Mathematics Standard • Year 11 • Module 2 • Lesson 13
Errors and Limits of Accuracy — Past-Paper Style
HSC Mathematics Standard 2-style writing on errors — short answers and one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A length is measured as 8.5 cm using a ruler with 1 mm graduations.
(a) State the absolute error.
(b) State the upper and lower bounds for the true length.
(c) Calculate the percentage error, correct to 2 significant figures. 3 marks Band 3
1.2 The dimensions of a swimming pool are measured as 25 m × 12 m using a tape with precision 0.5 m.
(a) State the absolute error and the bounds for the 25 m measurement.
(b) Find the maximum and minimum possible area of the pool. 3 marks Band 3-4
1.3 Two students measure the same length. Student A records it as 6 m using a tape with precision 0.5 m. Student B records it as 6.00 m using a tape with precision 0.01 m.
(a) Calculate the percentage error of each measurement to 2 decimal places.
(b) Explain in one sentence which student's measurement is more reliable, with a numerical comparison. 4 marks Band 4
2. Extended response
2.1 A surveyor measures three sides of a triangular community-garden plot using a distance wheel with precision 1 m. The recorded measurements are 120 m, 85 m and 95 m. The local council needs to order fencing wire for the perimeter and wants a buffer to be sure they don't run short.
Fencing wire is sold in 50 m rolls only (no part rolls).
The wire costs $135 per roll.
(a) State the absolute error for each side and find the bounds for each.
(b) Find the maximum possible perimeter and the minimum possible perimeter, and state the range (max − min).
(c) Calculate how many full rolls of wire the council must order to guarantee enough wire even in the maximum case. State the total cost, and a one-sentence conclusion. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct AE = 0.5 m for each side.
• 1 mark — correct bounds for each of the three sides.
Part (b) — 2 marks
• 1 mark — correct max perimeter = 301.5 m.
• 1 mark — correct min perimeter and range stated.
Part (c) — 3 marks
• 1 mark — divides max perimeter by 50 and rounds UP to a whole roll count.
• 1 mark — correct total cost = (number of rolls) × $135.
• 1 mark — explicit conclusion sentence stating roll count and cost with units.
Your response:
Stuck on (c)? 6 rolls × 50 m = 300 m, which is just below the max perimeter — so 7 rolls is needed to guarantee enough.How did this worksheet feel?
What I'll revisit before next class:
1.1 — 8.5 cm to nearest mm (3 marks)
Sample response.
(a) AE = ½ × 1 mm = 0.5 mm.
(b) Bounds: [8.45 cm, 8.55 cm].
(c) % error = 0.05 ÷ 8.5 × 100 ≈ 0.59%.
Marking notes. 1 mark — correct AE with units. 1 mark — correct bounds with units. 1 mark — correct % error to 2 s.f. with the % sign. Common error: mixing units in the % error step (using 0.5 mm and 8.5 cm without converting) — must be the same unit before dividing.
1.2 — Pool 25 m × 12 m, precision 0.5 m (3 marks)
(a) Sample. AE = 0.25 m. Bounds for 25 m: [24.75, 25.25] m.
(b) Sample. Bounds for 12 m: [11.75, 12.25] m. Max area = 25.25 × 12.25 = 309.3125 m². Min area = 24.75 × 11.75 = 290.8125 m².
Marking notes. (a) 1 mark — AE = 0.25 m and bounds correct. (b) 1 mark — max area correct. 1 mark — min area correct (both with units m²). Common error: AE = 0.5 m (forgetting to halve the precision).
1.3 — Two students measuring the same length (4 marks)
(a) Sample. Student A: AE = 0.25 m; % error = 0.25 ÷ 6 × 100 ≈ 4.17%. Student B: AE = 0.005 m; % error = 0.005 ÷ 6 × 100 ≈ 0.08%.
(b) Sample. Student B's measurement is more reliable because its percentage error (≈0.08%) is about 50 times smaller than Student A's (≈4.17%).
Marking notes. (a) 1 mark — % error for A. 1 mark — % error for B (both with %). (b) 1 mark — identifies B as more reliable. 1 mark — supports the choice with a numerical comparison (ratio or comment that B's % error is much smaller). Common error: writing "B is more accurate because the tape is more precise" without quoting the numbers loses the comparison mark.
2.1 — Community-garden fence order (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) AE and bounds.
AE = ½ × 1 = 0.5 m for each side. [1 mark — AE.]
Bounds: 120 m → [119.5, 120.5] m; 85 m → [84.5, 85.5] m; 95 m → [94.5, 95.5] m. [1 mark — all three bounds.]
(b) Max, min, range.
Max P = 120.5 + 85.5 + 95.5 = 301.5 m. [1 mark — max.]
Min P = 119.5 + 84.5 + 94.5 = 298.5 m. Range = 301.5 − 298.5 = 3 m. [1 mark — min and range.]
(c) Rolls needed and cost.
Max perimeter = 301.5 m. Rolls = 301.5 ÷ 50 = 6.03 → 7 rolls (round up, because 6 rolls give only 300 m which is below the max). [1 mark — rounded-up roll count.]
Cost = 7 × $135 = $945.00. [1 mark — multiplication and total cost.]
Conclusion: the council must order 7 rolls of fencing wire at a total cost of $945.00 to guarantee enough wire even in the maximum-perimeter case. [1 mark — explicit conclusion naming rolls, cost and the safety reasoning.]
Total: 7/7.
Band descriptors for marker.
Band 3: AE = 0.5 m stated and perimeter calculated using only the recorded values (300 m), with no use of bounds. ≈ 2-3 marks.
Band 4: Max and min perimeter computed correctly; tries the roll calculation but rounds down or to the nearest whole, missing the safety logic. ≈ 4-5 marks.
Band 5: Full numerical solution with 7 rolls and $945, but conclusion sentence missing the "guarantee enough" reasoning. ≈ 6 marks.
Band 6: Complete, correct, and conclusion explicitly justifies rounding UP (because 6 rolls would be insufficient at the max perimeter). 7/7.