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HKDSE Design and Applied Technology Practice Paper: Structures, Mechanisms, Electronics and Manufacturing
HKDSE DAT – 12 questions – 100 marks – 75 minutes
HKDSE Design and Applied Technology practice paper: beam reactions, gears, circuits, stress, energy cost, tolerance fits and quality control, with full solutions.
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Paper details HKDSE Design and Applied Technology practice paper on beam reactions, gears, material density, circuits, belt drives, stress, energy cost, tolerance fits and quality control, with a full marking scheme.
Section A: Structures and Mechanisms (Q1-Q4)
State the formula and all working; answers alone earn no method marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A horizontal beam 2.4 m long has negligible weight. Support A is at the left end and support B is 1.8 m from the left end. A load of 600 N hangs 1.2 m from the left end. (a) Taking moment about A, find the reaction at B. (4 marks) (b) Find the reaction at A and check it by vertical force equilibrium. (4 marks) |
(a) Moment about A: 600 x 1.2 = 720 N m clockwise. R_B x 1.8 = 720, so R_B = 400 N. (2 marks formula, 2 marks answer) (b) Vertical equilibrium: R_A + R_B = 600 N, so R_A = 200 N; 200 + 400 = 600 N, which matches the load. (2 marks + 2 marks check) |
| 2 | In a spur gear pair the driver has 18 teeth and rotates at 900 rpm, driving a 54-tooth driven gear. (a) Find the gear ratio and the speed of the driven gear. (4 marks) (b) If the input torque is 12 N m and losses are neglected, find the output torque and explain the relation between speed and torque. (4 marks) |
(a) Gear ratio = 54/18 = 3; driven speed = 900/3 = 300 rpm. (2 marks + 2 marks) (b) Output torque = 12 x 3 = 36 N m. Speed falls to one third while torque rises three times, so power is nearly unchanged (P = 2 pi n T / 60). (4 marks) |
| 3 | An aluminium plate measures 1.2 m by 0.8 m by 6 mm. The density of aluminium is 2700 kg/m3. (a) Find the volume of the plate in m3. (3 marks) (b) Find its mass in kg. (3 marks) (c) If a steel plate of density 7850 kg/m3 is used instead, by what percentage does the mass increase? (2 marks) |
(a) Volume = 1.2 x 0.8 x 0.006 = 0.00576 m3. (3 marks) (b) Mass = 0.00576 x 2700 = 15.55 kg. (3 marks) (c) Steel mass = 0.00576 x 7850 = 45.22 kg; increase = (45.22 – 15.55) / 15.55 x 100% = 190.7%. (2 marks) |
| 4 | A 12 V DC supply feeds two 220 ohm resistors in parallel, which are in series with a 100 ohm resistor. (a) Find the equivalent resistance of the parallel section. (3 marks) (b) Find the total resistance and the total current in mA. (3 marks) (c) Find the total power delivered by the supply. (2 marks) |
(a) Parallel equivalent = (220 x 220) / (220 + 220) = 110 ohm. (3 marks) (b) Total resistance = 110 + 100 = 210 ohm; current = 12 / 210 = 0.0571 A = 57.1 mA. (3 marks) (c) Power = 12 x 0.0571 = 0.69 W. (2 marks) |
Section B: Electronics, Energy and Drives (Q5-Q8)
Use g = 9.81 m/s2 and pi = 3.142 throughout.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | In a belt drive the driver pulley has a diameter of 80 mm and runs at 1440 rpm, driving a pulley of 240 mm diameter. (a) Find the speed of the driven pulley. (3 marks) (b) Find the belt linear speed in m/s. (3 marks) (c) If the belt slips by 4%, find the actual driven speed. (3 marks) |
(a) n2 = n1 x D1 / D2 = 1440 x 0.080 / 0.240 = 480 rpm. (3 marks) (b) v = pi D1 n1 / 60 = 3.142 x 0.080 x 1440 / 60 = 6.03 m/s. (3 marks) (c) Actual speed = 480 x 0.96 = 460.8 rpm. (3 marks) |
| 6 | A steel rod of 10 mm diameter carries an axial tensile load of 15 kN. The yield strength of the steel is 250 MPa. (a) Find the cross-sectional area in mm2. (3 marks) (b) Find the tensile stress in MPa. (3 marks) (c) Find the safety factor and state whether the design is safe (a factor of at least 1.5 is required). (3 marks) |
(a) A = pi d^2 / 4 = 3.142 x 10^2 / 4 = 78.54 mm2. (3 marks) (b) Stress = F / A = 15000 N / 78.54 mm2 = 191.0 MPa. (3 marks) (c) Safety factor = 250 / 191.0 = 1.31, below 1.5, so the design is unsafe; use a larger diameter or stronger steel. (3 marks) |
| 7 | A 1500 W electric kettle is used for 12 minutes a day for 30 days. Electricity costs $1.20 per kWh. (a) Find the daily energy use in kWh. (3 marks) (b) Find the total energy and the electricity cost for 30 days. (3 marks) (c) If a 1200 W model is used for the same time, how much money is saved in 30 days? (3 marks) |
(a) 12 minutes = 0.20 h; daily energy = 1.5 x 0.20 = 0.30 kWh. (3 marks) (b) Energy in 30 days = 0.30 x 30 = 9.0 kWh; cost = 9.0 x 1.20 = $10.80. (3 marks) (c) The 1200 W model uses 0.24 kWh a day, 7.2 kWh in 30 days, costing $8.64; saving = $2.16. (3 marks) |
| 8 | A pulley system lifts a 450 N load with an effort of 150 N. When the load rises 0.30 m the rope is pulled 1.20 m. (a) Find the mechanical advantage (MA). (3 marks) (b) Find the velocity ratio (VR). (3 marks) (c) Find the efficiency and state where the remaining energy goes. (3 marks) |
(a) MA = load / effort = 450 / 150 = 3.0. (3 marks) (b) VR = rope distance / load distance = 1.20 / 0.30 = 4.0. (3 marks) (c) Efficiency = MA / VR x 100% = 3.0 / 4.0 x 100% = 75%; the remaining 25% is lost to bearing friction and rope bending, becoming heat. (3 marks) |
Section C: Manufacturing, Cost and Quality (Q9-Q12)
Money is in Hong Kong dollars; round answers to two decimal places.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A hole and shaft fit is specified as: hole 25.00 (+0.02/0), shaft 24.98 (+/-0.01). (a) State the maximum and minimum hole sizes. (2 marks) (b) State the maximum and minimum shaft sizes. (2 marks) (c) Find the maximum and minimum clearance and state whether this is a clearance fit or an interference fit. (4 marks) |
(a) Hole: maximum 25.02 mm, minimum 25.00 mm. (2 marks) (b) Shaft: maximum 24.99 mm, minimum 24.97 mm. (2 marks) (c) Maximum clearance = 25.02 – 24.97 = 0.05 mm; minimum clearance = 25.00 – 24.99 = 0.01 mm. The minimum clearance is positive, so the shaft always turns freely: this is a clearance fit. (4 marks) |
| 10 | A 3D printed model has a volume of 45 cm3 and is printed at 20% infill. PLA has a density of 1.24 g/cm3 and filament costs $180 per kg. (a) Find the volume of material actually used. (3 marks) (b) Find the mass of material in g. (3 marks) (c) Find the material cost of the part. (2 marks) |
(a) Material volume = 45 x 0.20 = 9.0 cm3. (3 marks) (b) Mass = 9.0 x 1.24 = 11.16 g. (3 marks) (c) Cost = 11.16 g / 1000 x $180 per kg = $2.01. (2 marks) |
| 11 | A new product needs $120,000 of development cost. Variable cost is $35 per unit, the selling price is $95, and 2,500 units are expected to sell. (a) Find the contribution margin per unit. (2 marks) (b) Find the break-even quantity. (3 marks) (c) Find the profit when 2,500 units are sold. (3 marks) |
(a) Contribution margin = $95 – $35 = $60 per unit. (2 marks) (b) Break-even = $120,000 / $60 = 2,000 units. (3 marks) (c) Revenue = $95 x 2,500 = $237,500; total cost = $120,000 + $35 x 2,500 = $207,500; profit = $30,000. (3 marks) |
| 12 | A production line inspects 200 units and finds 8 defective. (a) Find the defect rate. (2 marks) (b) If the defect rate falls to 2.5%, how many fewer defective units are produced per 10 000 units? (3 marks) (c) Suggest one practical measure to reduce the defect rate and explain how its effect would be verified. (3 marks) |
(a) Defect rate = 8 / 200 x 100% = 4.0%. (2 marks) (b) At present 400 defects per 10 000 units; at 2.5% it is 250; so 150 fewer defective units. (3 marks) (c) Measure: add a first-off inspection at the critical station and monitor dimensions on a control chart, stopping the line when a trend or out-of-limit point appears. Verification: five consecutive lots must each show a defect rate of 2.5% or lower with no out-of-control pattern. (3 marks) |
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