IB Maths AI HLNumber ToolkitPaper 1 & 2LB ≤ x < UB~6 min read
Upper & Lower Bounds
A rounded number hides a range of values it could really have been. The smallest is the lower bound, the largest is the upper bound. This note shows how to find those bounds with the “half up, half down” rule, and how to combine them when rounded numbers are added, subtracted, multiplied or divided.
📘 What you need to know
The bounds of a rounded number x are the smallest and largest values it could have been: LB ≤ x < UB.
The lower bound is included (≤); the upper bound is not (<) — it would round up instead.
Find bounds with “half up, half down”: add or subtract half the degree of accuracy.
The degree of accuracy is the rounding unit — 0.1, 0.01, 1, 10, and so on.
Subtracting and dividing mix the bounds — biggest possible result comes from a large value over (or minus) a small one.
What are bounds?
When a number is rounded, the true value is lost — but it must lie within a fixed range. The lower bound (LB) is the smallest value that rounds to it; the upper bound (UB) is the cut-off above which it would round to something larger.
For example, 2.6 written to 1 decimal place could have been anything from 2.55 up to (but not including) 2.65, so 2.55 ≤ x < 2.65.
2.6 came from anywhere in 2.55 ≤ x < 2.65. Each bound sits half the degree of accuracy (½ × 0.1 = 0.05) from the value.
Finding bounds
To find bounds, identify the degree of accuracy — the unit the number was rounded to — and halve it. The rule is half up, half down.
Half up, half down
UB = value + ½ × (degree of accuracy)
LB = value − ½ × (degree of accuracy)degree of accuracy: 1 d.p. → 0.1, 2 d.p. → 0.01, nearest 10 → 10
Combining bounds in calculations
When rounded numbers are combined, choose the bound of each input that pushes the result to its extreme. Adding and multiplying are straightforward; subtracting and dividing mix the bounds.
Logic check: the largest result of a ÷ b comes from the biggest a divided by the smallest b — dividing by a smaller number makes a bigger answer.
🧠Recipe — bounds in five steps
Identify the degree of accuracy each number is rounded to.
Halve it — this is how far each bound sits from the value.
Find each bound: LB = value − half, UB = value + half; write LB ≤ x < UB.
For a calculation, pick the bound of each input that drives the result to its extreme (use the combine rules).
State the answer as LB ≤ result < UB, rounding sensibly if the question asks.
Worked examples
WE 1
Bounds of a single value
A parcel has a mass of 6.4 kg, correct to 1 decimal place. Write down the lower and upper bounds for its mass m.
degree of accuracy is 0.1half of 0.1 = 0.05half up, half downLB = 6.4 − 0.05 = 6.35UB = 6.4 + 0.05 = 6.456.35 ≤ m < 6.45the lower bound uses ≤, the upper bound uses <.
WE 2
Rounded to the nearest 10
The distance to a town is 80 km, correct to the nearest 10 km. Write down the bounds for the distance d.
degree of accuracy is 10half of 10 = 5half up, half downLB = 80 − 5 = 75UB = 80 + 5 = 8575 km ≤ d < 85 kmthe degree of accuracy isn’t always a decimal — here it is 10.
WE 3
Adding bounds
Two ribbons measure 25 cm and 18 cm, each to the nearest cm. They are laid end to end. Find the bounds for the total length T.
bounds of each ribbon (accuracy 1, half 0.5)25 cm: 24.5 ≤ a < 25.518 cm: 17.5 ≤ b < 18.5adding: UB = UB + UB, LB = LB + LBLB = 24.5 + 17.5 = 42UB = 25.5 + 18.5 = 4442 cm ≤ T < 44 cmfor a sum, both extremes line up — low with low, high with high.
WE 4
Subtracting bounds
A tank holds 50 litres of water and 12 litres are drained out, both to the nearest litre. Find the bounds for the volume R remaining.
bounds (accuracy 1, half 0.5)50 L: 49.5 ≤ a < 50.512 L: 11.5 ≤ b < 12.5subtracting: UB = UBa − LBb, LB = LBa − UBbUB = 50.5 − 11.5 = 39LB = 49.5 − 12.5 = 3737 L ≤ R < 39 Lmost left over = most poured in minus least drained out — the bounds mix.
WE 5
Dividing bounds
A cyclist travels 150 m, to the nearest 10 m, in a time of 12 s, to the nearest second. Find the bounds for the average speed, in m s−1, to 3 s.f.
bounds of distance and timedistance: 145 ≤ d < 155time: 11.5 ≤ t < 12.5speed = d ÷ t; dividing mixes the boundsUB = 15511.5 = 13.478…LB = 14512.5 = 11.611.6 m s−1 ≤ speed < 13.5 m s−1 (3 s.f.)fastest = furthest distance ÷ shortest time.
WE 6
Full question: perimeter and area
A rectangular tabletop has length 1.2 m (1 d.p.) and width 0.85 m (2 d.p.). (a) Find the bounds for the length and the width. (b) Find the bounds for the perimeter P and the area A, giving the area to 3 s.f.
(a) half up, half downlength: 1.15 ≤ L < 1.25width: 0.845 ≤ W < 0.855(b) perimeter P = 2L + 2W (a sum — bounds line up)LB = 2(1.15) + 2(0.845) = 3.99UB = 2(1.25) + 2(0.855) = 4.21area A = L × W (multiply: low×low, high×high)LB = 1.15 × 0.845 = 0.97175UB = 1.25 × 0.855 = 1.06875(a) above · (b) 3.99 m ≤ P < 4.21 m · 0.972 m2 ≤ A < 1.07 m2use the bound values, never the rounded ones, in every calculation.
💡 Top tips
Always halve the degree of accuracy — not the value itself.
Write bounds as LB ≤ x < UB: lower bound included, upper bound excluded.
For add and multiply, pair like with like (low–low, high–high).
For subtract and divide, the bounds mix — biggest result from a large value over (or minus) a small one.
Use logic to sanity-check: the largest area comes from the largest length and largest width.
âš Common mistakes
Adding or subtracting the whole degree of accuracy instead of half of it.
Using UB = UB − UB for subtraction — subtraction mixes the bounds: UB = UBa − LBb.
Dividing UB by UB — the upper bound of a quotient is UBa ÷ LBb.
Using the rounded value in the calculation instead of the bound value.
Writing ≤ for the upper bound — it is strictly less than (<).
Next up: Percentage Error — measuring how far an estimate sits from the exact value, as a percentage. Bounds and percentage error are close cousins: both ask how much a rounded or estimated number can differ from the truth.
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