first-order maximum relative uncertainty is approximately $Δ A / A = Δ l / l + Δ b / b$ .
3.The sum is $0.1/16.2 + 0.1/10.1 \approx 0.0161$ , or about 1.6%. Thus $Δ A \approx 2.6 c m^{2}$ .
4.Round the uncertainty and match the area’s place value: $A = (164 \pm 3) c m^{2}$ . This is a first-order bound, not a statistical standard-deviation calculation.

Why fractional uncertainties add in a product

Expanding the maximum productLaTeX

(l + Δ l) (b + Δ b) = l b + b Δ l + l Δ b + Δ l Δ b

For positive lengths, simultaneous positive deviations give the upper product bound.

If each uncertainty is small compared with its measured length, the final term is much smaller than the terms containing just one uncertainty. Neglect it to first order. The change in area is then approximately bΔl + lΔb. Dividing by the central area lb gives the sum of relative uncertainties. This explains the rectangle example rather than treating the rule as an unexplained prescription.

First-order relative area uncertaintyLaTeX

\frac{Δ A}{A} \approx \frac{Δ l}{l} + \frac{Δ b}{b}

This is a maximum-uncertainty estimate. It is not the root-sum-square rule for independent statistical standard uncertainties.

Surface area and a squared-unit conversion

A solid cylinder

Problem
Find the total surface area in mm² for a closed cylinder with radius 2.0 cm and height 10.0 cm.

1.The two circular ends contribute $2 π r^{2}$ . Unrolling the curved surface gives a rectangle of width $2 π r$ and height h, contributing $2 π r h$ .
2.Thus $A = 2 π r (r + h) = 2 π (2.0) (12.0) \approx 150.796 c m^{2}$ . Keep the extra digits temporarily.
3.Since $1 c m^{2} = 100 m m^{2}$ , the area is about $15079.6 m m^{2}$ .
4.The radius has two significant figures, so report $1.5 \times 1 0^{4} m m^{2}$ .

Calculating before rounding

Area and volume of a metal sheet

Problem
A sheet measures 4.234 m by 1.005 m and is 2.01 cm thick.

1.Convert thickness exactly to $0.0201 m$ .
2.Face area is $4.234 \times 1.005 = 4.25517 m^{2}$ ; report $4.255 m^{2}$ to four significant figures.
3.Use the unrounded area for volume: $4.25517 \times 0.0201 = 0.085528917 m^{3}$ .
4.Thickness limits the result to three significant figures: $0.0855 m^{3}$ .

Adding masses expressed in different units

Problem
A 2.30 kg box receives gold pieces of 20.15 g and 20.17 g. Find the total and the difference between the pieces.

1.Convert the pieces to $0.02015 kg$ and $0.02017 kg$ . The unrounded total is $2.34032 kg$ .
2.The box is reported to $0.01 kg$ , so the total is $2.34 kg$ .
3.The difference between the pieces is $20.17 - 20.15 = 0.02 g$ , reported at the hundredths place. The box’s precision does not affect this separate subtraction.

Quick check

Which rule applies when adding measured lengths?

Key Takeaways

• Products and ratios use significant figures; sums and differences use decimal places. • Exact counts do not limit measurement precision. • Keep guard digits and round the final answer. • An uncertainty must be interpreted relative to the measured value.

Previous · Lesson 2

Precision and Significant Figures

Next · Lesson 4

Dimensions and Dimensional Formulae