Upper & Lower Bounds

SPEC N16 · NUMBER — BOUNDS

Bounds of a single measurement · Combining bounds in calculations

Addition · Subtraction · Multiplication · Division · Appropriate accuracy

Upper & Lower Bounds H

A measurement stated as a rounded value could have come from a range of true values. The extremes of that range are the bounds.

UPPER BOUND (UB)

The largest value a measurement could have been before rounding.

For a value x given to the nearest unit u: UB = x + u/2

LOWER BOUND (LB)

The smallest value a measurement could have been before rounding.

For a value x given to the nearest unit u: LB = x − u/2

Note on notation

In exam questions asking for bounds, write them as a decimal (e.g. "LB = 8.25"), not as part of an inequality. The error-interval notation (a ≤ x < b) from Lesson 1.10 uses the same numbers, but bounds questions usually ask for the two values directly.

Bounds of a Single Measurement

UB = x + u/2 · LB = x − u/2

where u = the rounding unit.

Example — 5.6 m to 1 d.p.

Rounding unit u = 0.1 m.

Half-unit = 0.05.

LB = 5.55 m

UB = 5.65 m

Example — 20 s to 2 s.f.

20 to 2 s.f.: last sig fig in the units place → unit u = 1.

Half-unit = 0.5.

LB = 19.5 s

UB = 20.5 s

⚠️ EXAMINER ALERT — wrong half-unit.

Students add 0.1 instead of 0.05 for a value given to 1 d.p.:

UB of 8.3 = 8.3 + 0.1 = 8.4 ✗

Correct:UB = 8.3 + 0.05 = 8.35 ✓

The shift is half the rounding unit, not the full unit.

Practice Question

A time t is recorded as 14.7 seconds, correct to 1 d.p.

Write down the upper bound and the lower bound of t. 2 marks

M:

Rounding unit = 0.1 s. Half-unit = 0.05 s.

C:

LB = 14.7 − 0.05 = 14.65 s

UB = 14.7 + 0.05 = 14.75 s

E:

LB = 14.65 s | UB = 14.75 s ✓

B1 each — 2 marks for both correct.

Combining Bounds in Calculations

When a quantity is calculated from rounded measurements, the result is also uncertain. The bounds of the result depend on which combination of input bounds you use — and the rule is different for each of the four operations.

The Golden Rule

To find the UB of a result, ask: which combination of input bounds makes the result as large as possible?

To find the LB of a result, ask: which combination makes the result as small as possible?

Addition: A + B

UB(A + B) = UB(A) + UB(B) · LB(A + B) = LB(A) + LB(B)

Intuition: To make the sum as large as possible, take the largest of each. To make it as small as possible, take the smallest of each. "Same-sign" combination.

Example

Two pieces of string, each measured to the nearest cm.

String A = 23 cm, String B = 47 cm. Find the bounds of the total length A + B.

Bounds of A: LB = 22.5, UB = 23.5.

Bounds of B: LB = 46.5, UB = 47.5.

UB(A + B) = 23.5 + 47.5 = 71 cm

LB(A + B) = 22.5 + 46.5 = 69 cm

Extension

The addition pattern extends naturally to sums of more than two values. For A + B + C: UB of total = UB(A) + UB(B) + UB(C); LB of total = LB(A) + LB(B) + LB(C). Same idea for multi-factor products like A × B × C.

Subtraction: A − B

UB(A − B) = UB(A) − LB(B) · LB(A − B) = LB(A) − UB(B)

Intuition: To make A − B as large as possible, make A as large as possible AND B as small as possible (subtracting less leaves more). To make A − B as small as possible, make A small AND B large.

Example

A wooden plank is 180 cm long (to the nearest cm). A piece of length 45 cm (to the nearest cm) is cut off. Find the bounds of the remaining length.

Bounds of plank P: LB = 179.5, UB = 180.5.

Bounds of piece Q: LB = 44.5, UB = 45.5.

UB(P − Q) = UB(P) − LB(Q) = 180.5 − 44.5 = 136 cm

LB(P − Q) = LB(P) − UB(Q) = 179.5 − 45.5 = 134 cm

⚠️ EXAMINER ALERT — subtraction bounds flipped. Students compute LB of a difference as LB(A) − LB(B), mirroring the addition rule. This gives the wrong answer. Rule: for the smallest A − B, take the smallest A and the largest B.

Multiplication: A × B

UB(A × B) = UB(A) × UB(B) · LB(A × B) = LB(A) × LB(B)

Intuition: When both values are positive, the product is largest when both factors are largest, and smallest when both are smallest. "Same-sign" combination — like addition.

Example

A rectangle has length L = 8.3 cm and width W = 5.6 cm, both to 1 d.p. Find the bounds of the area.

L: LB = 8.25, UB = 8.35.

W: LB = 5.55, UB = 5.65.

UB(Area) = 8.35 × 5.65 = 47.1775 cm²

LB(Area) = 8.25 × 5.55 = 45.7875 cm²

Key assumption

The rules on this slide assume both quantities are positive — which is always true for physical measurements like length, time, mass. For signed quantities, the pattern can change, but at GCSE you will only be asked about positive measurements.

Division: A ÷ B

UB(A ÷ B) = UB(A) ÷ LB(B) · LB(A ÷ B) = LB(A) ÷ UB(B)

Intuition: To make a fraction as large as possible, you want the largest top and the smallest bottom. To make it as small as possible, the smallest top and the largest bottom. Dividing by a smaller number gives a larger result.

Example — Speed

A car travels 120 km (to the nearest km) in 2.4 hours (to 1 d.p.). Find the bounds of its average speed.

Distance: LB = 119.5, UB = 120.5.

Time: LB = 2.35, UB = 2.45.

UB(Speed) = UB(dist) ÷ LB(time) = 120.5 ÷ 2.35 = 51.28 km/h (2 d.p.)

LB(Speed) = LB(dist) ÷ UB(time) = 119.5 ÷ 2.45 = 48.78 km/h (2 d.p.)

⚠️ EXAMINER ALERT — division bounds flipped. Students compute LB of a quotient as LB(A) ÷ LB(B), mirroring the multiplication rule. This gives the wrong answer. Rule: for division, to make the result smallest, take the smallest top and the largest bottom.

Summary — All Four Operations

OperationUpper bound of resultLower bound of resultPattern

A + BUB(A) + UB(B)LB(A) + LB(B)Same-sign

A − BUB(A) − LB(B)LB(A) − UB(B)Flip B

A × BUB(A) × UB(B)LB(A) × LB(B)Same-sign

A ÷ BUB(A) ÷ LB(B)LB(A) ÷ UB(B)Flip B

Memory device

"Plus and times — same. Minus and divide — flip."

Addition and multiplication preserve the sign of the bound you pick: to make the result biggest, pick biggest. Subtraction and division reverse which bound of the second number you want.

MCE Worked Example

A rectangle has length L = 8.3 cm and width W = 5.6 cm, both correct to 1 d.p.

(a) Find the upper and lower bounds of the area.

(b) Give the area to a suitable degree of accuracy, justifying your answer. 5 marks

M Find bounds for L and W.

Area = L × W.

UB × UB for UB of area.

LB × LB for LB of area.

C L: 8.25 ≤ L < 8.35

W: 5.55 ≤ W < 5.65

LB = 8.25 × 5.55 = 45.7875 cm²

UB = 8.35 × 5.65 = 47.1775 cm²

E (a) LB = 45.7875 cm²

UB = 47.1775 cm²

(b) To 2 s.f.: 46 vs 47 — differ.

To 1 s.f.: both = 50. ✓

Area ≈ 50 cm² to 1 s.f.

Justifying the accuracy

The highest precision at which both bounds agree is the most precise justifiable value. Here: at 1 s.f. both LB (45.7875) and UB (47.1775) round to 50. At 2 s.f. they differ (46 vs 47) — so 2 s.f. is not justifiable.

Practice Question

A square has side length s = 12 cm, correct to the nearest cm.

Find the upper bound of the area of the square. 2 marks

M:

Area = s². Upper bound of s² uses UB of s (both factors are s).

C:

Rounding unit = 1 cm, half-unit = 0.5.

UB(s) = 12 + 0.5 = 12.5 cm.

UB(Area) = 12.5 × 12.5 = 156.25 cm²

E:

UB(Area) = 156.25 cm² ✓

Practice Question

A runner covers a distance d = 400 m (nearest metre) in a time t = 52 seconds (nearest second).

Find the lower bound of the runner's average speed in m/s.

Give your answer to 2 d.p. 3 marks

M:

Speed = distance ÷ time. For LB of speed: smallest distance, largest time.

LB(Speed) = LB(d) ÷ UB(t)

C:

LB(d) = 400 − 0.5 = 399.5 m

UB(t) = 52 + 0.5 = 52.5 s

LB(Speed) = 399.5 ÷ 52.5 = 7.6095... m/s

E:

LB(Speed) = 7.61 m/s (to 2 d.p.) ✓

Appropriate Degree of Accuracy

Exam questions often ask: "Give your answer to a suitable degree of accuracy and justify." The standard method:

Method

1. Compute LB and UB of the result.

2. Round both to progressively lower precision (1 more s.f., then 1 less, etc.).

3. The highest precision at which both bounds round to the same value is the answer.

4. State both bounds, show both round to the same value, then quote the result at that precision.

Example — density calculation

LB of density = 2.3418 g/cm³, UB = 2.3472 g/cm³.

To 3 s.f.: 2.34 vs 2.35 — differ.

To 2 s.f.: 2.3 vs 2.3 — same! ✓

Answer: density ≈ 2.3 g/cm³ (to 2 s.f.).

Justification: both bounds round to 2.3 to 2 s.f.

Mark scheme

"Appropriate accuracy" questions are worth 4–5 marks. Typical 4-mark breakdown: M1 for correct bound choices; M1 for correct operation; A1 for the two bounds; B1 for stating the final answer at an appropriately justified precision. Higher-mark versions add a separate M mark for arithmetic.

Common Errors — Don't Lose These Marks

Students often writeCorrect versionError type

UB of 8.3 (1 d.p.) = 8.3 + 0.1 = 8.4UB = 8.3 + 0.05 = 8.35Wrong half-unit

LB(A − B) = LB(A) − LB(B)LB(A) − UB(B)Subtraction flip

LB(A ÷ B) = LB(A) ÷ LB(B)LB(A) ÷ UB(B)Division flip

Quote single bound as answerRound both bounds, pick matching precisionAppropriate accuracy method

UB = x + u (full unit)UB = x + u/2 (half unit)Bounds are half-unit shifts

Memory shortcut

"Plus and times — same. Minus and divide — flip." Write this at the top of your paper at the start of the exam if bounds questions make you nervous.

Summary

Key Points — Lesson 1.11 H

Single value: UB = x + u/2, LB = x − u/2 (u = rounding unit).

A + B: UB = UB(A) + UB(B), LB = LB(A) + LB(B). Same-sign.

A − B: UB = UB(A) − LB(B), LB = LB(A) − UB(B). Flip B.

A × B: UB = UB(A) × UB(B), LB = LB(A) × LB(B). Same-sign.

A ÷ B: UB = UB(A) ÷ LB(B), LB = LB(A) ÷ UB(B). Flip B.

Appropriate accuracy: highest precision at which both bounds match.

The two flips

Only subtraction and division flip the second number's bound.

Addition and multiplication are "obvious" — same-sign.

Core intuition

Always ask: which combination makes the result biggest/smallest?

Then pick the input bounds that achieve it.

Exam Tip

Always write out the bounds of each input value before combining. Label them LB(A), UB(A), LB(B), UB(B) clearly — examiners award method marks for showing correct bound choices even if the arithmetic goes wrong.