📚 GCSE Science: Worked Example Questions | GCSE 科学：典型例题详解
Mastering GCSE Science requires more than just memorising facts — you need to apply your knowledge to unfamiliar scenarios. This article walks you through ten carefully chosen worked examples covering Biology, Chemistry and Physics. Each question is broken down step by step to show you how to structure your answers, avoid common pitfalls and gain full marks.
掌握 GCSE 科学不仅需要记忆事实，还需要将知识应用到不熟悉的场景中。本文详细讲解十个精心挑选的典型例题，涵盖生物、化学和物理。每个题目都逐步分解，展示如何组织答案、避开常见错误并获得满分。
1. Biology: Calculating Magnification | 生物：计算放大倍数
s ×10 and the objective lens magnification is ×40. Calculate the total magnification used.
例题：一名学生用光学显微镜观察植物细胞。目镜放大倍数为 ×10，物镜放大倍数为 ×40。计算使用的总放大倍数。
Step 1: Recall the formula.
步骤1：回忆公式。
Total magnification = Eyepiece magnification × Objective magnification
Step 2: Substitute the values into the formula. Total magnification = 10 × 40.
步骤2：将数值代入公式。总放大倍数 = 10 × 40。
Step 3: Perform the multiplication to obtain 400. Write the answer as '×400' or '400 times'. Magnification has no unit.
步骤3：进行乘法运算得到 400。答案写为 '×400' 或 '400 倍'。放大倍数没有单位。
Common mistake: Some students add the magnifications instead of multiplying them. Always multiply the eyepiece and objective magnifications.
常见错误：部分学生会将放大倍数相加而不是相乘。务必记住是将目镜与物镜的放大倍数相乘。
2. Biology: Osmosis and Percentage Change in Mass | 生物：渗透作用与质量变化百分比
Example question: A student placed potato cylinders in different sugar solutions. The initial mass of a cylinder was 2.5 g. After 30 minutes in a 0.4 mol/dm³ sucrose solution, the final mass was 2.3 g. Calculate the percentage change in mass and explain what happened.
例题：一名学生将土豆圆柱体放入不同蔗糖溶液中。某圆柱体初始质量为 2.5 g。在 0.4 mol/dm³ 蔗糖溶液中放置 30 分钟后，最终质量为 2.3 g。计算质量变化百分比并解释发生了什么。
Step 1: Calculate the change in mass: final mass – initial mass = 2.3 – 2.5 = –0.2 g.
步骤1：计算质量变化：最终质量 – 初始质量 = 2.3 – 2.5 = –0.2 g。
Step 2: Work out the percentage change using the formula:
步骤2：使用公式计算变化百分比：
Percentage change = (Change in mass ÷ Initial mass) × 100
Step 3: Substitute: (–0.2 ÷ 2.5) × 100 = –8.0% . The negative sign indicates a decrease in mass.
步骤3：代入数值：(–0.2 ÷ 2.5) × 100 = –8.0% 。负号表示质量减少。
Step 4: Explain using osmosis. The sucrose solution had a lower water potential than the potato cells. Water moved out of the potato cells by osmosis, so the cylinder lost mass.
步骤4：用渗透作用解释。蔗糖溶液的水势低于土豆细胞。水通过渗透作用从土豆细胞中流出，因此圆柱体质量减少。
Common mistake: Forgetting to divide by the initial mass, or misreading a negative percentage as an error. Always state 'decrease' and link to water movement.
常见错误：忘记除以初始质量，或误认为负百分比是计算错误。务必说明“减少”并将其与水分移动联系起来。
3. Chemistry: Balancing Chemical Equations | 化学：配平化学方程式
tion: Balance the equation for the combustion of methane: __ CH₄ + __ O₂ → __ CO₂ + __ H₂O
例题：配平甲烷燃烧的方程式：__ CH₄ + __ O₂ → __ CO₂ + __ H₂O
Step 1: Count the atoms on each side. Left: C=1, H=4, O=2. Right: C=1, H=2, O=3.
步骤1：数出两侧的原子个数。左边：C=1，H=4，O=2。右边：C=1，H=2，O=3。
placing a coefficient 2 in front of H₂O: CH₄ + O₂ → CO₂ + 2H₂O. Now right side: H=4, O=4.
步骤2：先配平氢，在 H₂O 前添加系数 2：CH₄ + O₂ → CO₂ + 2H₂O。此时右边：H=4，O=4。
ide needs 4 oxygen atoms, so place a coefficient 2 in front of O₂: CH₄ + 2O₂ → CO₂ + 2H₂O.
步骤3：配平氧。左边需要 4 个氧原子，因此在 O₂ 前添系数 2：CH₄ + 2O₂ → CO₂ + 2H₂O。
left C=1, H=4, O=4; right C=1, H=4, O=4. The balanced equation is: CH₄ + 2O₂ → CO₂ + 2H₂O.
步骤4：最终核对：左边 C=1, H=4, O=4；右边 C=1, H=4, O=4。配平的方程式为：CH₄ + 2O₂ → CO₂ + 2H₂O。
Common mistake: Changing the subscript numbers inside a formula instead of adding coefficients. Never alter the chemical formulas themselves.
常见错误：改变化学式内部的下标数字而非添加系数。切勿改动化学式本身。
4. Chemistry: Moles and Mass Calculations | 化学：摩尔与质量计算
number of moles in 8.0 g of magnesium oxide, MgO. Relative formula mass (Mᵣ) of MgO = 40.
例题：计算 8.0 g 氧化镁 (MgO) 的物质的量。MgO 的相对式量 (Mᵣ) = 40。
Step 1: Write down the linking formula.
步骤1：写出关联公式。
Moles = Mass ÷ Relative formula mass (or Mᵣ)
Step 2: Substitute the numbers: Moles = 8.0 ÷ 40 .
步骤2：代入数值：物质的量 = 8.0 ÷ 40。
Step 3: Complete the division: 8.0 ÷ 40 = 0.20 mol.
步骤3：完成除法运算：8.0 ÷ 40 = 0.20 mol。
Step 4: Always give the unit 'mol' and round to an appropriate number of significant figures (here 2 sf).
步骤4：务必给出单位 'mol'，并四舍五入到合适有效数字位数（此处为 2 位有效数字）。
Common mistake: Confusing Mᵣ with atomic number, or using the wrong formula mass. Also, remember to convert grams to kilograms only for specific physics equations – in chemistry, mass stays in grams.
常见错误：混淆 Mᵣ 与原子序数，或使用了错误的式量。此外请注意，仅在特定物理公式中需将克转换为千克，化学计算中质量以克为单位。
5. Chemistry: Rate of Reaction from a Graph | 化学：由曲线图求反应速率
Example question: The graph shows the volume of carbon dioxide gas produced when excess calcium carbonate reacts with hydrochloric acid. After 60 seconds, 48 cm³ of gas had been collected. After 120 seconds, the total volume was 72 cm³. Calculate the mean rate of reaction between 60 s and 120 s, and explain why the rate decreases over time.
例题：图表显示了过量碳酸钙与盐酸反应时产生的二氧化碳气体体积。60 秒后收集到 48 cm³ 气体，120 秒后总体积为 72 cm³。计算 60 s 至 120 s 间的平均反应速率，并解释为何速率随时间下降。
Step 1: Calculate the change in volume: 72 – 48 = 24 cm³.
步骤1：计算气体体积的变化量：72 – 48 = 24 cm³。
Step 2: Determine the time interval: 120 – 60 = 60 s.
步骤2：确定时间间隔：120 – 60 = 60 s。
Step 3: Use the rate formula:
步骤3：使用速率公式：
Mean rate = Change in volume ÷ Change in time
Step 4: Substitute values: 24 cm³ ÷ 60 s = 0.40 cm³/s.
步骤4：代入数值：24 cm³ ÷ 60 s = 0.40 cm³/s。
Step 5: Explain the decreasing rate. As the reaction progresses, the concentration of hydrochloric acid falls, and the surface area of calcium carbonate decreases. Fewer successful collisions occur per second, so the rate slows down.
步骤5：解释速率下降的原因。随着反应进行，盐酸浓度降低，碳酸钙表面积减小。每秒发生的有效碰撞减少，因此速率减慢。
Common mistake: Using the total volume for the whole reaction instead of the change. Always read the question carefully to identify the correct time interval.
常见错误：将整个反应的总气体量当作变化量。务必仔细审题，确定正确的时间间隔。
6. Physics: Ohm's Law and Resistance | 物理：欧姆定律与电阻
Example question: A student measures the current through a resistor as 0.25 A when the potential difference across it is 5.0 V. Calculate the resistance. State whether the resistor is an ohmic conductor and explain your answer.
例题：一名学生测得通过某电阻器的电流为 0.25 A，其两端的电势差为 5.0 V。计算电阻值。判断该电阻器是否为欧姆导体并解释。
Step 1: Recall Ohm's law: V = I × R, which rearranges to R = V ÷ I.
步骤1：回忆欧姆定律：V = I × R，可变形为 R = V ÷ I。
Step 2: Substitute the values: R = 5.0 V ÷ 0.25 A = 20 Ω.
步骤2：代入数值：R = 5.0 V ÷ 0.25 A = 20 Ω。
Step 3: An ohmic conductor has a constant resistance provided the temperature is constant. In the question, only one pair of V–I readings is given, so we cannot check if the ratio V/I is constant. Therefore, we cannot conclude it is an ohmic conductor without more data.
步骤3：在温度恒定的条件下，欧姆导体的电阻是恒定的。题干中只给出一组 V–I 数据，无法验证 V/I 比值是否恒定。因此，在缺少更多数据时无法得出它是欧姆导体的结论。
Step 4: If the question provided a graph, you would expect a straight line through the origin for an ohmic resistor. A curve would indicate a non-ohmic device, such as a filament lamp or diode.
步骤4：若题目提供了图表，欧姆电阻的图线应为一条通过原点的直线。若为曲线，则表明该器件不是欧姆导体，如灯丝灯泡或二极管。
Common mistake: Assuming any resistor is automatically ohmic. Always check the temperature or the shape of the I–V graph.
常见错误：想当然认为任何电阻器都是欧姆导体。务必检查温度条件或 I–V 图线的形状。
7. Physics: Energy, Power and Time | 物理：能量、功率与时间
Example question: An electric kettle has a power rating of 2200 W. When it is switched on for 150 seconds, calculate the energy transferred. Give your answer in kilojoules.
例题：一个电热水壶的额定功率为 2200 W。当它接通电源 150 秒时，计算所传递的能量。以千焦为单位给出答案。
Step 1: Use the equation linking energy, power and time:
步骤1：使用关联能量、功率和时间的公式：
E = P × t
where E is energy in joules, P is power in watts, and t is time in seconds.
其中 E 为能量，单位焦耳；P 为功率，单位瓦特；t 为时间，单位秒。
Step 2: Substitute: E = 2200 W × 150 s = 330 000 J.
步骤2：代入数值：E = 2200 W × 150 s = 330 000 J。
Step 3: Convert joules to kilojoules by dividing by 1000: 330 000 J ÷ 1000 = 330 kJ.
步骤3：将焦耳转换为千焦，除以 1000：330 000 J ÷ 1000 = 330 kJ。
Step 4: Check significant figures. The power and time are given as 2200 W (2 sf) and 150 s (2 or 3 sf), so an answer of 330 kJ is appropriate.
步骤4：检查有效数字。功率为 2200 W（2 位有效数字），时间为 150 s（2 或 3 位），故答案 330 kJ 是合适的。
e in minutes instead of seconds. Always convert time to seconds before applying E = P × t.
常见错误：忘记转换为千焦，或时间以分钟代入却未转换。使用 E = P × t 前务必将时间转换为秒。
8. Physics: Resultant Force and Acceleration | 物理：合力与加速度
Example question: A car of mass 1200 kg accelerates uniformly. The driving force from the engine is 2500 N, and the resistive force (friction and air resistance) is 700 N. Calculate the acceleration of the car.
例题：一辆质量为 1200 kg 的汽车匀加速行驶。发动机的驱动力为 2500 N，阻力（摩擦和空气阻力）为 700 N。计算汽车的加速度。
Step 1: Determine the resultant force. Resultant force = driving force – resistive force = 2500 N – 700 N = 1800 N.
步骤1：确定合力。合力 = 驱动力 – 阻力 = 2500 N – 700 N = 1800 N。
Step 2: Apply Newton's second law:
步骤2：应用牛顿第二定律：
F = m × a → a = F ÷ m
Step 3: Substitute the values: a = 1800 N ÷ 1200 kg = 1.5 m/s².
步骤3：代入数值：a = 1800 N ÷ 1200 kg = 1.5 m/s²。
Step 4: State the direction. The acceleration is in the direction of the resultant force (forward). Always give the unit m/s².
步骤4：说明方向。加速度方向与合力方向相同（向前）。务必给出单位 m/s²。
Common mistake: Using the driving force directly without subtracting resistive forces, or forgetting to rearrange the formula correctly. Always calculate the net force first.
常见错误：直接用驱动力计算而忘记减去阻力，或公式变形错误。务必先计算净合力。
9. Biology: Monohybrid Genetic Cross | 生物：单因子遗传杂交
Example question: In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). A heterozygous tall plant is crossed with a homozygous recessive short plant. Determine the expected ratio of tall to short offspring using a Punnett square.
例题：在豌豆植株中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。一株杂合高茎植株与一株隐性纯合矮茎植株杂交。使用庞纳特方格确定子代表现型的预期比例。
Step 1: Identify the genotypes of the parents. Heterozygous tall is Tt; homozygous recessive is tt.
步骤1：确定亲本的基因型。杂合高茎为 Tt；隐性纯合为 tt。
Step 2: Determine the gametes. Parent 1 (Tt) produces gametes T and t. Parent 2 (tt) produces gametes t and t.
步骤2：确定配子。亲本 1 (Tt) 产生的配子为 T 和 t。亲本 2 (tt) 产生的配子为 t 和 t。
Step 3: Draw a Punnett square (two-by-two grid). Combine gametes: offspring genotypes are Tt, Tt, tt, tt.
步骤3：绘制庞纳特方格（2×2 网格）。配子组合：子代基因型为 Tt、Tt、tt、tt。
Step 4: Translate genotypes to phenotypes. Tt produces tall stems because T is dominant. tt produces short stems. Therefore, two tall : two short, simplified to a 1 : 1 ratio.
步骤4：将基因型转化为表现型。Tt 表现为高茎，因为 T 是显性。tt 表现为矮茎。因此，两株高茎 : 两株矮茎，简化为 1 : 1 比例。
Common mistake: Writing the gametes incorrectly (e.g. forgetting to separate alleles), or confusing heterozygous and homozygous. Remember each gamete carries only one allele for each gene.
常见错误：配子书写错误（例如忘记等位基因分离），或混淆杂合与纯合。记住每个配子对每个基因只携带一个等位基因。
10. Scientific Skills: Interpreting Data and Drawing Conclusions | 科学技能：解读数据与得出结论
Example question: An investigation into enzyme activity produced the following data. At pH 5, the time taken for starch to break down was 120 s; at pH 6 it was 60 s; at pH 7 it was 30 s; at pH 8 it was 50 s; and at pH 9 it was 140 s. Describe the trend and identify the optimum pH.
例题：一项酶活性研究得到如下数据：pH 5 时，淀粉完全分解所需时间为 120 s；pH 6 时为 60 s；pH 7 时为 30 s；pH 8 时为 50 s；pH 9 时为 140 s。描述变化趋势并确定最适 pH。
e time decreases (120 s → 60 s → 30 s). This indicates the rate of reaction is increasing.
步骤1：审视 pH 与时间的关系。当 pH 从 5 升至 7 时，时间减少（120 s → 60 s → 30 s）。这表明反应速率在增加。
From pH 7 to 9, the time increases (30 s → 50 s → 140 s), showing the reaction slows down.
步骤2：分析 pH 高于 7 的数据。从 pH 7 到 9，时间增加（30 s → 50 s → 140 s），说明反应速率减慢。
Step 3: Identify the optimum. The shortest time (30 s) occurred at pH 7, which means the enzyme works fastest at this pH. So the optimum pH is 7.
步骤3：确定最适值。最短时间 (30 s) 出现在 pH 7，意味着酶在此 pH 下作用最快。因此最适 pH 为 7。
Step 4: Explain using enzyme theory. At pH values away from the optimum, the enzyme's active site changes shape, reducing its ability to bind to the substrate, so the rate decreases.
步骤4：用酶学理论解释。当 pH 偏离最适值时，酶的活性位点形状发生变化，与底物结合的能力下降，因此速率降低。
Common mistake: Confusing time taken with rate of reaction. Shorter time means faster rate. Also, always use data values to support your description, not just 'it goes up and down'.
常见错误：将分解所需时间与反应速率混淆。时间越短表示速率越快。此外，要始终用数据值来支撑描述，而不能只说“有升有降”。
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