- When a value is reported as a rounded measurement (for example, “16 m to the nearest metre”), the true value is not exactly known.
- Instead, it lies in an interval of possible values.
- The lower bound is the smallest possible value.
- The upper bound is the largest possible value (not included for continuous data).
- Together, these are called the limits of accuracy.
- Many real measurements are continuous: they can take any value in an interval (for example, height, mass, time, temperature).
- This is different from discrete quantities, which come in fixed steps (like number of students in a class).
- Because continuous measurements are recorded to a stated accuracy (nearest metre, nearest 0.1 kg, 2 significant figures, etc.), you should treat the recorded number as a label for an interval of possible true values, not as an exact value.
Think of the rounded value as the name of a whole neighbourhood of numbers.
- The lower bound is the first house in the neighbourhood.
- The upper bound is just before the last house.
- Rounding works by grouping values around a midpoint.
- Suppose a measurement is rounded to the nearest unit of size $a$.
- Examples of step size $a$:
- Nearest 10 → $a=10$
- Nearest 0.1 → $a=0.1$
- Nearest 0.01 → $a=0.01$
- Then the half-step is $\frac{a}{2}$.
- If the recorded (rounded) value is $x$, then
- Lower bound = $x−\frac{a}{2}$
- Upper bound = $x+\frac{a}{2}$
- So the true value TT satisfies $$x-\frac{a}{2} \leq T<x+\frac{a}{2}$$
- We usually include the lower bound but exclude the upper bound.
- This matches the rule that values exactly halfway ($\frac{a}{2}$ above the midpoint) round up.
A suitcase weighs 16 kg to the nearest kilogram.
- Step size: $a=1 \text{ kg}$
- Half-step: $\frac{a}{2}=0.5 \text{ kg}$
So
- Lower bound: $16−0.5=15.5 \text{ kg}$
- Upper bound: $16+0.5=16.5 \text{ kg}$
The actual mass $w$ satisfies $$15.5 \leq w<16.5$$
- Do not write $15.5 \leq w \leq 16.5$.
- If $w=16.5$, it would round to 17 kg , not 16 kg .
The inequality signs matter because of how rounding behaves at the midpoint.
A lion’s mass is reported as 300 kg to the nearest 100 kg.
- Step size: $a=100 \text{ kg}$
- Half-step: $\frac{a}{2}=50 \text{ kg}$
- Lower bound: $300−50=250 \text{ kg}$
- Upper bound: $300+50=350 \text{ kg}$
So the true mass mm satisfies $$250 \leq m < 350$$
If the lion actually had mass $m=350 \text{ kg}$, rounding to the nearest 100 kg would give 400 kg (values exactly halfway round up).
When you see “to the nearest …”, immediately:
- Identify the rounding step aa (for example, nearest 5 → a=5a=5).
- Use the structure
$$x-\frac{a}{2} \leq \text{ true value } < x+\frac{a}{2}$$
Even if you later make a small arithmetic slip, this shows a correct method.
- If a length is 1.6 m to the nearest 0.1 m, then
- Step size: $a=0.1$
- Half-step: $\frac{a}{2}=0.05$
- So the true length $L$ satisfies $1.55 \leq L < 1.65$$
- The lower bound is $1.55 \text{ m}$.
With significant figures, the step size depends on the place value of the last significant digit.
$22 \text{ kg}$ to 2 significant figures
- The second significant figure is in the ones place.
- So the rounding step is $a=1 \text{ kg}$
- Half-step: $\frac{a}{2}=0.5 \text{ kg}$
Therefore: $$21.5\leq M<22.5$$
$0.7 text{ t}$ to 1 significant figure
- The first significant figure is in the tenths place.
- The rounding step is $a=0.1 \text{ t}$
- Half-step: $\frac{a}{2}=0.05 \text{ t}$
Therefore: $$0.65 \leq T<0.75$$
To find the rounding step for significant figures:
- Identify the place value of the last kept digit.
- Use that place value as the step $a$.
- “Nearest thousand” → $a=1000$, half-step $\frac{a}{2}=500$
- “Nearest 0.01” → $a=0.01$, half-step $\frac{a}{2}=0.005$
- Once you have lower and upper bounds for quantities, you can use them to find maximum and minimum possible results of calculations.
- Let $A$ and $B$ be quantities with bounds. To:
- $A+B$: use upper bound + upper bound
- $A−B$: use upper bound − lower bound
- $A\times B$: use upper bound × upper bound (if both are positive)
- $\frac{A}{B}$: use upper bound ÷ lower bound (if both are positive)
- $A+B$: use lower bound + lower bound
- $A−B$: use lower bound − upper bound
- $A\times B$: use lower bound × lower bound (if both are positive)
- $\frac{A}{B}$: use lower bound ÷ upper bound (if both are positive)
These rules match the summary table: to make an answer as big as possible, combine the bounds in a way that pushes the result up; to make it as small as possible, combine them to push the result down.
- A lower bound is often the safest guaranteed minimum you can claim.
- If a crate is labelled 40 kg to the nearest 0.5 kg, the minimum possible mass is the lower bound, $39.75 \text{ kg}$.
- When measurements are used for safety limits, capacity planning or costs, working with lower and upper bounds helps avoid over-estimating what is guaranteed.
- Card length: 12.0 cm to the nearest 0.1 cm → upper bound $=12.05 \text{ cm}$
- Envelope length: 12.0 cm to the nearest 0.5 cm → lower bound $=11.75 \text{ cm}$
The largest possible card $(12.05 \text{ cm})$ is bigger than the smallest possible envelope $(11.75 \text{ cm})$, so you cannot guarantee that the card will fit in the envelope.
- If a manufacturer designs packaging using only rounded measurements, some items may not fit, causing waste and rework.
- Using lower and upper bounds leads to better decisions and more efficient use of materials.
- A number line helps you visualize the interval idea.
- Mark the stated rounded value in the centre.
- Move left and right by half the rounding step.
- Mark:
- a solid (closed) point at the lower bound (included),
- an open point at the upper bound (excluded).
- A rope is 8 m to the nearest metre. Write the lower bound.
- Mauritius is 2000 km to the nearest 10 km. Write the lower bound.
- A temperature is 23.4°C to 1 decimal place. Write the lower bound.