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HKDSE Technology and Living Practice Paper: Food Science and Nutrition, Clothing and Textiles
HKDSE Technology and Living – 12 questions – 100 marks – 75 minutes
DSE Technology and Living practice paper: 12 questions, 100 marks, 75 minutes on food science, nutrition, clothing and textiles, with marking points and printable PDFs.
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Paper details HKDSE Technology and Living practice paper: 12 questions, 100 marks, 75 minutes, covering food science, nutrition, clothing and textiles, with marking points and printable PDFs.
Section A: Food Science and Nutrition Basics
Answer all questions. Show your working and state the units.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A student lunch contains 30 g of protein, 55 g of carbohydrate and 18 g of fat. The energy values of protein, carbohydrate and fat are 17 kJ, 17 kJ and 37 kJ per gram respectively. (a) Calculate the total energy of the meal in kJ. (b) Convert the answer to kcal (1 kcal = 4.184 kJ, give a whole number). | Answer: (a) 30 x 17 + 55 x 17 + 18 x 37 = 510 + 935 + 666 = 2,111 kJ. (b) 2,111 / 4.184 = 504.5, about 505 kcal. Marking: 2 marks for each energy term, 1 mark for the total and 1 mark for the conversion; no marks if the conversion omits 4.184. |
| 2 | A 60 kg athlete needs 1.1 g of protein per kilogram of body weight per day. On that day the athlete takes 45 g of protein from meals and plans to make up the difference with eggs, each providing 6.5 g of protein. (a) Calculate the recommended daily protein intake. (b) Find the protein still required. (c) What is the minimum number of eggs needed? | Answer: (a) 1.1 x 60 = 66 g. (b) 66 – 45 = 21 g. (c) 21 / 6.5 = 3.23, so at least 4 eggs (3 eggs supply only 19.5 g, which is short). Marking: 1 mark for each correct expression with units; no marks for an answer of 3.23 eggs left unrounded. |
| 3 | A drink label states that 100 g contains 12.5 g of sugar and one pack weighs 40 g. The World Health Organization advises that free sugar intake stay below 50 g per day, and ideally below 25 g. (a) Calculate the sugar in one pack. (b) Using 50 g and 25 g as limits, calculate the percentage of the daily limit provided by one pack. | Answer: (a) 12.5 x 40 / 100 = 5.0 g. (b) Against 50 g: 5.0 / 50 x 100% = 10%; against 25 g: 5.0 / 25 x 100% = 20%. Marking: 2 marks for the proportion, 2 marks for each percentage; deduct 1 mark where units or the percent sign are missing. |
| 4 | The Centre for Food Safety states that bacteria multiply rapidly between 4 and 60 degrees Celsius, cooked food kept at room temperature for more than 2 hours should not be eaten, and reheated leftovers must reach a core temperature above 75 degrees. (a) Is a room temperature of 25 degrees inside the danger zone? (b) How many degrees above the top of the danger zone is the reheating temperature? (c) Give one practical step in a home kitchen that slows bacterial growth. | Answer: (a) Yes, 25 degrees lies between 4 and 60 degrees, so it is inside the danger zone. (b) 75 – 60 = 15 degrees. (c) For example, refrigerate cooked food at 4 degrees within 2 hours, or divide food into small portions so that it cools faster. Marking: (a) 1 mark for the judgement and 1 mark for the reason; (b) 2 marks for the working; (c) 2 marks for a specific step, no marks for a slogan such as keep it clean. |
Section B: Food Technology and Nutrition Applications
Answer all questions. Explain the scientific principles involved.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A glycaemic index (GI) below 55 is low, 56 to 69 is medium and 70 or above is high. White rice has a GI of 83 and 45 g of carbohydrate per portion; wholemeal pasta has a GI of 45 and 40 g of carbohydrate per portion. Glycaemic load (GL) = GI x grams of carbohydrate / 100. (a) Calculate the GL of both foods to one decimal place. (b) State which food affects blood glucose more and explain why. | Answer: (a) White rice: 83 x 45 / 100 = 37.35, that is 37.4; wholemeal pasta: 45 x 40 / 100 = 18.0. (b) White rice has the larger effect: its GI is in the high range and its GL is more than twice that of the pasta. Marking: 2 marks for each GL; 1 mark for the choice and 2 marks for the reason. |
| 6 | Vinaigrette is usually made with a volume ratio of oil to vinegar of 3 to 1, and the two liquids form an emulsion after whisking. (a) If 30 mL of vinegar is used, how much oil is required? (b) What is the total volume of the dressing? (c) Explain why the emulsion separates on standing. | Answer: (a) 30 x 3 = 90 mL. (b) 90 + 30 = 120 mL. (c) Oil and vinegar do not mix; whisking only disperses oil droplets through the vinegar. On standing the droplets, being less dense, rise and coalesce under interfacial tension, so the emulsion separates. Marking: 2 marks each for (a) and (b); (c) 2 marks for immiscibility, 2 marks for droplets rising and coalescing, 1 mark for breakdown of the emulsifier. |
| 7 | Cooked food at room temperature is contaminated with bacteria, initially 100 cells per gram. At 25 degrees the generation time of the bacterium is 20 minutes. (a) How many generations occur in 3 hours? (b) Calculate the number of bacteria per gram after 3 hours. (c) Explain why refrigeration slows food spoilage and microbial growth. | Answer: (a) 180 / 20 = 9 generations. (b) 100 x 2^9 = 51,200 cells. (c) Refrigeration lowers the temperature to 4 degrees, below the optimum for microbial growth, so enzyme activity and metabolic rate fall and the generation time lengthens; growth slows but the bacteria are not killed. Marking: (a) 2 marks, (b) 3 marks, (c) 1 mark for temperature, 1 mark for metabolism, 1 mark for noting that bacteria survive. |
| 8 | Blanching and steaming keep different amounts of vitamin C. A vegetable contains 45 mg of vitamin C per 100 g; blanching in water for 5 minutes loses 55% while steaming for 5 minutes loses 25%. (a) Calculate the vitamin C retained per 100 g after each method. (b) Find the difference between the two. (c) Suggest one cooking practice that preserves water-soluble vitamins and explain why. | Answer: (a) Blanching: 45 x (1 – 0.55) = 20.25 mg; steaming: 45 x (1 – 0.25) = 33.75 mg. (b) 33.75 – 20.25 = 13.5 mg. (c) Steam or stir-fry briefly with little water instead of boiling for a long time, because vitamin C is water-soluble and is also destroyed by heat, so less water and shorter heating reduce the loss. Marking: 2 marks for each retained amount, 2 marks for the difference, 3 marks for the practice and the reason. |
Section C: Clothing, Textiles and Design
Answer all questions. Use textile and sewing terms accurately.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | Moisture regain is the percentage of water a textile fibre absorbs in a standard atmosphere. Cotton has a moisture regain of about 8.5% and polyester about 0.4%. (a) Find the difference in moisture regain. (b) How many times greater is the moisture regain of cotton than that of polyester, to one decimal place? (c) Explain why cotton is chosen for close-fitting summer clothing. | Answer: (a) 8.5 – 0.4 = 8.1 percentage points. (b) 8.5 / 0.4 = 21.25, about 21.3 times. (c) Cotton has high moisture regain, so it absorbs sweat and carries moisture to the fabric surface for evaporation; it is breathable and feels dry. Polyester absorbs little, so sweat stays between skin and fabric and feels clammy. Marking: 2 marks each for (a) and (b); (c) 1 mark for absorption, 1 mark for evaporation, 2 marks for the comparison with polyester. |
| 10 | One dress requires 1.6 m of fabric. The fabric costs 48 dollars per metre and 10% must be allowed for waste. (a) How many metres are needed for 10 dresses before waste? (b) How many metres are needed after allowing for waste? (c) Calculate the fabric cost for 10 dresses. | Answer: (a) 1.6 x 10 = 16 m. (b) 16 x 1.1 = 17.6 m. (c) 17.6 x 48 = 844.8 dollars. Marking: (a) 2 marks, (b) 3 marks, (c) 3 marks; no marks in (c) if the cost is based on 16 m. |
| 11 | A cotton fabric shrinks from 100 cm to 96 cm after washing. (a) Calculate the shrinkage. (b) To obtain a finished length of 120 cm after washing, what length must be cut? (c) Explain why knitted cotton shows more shrinkage than woven cotton. | Answer: (a) (100 – 96) / 100 x 100% = 4%. (b) 120 / 0.96 = 125 cm. (c) In knitted fabric the loops are held under tension during spinning and finishing, and washing relaxes the fibres so the loops return to their curved shape, giving greater shrinkage; woven fabric has tightly interlaced yarns and recovers less. Marking: (a) 2 marks, (b) 3 marks, (c) 3 marks. |
| 12 | A sewing sample shows 12 stitches in every 3 cm and the seam is 1.5 m long. (a) Find the number of stitches per cm. (b) Calculate the total number of stitches in the seam. (c) If the stitch length is changed from 2.5 mm to 3.5 mm, describe the effect on seam strength and appearance. | Answer: (a) 12 / 3 = 4 stitches per cm. (b) 4 x 150 = 600 stitches. (c) A longer stitch length gives fewer stitches per cm and fewer contact points between thread and fabric, so seam strength falls; the stitching looks coarser but the machine sews faster. Marking: (a) 2 marks, (b) 3 marks, (c) 1.5 marks each for strength and appearance. |
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