Secondary 4 Design And Applied Technology Practice Paper: Design Process And Creativity, Materials And Manufacturing, Structures And Mechanics, Electricity And Electronics, Systems And Control, Product Evaluation And Sustainable Design

Secondary 4 Design and Applied Technology Practice Paper: Design Process and Creativity, Materials and Manufacturing, Structures and Mechanics, Electricity and Electronics, Systems and Control, Product Evaluation and Sustainable Design

2026-11-14

Design and Applied Technology – Secondary 4 – Practice Paper

S4 Design and Applied Technology practice paper: 24 parts in 60 minutes on design process, material mass, beam reactions, axial stress, Ohm law and carbon footprint.

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Paper details　S4 DAT practice paper, 24 parts in 60 minutes, covering questionnaire response, weighted decision matrix, aluminium plate mass, injection shrinkage, simply supported beam reactions, axial stress and safety factor, Ohm law, parallel resistors, ADC resolution, PWM, micro-step travel and carbon footprint, with marking points for every part.

Section A: Design Process and Creativity (16 marks)

A school design team develops a foldable shopping trolley for elderly users, collecting user views by questionnaire and comparing three concepts with a weighted scoring matrix.
Of 400 questionnaires issued, 92.5% are valid returns, and 40.0% of respondents prefer the foldable design.
The team also fixes ergonomic dimensions and a target cost before building models and prototypes.

#	Question	Answer and marking notes
1	(a) Calculate the number of valid questionnaires and the number of respondents who prefer the foldable design.	Answer: 400 x 92.5% = 370 valid returns, and 370 x 40.0% = 148 respondents. Marking points: 2 marks for the valid returns and 2 for the respondent count; using 400 as the base loses 1 mark.
2	(b) The target retail price is HK$480 with a gross margin of 25%; calculate the maximum manufacturing cost and explain why it must be fixed at the design stage.	Answer: 480 x 0.75 = HK$360. Marking points: 2 marks for the working, 1 for the answer and 1 for noting that the cost ceiling drives material and process choice, and that changing drawings is cheaper than changing tooling.
3	(c) The user is 170 cm tall; desk height is 0.43 of stature and seat height 0.25 of stature. Calculate both dimensions to one decimal place and state two problems caused by a desk that is too high.	Answer: desk 170 x 0.43 = 73.1 cm and seat 170 x 0.25 = 42.5 cm. Problems: the forearms cannot rest flat so the shoulders are raised, and the neck leans forward, loading the lower back. Marking points: 1 mark per dimension and 1 per problem.
4	(d) A weighted matrix uses function 0.4, cost 0.35 and aesthetics 0.25 (total 1.0). On a 10 point scale, concept A scores 8, 6, 7; concept B scores 6, 9, 8; concept C scores 7, 7, 9. Calculate the three weighted totals and select the best concept.	Answer: A = 0.4 x 8 + 0.35 x 6 + 0.25 x 7 = 7.05; B = 0.4 x 6 + 0.35 x 9 + 0.25 x 8 = 7.55; C = 0.4 x 7 + 0.35 x 7 + 0.25 x 9 = 7.5. B is highest, so concept B is chosen. Marking points: 1 mark per weighted total and 1 for the conclusion; comparing raw totals without weights scores nothing.

Section B: Materials and Manufacturing (17 marks)

A product with an aluminium alloy shell and a steel bracket requires material, tooling and finishing decisions at the design stage.
The team compares the density, scrap rate and processing cost of aluminium and steel, and selects injection moulding plus spray coating as the main production route.
Every quantity must be estimated in advance so that cost overruns are not discovered after mass production.

#	Question	Answer and marking notes
5	(a) An aluminium plate measures 1.2 m by 0.8 m by 3 mm; the density of aluminium is 2 700 kg/m3. Calculate its volume in m3 and its mass in kg to three decimal places.	Answer: 3 mm = 0.003 m; volume 1.2 x 0.8 x 0.003 = 0.00288 m3; mass 2 700 x 0.00288 = 7.776 kg. Marking points: 1 for unit conversion, 1 for volume, 1 for the density formula, 1 for mass and 1 for three decimal places.
6	(b) The steel bracket needs 12.5 kg of net material; the scrap rate is 8% and steel costs HK$18 per kg. Calculate the purchase quantity in kg and the material cost in HK$.	Answer: 12.5 x 1.08 = 13.5 kg, and 13.5 x 18 = HK$243.0. Marking points: 2 for handling the scrap rate (pricing the net mass alone earns 1), 1 for the purchase quantity and 1 for the cost.
7	(c) An injection mould cavity measures 100.0 mm and the material shrinks by 0.6%. Calculate the moulded part size in mm to one decimal place.	Answer: 100.0 x 0.994 = 99.4 mm. Marking points: 2 for taking the shrinkage in the correct direction (enlarging the part is wrong), 1 for the working and 1 for one decimal place.
8	(d) The shell has a surface area of 0.9 m2; the coating covers 8 m2 per litre and 2 layers are sprayed. Calculate the coating volume needed in litres to three decimal places.	Answer: 0.9 x 2 / 8 = 0.225 L. Marking points: 1 for including both layers, 1 for dividing by the coverage rate, 1 for the value and 1 for three decimal places.

Section C: Structures and Mechanics (17 marks)

A test rig contains a simply supported beam, a wrench and a gear train, together with a square post carrying an axial compressive load.
Engineers check each member with static equilibrium and stress formulae, using a safety factor to compare working stress with allowable stress.
All working must show units: length in millimetres or metres and force in newtons.

#	Question	Answer and marking notes
9	(a) A simply supported beam of span 2.4 m carries a central point load of 480 N; the self weight is neglected. Calculate the reactions at both supports and explain your reasoning.	Answer: the load is central, so the arrangement is symmetrical and each reaction is 480 / 2 = 240 N. Marking points: 1 for the symmetry argument, 2 for the calculation and 1 for stating 240 N upwards at each support.
10	(b) A wrench has an effective arm of 0.25 m and a force of 120 N is applied. Calculate the torque in N m; if the arm shortens to 0.2 m with the same torque, what force is needed?	Answer: torque = 120 x 0.25 = 30.0 N m; with a 0.2 m arm the force is 30.0 / 0.2 = 150.0 N. Marking points: 2 for the torque, 1 for the reverse calculation and 1 for noting that a shorter arm needs a larger force.
11	(c) A driving gear has 12 teeth at 1200 rpm and the driven gear has 36 teeth. Calculate the driven speed in rpm and the reduction ratio.	Answer: speed = 1200 x 12 / 36 = 400 rpm; reduction ratio 36 to 12, that is 3 to 1. Marking points: 2 for the speed, 1 for the ratio and 1 for noting that speed falls while torque rises.
12	(d) A square post has a 20 mm by 10 mm section and carries an axial compressive load of 4000 N. The ultimate strength is 100 N/mm2 and the safety factor is 4. Calculate the cross sectional area, the working stress and the allowable stress, then state whether the post is safe.	Answer: area 20 x 10 = 200 mm2; working stress 4000 / 200 = 20 N/mm2; allowable stress 100 / 4 = 25 N/mm2. Since 20 N/mm2 is below 25 N/mm2 the post is safe. Marking points: 1 for the area, 2 for the working stress, 1 for the allowable stress and 1 for the comparison and conclusion.

Section D: Electricity and Electronics (17 marks)

A laboratory test uses a 12 V direct current supply, resistors, a light emitting diode and a potential divider, with meters reading current and voltage.
Another group calculates monthly energy use and electricity cost to compare lighting loads.
Working must distinguish series and parallel equivalent resistance and the units of power and energy.

#	Question	Answer and marking notes
13	(a) A 12 V direct current supply is connected across a 240 ohm resistor. Calculate the current in A and in mA.	Answer: 12 / 240 = 0.05 A = 50.0 mA. Marking points: 2 for the Ohm law formula, 1 for the value and 1 for the unit conversion.
14	(b) A 9 W light emitting diode lamp runs 5 hours a day for 30 days a month; electricity costs HK$1.2 per kWh. Calculate the monthly energy use in kWh and the cost in HK$.	Answer: 9 x 5 x 30 / 1000 = 1.35 kWh, and 1.35 x 1.2 = HK$1.62. Marking points: 2 for dividing by 1000 to obtain kWh, 1 for the energy and 1 for the cost.
15	(c) Resistors of 100 ohm and 150 ohm are connected in parallel. Calculate the equivalent resistance in ohms.	Answer: 1/R = 1/100 + 1/150, so R = 100 x 150 / (100 + 150) = 60.0 ohm. Marking points: 2 for the parallel formula, 1 for the value and 1 for noting that the equivalent value is smaller than either resistor.
16	(d) A 9 V supply drives 1000 ohm and 2000 ohm resistors in series. Calculate the voltage across the 2000 ohm resistor in V to one decimal place and the circuit current in mA to one decimal place.	Answer: total resistance 1000 + 2000 = 3000 ohm; current 9 / 3000 = 3.0 mA; divider output 9 x 2000 / 3000 = 6.0 V. Marking points: 2 for the current, 2 for the divided voltage and 1 for one decimal place.

Section E: Systems and Control (17 marks)

An automated line has a safety gate and sensors; a microcontroller drives a motor by pulse width modulation and a lead screw moves a slide.
The system may start only when the gate is closed and the zone is clear; if either condition fails it must stop.
Engineers assess control accuracy through resolution, period and displacement per step.

#	Question	Answer and marking notes
17	(a) An 8 bit analogue to digital converter has a reference voltage of 5.0 V. Calculate the number of levels and the resolution in mV to one decimal place.	Answer: levels 2 to the power 8 = 256; resolution 5.0 / 256 = 0.01953125 V = 19.5 mV. Marking points: 1 for the level count, 2 for the division and 1 for the mV conversion and one decimal place.
18	(b) A pulse width modulation signal has a duty cycle of 40% at 500 Hz. Calculate the period in ms to two decimal places and the on time in ms.	Answer: period 1 / 500 = 0.002 s = 2.0 ms; on time 2.0 x 40% = 0.8 ms. Marking points: 2 for the period, 1 for the on time and 1 for noting that at fixed duty cycle a higher frequency shortens each on pulse.
19	(c) A stepper motor has 200 steps per revolution, the driver uses 16 micro steps and the lead screw has a 4.0 mm lead. Calculate the displacement per micro step in micrometres and the number of steps to move the slide 20 mm.	Answer: 200 x 16 = 3200 micro steps per revolution; 4.0 / 3200 = 0.00125 mm = 1.25 micrometres per step; 20 / 0.00125 = 16000 steps. Marking points: 1 for the micro step count, 2 for the displacement per step, 1 for the step count and 1 for noting that more micro steps give finer accuracy.
20	(d) Signal D means the gate is closed, C means the zone is clear, S means the start button is pressed and output Y means starting is permitted. Write the Boolean expression for Y and explain why the start button alone must not drive the motor.	Answer: Y = D AND C AND S. Marking points: 2 for combining the three signals with AND, 1 for a truth table with 8 rows, and 2 for explaining the interlock: if the gate is open or the zone is occupied the output is 0, so moving parts cannot injure anyone.

Section F: Product Evaluation and Sustainable Design (16 marks)

Product evaluation covers not only function and cost but also carbon footprint, recyclability and energy use in the use phase.
The team compares two designs with a life cycle view and selects the more sustainable option by weighted score.
All values must be given to the stated decimal place with units.

#	Question	Answer and marking notes
21	(a) A product uses 12 kg of material with a carbon intensity of 8.5 kg carbon dioxide equivalent per kg; it travels 600 km by lorry at 0.09 kg carbon dioxide equivalent per tonne kilometre. Calculate the combined material and transport footprint in kg carbon dioxide equivalent to one decimal place.	Answer: material 12 x 8.5 = 102.0 kg; transport 0.012 tonne x 600 x 0.09 = 0.648 kg; total 102.0 + 0.648 = 102.6 kg carbon dioxide equivalent. Marking points: 1 for the material, 2 for the transport (the mass must first be converted to tonnes) and 1 for the total.
22	(b) A product weighs 2.4 kg and contains 1.92 kg of recyclable material. Calculate the recycling rate as a percentage and suggest two design measures that raise it.	Answer: 1.92 / 2.4 x 100 = 80.0%. Measures: reduce adhesive bonding and inserts between different materials, use snap fits or screws instead of glue, and label material codes for easier sorting. Marking points: 2 for the rate and 1 for each measure.
23	(c) An older device draws 60 W and the new model 18 W, used 4 hours a day for 365 days a year, with electricity at HK$1.2 per kWh. Calculate the annual energy saving in kWh and the money saved in HK$ to two decimal places.	Answer: power difference 60 – 18 = 42 W; 42 x 4 x 365 / 1000 = 61.32 kWh; 61.32 x 1.2 = HK$73.58. Marking points: 1 for the power difference, 2 for the kWh and 1 for the money saved.
24	(d) A sustainability matrix weights materials 0.3, energy 0.3, recycling 0.2 and packaging 0.2 (total 1.0). Product A scores 7, 6, 8, 5 and product B scores 6, 9, 7, 8 on a 10 point scale. Calculate both weighted totals and select the more sustainable option.	Answer: A = 0.3 x 7 + 0.3 x 6 + 0.2 x 8 + 0.2 x 5 = 6.5; B = 0.3 x 6 + 0.3 x 9 + 0.2 x 7 + 0.2 x 8 = 7.5. B is higher, so product B is selected. Marking points: 1 per weighted total, 1 for the conclusion and 1 for noting that the weights reflect the evaluation priorities and can change the outcome.

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