Mathematics Standard • Year 11 • Module 2 • Lesson 13
Errors and Limits of Accuracy — Problem Set
Apply absolute error, bounds and percentage error to swimming pool, paddock, timber-cutting and pharmacy scenarios where uncertainty actually matters.
Problem 1 — Swimming pool surface area (bounds for an area)
A rectangular swimming pool is measured as 25 m × 12 m using a tape measure with precision 0.5 m. The owner wants to buy a pool cover and needs to know the maximum and minimum possible surface area.
Set up: What are we solving for?
(i) State the absolute error and the bounds for each measurement. 1 mark
(ii) Find the maximum possible area of the pool. 1 mark
(iii) Find the minimum possible area of the pool. 1 mark
(iv) Pool-cover material costs $24/m². State the difference in cost between buying for the maximum area and the minimum area. 2 marks
Stuck? Revisit lesson § Worked Example 3 — Bounds for Area. Multiply uppers for max, lowers for min.Problem 2 — Surveying a paddock (bounds for a perimeter)
A farmer measures the three sides of a triangular paddock using a distance wheel with precision 1 m. The recorded sides are 120 m, 85 m, and 95 m. She needs to order fencing wire and wants a guaranteed-enough length.
Set up: What are we solving for?
(i) State the absolute error for each measurement. 1 mark
(ii) Find the maximum possible perimeter. 1 mark
(iii) Find the minimum possible perimeter and state the range (max − min). 2 marks
(iv) Fencing wire is sold in 50 m rolls. How many rolls must the farmer buy to guarantee she has enough for the maximum-possible perimeter? 1 mark
Stuck on (iv)? Divide the maximum perimeter by 50 and round UP — partial rolls are not sold.Problem 3 — Cutting 20 pieces from one plank
A carpenter needs to cut 20 identical pieces of timber, each marked as 45 cm long, from a single plank. He measures using a tape with millimetre graduations (precision 1 mm).
Set up: What are we solving for?
(i) State the absolute error for a single 45 cm measurement and the bounds for one piece. 1 mark
(ii) Find the maximum possible total length of the 20 pieces. 2 marks
(iii) Find the minimum possible total length, and the range (max − min) in cm. 2 marks
(iv) The plank he plans to use is exactly 9.05 m long. Will it be enough to guarantee 20 pieces, even in the maximum case? Justify in one sentence. 2 marks
Stuck? Errors compound when measurements are added — max total = 20 × upper bound; min total = 20 × lower bound.Problem 4 — Pharmacy weighing (small measurement, large % error)
A pharmacy assistant weighs out a tablet ingredient as 0.4 g using digital scales with precision 0.1 g. A second batch is weighed as 4.0 g on the same scales.
Set up: What are we solving for?
(i) Find the absolute error for both measurements (it is the same). 1 mark
(ii) Find the percentage error of each measurement to 2 d.p. 2 marks
(iii) Explain in one sentence why the smaller mass has a much larger percentage error, even though both measurements were taken on the same scales. 2 marks
Stuck on (iii)? The AE is fixed by the instrument; the % error depends on how big the measured value is compared with that AE.Problem 5 — Tile order for a courtyard (compounding errors)
A renovator is tiling a rectangular courtyard. The dimensions are measured using a tape (precision 1 cm) as 4.20 m × 3.50 m. Tiles cost $58/m² supplied and laid, and the supplier requires a 10% wastage allowance on top of the floor area.
Set up: What are we solving for?
(i) State the absolute error and the bounds for each dimension. 1 mark
(ii) Find the nominal area (using the recorded values 4.20 × 3.50). 1 mark
(iii) Find the maximum possible area in m², to 3 d.p. 2 marks
(iv) Using the maximum possible area, find the cost of tiles to the nearest dollar (include the 10% wastage). State a one-sentence conclusion about whether ordering based on the nominal area would be enough. 3 marks
Stuck on (iv)? Multiply your max area by 1.10 (for wastage), then by $58, then round to the nearest dollar.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Pool surface area
Set up. Bounds for each dim, then upper×upper and lower×lower.
(i) AE = 0.25 m. Bounds: 25 m → [24.75, 25.25] m; 12 m → [11.75, 12.25] m.
(ii) Max area = 25.25 × 12.25 = 309.3125 m².
(iii) Min area = 24.75 × 11.75 = 290.8125 m².
(iv) Difference in m² = 309.3125 − 290.8125 = 18.5 m². Cost difference = 18.5 × $24 = $444.00.
Problem 2 — Triangular paddock
Set up. AE for each side, then sum uppers / sum lowers for perimeter bounds; round up rolls.
(i) AE = ½ × 1 = 0.5 m for each side.
(ii) Max P = 120.5 + 85.5 + 95.5 = 301.5 m.
(iii) Min P = 119.5 + 84.5 + 94.5 = 298.5 m. Range = 301.5 − 298.5 = 3 m.
(iv) Rolls needed = 301.5 ÷ 50 = 6.03 → round up to 7 rolls (6 rolls give only 300 m, not enough).
Problem 3 — 20 pieces of timber
Set up. Single-piece bounds first, then ×20 for compounded total.
(i) AE = 0.5 mm = 0.05 cm. One piece in [44.95, 45.05] cm.
(ii) Max total = 20 × 45.05 = 901 cm = 9.01 m.
(iii) Min total = 20 × 44.95 = 899 cm = 8.99 m. Range = 901 − 899 = 2 cm.
(iv) Plank = 9.05 m = 905 cm. Even at the maximum case (901 cm), the plank has 905 − 901 = 4 cm spare, so yes, the 9.05 m plank guarantees 20 pieces with 4 cm to spare.
Problem 4 — Pharmacy scales
Set up. Same AE both times because same instrument; % error scales with how big the measurement is.
(i) AE = ½ × 0.1 = 0.05 g for both.
(ii) 0.4 g: % error = 0.05 ÷ 0.4 × 100 = 12.50%. 4.0 g: % error = 0.05 ÷ 4.0 × 100 = 1.25%.
(iii) The absolute error (0.05 g) is set by the instrument and stays the same, but it is a much larger fraction of 0.4 g than of 4.0 g — so the smaller mass has a much larger percentage error.
Problem 5 — Courtyard tiling
Set up. Bounds for each side, max area (uppers), then × 1.10 wastage, then × $58.
(i) AE = ½ × 1 cm = 0.5 cm = 0.005 m. Bounds: 4.20 m → [4.195, 4.205] m; 3.50 m → [3.495, 3.505] m.
(ii) Nominal area = 4.20 × 3.50 = 14.7 m².
(iii) Max area = 4.205 × 3.505 = 14.739 m² (to 3 d.p.).
(iv) With 10% wastage: 14.739 × 1.10 = 16.2129 m². Cost = 16.2129 × $58 = $940.35 ≈ $940. Conclusion: ordering on the nominal 14.7 m² (cost ≈ $937 with wastage) would only be about $3 short, but using the max area gives a safety buffer — order based on the max area to guarantee enough tile.