📚 Year 10 CIE Mathematics: Interdisciplinary Problem-Solving Training | Year 10 CIE 数学：跨学科综合题型训练
Interdisciplinary problems are a key feature of CIE IGCSE Mathematics. They require you to apply your mathematical skills to real-world contexts from physics, chemistry, biology, geography and economics. This revision article provides targeted training on such cross-curricular question types, helping you bridge the gap between abstract maths and practical scenarios.
跨学科问题是 CIE IGCSE 数学的一个重要特色。这类题目要求你将数学技能应用到物理、化学、生物、地理和经济学等现实情境中。这篇复习文章针对跨学科综合题型进行专项训练，帮助你衔接抽象数学与实践场景。
1. Motion and Speed Graphs | 运动与速度图像
tance, speed and time. The formula is Distance = Speed × Time, so Speed = Distance ÷ Time.
一名骑自行车的人以恒定速度行驶了 2 小时，行程 30 公里。要计算速度，可以利用距离、速度和时间的关系。公式为 距离 = 速度 × 时间，因此 速度 = 距离 ÷ 时间。
Speed = Distance / Time
In a speed–time graph, the area under the graph represents the distance travelled. If the graph is a horizontal straight line, the area is a rectangle; if the speed changes at a constant rate, the area is a triangle or a trapezium.
在速度–时间图像中，图像下方的面积代表所行驶的距离。如果图像是一条水平直线，面积就是一个矩形；如果速度以恒定速率变化，面积就是一个三角形或梯形。
2. Density and Buoyancy | 密度与浮力
metal block with a density of 8 g/cm³ and a volume of 15 cm³ has a mass of 8 × 15 = 120 g.
材料的密度通过公式 质量 = 密度 × 体积 将质量和体积联系起来。例如，一块密度为 8 g/cm³、体积为 15 cm³ 的金属块，其质量为 8 × 15 = 120 g。
Mass = Density × Volume
When an object floats, the mass of water displaced equals the mass of the object. This principle often leads to problems where you compare densities or calculate submerged volumes using ratios.
当物体漂浮时，排开水的质量等于物体的质量。这一原理常常引出比较密度或用比例计算浸没体积的问题。
3. Concentration in Chemistry | 化学中的浓度计算
la is Concentration = Mass of solute ÷ Volume of solution. Units are often g/dm³ or g/cm³.
浓度表示在一定体积的溶剂中溶解了多少溶质。最常用的公式是 浓度 = 溶质质量 ÷ 溶液体积。单位通常是 g/dm³ 或 g/cm³。
Concentration = Mass / Volume
g/cm³. Changing the volume or mass of solute will change the concentration proportionally.
如果将 0.4 g 盐溶解在 200 cm³ 水中，浓度（g/cm³）为 0.4 ÷ 200 = 0.002 g/cm³。改变溶液的体积或溶质的质量会按比例改变浓度。
4. Map Scales and Real Distances | 地图比例尺与实际距离
A map scale such as 1 : 50 000 means that 1 cm on the map represents 50 000 cm (or 0.5 km) on the ground. To find a real distance, multiply the map distance by the scale factor.
地图比例尺如 1:50 000 表示地图上 1 cm 代表实际地面上 50 000 cm（即 0.5 km）。要计算实际距离，可将地图上的距离乘以比例尺因子。
Actual distance = Map distance × Scale factor
For area, the scale factor for lengths must be squared. If the scale is 1 : n, then the area scale factor is n². This is vital when calculating the real area of a field or a nature reserve from a map.
对于面积，长度比例尺因子必须平方。如果比例尺为 1 : n，那么面积比例因子就是 n²。这在根据地图计算农田或自然保护区的实际面积时至关重要。
5. Population Growth and Exponential Functions | 人口增长与指数函数
tage each year. The formula for compound growth is Final amount = Initial amount × (1 + r)n, where r is the growth rate per period and n is the number of periods.
在生物学中，种群数量每年会按固定百分比增长。复合增长公式为 最终数量 = 初始数量 × (1 + r)n，其中 r 是每阶段的增长率，n 是阶段数。
P = P0 (1 + r)n
ony of 1000 doubles every 20 minutes, after 2 hours (6 periods) the population is 1000 × 26 calculate the size after several growth intervals or to find the time needed to reach a target population.
如果一个 1000 个细菌的菌落每 20 分钟翻倍一次，2 小时后（6 个周期）细菌数量为 1000 × 26 = 64 000。CIE 的题目常要求计算若干个生长间隔后的数量，或求出达到目标数量所需的时间。
6. Profit, Loss and Percentage Change | 利润、亏损与百分比变化
ntexts, profit percentage is calculated as (Selling price − Cost price) ÷ Cost price × 100%. Loss percentage uses the same structure when the selling price is lower.
在商业和经济情境中，利润百分比的计算公式为 （售价 − 成本价）÷ 成本价 × 100%。当售价较低时，亏损百分比采用同样的结构。
Profit % = (S.P. − C.P.) / C.P. × 100%
An item bought for $80 and sold for $100 gives a profit of $20 and a profit percentage of 25%. If a discount of 15% is applied to the selling price, you need to adjust the calculations step by step, which tests your ability to chain percentage changes.
一件商品以 80 美元买入，100 美元卖出，获利 20 美元，利润百分比为 25%。如果售价再打 15% 的折扣，就需要逐步调整计算，这考查了你串联百分比变化的能力。
7. Experimental Data and Averages | 实验数据与平均值
Scientists repeat measurements to improve reliability. The mean (average) of a data set is sum of values ÷ number of values. The median is the middle value when data is ordered, and the mode is the most frequent value. The range gives the spread: highest minus lowest.
科学家会重复测量以提高可靠性。一组数据的平均数（平均值）是 数值之和 ÷ 数值个数。中位数是排序后中间的值，众数是出现最频繁的值。极差表示离散程度：最大值减最小值。
5.3+5.4+5.1) ÷ 5 = 5.28 g, the median is 5.3 g, and the mode is 5.4 g. The range is 0.3 g.
例如，某实验得到质量数据 5.2 g, 5.4 g, 5.3 g, 5.4 g, 5.1 g。平均值为 (5.2+5.4+5.3+5.4+5.1) ÷ 5 = 5.28 g，中位数为 5.3 g，众数为 5.4 g。极差为 0.3 g。
8. Geometric Design and Area Calculation | 几何设计与面积计算
Many design and architecture questions involve composite shapes. You may need to find the area of a rectangular garden with a semicircular pond, then calculate the cost of paving or planting. Break the shape into simpler parts: rectangles, triangles, circles.
许多设计与建筑问题涉及组合图形。你可能需要计算一个带有半圆形池塘的矩形花园的面积，再计算铺砌或种植的成本。将图形分解为更简单的部分：矩形、三角形、圆形。
Area of circle = π × r²
1.5 m cut out. The grass area is 96 − π × 1.5² ≈ 96 − 7.07 = 88.93 m². If grass seed costs £2.50 per square metre, total cost is £222.33 (to the nearest penny).
一块 12 m × 8 m 的矩形草坪中挖去一个半径 1.5 m 的圆形花坛。草地面积为 96 − π × 1.5² ≈ 96 − 7.07 = 88.93 m²。如果草籽每平方米售价 2.50 英镑，总费用为 222.33 英镑（精确到便士）。
9. Scatter Graphs and Line of Best Fit | 散点图与最佳拟合线
A scatter graph is used to investigate the relationship between two variables, such as temperature and rate of reaction in chemistry. If points roughly follow a straight line, we say there is a linear correlation. You may be asked to draw a line of best fit by eye.
散点图用于研究两个变量之间的关系，例如化学中的温度和反应速率。如果点大致沿一条直线分布，我们称之为线性相关。题目可能会要求你用目测法画出一条最佳拟合线。
The line of best fit passes through the mean point of the data and has roughly equal numbers of points above and below it. You can then use the line to estimate an unknown value, either by interpolation (within the data range) or extrapolation (outside the range).
最佳拟合线通过数据的平均点，且线上方和下方的点数大致相等。然后你可以用这条直线来估计未知值，无论是在数据范围内的插值，还是超出范围的推估。
10. Financial Mathematics: Compound Interest | 金融数学：复利
Compound interest is a classic real-world application of percentage growth. The amount A after n years is given by A = P × (1 + r/100)n, where P is the principal and r is the annual interest rate.
复利是百分比增长在现实中的经典应用。n 年后的金额 A 由 A = P × (1 + r/100)n 给出，其中 P 为本金，r 为年利率。
A = P (1 + r/100)n
If £5000 is invested at 3% per annum compound interest, after 4 years it grows to 5000 × (1.03)4 = £ ompound interest over the same period, or to work backwards to find the principal or rate.
如果 5000 英镑按年利率 3% 的复利投资，4 年后将增长到 5000 × (1.03)4 = 5627.54 英镑。部分 CIE 题目还会要求比较同期的单利和复利，或者反向计算以求出本金或利率。
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