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Pearson Edexcel 4MA1 IGCSE  ·  Topic G7.2  ·  Higher tier

G7.2 Bounds and Limits of Accuracy

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01 — Explanation

Why does a rounded measurement have "bounds"?

Every measurement is really an approximation. If a length is recorded as 2121 cm to the nearest cm, the true value could be anything from 20.520.5 cm up to (but not including) 21.521.5 cm — any of those values round back to 2121. These two edges are the lower bound (LB) and upper bound (UB), written as an error interval:

20.5≤L<21.520.5 \le L < 21.5

The upper bound uses a strict << because 21.521.5 itself would round up to 2222, not 2121.

02 — Explanation

How do I find the bounds of a rounded value?

Use the half-unit rule: halve the rounding unit and add/subtract it from the given value.

  • Nearest cm (unit =1=1): half-unit =0.5=0.5 → 21±0.521\pm0.5.
  • Nearest 5 m (unit =5=5): half-unit =2.5=2.5 → 105±2.5⇒102.5≤L<107.5105\pm2.5 \Rightarrow 102.5\le L<107.5.
  • 2 d.p. (unit =0.01=0.01): half-unit =0.005=0.005 → 5.36±0.0055.36\pm0.005.
  • 2 s.f., e.g. x=240x=240: the last significant digit (4) is in the tens column, so the unit is 1010, half-unit 55 → 235≤x<245235\le x<245.

Exam Tip: For significant figures, find the place value of the last significant digit first — that tells you the rounding unit. Don't assume it's always 0.5.

03 — Explanation

How are bounds different for a truncated number?

Truncation just chops digits off without rounding — 5.875.87 truncated to 1 d.p. is 5.85.8, not 5.95.9. The truncated value is always the lower bound; the upper bound is one unit of the last digit's place value higher.

7.3 (truncated to 1 d.p.)⇒7.3≤x<7.47.3 \text{ (truncated to 1 d.p.)} \Rightarrow 7.3 \le x < 7.4

04 — Explanation

How do bounds combine when I add, subtract, multiply, or divide?

Addition and multiplication are intuitive — to get the biggest result, use the biggest inputs. Subtraction and division are not intuitive, because making the result as large as possible means pairing a big value with a small one.

OperationLower BoundUpper Bound
A+BA+BLB(A)+LB(B)LB(A)+LB(B)UB(A)+UB(B)UB(A)+UB(B)
A−BA-BLB(A)−UB(B)LB(A)-UB(B)UB(A)−LB(B)UB(A)-LB(B)
A×BA\times BLB(A)×LB(B)LB(A)\times LB(B)UB(A)×UB(B)UB(A)\times UB(B)
A÷BA\div BLB(A)÷UB(B)LB(A)\div UB(B)UB(A)÷LB(B)UB(A)\div LB(B)

Exam Tip: For subtraction and division, say the rule out loud before substituting: "biggest result needs the biggest AA and the smallest BB." Writing it wrong is the single most common bounds error.

Worked check (subtraction): b=c2−a2b=\sqrt{c^2-a^2}, with c=25c=25 cm, a=10a=10 cm (both nearest cm). Bounds: 24.5≤c<25.524.5\le c<25.5, 9.5≤a<10.59.5\le a<10.5. For LB(b)LB(b) we need LB(c2)−UB(a2)=24.52−10.52=600.25−110.25=490LB(c^2)-UB(a^2) = 24.5^2-10.5^2 = 600.25-110.25=490, so LB(b)=490≈22.14LB(b)=\sqrt{490}\approx22.14 cm.

05 — Explanation

How do I justify an answer "to a suitable degree of accuracy"?

Calculate both the final LB and UB, then find the most precise rounding that gives the same value for both — any digit they agree on, you're certain of.

Worked exam-style question: P=F/AP = F/A where F=28.5F=28.5 (3 s.f.) and A=0.195A=0.195 (3 s.f.). Find PP to a suitable degree of accuracy, with justification.

  1. Bounds: 28.45≤F<28.5528.45\le F<28.55 and 0.1945≤A<0.19550.1945\le A<0.1955.
  2. LB(P)=LB(F)÷UB(A)=28.45÷0.1955≈145.52LB(P)=LB(F)\div UB(A) = 28.45\div0.1955\approx145.52.
  3. UB(P)=UB(F)÷LB(A)=28.55÷0.1945≈146.79UB(P)=UB(F)\div LB(A)=28.55\div0.1945\approx146.79.
  4. Round both to increasing accuracy: at 1 s.f. both give 100100; at 2 s.f. both give 150150; at 3 s.f. they disagree (146146 vs 147147).
  5. Answer: P≈150P\approx150 (2 s.f.), justified because both the LB and UB round to 150150 at that accuracy — rounding further is not supported by the data.

06 — Common mistakes

Common mistakes to avoid

  • Using ≤\le on both sides of the error interval — the upper bound is always a strict << for continuous data.
  • Flipping the subtraction/division rule and pairing LB with LB, UB with UB.
  • Treating π\pi or other exact constants as having bounds — only measured quantities do.
  • Confusing truncation with rounding — truncated numbers don't have a symmetric half-unit either side.