📚 Year 10 OCR Mathematics: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练
In Year 10 OCR Mathematics, students are expected to apply their mathematical skills to real-world contexts, often blending concepts from physics, chemistry, biology, geography, and economics. This article provides a collection of interdisciplinary problem-solving exercises designed to strengthen your ability to transfer mathematical techniques across subjects. Each section presents a scenario, key mathematical ideas, and a step-by-step approach.
在OCR Year 10数学中，学生需要将数学技能应用于现实世界情境，常常融合物理、化学、生物、地理和经济等学科的概念。本文提供了一系列跨学科问题解决训练，旨在增强你将数学方法迁移到其他学科的能力。每节呈现一个情境、关键数学思想以及分步解决方法。
1. Physics: Interpreting Distance-Time Graphs | 物理：距离–时间图的解读
In physics, the motion of an object is often described using a distance-time graph. The gradient (slope) of the graph represents the speed. A horizontal line means the object is stationary. Understanding gradients and interpreting straight-line segments is a core mathematical skill that directly applies to mechanics. Consider a car journey: from 0 s to 10 s it travels 800 m in a straight line, then remains stopped from 10 s to 30 s, and finally returns to the start over the next 20 s.
在物理学中，物体的运动常用距离–时间图描述。图线的梯度（斜率）代表速度。水平线段表示物体静止。理解梯度并解读直线段是直接应用于力学的核心数学技能。考虑一段汽车行程：从0秒到10秒沿直线行驶800米，之后在10秒到30秒间停止不动，最后在接下来的20秒内返回起点。
The graph consists of three straight segments. For the first segment, the coordinates are (0,0) to (10, 800). The speed is the gradient: (800 − 0) ÷ (10 − 0) = 80 m/s. The second segment from (10, 800) to (30, 800) is horizontal, so gradient = 0, indicating the car is stationary. The third segment from (30, 800) to (50, 0) gives a return speed of (0 − 800) ÷ (50 − 30) = −800 ÷ 20 = −40 m/s; the negative sign shows movement back toward the start, so the speed is 40 m/s. Average speed is total distance divided by total time: (800 + 0 + 800) m ÷ 50 s = 1600 ÷ 50 = 32 m/s.
该图由三条直线段组成。第一段，坐标为(0,0)到(10,800)。速度即梯度：(800 − 0) ÷ (10 − 0) = 80 米/秒。第二段从(10,800)到(30,800)为水平线，梯度为0，表明汽车静止。第三段从(30,800)到(50,0)，返回速度为 (0 − 800) ÷ (50 − 30) = −800 ÷ 20 = −40 米/秒；负号表示返回起点，因此速率为40米/秒。平均速度 = 总距离 ÷ 总时间 = (800 + 0 + 800) 米 ÷ 50 秒 = 1600 ÷ 50 = 32 米/秒。
Speed = Δdistance ÷ Δtime
This example reinforces how gradient calculations from coordinate geometry are used to analyse everyday motion. Interdisciplinary problems often require unit conversions and careful interpretation of graphical information, exactly the skills assessed in OCR examinations.
这个例子强化了坐标几何中梯度计算如何用于分析日常运动。跨学科问题通常要求进行单位换算并仔细解读图形信息，这正是OCR考试所评估的技能。
2. Chemistry: Proportions and Scaling in Stoichiometry | 化学：计量学中的比例与配方缩放
Chemical equations provide fixed ratios between reactants and products. These ratios allow us to scale up masses and volumes using direct proportion – a fundamental Year 10 topic. Consider the production of water: 2H₂ + O₂ → 2H₂O. If we wish to prepare 36 g of water, how many grams of hydrogen gas (H₂) are required? First, note the molar masses: H₂ = 2 g/mol, O₂ = 32 g/mol, H₂O = 18 g/mol.
化学反应方程式给出了反应物与生成物之间的固定比例。利用这些比例，我们可以通过正比例关系放大质量与体积——这是Year 10的基础课题。考虑水的生成：2H₂ + O₂ → 2H₂O。如果我们想制取36克水，需要多少克氢气(H₂)？首先，注意摩尔质量：H₂ = 2 克/摩尔，O₂ = 32 克/摩尔，H₂O = 18 克/摩尔。
From the balanced equation, 2 moles of H₂ produce 2 moles of H₂O, so the mole ratio is 1 : 1. Mass-wise, 2 mol × 2 g/mol = 4 g of H₂ produce 2 mol × 18 g/mol = 36 g of H₂O. Thus, 4 g of hydrogen are needed. This can be set up as a proportion: mass of H₂ / mass of H₂O = 4/36 = 1/9. For any desired mass of water, you multiply by 1/9 to get the mass of hydrogen.
由配平的方程式，2摩尔H₂生成2摩尔H₂O，摩尔比为1:1。质量方面，2 mol × 2 克/摩尔 = 4 克 H₂ 生成 2 mol × 18 克/摩尔 = 36 克 H₂O。因此需要4克氢气。这可以设为比例：H₂质量 / H₂O质量 = 4/36 = 1/9。对于任何所需水的质量，乘以1/9即可得到氢气质量。
| Substance 物质 | Molar mass (g/mol) 摩尔质量 | Moles 摩尔数 | Mass (g) 质量 |
|---|---|---|---|
| 2H₂ | 2 | 2 | 4 |
| O₂ | 32 | 1 | 32 |
| 2H₂O | 18 | 2 | 36 |
Proportional reasoning also helps if we need to scale up: to produce 180 g of water, the mass of hydrogen required is 180 × (4/36) = 20 g. This method avoids mole calculations when they are not required, linking directly to the ratio and proportion strand of the OCR curriculum.
比例推理同样适用于放大生产：要生产180克水，所需氢气质量为 180 × (4/36) = 20 克。这种方法在不需要摩尔计算时直接使用，与OCR课程中比和比例知识紧密相连。
3. Biology: Exponential Growth of Bacteria | 生物：细菌的指数增长
Many biological populations grow exponentially when resources are unlimited. A bacterial colony that doubles every 20 minutes can be modelled using powers of 2. Starting with 100 bacteria, after 3 hours (180 minutes) how many bacteria are present? The number of doubling periods is 180 ÷ 20 = 9. Hence the population = 100 × 2⁹.
许多生物种群在资源无限时呈指数增长。一个每20分钟翻倍的细菌菌落可以用2的幂来建模。初始有100个细菌，3小时（180分钟）后有多少细菌？翻倍次数为 180 ÷ 20 = 9。因此数量 = 100 × 2⁹。
Calculate 2⁹ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512. So the final count is 100 × 512 = 51,200 bacteria. The general formula is N = N₀ × 2^(t/T), where N₀ is the initial population, t is total time, and T is the doubling time. This is an example of a geometric sequence, a key part of the OCR algebra topic.
计算 2⁹ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512。因此最终数量为 100 × 512 = 51,200 个细菌。通用公式为 N = N₀ × 2^(t/T)，其中 N₀ 为初始数量，t 为总时间，T 为翻倍时间。这是一个几何数列例子，属于OCR代数部分的重要内容。
We can also reverse the process: if after 2 hours the count is 6,400, starting from 100, how many doubling periods have passed? 6,400 ÷ 100 = 64 = 2⁶, so n = 6, meaning each period is 120 ÷ 6 = 20 minutes. These exercises strengthen understanding of indices and growth modelling.
我们也可以逆向求解：如果2小时后数量为6,400，开始为100，翻倍次数是多少？6,400 ÷ 100 = 64 = 2⁶，因此 n = 6，即每次翻倍时间为 120 ÷ 6 = 20 分钟。这些练习增强对指数和增长建模的理解。
4. Geography: Map Scales and Bearings | 地理：地图比例尺与方位
Map reading combines ratio, proportion, and angle measurement. A map with scale 1:25,000 means 1 cm on the map represents 25,000 cm in reality. If two points are 4.5 cm apart on the map, the actual distance is 4.5 × 25,000 = 112,500 cm. Convert to metres by dividing by 100: 1,125 m, or 1.125 km. This requires fluency with metric conversions and multiplying by scale factors.
地图阅读结合了比、比例和角度测量。一张比例尺为1:25,000的地图表示图上1厘米代表实际25,000厘米。若图上两点距离为4.5厘米，实际距离为 4.5 × 25,000 = 112,500 厘米。除以100转换为米：1,125米，或1.125公里。这需要熟练掌握公制单位换算和比例尺的乘法。
Bearings are measured clockwise from north. If the bearing from point A to point B is 065°, then the return bearing from B to A is found by adding or subtracting 180° (as long as it stays within 0°–360°). 065° + 180° = 245°. This uses angle properties on a straight line and reinforces three-figure bearings. Interdisciplinary tasks often combine distance and bearing to describe positions, linking to geometry and angle facts.
方位角从正北起顺时针测量。若从A点到B点的方位角为065°，则从B到A的返回方位角加或减180°（保持在0°–360°内）。065° + 180° = 245°。这利用了直线上的角度性质并强化了三方位角度制。跨学科任务常结合距离与方位来描述位置，联系几何与角度知识。
5. Economics: Compound Interest and Percentage Change | 经济学：复利与百分比变化
Compound interest is a direct application of repeated percentage increase, a topic firmly within the OCR specification. Suppose you deposit £2000 in a savings account offering 5% annual compound interest. After 3 years, the amount A is calculated using the multiplier 1.05: A = 2000 × (1.05)³.
复利是重复百分比增长的直接应用，是OCR大纲中的核心内容。假设你将2000英镑存入年利率5%的复利账户。3年后，金额A使用乘数1.05计算：A = 2000 × (1.05)³。
First compute (1.05)³ = 1.05 × 1.05 × 1.05. 1.05² = 1.1025, then 1.1025 × 1.05 = 1.157625. Multiply by 2000: £2315.25. This is greater than simple interest (where you would earn 5% of £2000 = £100 each year, totalling £2300), demonstrating the power of compounding. Percentage decrease problems, such as depreciation of a car by 15% per year, use a multiplier of 0.85 in the same formula.
先计算 (1.05)³ = 1.05 × 1.05 × 1.05。1.05² = 1.1025，然后 1.1025 × 1.05 = 1.157625。乘以2000等于2315.25英镑。这高于单利（每年获5% × £2000 = £100，3年共£2300），展示了复利的威力。百分比减少问题，如汽车每年贬值15%，使用乘数0.85套用同一公式。
Understanding the difference between simple and compound interest requires careful construction of formulas and calculation. Such skills are tested in worded problems that mix financial contexts with algebraic expressions.
理解单利与复利的区别需要仔细构建公式并计算。这类技能会在融合金融情境与代数表达式的应用题中进行考查。
6. Design & Technology: Optimising Surface Area and Volume | 设计技术：优化表面积与体积
In product design, manufacturers often need to minimise material for a given volume. Consider a cylindrical can that must hold exactly 500 cm³ of liquid. The volume is V = πr²h, and the surface area (total material) is S = 2πr² + 2πrh. If you choose a radius r, you can find the height h = V / (πr²) and
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