Engineering Maths Skills ProgrammeTutor answer sheet · Modules 01–05
GCSE Maths in Engineering · Not for learner issue
Answer sheet
Full answers, methods and mark allocations for all five modules, with the accepted tolerance for each auto-marked box and the errors that cost learners marks most often. Each module carries 12 marks; the programme totals 60.
Modules5
Total marks60
Questions20
PrintA4 portrait, 2 pages
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Answers by module
Q1–Q3 practice · Q4 exam-style01
Sizing a Footbridge Truss
GM-01/TRUSSDesign window: Every panel count from 4 to 9 has a workable depth band; 10 panels has none. 4 panels needs 2.15–2.25 m, 6 needs 1.45–1.55 m, 9 needs 1.00–1.05 m. The taught answer is 6 panels at 1.50 m (36.9°, £1767.39). Shallow fails the angle rule, deep breaks the budget.
| Answer | Method | Marks | |
|---|---|---|---|
| Q1 | 3.35 m | Panel width 12 ÷ 4 = 3.00 m. d = √(3.00² + 1.50²) = √11.25tolerance accepted ±0.03 | 2 mk |
| Q2 | 26.6° | tanθ = 1.50 ÷ 3.00 = 0.5, θ = tan¹(0.5). Fails the 35°–55° rule — too flat.tolerance accepted ±0.2 | 2 mk |
| Q3 | 2.14 m | Panel width 1.50 m. d = 1.50 × tan 55° = 1.50 × 1.4281tolerance accepted ±0.05 | 3 mk |
| Q4 | £1514 | Diagonal √(2.00²+1.60²) = 2.5613. One frame 20 + 6(1.60) + 5(2.5613) = 42.406 m. Two frames 84.813 m → 966.87 kg = 0.96687 t → ×1450×1.08tolerance accepted ±12 | 5 mk |
Commonest errors5 panels means 6 verticals, not 5. Also watch learners rounding the diagonal to 2.56 before multiplying by five.
02
Batching a Concrete Base
GM-02/CONCDesign window: Depth ≥ 0.40 m (footing) and ≤ 0.48 m in a 1:2:4 mix (budget); the C30 mix pulls that ceiling to about 0.43 m. 1:3:6 fails on strength whatever the depth. Note the drum accepts equivalent ratios — 2:4:8 and 3:6:12 are both read as C25.
| Answer | Method | Marks | |
|---|---|---|---|
| Q1 | 2.16 m³ | V = 2.40 × 1.80 × 0.50tolerance accepted ±0.02 | 2 mk |
| Q2 | 740.6 kg | Mass = 2.16 × 2400 = 5184 kg. Parts 1+2+4 = 7, cement = 5184 ÷ 7tolerance accepted ±1.2 | 2 mk |
| Q3 | 30 bags | 740.57 ÷ 25 = 29.6, rounded up — part bags cannot be boughttolerance accepted ±0.4 | 3 mk |
| Q4 | £414 | V = 3.50×2.20×0.40 = 3.08 m³. Order 3.08×1.05 = 3.234 m³. Cost ×128 = £413.95. Cement: 3.08×2400 = 7392 kg ÷ 5.5 parts = 1344 kgtolerance accepted ±12 | 5 mk |
Commonest errorsRounding 29.6 down to 29 bags in Q3. In Q4, applying the 5% to the cost rather than the volume.
03
Sizing a Cable Run
GM-03/CABLEDesign window: 2.5 mm² fails volt drop beyond 31.9 m; 6 mm² passes electrically but costs £154 at 35 m. 4 mm² is the only size that clears both (7.70 V, 3.35%, £110.25).
| Answer | Method | Marks | |
|---|---|---|---|
| Q1 | 10.08 V | Vd = (18 × 20 × 28) ÷ 1000tolerance accepted ±0.06 | 2 mk |
| Q2 | 4.4% | 10.08 ÷ 230 × 100 = 4.383… Passes, but only justtolerance accepted ±0.1 | 2 mk |
| Q3 | 49.2 m | 11.5 × 1000 = 11500; 7.3 × 32 = 233.6; L = 11500 ÷ 233.6tolerance accepted ±0.3 | 3 mk |
| Q4 | 9.29 V | A = 4×10⁻⁶ m², L = 2×45 = 90 m. R = (1.72×10⁻⁸ × 90) ÷ (4×10⁻⁶) = 0.387 Ω. V = IR = 24 × 0.387tolerance accepted ±0.06 | 5 mk |
Commonest errorsForgetting to double the cable length for the return conductor, and converting mm² to m² by dividing by 1000 instead of 1 000 000.
04
Setting CNC Cutting Speeds
GM-04/CNCDesign window: At Vc = 120 m/min the 6 mm cutter needs 6366 rpm — beyond the machine. Dropping to 75.4 m/min would fit the spindle but falls below the 100 m/min tool-life minimum, so the 6 mm cutter is unusable here. 12 mm and 25 mm both work.
| Answer | Method | Marks | |
|---|---|---|---|
| Q1 | 1432 rev/min | N = (1000 × 90) ÷ (π × 20) = 90000 ÷ 62.83tolerance accepted ±4 | 2 mk |
| Q2 | 430 mm/min | Vf = 1432 × 0.10 × 3 (429.7 from the unrounded speed)tolerance accepted ±4 | 2 mk |
| Q3 | 100.5 m/min | Vc = Nπd ÷ 1000 = (4000 × π × 8) ÷ 1000tolerance accepted ±1 | 3 mk |
| Q4 | 22.4 s | N = 140000 ÷ (π×16) = 2785 rpm. Vf = 2785.2 × 0.06 × 4 = 668.4 mm/min. t = 250 ÷ 668.4 = 0.374 min × 60tolerance accepted ±0.35 | 5 mk |
Commonest errorsOmitting the number of teeth from the feed rate, and reading 0.374 min as 37 seconds instead of converting properly.
05
Sizing a Hydraulic Tank
GM-05/TANKDesign window: Only 3 mm plate can pass at all — at £28 and £34 per m² even the smallest compliant tank breaks the £45 budget. In 3 mm the window is roughly 0.56–0.89 m long by 0.38–0.60 m high. The taught answer, 0.80 × 0.45 m, gives 162 L at £40.59.
| Answer | Method | Marks | |
|---|---|---|---|
| Q1 | 150 litres | V = 0.75 × 0.40 × 0.50 = 0.15 m³, ×1000tolerance accepted ±1.5 | 2 mk |
| Q2 | 117.5 kg | Oil volume 0.15 × 0.9 = 0.135 m³, mass = ×870tolerance accepted ±1 | 2 mk |
| Q3 | 1.75 m² | SA = 2(0.30 + 0.375 + 0.20)tolerance accepted ±0.03 | 3 mk |
| Q4 | 226 litres | r = 0.30 m. V = πr²h = π×0.09×0.80 = 0.2262 m³ → 226 L. Plate: SA = 2πr² + 2πrh = 2.0735 m², mass = ×0.003×7850 = 48.8 kgtolerance accepted ±5 | 5 mk |
Commonest errorsUsing the diameter in place of the radius in Q4 — it makes the capacity four times too big, around 900 litres.
Produced for EAL engineering learners.
Tutor copy — withhold from learners until the modules are complete.