📚 Interdisciplinary Applied Mathematics Revision for Year 10 WJEC | Year 10 WJEC 数学：跨学科综合题型训练
This revision guide bridges Year 10 WJEC Mathematics with real-world applications across science, geography, economics, and design. You will practise interpreting graphs, solving equations, using ratios, and analysing data exactly as they appear in cross-curricular exam questions. Each section provides paired English–Chinese explanations followed by worked examples that mirror the style of WJEC assessments.
本复习指南将 Year 10 WJEC 数学与科学、地理、经济和设计等学科的实际应用联系起来。你将练习解读图像、解方程、使用比例、分析数据，这些正是跨学科考题中常见的形式。每个小节先提供中英文对照解释，再给出符合 WJEC 评估风格的例题及解答。
1. Physics: Interpreting Distance–Time Graphs | 物理：距离–时间图的解读
A distance–time graph shows how far an object has travelled over time. The gradient (steepness) of the line represents speed. A horizontal line means the object is stationary. In WJEC questions, you may need to calculate average speed between two points or find the speed at a particular instant by drawing a tangent.
距离–时间图展示物体在一段时间内移动的距离。线的斜率（陡峭程度）代表速度。水平线表示物体静止。在 WJEC 考题中，可能需要你计算两点之间的平均速度，或者通过画切线求某一时刻的瞬时速度。
Example: A cyclist travels 150 m in 10 s, then stops for 5 s, and finally cycles 90 m in 15 s. Plot the graph and calculate the average speed for the whole journey.
例题：一名自行车手 10 秒内骑了 150 米，然后停了 5 秒，最后 15 秒骑了 90 米。画出图像并计算全程平均速度。
= 150 + 0 + 90 = 240 m, total time = 10 + 5 + 15 = 30 s, average speed = 240 ÷ 30 = 8 m/s.
解：总距离 = 150 + 0 + 90 = 240 米，总时间 = 10 + 5 + 15 = 30 秒，平均速度 = 240 ÷ 30 = 8 米/秒。
2. Chemistry: Molar Ratios and Proportional Reasoning | 化学：摩尔比与比例推理
In chemistry, balanced equations give mole ratios. You can use direct proportion to scale up or down. For instance, if 2 moles of hydrogen react with 1 mole of oxygen to produce water, then 5 moles of hydrogen need 2.5 moles of oxygen. WJEC often embeds these ratios in a mathematical context—unitary method or ratio tables are expected.
在化学中，配平的化学方程式提供摩尔比。你可以用正比例进行放大或缩小。例如，如果 2 摩尔氢气与 1 摩尔氧气反应生成水，那么 5 摩尔氢气需要 2.5 摩尔氧气。WJEC 经常将这些比例嵌入数学情境——要求你使用归一法或比例表。
xample: The reaction is 2Mg + O₂ → 2MgO. How many grams of MgO are produced from 6 g of Mg? (Aᵣ: Mg = 24, O = 16)
例题：反应方程式为 2Mg + O₂ → 2MgO。用 6 克 Mg 能生成多少克 MgO？（相对原子质量：Mg = 24，O = 16）
Mg : MgO = 2:2 = 1:1, so 0.25 mol MgO. Mᵣ of MgO = 24 + 16 = 40. Mass = 0.25 × 40 = 10 g.
步骤：Mg 的物质的量 = 质量 ÷ 相对原子质量 = 6 ÷ 24 = 0.25 摩尔。摩尔比 Mg:MgO = 2:2 = 1:1，所以生成 0.25 摩尔 MgO。MgO 的相对分子质量 = 24 + 16 = 40。质量 = 0.25 × 40 = 10 克。
3. Biology: Exponential Growth and Decay Models | 生物：指数增长与衰减模型
ations of bacteria often grow exponentially under ideal conditions. The formula N = N₀ × 2t/d is common, where N₀ is the initial number, t is time, and d is doubling period. WJEC questions may ask you to complete a table, plot a growth curve, or use trial and improvement for a related equation.
在理想条件下，细菌数量常呈指数增长。常见公式为 N = N₀ × 2t/d，其中 N₀ 为初始数量，t 为时间，d 为复制周期。WJEC 考题可能要求你完成表格、绘制增长曲线或使用试错法解相关方程。
Example: A culture has 500 bacteria at 8 am. The number doubles every 3 hours. Find the number of bacteria at 8 pm on the same day.
例题：上午 8 点某培养物中有 500 个细菌。每 3 小时数量翻倍。求当天晚上 8 点的细菌数量。
ed = 12 hours, d = 3, so number of doublings = 12 ÷ 3 = 4. N = 500 × 2⁴ = 500 × 16 = 8000.
经过时间 = 12 小时，复制周期 d = 3，所以翻倍次数 = 12 ÷ 3 = 4。N = 500 × 2⁴ = 500 × 16 = 8000。
4. Geography: Map Scales and Bearings | 地理：地图比例尺与方位角
250 m in reality. Bearings are measured clockwise from north, expressed as three-digit angles. WJEC exams test conversion between units, area scale factors, and finding distances using bearings and trigonometry.
地图比例尺以比的形式表示，例如 1:25 000。图上 5 厘米代表实际 5 × 25 000 = 125 000 厘米 = 1250 米。方位角从正北顺时针测量，用三位数表示。WJEC 考试考查单位换算、面积比例尺以及利用方位角和三角函数求距离。
Task: A map has scale 1:50 000. Two villages are 8.4 km apart. What is the distance on the map in centimetres? Also, the bearing of B from A is 125°. If C is 5 km due east of A, find the bearing of C from B (using the sine rule or accurate drawing).
任务：地图比例尺为 1:50 000。两村庄相距 8.4 公里。图上距离是多少厘米？此外，B 在 A 的 125° 方位。若 C 在 A 正东 5 公里处，求 C 在 B 的方位（可用正弦定理或精确作图）。
he bearing problem, use cosine rule first to find BC, then sine rule to find angle ABC, finally compute required bearing. (Full working would be shown in class.)
图上距离 = (8.4 公里 × 100 000 厘米/公里) ÷ 50 000 = 840 000 ÷ 50 000 = 16.8 厘米。方位问题先用余弦定理求 BC，再用正弦定理求角 ABC，最后计算所需方位。（完整步骤将在课堂展示。）
5. Economics: Simple and Compound Interest | 经济：单利与复利
= P × r × t. Compound interest adds interest to the principal each period: A = P(1 + r/n)nt. WJEC often embeds these in multi-step word problems involving savings, loans, or depreciation.
单利仅基于本金计算：I = P × r × t。复利每期将利息加入本金：A = P(1 + r/n)nt。WJEC 常将其嵌入涉及储蓄、贷款或折旧的多步骤应用题中。
Example: £2000 is invested at 3.5% per annum compound interest. What is the total amount after 4 years? How much more interest would be earned compared to simple interest at the same rate?
例题：将 £2000 以年利率 3.5% 的复利进行投资。4 年后总金额是多少？与同利率下的单利相比，多赚多少利息？
Compound: A = 2000 × (1.035)4 = 2000 × 1.1475... ≈ £2295.05. Simple interest: I = 2000 × 0.035 × 4 = £280, total = £2280. Difference = £2295.05 - £2280 = £15.05.
复利：A = 2000 × (1.035)4 = 2000 × 1.1475... ≈ £2295.05。单利：I = 2000 × 0.035 × 4 = £280，总金额 = £2280。差额 = £2295.05 - £2280 = £15.05。
6. Design & Technology: Surface Area and Volume of Composite Solids | 设计与技术：组合体的表面积与体积
Real-world products often combine cuboids, cylinders, and prisms. You need to split the shape into simpler parts, calculate the area or volume of each, and then add or subtract. WJEC exams frequently give nets or 3D drawings and ask for the amount of material needed or capacity.
现实中的产品常由长方体、圆柱体和棱柱组合而成。你需要将图形拆分成简单部分，分别计算面积或体积，然后进行加减。WJEC 考试常给出展开图或三维图，要求计算所需材料量或容积。
late the total surface area of the block, including the inside of the hole. (Use π = 3.14)
问题：一个积木由一个棱长 4 厘米的立方体和一个从中心钻穿的半径 1 厘米的圆柱孔组成。计算该积木的总表面积，包括孔的内表面。（使用 π = 3.14）
cylinder = 2πrh = 2 × 3.14 × 1 × 4 = 25.12 cm². So total = 96 - 6.28 + 25.12 = 114.84 cm².
表面积：立方体表面积 = 6 × 4 × 4 = 96 平方厘米。两个圆孔减去面积 = 2 × π × 1² = 6.28 平方厘米，加上圆柱侧面积 = 2πrh = 2 × 3.14 × 1 × 4 = 25.12 平方厘米。总表面积 = 96 - 6.28 + 25.12 = 114.84 平方厘米。
7. Psychology: Descriptive Statistics and Interpretation | 心理学：描述性统计与解读
Psychology experiments produce data sets requiring mean, median, mode, range, and interquartile range. WJEC may present a table of reaction times or scores and ask you to compare two groups using measures of central tendency and spread. You must comment on what the statistics suggest about behaviour.
心理学实验会产生需要计算平均数、中位数、众数、极差和四分位距的数据集。WJEC 可能给出反应时间或得分表格，要求你使用集中趋势和离散程度的度量比较两组数据。你必须根据统计结果对行为做出推断。
Example: Group A scores: 12, 15, 14, 13, 18. Group B: 9, 20, 10, 22, 14. Calculate mean and range for each. Which group is more consistent? Explain.
例题：A 组得分：12, 15, 14, 13, 18。B 组：9, 20, 10, 22, 14。计算每组的平均数和极差。哪组更稳定？请解释。
is more consistent because its range is much smaller, and the mean is stable; group B has a higher mean but greater variability.
平均数 A = (12+15+14+13+18)/5 = 14.4；极差 A = 18 - 12 = 6。平均数 B = (9+20+10+22+14)/5 = 15；极差 B = 22 - 9 = 13。A 组更稳定，因为其极差小很多，且平均数较稳定；B 组平均数更高但变异性大。
8. Sports Science: Scatter Graphs and Lines of Best Fit | 运动科学：散点图与最佳拟合线
Scatter graphs show the relationship between two variables, such as hours of training and race time. A line of best fit (drawn by eye or mathematically) helps predict values. WJEC questions require plotting points, describing correlation (positive, negative, none), and using the line to interpolate or extrapolate.
散点图显示两个变量之间的关系，如训练时长与比赛用时。最佳拟合线（目测或数学方法画出）有助于预测数值。WJEC 考题要求描点、描述相关性（正、负、无相关），并用拟合线进行内插或外推。
Data: (hours training, time in seconds): (5, 58), (6, 54), (8, 48), (10, 44), (12, 40). Plot the points, draw the line of best fit, and estimate the time for 9 hours of training. Describe the correlation.
数据：（训练小时数，秒数）：(5, 58), (6, 54), (8, 48), (10, 44), (12, 40)。描点，画出最佳拟合线，估计训练 9 小时对应的用时。描述相关性。
e: about 45 seconds. (Using two points on line, e.g. (5, 58) and (12, 40), gradient = (40-58)/(12-5) = -18/7 = -2.57; equation y = -2.57x + 70.85; for x=9, y ≈ 47.7 s, rounding to 48 s.)
描点可得到明显的负相关：时间随小时数增加而减少。拟合线大致经过 (7, 50)。对于 9 小时，从线读出约 45 秒。（若取线上两点 (5, 58) 和 (12, 40)，斜率 = (40-58)/(12-5) = -18/7 = -2.57；方程 y = -2.57x + 70.85；当 x=9，y ≈ 47.7 秒，四舍五入为 48 秒。）
9. Business: Break-even Analysis with Simultaneous Equations | 商业：用联立方程做盈亏平衡分析
A company' . 5n = 2n + 500. This is a linear simultaneous equation concept. WJEC may ask students to solve algebraically and interpret the solution in context.
某公司的总成本可表示为 C = 2n + 500，收入为 R = 5n，其中 n 为销售数量。盈亏平衡点是成本等于收入之处，即 5n = 2n + 500。这采用了线性联立方程的概念。WJEC 可能要求学生用代数方法求解并在上下文中解释解的意义。
167 items to start making profit. Check: C = 2(167)+500 = 834, R = 5(167) = 835, profit = £1.
求解：5n = 2n + 500 → 3n = 500 → n = 166.67。由于产品须为整数，必须卖出 167 件才能开始盈利。检验：C = 2(167)+500 = 834，R = 5(167) = 835，利润 = £1。
10. Environmental Science: Rate of Water Flow and Decay of Pollutants | 环境科学：水流速率与污染物衰减
Environmental data often involve rates of change. A river' led with a formula. The concentration of a pollutant may decay exponentially, e.g. C = C₀e-kt (though at GCSE, simpler decay using halving time is common). WJEC may present a graph of concentration over time and ask for gradient at a point (rate of decay) or total volume calculations using area under a graph.
环境数据常涉及变化率。河流流量（立方米/秒）可用公式建模。污染物浓度可能呈指数衰减，如 C = C₀e-kt（不过在 GCSE 水平，通常用半衰期进行简单衰减计算）。WJEC 可能给出浓度随时间变化的图像，要求你求某点的梯度（衰减速率）或利用图像下方面积计算总量。
Scenario: A factory spill causes initial pollutant concentration of 80 mg/L. The concentration halves every 2 days. Write a formula for concentration after t days. Find when concentration drops below 5 mg/L.
情境：工厂泄漏导致初始污染物浓度为 80 毫克/升。浓度每 2 天减半。写出 t 天后的浓度公式。求浓度何时降至 5 毫克/升以下。
Formula: C = 80 × (1/2)t/2. Set C = 5: 5 = 80 × (0.5)t/2 → (0.5)t/2 8 days. After 8 days, concentration is 5 mg/L, so it drops below 5 mg/L just after 8 days.
公式：C = 80 × (1/2)t/2。令 C = 5：5 = 80 × (0.5)t/2 → (0.5)t/2 = 5/80 = 1/16 = 0.0625。由于 0.5⁴ = 0.0625，t/2 = 4 → t = 8 天。8 天后浓度为 5 毫克/升，因此刚好 8 天后浓度降至 5 毫克/升以下。
11. Nutrition: Proportions in Recipes and Scaling | 营养学：食谱比例与缩放
ts by 10/4 = 2.5. WJEC tests ratio and proportion in real-life contexts, including best buys and conversion between units like grams and ounces.
营养学家使用比例调整食材用量。若一份 4 人食用的食谱需用 300 克面粉，供 10 人食用则需将所有配料乘以 10/4 = 2.5。WJEC 测试实际生活中的比例问题，包括最佳购买以及克与盎司等单位的换算。
Problem: A smoothie for 3 uses 2 bananas, 150 ml yoghurt, and 200 ml juice. How much of each ingredient is needed for 7 people? If bananas cost £0.15 each, what is the total cost for 7 smoothies?
问题：一份 3 人份的奶昔需用 2 根香蕉、150 毫升酸奶、200 毫升果汁。7 人份需要多少每种配料？若香蕉每根 £0.15，7 份奶昔的总成本是多少？
Scale factor = 7/3. Bananas = 2 × 7/3 = 14/3 ≈ 4.67, you' ically. Yoghurt = 150 × 7/3 = 350 ml. Juice = 200 × 7/3 = 466.7 ml. Cost for bananas = 5 × £0.15 = £0.75. (Other costs would be added if given.)
缩放因子 = 7/3。香蕉 = 2 × 7/3 = 14/3 ≈ 4.67，实际需要 5 根。酸奶 = 150 × 7/3 = 350 毫升。果汁 = 200 × 7/3 = 466.7 毫升。香蕉成本 = 5 × £0.15 = £0.75。（其他配料成本若有给出可另加。）
12. Astronomy: Scientific Notation and Light Years | 天文学：科学记数法与光年
³ km. WJEC questions may involve multiplying and dividing numbers in standard form, converting units (e.g. km to light years), and understanding orders of magnitude.
太空中的距离极其巨大，常以标准形式（科学记数法）书写，如比邻星的距离约 4.01 × 10¹³ 公里。WJEC 考题可能涉及科学记数法的乘除运算、单位换算（如公里与光年），以及对数量级的理解。
wer in standard form. Then, if a star is 3.78 × 10¹³ km away, how many light-years is that?
光速为 3 × 10⁵ 公里/秒。一年有 3.156 × 10⁷ 秒。计算 1 光年是多少公里，答案用标准形式表示。若某恒星距离地球 3.78 × 10¹³ 公里，这相当于多少光年？
68 × 10¹² km. Distance in light-years = 3.78 × 10¹³ ÷ (9.468 × 10¹²) = (3.78 ÷ 9.468) × 1013-12 = 0.3992 × 10¹ = 3.992 ≈ 4.0 light-years.
1 光年 = (3 × 10⁵) × (3.156 × 10⁷) = 9.468 × 10¹² 公里。距离的光年数 = 3.78 × 10¹³ ÷ (9.468 × 10¹²) = (3.78 ÷ 9.468) × 1013-12 = 0.3992 × 10¹ = 3.992 ≈ 4.0 光年。
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