Upper and lower bounds — limits of accuracy [Higher tier]
Every measurement is rounded. If a length is given as 4.5 cm "to 1 d.p.", the actual length could be anything from 4.45 cm up to (just below) 4.55 cm. Bounds are how we make this precise.
The half-up, half-down rule
For a value rounded to a particular precision (1 d.p., 1 s.f., nearest 10, etc.):
- Lower bound (LB) = stated value − half the rounding interval.
- Upper bound (UB) = stated value + half the rounding interval.
Worked example: a length is 4.5 cm to 1 d.p.
- Rounding interval = 0.1 cm.
- Half of that = 0.05 cm.
- LB = 4.5 − 0.05 = 4.45 cm.
- UB = 4.5 + 0.05 = 4.55 cm.
Strictly, the upper bound is not attainable (it would round up to 4.6), so the inequality is 4.45 ≤ length < 4.55. AQA accept either statement of the UB; just be consistent.
Other examples
- 27 cm to the nearest cm: LB = 26.5; UB = 27.5.
- 2,300 to 2 s.f.: rounding interval = 100; LB = 2,250; UB = 2,350.
- 0.06 g to 1 s.f.: rounding interval = 0.01 (0.05 to 0.15); but since the leading digit is 6, the interval is 0.01 → LB = 0.055; UB = 0.065.
Bounds in calculations
When combining bounds, choose the LB and UB of each input to maximise or minimise the result.
Sum (a + b)
- Max sum = UB(a) + UB(b).
- Min sum = LB(a) + LB(b).
Difference (a − b)
- Max difference = UB(a) − LB(b).
- Min difference = LB(a) − UB(b).
- Note the cross-pairing: maximising a − b means biggest a, smallest b.
Product (a × b)
- Max product = UB(a) × UB(b) (assuming both positive).
- Min product = LB(a) × LB(b).
Quotient (a ÷ b)
- Max quotient = UB(a) ÷ LB(b).
- Min quotient = LB(a) ÷ UB(b).
- Cross-pairing again: dividing by a smaller denominator gives a bigger quotient.
✦Worked example— Worked example — speed
A car covers 240 km (to the nearest 10 km) in 3 hours (to the nearest hour).
- Distance: LB = 235, UB = 245.
- Time: LB = 2.5, UB = 3.5.
For the maximum possible average speed:
- Speed = D / T → max when D is biggest, T is smallest.
- Max speed = 245 ÷ 2.5 = 98 km/h.
For the minimum possible average speed:
- Min speed = 235 ÷ 3.5 ≈ 67.14 km/h.
"Truncation" vs "rounding"
If a measurement is given to a stated accuracy by truncation (chopping off rather than rounding), the bounds are different. AQA defaults to rounding, so use the half-up/half-down rule unless told "truncated".
Stating answers "to a suitable degree of accuracy"
A favourite Higher question: "A length is given as 7.43 m. Calculate the area of a square of this side, and state your answer to a suitable degree of accuracy."
- Find LB and UB of the side: 7.425 and 7.435.
- LB(area) = 7.425² = 55.130625.
- UB(area) = 7.435² = 55.279225.
- Both round to 55.2 to 3 s.f., so the area is 55.2 m² to 3 s.f..
- Both round to 55 to 2 s.f., so 55 m² to 2 s.f. is also justified.
- The "suitable" answer is the most precise figure that LB and UB agree on. So: 55.2 m² (3 s.f.).
⚠Common mistakes— Common mistakes (examiner traps)
- Adding/subtracting halves of the wrong interval. A measurement to 2 d.p. has interval 0.01 → halve to 0.005.
- Min difference paired wrong. For a − b, MIN uses LB(a) − UB(b), not LB(a) − LB(b).
- Ignoring units consistency (e.g. minutes vs hours).
- Stating UB as the rounded boundary as a strict equality.
UB = 4.55is fine for AQA, but technically it's the supremum. - Writing the final answer with too many digits when LB and UB diverge. Truncate to the precision they agree on.
➜Try this— Quick check
A rectangle has length 9.4 cm and width 5.7 cm, each to 1 d.p. Find the bounds of the area.
- L: 9.35 ≤ L < 9.45.
- W: 5.65 ≤ W < 5.75.
- Min area = 9.35 × 5.65 = 52.8275 cm².
- Max area = 9.45 × 5.75 = 54.3375 cm².
AI-generated · claude-opus-4-7 · v3-deep-number