📚 CAIE IGCSE Co-ordinated Sciences (Double Award) 0654: Mastering Past Papers for Year 11 Success | CAIE IGCSE 综合科学（双奖）0654：Year 11 历年真题致胜解析
Past papers are the single most valuable revision resource for the CAIE IGCSE Co-ordinated Sciences (0654) examination. They reveal the precise command words, mark allocation patterns, and depth of understanding that examiners consistently demand. This article provides a deep-dive analysis of real past paper questions across Biology, Chemistry, and Physics, extracting the recurring themes, common pitfalls, and strategic approaches that distinguish A* candidates from the rest.
历年真题是 CAIE IGCSE 综合科学（0654）考试中最宝贵的复习资源。它们揭示了考官一贯要求的精确指令词、分值分配模式和理解深度。本文深入分析了生物、化学和物理三科的真实历年考题，提炼出反复出现的主题、常见陷阱以及将 A* 考生与其他人区分开来的战略方法。
1. Understanding the 0654 Paper Structure and Command Words | 理解 0654 试卷结构与指令词
The 0654 syllabus sandwiches three sciences into a single qualification, examined across three papers. Paper 2 and Paper 4 are the core theory papers for the Core and Extended tiers respectively, while Paper 6 tests experimental skills through alternative-to-practical questions. Every question is rooted in an Assessment Objective: AO1 (Knowledge with understanding), AO2 (Handling information and problem-solving), or AO3 (Experimental skills and investigations). The most common command words include 'State', 'Describe', 'Explain', 'Suggest', 'Calculate', and 'Compare'. 'State' demands a brief factual answer without justification, often worth 1 mark; 'Describe' requires a detailed account of what happens, focusing on the sequence of observations; 'Explain' is the A* discriminator, demanding a scientific reason using the 'because' or 'therefore' logic chain. Mark schemes consistently award marks for key terms: mentioning 'random movement' when defining diffusion, specifying 'activation energy' when explaining the role of enzymes, and identifying 'resultant force' when describing motion changes.
0654 教学大纲将三门科学合并为一个资格认证，通过三份试卷进行考查。试卷二和试卷四分别是核心课程和拓展课程的核心理论试卷，而试卷六通过实验替代题测试实验技能。每道题都基于一个评估目标：AO1（知识与理解）、AO2（处理信息与解决问题）或 AO3（实验技能与调查）。最常见的指令词包括 "陈述"、"描述"、"解释"、"建议"、"计算" 和 "比较"。"陈述" 要求给出简短的事实性答案，无需论证，通常值 1 分；"描述" 要求详细说明发生了什么，侧重观察的顺序；"解释" 是 A* 区分器，要求使用 "因为" 或 "因此" 逻辑链给出科学原因。评分标准始终为关键词给分：定义扩散时提到 "随机运动"，解释酶的作用时明确 "活化能"，描述运动变化时指出 "合力"。
2. Biology: Respiration and Gas Exchange – A Recurring Challenger | 生物：呼吸作用与气体交换 – 反复出现的难点
mical equation for aerobic respiration. The examiner expects: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, with energy released explicitly mentioned. Many candidates lose marks by writing 'heat' instead of 'energy' ight present a graph showing the effect of exercise on breathing rate and depth, then ask: 'Explain why breathing rate remains elevated after exercise has stopped.' nd must be oxidised; extra oxygen is required to break down lactic acid into carbon dioxide and water; this post-exercise oxygen consumption is the oxygen debt, keeping the breathing rate high until the debt is repaid. A* candidates link the elevated post-exercise oxygen demand (EPOC) to the breakdown of lactic acid in the liver, demonstrating integrated understanding of anaerobic respiration and circulatory transport.
关于有氧呼吸和呼吸机制的题目几乎每年都会出现。一个常见问题是要求考生陈述有氧呼吸的平衡化学方程式。考官期望的答案是：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O，并明确提及能量释放。许多考生因写成 "热" 而非 "能量"，或忘记平衡方程式而失分。一道高阶题目可能给出展示运动对呼吸频率和深度影响的图表，然后问："解释为什么运动停止后呼吸频率仍然升高。" 正确答案涉及氧债：剧烈运动时，肌肉进行无氧呼吸，产生乳酸；乳酸有毒，必须被氧化；需要额外的氧气将乳酸分解为二氧化碳和水；这种运动后的氧气消耗就是氧债，使呼吸频率保持在高位，直到氧债偿清。A* 考生将运动后过量氧耗（EPOC）与肝脏中乳酸的分解联系起来，展示了对无氧呼吸和循环运输的综合理解。
3. Biology: Inheritance and Genetic Crosses – The Logic of Monohybrid Diagrams | 生物：遗传与基因杂交 – 单基因杂交图的逻辑
Genetic pedigree diagrams and monohybrid crosses test a candidate's ability to apply logical reasoning to biological principles. A classic question provides the genotypes of parents for a condition like cystic fibrosis (recessive allele f, normal allele F) and asks for the probability of an affected child. The expected working includes a Punnett square with parental gametes (F and f from each heterozygous parent), producing offspring genotypes FF, Ff, Ff, and ff. The phenotype ratio is 3 normal : 1 affected, giving a 25% probability. The trap is incomplete gamete labelling or confusing genotype with phenotype in the final answer. Extended questions often ask: 'Explain what is meant by the term "carrier".' A carrier is a heterozygous individual (Ff) who possesses one recessive allele for a genetic disorder but does not show symptoms; they can pass the allele to their offspring. Some papers embed inheritance questions within a context of selective breeding: 'Explain how a farmer could use selective breeding to produce cattle with high milk yield.' The answer must include: select cows with highest milk yield; breed them with a bull whose mother had high milk yield; select best offspring and repeat over many generations; this increases the frequency of the desirable alleles for high milk yield in the population.
遗传谱系图和单基因杂交题测试考生将逻辑推理应用于生物学原理的能力。一道经典题目给出某种疾病（如囊性纤维化，隐性等位基因 f，正常等位基因 F）的亲本基因型，并要求计算患病孩子的概率。期望的解题过程包括绘制庞纳特方格，配子为每个杂合亲本的 F 和 f，产生子代基因型 FF、Ff、Ff 和 ff。表型比例为 3 正常 : 1 患病，概率为 25%。陷阱在于配子标记不完整或在最终答案中混淆基因型与表型。拓展题常问："解释 '携带者' 一词的含义。" 携带者是杂合个体（Ff），拥有一个遗传病的隐性等位基因但不表现出症状；他们可以将该等位基因传给后代。有些试卷将遗传问题嵌入选择性育种的情境中："解释农民如何利用选择性育种培育高产奶量的奶牛。" 答案必须包括：选择产奶量最高的奶牛；与一头母亲产奶量高的公牛交配；选择最好的后代，经过多代重复；这增加了群体中高产奶量理想等位基因的频率。
4. Chemistry: Moles, Concentration, and Titration Calculations – Precision is Key | 化学：摩尔、浓度与滴定计算 – 精准是关键
ume form the quantitative backbone of the Chemistry component. A typical Paper 4 question: ' cm³ of dilute sulfuric acid. Calculate the concentration of the sulfuric acid in mol/dm³.' mol H₂SO₄, so moles H₂SO₄ = 0.00250 ÷ 2 = 0.00125 mol. Step 3: concentration H₂SO₄ = (0.00125 ÷ (30.0/1000)) = 0.0417 mol/dm³. The mark scheme awards individual marks for each step, so even if the final answer is wrong due to an arithmetic slip, method marks are available. The most frequent error is forgetting the 2:1 mole ratio from the balanced equation. Examiners also test understanding through 'back-titration' hese require candidates to think in reverse, calculating the amount that actually reacted.
涉及摩尔、浓度和体积的计算构成了化学部分的定量基础。一道典型的试卷四题目："25.0 cm³ 浓度为 0.100 mol/dm³ 的氢氧化钠溶液被 30.0 cm³ 的稀硫酸中和。计算硫酸的浓度，单位为 mol/dm³。" 方程式为 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。第 1 步：NaOH 的摩尔数 = (25.0/1000) × 0.100 = 0.00250 mol。第 2 步：从方程式可知，2 mol NaOH 与 1 mol H₂SO₄ 反应，因此 H₂SO₄ 摩尔数 = 0.00250 ÷ 2 = 0.00125 mol。第 3 步：H₂SO₄ 浓度 = (0.00125 ÷ (30.0/1000)) = 0.0417 mol/dm³。评分标准为每一步单独给分，因此即使最终因算术错误导致答案错误，也能获得方法分。最常见的错误是忘记了平衡方程式中的 2:1 摩尔比。考官还通过 "返滴定" 题目测试理解能力，即加入过量的一种试剂，然后滴定剩余部分；这要求考生反向思考，计算实际反应了的量。
5. Chemistry: Rates of Reaction and Collision Theory – The Particle Model in Action | 化学：反应速率与碰撞理论 – 粒子模型的实际应用
A graph showing the volume of gas produced against time is a staple of Paper 6 and Paper 4. Questions ask candidates to explain, using collision theory, why increasing the concentration of hydrochloric acid increases the rate of its reaction with marble chips. The 4-mark answer structure: (1) Increasing concentration means more particles of hydrochloric acid per unit volume; (2) The particles are closer together; (3) Therefore, the frequency of successful collisions between hydrogen ions and calcium carbonate particles increases; (4) More successful collisions per unit time lead to a faster rate of reaction. Crucially, examiners reject answers that simply state 'there are more collisions' without specifying 'per unit time' or 'frequency'. A follow-up question might present two curves: one at a lower temperature and one at a higher temperature, and ask why the higher temperature curve is steeper and reaches the same final volume. The explanation requires two parts: higher temperature increases the kinetic energy of particles, so a greater proportion of particles have energy greater than or equal to the activation energy; this increases the frequency of successful collisions. The final volume is the same because the same mass of marble chips and volume of acid are used, producing the same total amount of carbon dioxide.
展示气体产生体积随时间变化的图表是试卷六和试卷四的常见内容。题目要求考生利用碰撞理论解释为什么增加盐酸浓度会提高其与大理石碎片反应的速率。4 分的答案结构：(1) 增加浓度意味着单位体积内有更多的盐酸粒子；(2) 粒子彼此更靠近；(3) 因此，氢离子与碳酸钙粒子之间成功碰撞的频率增加；(4) 单位时间内更多的成功碰撞导致更快的反应速率。关键是，考官会拒绝那些只陈述 "有更多碰撞" 而不指明 "单位时间" 或 "频率" 的答案。后续题目可能给出两条曲线：一条在较低温度，一条在较高温度，并问为什么较高温度的曲线更陡且达到相同的最终体积。解释需要两部分：较高温度增加了粒子的动能，因此更大比例的粒子具有大于或等于活化能的能量；这增加了成功碰撞的频率。最终体积相同是因为使用了相同质量的大理石碎片和相同体积的酸，产生的二氧化碳总量相同。
6. Physics: Forces, Resultant Force, and Motion Graphs – Interpreting the Story | 物理：力、合力与运动图像 – 解读故事
car accelerating, travelling at constant velocity, then decelerating. Questions progress: 'Describe the motion of the car between 0 and 5 seconds.' Answer: The car accelerates uniformly; the velocity increases at a constant rate. 'Calculate the acceleration during the first 5 seconds.' a = Δv ÷ Δt, if the velocity increases from 0 m/s to 20 m/s, a = (20 - 0) ÷ 5 = 4.0 m/s². 'Calculate the distance travelled during the entire journey.' quals the sum of these individual areas. The distance is measured in metres, not metres per second. An Extended tier question may ask: 'Explain, in terms of forces, why the car reaches a constant velocity.' The answer: As the car' drag force equals the driving force, the resultant force becomes zero; according to Newton's first law, the car then continues at constant velocity. The term 'resultant force' is the key to securing both marks.
速度-时间图像讲述了运动的详细故事，考官要求考生流畅地解读这个故事。一幅图像可能展示汽车加速、匀速行驶、然后减速。问题逐步深入："描述汽车在 0 至 5 秒内的运动。" 答案：汽车匀加速；速度以恒定速率增加。"计算前 5 秒的加速度。" 使用 a = Δv ÷ Δt，如果速度从 0 m/s 增加到 20 m/s，a = (20 - 0) ÷ 5 = 4.0 m/s²。"计算整个行程中行驶的距离。" 这需要求出速度-时间图像下的面积。对于由三角形、矩形和另一个三角形组成的图形，总面积等于这些单独面积之和。距离的单位是米，不是米每秒。一道拓展层级的题目可能问："从力的角度解释为什么汽车达到恒定速度。" 答案：随着汽车速度增加，空气阻力（阻力）增加；发动机的驱动力最初保持不变；当增加的阻力等于驱动力时，合力变为零；根据牛顿第一定律，汽车随后以恒定速度继续行驶。"合力" 一词是获得两分的关键。
7. Physics: Electrical Circuits – Series, Parallel, and Potential Dividers | 物理：电路 – 串联、并联与分压器
y, a fixed resistor, and a variable resistor in series, with a voltmeter across the fixed resistor. The question asks: ' s to the reading on the voltmeter as the resistance of the variable resistor is increased.' cuit increases; current through the circuit decreases (from I = V ÷ R); the fixed resistor' (from V = I × R). Many candidates incorrectly state that increasing the variable resistor' ut showing the intermediate steps involving current. Parallel circuit questions often ask: ' of resistors in parallel is less than the resistance of the smallest individual resistor.' w through the circuit for the same potential difference; since R = V ÷ I, if I increases for a constant V, total R must decrease. A potential divider question might ask for the output voltage Vout from two resistors in series, and then ask what happens if one resistor is replaced with a light-dependent resistor (LDR) and light intensity increases. The answer links the decreasing LDR resistance to a decreasing share of the total voltage, thus Vout decreases.
电路分析题在核心和拓展试卷中都持续出现。一道典型题目提供一张电路图，包含电池、一个定值电阻和一个可变电阻串联，电压表接在定值电阻两端。题目问："解释当可变电阻的阻值增加时，电压表读数会发生什么变化。" 正确的推理链：电路总电阻增加；通过电路的电流减小（根据 I = V ÷ R）；定值电阻的阻值不变；因此，定值电阻两端的电势差减小（根据 V = I × R）。许多考生错误地陈述增加可变电阻的阻值会直接使电压表读数减小，而没有展示涉及电流的中间步骤。并联电路题常问："解释为什么并联电阻的总电阻小于最小单个电阻的阻值。" 解释需要概念性理解：并联增加一个电阻为电流提供了额外路径；对于相同的电势差，更多电流可以通过电路；由于 R = V ÷ I，如果在 V 恒定时 I 增加，总电阻 R 必然减小。一道分压器题目可能要求计算两个串联电阻的输出电压 Vout，然后问如果其中一个电阻更换为光敏电阻（LDR）且光照强度增加，会发生什么。答案将 LDR 电阻的减小与其占总电压份额的减小联系起来，因此 Vout 减小。
8. Physics: Energy, Work, and Power – Units and Conversions | 物理：能量、功和功率 – 单位与换算
uire fluency in unit conversions and formula selection. A question frequently encountered: ' through a vertical height of 12 m in 15 seconds. Calculate the power output of the crane.' ÷ time = 60,000 ÷ 15 = 4000 W. The answer must include correct units; writing 4 kW is acceptable but 4000 J is wrong because the unit of power is the watt. Efficiency questions connect energy input and useful output: ' t uses 250 J of electrical energy to lift a weight, doing 150 J of useful mechanical work.' Efficiency = (useful energy output ÷ total energy input) × 100% = (150 ÷ 250) × 100% = 60% gainst friction and energy dissipated as heat and sound. Extended questions sometimes ask: 'Explain how the principle of conservation of energy applies to a bouncing ball.' ergy is transferred to thermal energy in the ball and the floor, and some to sound; the remaining kinetic energy causes the ball to rebound to a lower height; total energy is conserved throughout but becomes more spread out.
0654 中的能量计算要求熟练掌握单位换算和公式选择。一个常遇到的问题："一台起重机在 15 秒内将 500 kg 的质量垂直提升 12 m。计算起重机的输出功率。" 第 1 步：重量 = 质量 × 重力场强度 = 500 × 10 = 5000 N。第 2 步：做功 = 力 × 距离 = 5000 × 12 = 60,000 J。第 3 步：功率 = 做功 ÷ 时间 = 60,000 ÷ 15 = 4000 W。答案必须包含正确的单位；写 4 kW 可以接受，但写 4000 J 是错误的，因为功率的单位是瓦特。效率问题将能量输入与有用输出联系起来："计算一台电动机的效率，它使用 250 J 的电能来提升重物，做了 150 J 的有用机械功。" 效率 = (有用能量输出 ÷ 总能量输入) × 100% = (150 ÷ 250) × 100% = 60%。效率低下的原因在于克服摩擦做功以及能量以热和声音的形式耗散。拓展题有时会问："解释能量守恒定律如何适用于一个弹跳的球。" 球在最高点具有最大的重力势能；下落时转化为动能；撞击时，部分动能转移为球和地板的热能，以及部分转化为声音；剩余的动能导致球反弹到较低的高度；总能量在整个过程中守恒，但变得更加分散。
9. Paper 6: Experimental Skills – Planning, Variables, and Graphical Analysis | 试卷六：实验技能 – 计划、变量与图像分析
Paper 6 requires candidates to design investigations, identify variables, and handle data. A typical planning question: 'A student wants to investigate how the concentration of sugar solution affects the mass of potato chips after 30 minutes. Describe how the student should carry out this investigation.' The answer must be a logical sequence: cut equal-sized potato chips using a cork borer and knife; measure initial mass of each chip using a balance; prepare sugar solutions of different concentrations (e.g., 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³); place one chip in each solution in separate boiling tubes; after 30 minutes, remove chips, blot dry, and reweigh; calculate percentage change in mass for each concentration. The independent variable is the concentration of sugar solution; the dependent variable is the percentage change in mass; control variables include temperature, volume of solution, time, and surface area of potato chips. A graph plotting question will provide data and ask candidates to plot points, draw a best-fit line, and identify anomalies. The line of best fit is a smooth curve or straight line that passes through as many points as possible, with roughly equal numbers of points above and below the line. Anomalies are data points that lie far from the line of best fit and should be circled. When calculating the gradient of a straight-line graph, candidates must use a large triangle (covering more than half the line) and state the coordinates clearly.
试卷六要求考生设计调查、识别变量和处理数据。一道典型的计划题："一名学生想研究糖溶液浓度如何影响 30 分钟后马铃薯条的质量。描述该学生应如何进行这项调查。" 答案必须是一个逻辑序列：使用打孔器和刀切出大小相等的马铃薯条；用天平测量每条马铃薯的初始质量；配制不同浓度的糖溶液（例如 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol/dm³）；将一条马铃薯放入不同试管内的每种溶液中；30 分钟后，取出马铃薯条，吸干水分并重新称重；计算每种浓度的质量变化百分比。自变量是糖溶液的浓度；因变量是质量变化百分比；控制变量包括温度、溶液体积、时间和马铃薯条的表面积。绘制图像的问题会提供数据，要求考生描点、画出最佳拟合线并识别异常值。最佳拟合线是一条平滑曲线或直线，尽可能多地穿过数据点，线上方和下方的点数大致相等。异常值是远离最佳拟合线的数据点，应圈出。计算直线图像的斜率时，考生必须使用一个大的三角形（覆盖线的一半以上），并清楚地说明坐标。
10. Paper 6: Drawing Conclusions and Evaluating Experiments | 试卷六：得出结论与评估实验
After plotting a graph, candidates are often asked to describe the relationship shown. The answer must match the shape of the graph: 'as the concentration of sugar solution increases, the percentage change in mass decreases; at low concentrations, the potato gains mass because water enters by osmosis; at high concentrations, the potato loses mass because water leaves by osmosis; at a specific concentration (where the graph crosses the x-axis), there is no net change in mass because the water potential inside the potato cells equals the water potential of the solution.' Evaluation questions ask for sources of error and improvements. Common sources of error include: difficulty in cutting potato chips to exactly the same size; incomplete blotting of surface water before weighing; temperature fluctuations affecting the rate of osmosis. Improvements correspond directly: use a cutter with a fixed blade to ensure uniform size; blot each chip the same number of times on filter paper; use a water bath to control temperature. A key mark is awarded for not just listing errors but explicitly describing how each error affects the result (e.g., 'if surface water is not fully removed, the final mass will be higher, making the calculated percentage change less negative' or 'more positive than it should be').
绘制图像后，考生常被要求描述所示的关系。答案必须与图像形状相符："随着糖溶液浓度增加，质量变化百分比减小；在低浓度时，马铃薯质量增加，因为水通过渗透进入；在高浓度时，马铃薯质量减少，因为水通过渗透离开；在特定浓度下（图像与 x 轴相交处），质量没有净变化，因为马铃薯细胞内的水势等于溶液的水势。" 评估题要求提出误差来源和改进措施。常见的误差来源包括：难以将马铃薯条切成完全相同的尺寸；称重前表面水分未完全吸干；温度波动影响渗透速率。改进措施直接对应：使用带有固定刀片的切割器确保尺寸均匀；每条马铃薯在滤纸上吸干相同次数；使用水浴控制温度。一个关键分数不仅在于列出误差，还在于明确描述每种误差如何影响结果（例如，"如果表面水分未完全去除，最终质量会偏高，使计算出的百分比变化负值减小" 或 "正值偏高"）。
11. Strategic Revision: Interleaving, Timed Practice, and Mark Scheme Analysis | 策略性复习：交叉练习、计时训练与评分标准分析
Inefficient revision often involves working through past papers one subject at a time. Research in cognitive science supports interleaving: switching between Biology, Chemistry, and Physics questions within a single revision session strengthens long-term retention by forcing the brain to repeatedly retrieve different types of problem-solving frameworks. A recommended approach is to print a complete Paper 4, set a timer for 2 hours, and attempt it under strict exam conditions. After finishing, mark your paper using the official mark scheme, but do not simply tick correct answers. For every mark lost, write a brief note explaining why: was the key term missing (e.g., 'frequency of collision' instead of 'frequency of successful collisions')? Was a step missed in a calculation (e.g., not dividing by 2 for the mole ratio)? Was the command word misinterpreted (e.g., describing when asked to explain)? Collating these errors into a personal 'error log' over 3–4 past papers reveals patterns. Many candidates discover they consistently lose marks on 'explain' questions in Physics because they fail to invoke resultant force, or on 'suggest' questions in Chemistry because they offer vague answers rather than referencing specific reactivity series trends or particle size effects.
低效的复习往往是一次只做一门科目的历年真题。认知科学研究支持交叉练习：在一次复习课程中切换生物、化学和物理问题，通过迫使大脑反复检索不同类型的问题解决框架，增强长期记忆。推荐的方法是打印一份完整的试卷四，设置 2 小时计时器，在严格考试条件下尝试完成。完成后，使用官方评分标准批改你的试卷，但不要只给正确答案打勾。对于每一个丢失的分数，写一条简短的笔记解释原因：是遗漏了关键术语（例如，写 "碰撞频率" 而非 "成功碰撞的频率"）？是计算中跳过了某一步（例如，没有根据摩尔比除以 2）？是误解了指令词（例如，要求解释时却做了描述）？将这些错误整理到个人的 "错误日志" 中，经过 3-4 份历年真题后，模式就会显现。许多考生发现他们在物理的 "解释" 题上持续失分，因为他们未能运用合力，或在化学的 "建议" 题上失分，因为他们给出的答案模糊，而没有参考具体的反应性顺序趋势或粒子尺寸效应。
12. The Night Before and Exam Hall Tactics | 考前之夜与考场策略
-negotiable; cognitive performance in analytical problem-solving drops significantly with sleep deprivation. In the exam hall, read the entire question before starting to write. For 0654, note that formulas for Physics are not provided on a separate sheet but must be memorised: recall F = ma, p = mv, P = IV, E = mcΔθ, v = fλ, and the transformer equation Vp/Vs = Np/Ns. Allocate time proportionally to marks: for a 40-mark question section in Paper 4, allow no more than 50 minutes. If a question asks for three answers but you can only recall two, write the two you know and make an intelligent guess for the third, drawing from the context of the question. For calculations, always write the formula first, substitute numbers with units, and present the final answer with correct units and to an appropriate number of significant figures (usually 2 or 3). Never leave a question blank; a blank guarantees zero marks, but a reasoned attempt may earn partial credit.
考前一晚应专注于复习个人错误日志和关键公式总结，而不是尝试新题目。睡眠是不可妥协的；在分析性解决问题时，认知表现会因睡眠不足而显著下降。在考场中，开始写答案前先阅读整个问题。对于 0654，请注意物理公式不提供在单独的纸上，必须记忆：回忆 F = ma、p = mv、P = IV、E = mcΔθ、v = fλ 和变压器方程 Vp/Vs = Np/Ns。按分数比例分配时间：对于试卷四中一个 40 分的问题部分，最多允许 50 分钟。如果一道题要求给出三个答案，但你只能回忆起两个，写出你知道的两个，并利用问题上下文对第三个做出合理猜测。对于计算，始终先写公式，代入带单位的数字，并以正确的单位和适当的有效数字位数（通常为 2 或 3 位）呈现最终答案。永远不要让任何问题空着；空白保证零分，但合理的尝试可能获得部分分数。
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