📚 Year 11 OCR Maths: Cross-Curricular Mixed Question Training | Year 11 OCR 数学：跨学科综合题型训练
In the OCR GCSE Mathematics exam, you will encounter questions that blend maths with science, geography, business and other real-world contexts. These cross-curricular problems test your ability to translate situations into numbers, formulas and logical steps – exactly the skills that top marks demand. This article provides targeted training across ten key interdisciplinary themes, showing you how to decode wordy questions and apply your mathematical toolkit with confidence.
在 OCR GCSE 数学考试中，你会遇到将数学与科学、地理、商业等现实情境相融合的题目。这些跨学科问题考验你将情境转化为数字、公式和逻辑步骤的能力——这正是获得高分所需的关键技能。本文围绕十大跨学科主题提供针对性训练，教会你如何拆解题干信息，自信地运用数学工具。
1. Understanding Cross-Curricular Questions | 理解跨学科题型
Cross-curricular questions in OCR Maths are not simply about harder calculations – they require you to identify which mathematical concept is being tested inside a scientific, geographical or financial shell. The first step is always to strip away the context and find the core mathematical relationship, such as direct proportion, percentage change or interpreting a graph.
OCR 数学中的跨学科题目并非只是计算更复杂——它们要求你在科学、地理或金融外壳下识别出正在考查的数学概念。第一步永远是剥去情境，找到核心的数学关系，例如正比例、百分比变化或解读图表。
You will often be given real data, unfamiliar units or complex diagrams. Stay calm, highlight the quantities you know and the quantity you need, then write an equation or draw a sketch. Practice with the examples below will build your fluency across these hybrid challenges.
你经常会遇到真实数据、不熟悉的单位或复杂图表。保持冷静，标出已知量和待求量，然后写出方程或画出示意图。通过下列例题练习，你将能熟练应对这些混合型挑战。
2. Physics: Speed, Distance and Time | 物理：速度、距离和时间
The relationship linking speed, distance and time appears frequently in both mechanics and graph interpretation questions. The formula triangle is worth memorising:
速度、距离与时间的关系经常出现在力学和图表解读题中。值得记住公式三角形：
speed = distance ÷ time
Worked example: A cyclist maintains a constant speed of 16 km/h for 2 hours and 30 minutes. Calculate the distance covered in kilometres.
例题：一位自行车手以 16 km/h 的恒定速度骑行了 2 小时 30 分钟。计算骑行的距离（公里）。
First convert the time entirely to hours: 2 hours 30 minutes = 2.5 h. Then apply the rearranged formula: distance = speed × time = 16 × 2.5 = 40 km. Always check that units are consistent – here both time and speed use hours, so no conversion is needed beyond the minutes.
首先将时间完全转换为小时：2 小时 30 分钟 = 2.5 小时。然后代入变形公式：距离 = 速度 × 时间 = 16 × 2.5 = 40 公里。务必检查单位一致——这里时间和速度都用小时，因此除分钟换算外无需额外转换。
When a graph of distance against time is given, the gradient represents speed. A straight line means constant speed, while a curved line indicates acceleration or deceleration.
当给出距离-时间图时，梯度代表速度。直线表示匀速，曲线则表示加速或减速。
3. Chemistry: Density and Concentration | 化学：密度与浓度
Density links mass and volume – a concept equally at home in chemistry and physics problems. The core equation is:
密度将质量与体积联系在一起——这个概念在化学和物理问题中同样常见。核心方程为：
density = mass ÷ volume
Example: A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density and determine whether it could be aluminium, given that the density of aluminium is 2.7 g/cm³.
例题：一块金属的质量为 540 g，体积为 200 cm³。计算其密度，并判断它是否可能是铝，已知铝的密度为 2.7 g/cm³。
Density = 540 ÷ 200 = 2.7 g/cm³. This matches the reference density for aluminium exactly, so it is likely aluminium. Notice how reading the units helps: g/cm³ tells you mass in grams divided by volume in cm³.
密度 = 540 ÷ 200 = 2.7 g/cm³。该数值与铝的参考密度完全吻合，因此很可能就是铝。注意解读单位的帮助：g/cm³ 表示以克为单位的质量除以以立方厘米为单位的体积。
In concentration problems, you use a similar structure: concentration = mass of solute ÷ volume of solution. For instance, if 5 g of salt is dissolved in 0.25 dm³ of water, the concentration is 5 ÷ 0.25 = 20 g/dm³.
在浓度问题中，结构类似：浓度 = 溶质质量 ÷ 溶液体积。例如，若将 5 g 食盐溶解于 0.25 dm³ 水中，浓度为 5 ÷ 0.25 = 20 g/dm³。
4. Geography: Map Scales and Bearings | 地理：地图比例尺与方位角
Map reading tasks require you to convert scaled distances to real lengths and to work with three-figure bearings. A scale of 1 : 25000 means that 1 cm on the map represents 25000 cm in reality.
地图辨读任务要求你将比例尺距离转换为实际长度，并处理三位数方位角。比例尺 1 : 25000 表示地图上 1 cm 代表实地 25000 cm。
Example: Two villages are 8 cm apart on a map with a scale of 1 : 50000. Find the real distance in kilometres.
例题：在比例尺为 1 : 50000 的地图上，两个村庄相距 8 cm。求实际距离（公里）。
Real distance = 8 × 50000 = 400000 cm. Convert to metres by dividing by 100 → 4000 m, then to kilometres by dividing by 1000 → 4 km. So the villages are 4 km apart.
实地距离 = 8 × 50000 = 400000 cm。除以 100 换算为米 → 4000 m，再除以 1000 换算为公里 → 4 km。因此两村相距 4 公里。
Bearings are always measured clockwise from north and expressed as three digits. If a lighthouse is on a bearing of 125° from a boat, you can use trigonometry or scale drawing to find distances back to the coast.
方位角始终从正北顺时针测量，并用三位数表示。如果从船上看到灯塔的方位角为 125°，你可以利用三角学或比例绘图求出返回海岸的距离。
5. Biology: Percentage Change and Growth | 生物学：百分比变化与增长
Biologists use percentage change to compare populations, cell counts or reaction rates over time. The formula is:
生物学家用百分比变化来比较随时间变化的种群数量、细胞计数或反应速率。公式为：
percentage change = (final value − original value) ÷ original value × 100
Example: A bacterial culture starts with 500 cells and grows to 850 cells in 4 hours. Calculate the percentage increase.
例题：一种细菌培养物初始有 500 个细胞，4 小时后增长至 850 个细胞。计算增长百分比。
Change = 850 − 500 = 350. Percentage increase = (350 ÷ 500) × 100 = 0.7 × 100 = 70%. The population increased by 70%.
变化量 = 850 − 500 = 350。增长百分比 = (350 ÷ 500) × 100 = 0.7 × 100 = 70%。数量增长了 70%。
A negative result indicates a decrease; for instance, if a population falls from 2000 to 1700, the change is −300, giving a 15% decrease.
结果为负表示减少；例如，若种群从 2000 降至 1700，变化量为 −300，即减少 15%。
6. Business: Profit, Loss and Discount | 商业：利润、亏损与折扣
Business scenarios test your ability to calculate profit margins, selling prices and percentage discounts. Profit is simply selling price minus cost price, but exam questions often ask for profit as a percentage of the cost price.
商业场景考查你计算利润率、售价和折扣百分比的能力。利润即售价减去成本价，但考试题常要求将利润表示为成本价的百分比。
Example: A shop buys a jacket for £40 and sells it for £54. Find the profit as a percentage of the cost price.
例题：一家商店以 £40 购入一件夹克，以 £54 售出。求利润占成本价的百分比。
Profit = £54 − £40 = £14. Percentage profit = (14 ÷ 40) × 100 = 35%. This means the shop makes a 35% return on the original cost.
利润 = £54 − £40 = £14。利润百分比 = (14 ÷ 40) × 100 = 35%。这意味着商店获得了原始成本 35% 的回报。
For discounts, a 20% off sale reduces an item’s price. If a TV originally costs £320, the sale price is 80% of £320 = 0.8 × 320 = £256. Always check whether the question asks for the discount amount or the new price.
对于折扣，打八折会降低商品价格。若一台电视原价 £320，促销价即为 £320 的 80% = 0.8 × 320 = £256。务必看清题目是要求折扣金额还是新价格。
7. Science Experiments: Interpreting Graphs | 科学实验：图表解读
Experimental data in physics, chemistry or biology is often presented as line graphs or scatter plots. You need to read axes, calculate gradients and describe trends.
物理、化学或生物学中的实验数据常以折线图或散点图呈现。你需要读取坐标轴、计算梯度并描述趋势。
For a distance–time graph, the gradient gives speed. Select two well-separated points, find the rise (change in distance) and run (change in time), then divide. If the graph shows a curve, draw a tangent at the desired point and find its gradient for the instantaneous rate.
对于距离–时间图，梯度给出速度。选择两个相距较远的点，计算纵轴增量（距离变化）和横轴增量（时间变化），然后相除。若图形为曲线，在所需点处作切线，求其梯度即为瞬时速率。
Example: The temperature of a liquid is recorded every minute and plotted. The gradient of the steepest part of the graph gives the maximum heating rate in °C per minute. Relate this to the power of the heater using the appropriate formula from physics, such as energy = power × time.
例题：每分钟记录一次液体温度并绘图。图形最陡部分的梯度给出了以 °C/分钟为单位的最大加热速率。结合物理学公式，如能量 = 功率 × 时间，可将此与加热器功率关联起来。
8. Genetics: Probability with Punnett Squares | 遗传学：庞纳特方格概率
Genetic crosses offer a perfect context for probability questions. A Punnett square shows all possible combinations of alleles from two parents, and you can express the chance of a certain genotype as a fraction or percentage.
遗传杂交为概率问题提供了绝佳情境。庞纳特方格展示来自双亲的所有等位基因组合，你可以用分数或百分比表示某种基因型出现的几率。
Consider a monohybrid cross between two heterozygous parents (Bb × Bb) where B is the dominant allele for brown eyes and b is recessive for blue eyes. The Punnett square yields genotypes BB, Bb, bB and bb.
考虑两个杂合亲本之间的单基因杂交（Bb × Bb），其中 B 为褐色眼睛的显性等位基因，b 为蓝色眼睛的隐性等位基因。庞纳特方格得出基因型 BB、Bb、bB 和 bb。
Probability of a brown-eyed offspring (BB, Bb or bB) = 3 out of 4, or 3/4 = 0.75 = 75%. Probability of a blue-eyed offspring (bb) = 1/4 = 25%. You can then extend this to predict numbers in a sample of 200 plants or animals by multiplying the probability by the total.
褐色眼睛后代（BB、Bb 或 bB）的概率 = 4 分之 3，即 3/4 = 0.75 = 75%。蓝色眼睛后代（bb）的概率 = 1/4 = 25%。随后可将概率乘以总数，预测在 200 株植物或动物样本中的数量。
9. Finance: Compound Interest and Depreciation | 金融：复利与折旧
Compound interest and depreciation are modelled using exponential growth or decay. The formula for compound interest is:
复利与折旧使用指数增长或衰减来建模。复利计算公式为：
total = P(1 + r)ⁿ
where P is the principal, r is the interest rate per period as a decimal, and n is the number of periods. For depreciation, use (1 − r)ⁿ instead.
其中 P 为本金，r 是每期利率（小数形式），n 为期数。对于折旧，则使用 (1 − r)ⁿ。
Example: £2000 is invested at 4% compound interest per annum for 3 years. Calculate the total amount.
例题：£2000 以年利率 4% 的复利投资 3 年。计算最终总金额。
r = 0.04, n = 3. Total = 2000 × (1.04)³ = 2000 × 1.124864 ≈ £2249.73. The interest earned is about £249.73.
r = 0.04，n = 3。总额 = 2000 × (1.04)³ = 2000 × 1.124864 ≈ £2249.73。赚取的利息约为 £249.73。
Depreciation works similarly: a car bought for £15000 depreciates at 20% per year. After 2 years, value = 15000 × (0.8)² = 15000 × 0.64 = £9600.
折旧类似：一辆以 £15000 购入的汽车每年贬值 20%。2 年后，价值 = 15000 × (0.8)² = 15000 × 0.64 = £9600。
10. Rates of Change: Gradient in Real Life | 变化率：实际生活中的梯度
Gradient is not just an abstract slope – it represents the rate at which one quantity changes relative to another. In physics, the gradient of a velocity–time graph gives acceleration. In geography, the gradient of a river profile shows how steeply the river falls.
梯度不仅是一个抽象的斜率——它表示一个量相对于另一个量的变化速率。在物理学中，速度–时间图的梯度代表加速度。在地理学中，河流纵剖面的梯度显示河流落差有多陡。
Example: A water tank is emptied, and the volume remaining is recorded every 10 seconds. The graph of volume (litres) against time (seconds) is a straight line from 100 L at 0 s to 0 L at 50 s. Find the rate of flow.
例题：一个水罐被排空，每 10 秒记录一次剩余体积。体积（升）与时间（秒）的关系图为一条直线，从 0 秒时的 100 升降至 50 秒时的 0 升。求流量速率。
Change in volume = 0 − 100 = −100 L, change in time = 50 − 0 = 50 s. Gradient = −100 ÷ 50 = −2 L/s. The negative sign indicates a decrease; the rate of flow is 2 litres per second out of the tank.
体积变化 = 0 − 100 = −100 升，时间变化 = 50 − 0 = 50 秒。梯度 = −100 ÷ 50 = −2 升/秒。负号表示减少；即水以 2 升/秒的速率流出水罐。
Interpreting the gradient with its units is crucial. Always write the rate as ‘unit of y per unit of x’, such as metres per second, people per year or kilograms per hour.
在单位层面解读梯度至关重要。始终将速率写成“y 轴单位 / x 轴单位”，例如米/秒、人/年或千克/小时。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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