IB Maths AI HL Number Toolkit Paper 1 & 2 LB ≤ 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

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 to 1 d.p. — the range it came from 2.5 2.7 rounds to 2.62.55 lower bound — included 2.65 upper bound — not included 0.05 0.05
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.

Combining bounds (calculating with a and b) add:  UB = UBa + UBb  ·  LB = LBa + LBb subtract:  UB = UBa − LBb  ·  LB = LBa − UBb multiply:  UB = UBa × UBb  ·  LB = LBa × LBb divide:  UB = UBa ÷ LBb  ·  LB = LBa ÷ UBb
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

  1. Identify the degree of accuracy each number is rounded to.
  2. Halve it — this is how far each bound sits from the value.
  3. Find each bound: LB = value − half, UB = value + half; write LB ≤ x < UB.
  4. For a calculation, pick the bound of each input that drives the result to its extreme (use the combine rules).
  5. 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 0.1 half of 0.1 = 0.05 half up, half down LB = 6.4 − 0.05 = 6.35 UB = 6.4 + 0.05 = 6.45 6.35 ≤ m < 6.45 the 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 10 half of 10 = 5 half up, half down LB = 80 − 5 = 75 UB = 80 + 5 = 85 75 km ≤ d < 85 km the 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 < 25.5 18 cm: 17.5 ≤ b < 18.5 adding: UB = UB + UB, LB = LB + LB LB = 24.5 + 17.5 = 42 UB = 25.5 + 18.5 = 44 42 cm ≤ T < 44 cm for 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 < 50.5 12 L: 11.5 ≤ b < 12.5 subtracting: UB = UBa − LBb, LB = LBa − UBb UB = 50.5 − 11.5 = 39 LB = 49.5 − 12.5 = 37 37 L ≤ R < 39 L most 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 time distance: 145 ≤ d < 155 time: 11.5 ≤ t < 12.5 speed = d ÷ t; dividing mixes the bounds UB = 15511.5 = 13.478… LB = 14512.5 = 11.6 11.6 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 down length: 1.15 ≤ L < 1.25 width: 0.845 ≤ W < 0.855 (b) perimeter P = 2L + 2W (a sum — bounds line up) LB = 2(1.15) + 2(0.845) = 3.99 UB = 2(1.25) + 2(0.855) = 4.21 area A = L × W (multiply: low×low, high×high) LB = 1.15 × 0.845 = 0.97175 UB = 1.25 × 0.855 = 1.06875 (a) above · (b) 3.99 m ≤ P < 4.21 m  ·  0.972 m2A < 1.07 m2 use the bound values, never the rounded ones, in every calculation.

💡 Top tips

âš  Common mistakes

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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