Interdisciplinary Problem-Solving Training | 跨学科综合题型训练

📚 Interdisciplinary Problem-Solving Training | 跨学科综合题型训练

Welcome to your Year 7 Edexcel Further Mathematics revision guide on interdisciplinary problems. Mathematics is not just a standalone subject — it is the language of science, engineering, economics and everyday life. In this article, you will practise applying your number, algebra, geometry and data skills to real-world contexts, from calculating speeds in physics to interpreting population density in geography. Each section presents a key topic with worked examples and explanations to build your confidence in solving unfamiliar problems.

欢迎来到七年级爱德思进阶数学跨学科问题复习指南。数学不仅是一门独立的学科——它更是科学、工程、经济学和日常生活的语言。在本文中，你将练习将数、代数、几何和数据处理技能应用于真实世界情境，从计算物理中的速度到解读地理中的人口密度。每一节都围绕一个关键主题，提供详细例题和解释，帮助你增强解决陌生问题的信心。

1. Speed, Distance and Time (Physics) | 速度、距离与时间（物理）

In physics, the link between speed, distance and time is captured by a simple formula. The average speed of an object is found by dividing the total distance travelled by the total time taken. We often write this as:

在物理学中，速度、距离和时间之间的关系可以用一个简单的公式表示。物体的平均速度等于总行驶距离除以总所用时间。我们通常写作：

Speed = Distance ÷ Time

速度 = 距离 ÷ 时间

The units must be consistent: if distance is measured in kilometres and time in hours, speed is in kilometres per hour (km/h).

单位必须保持一致：如果距离以千米为单位，时间以小时为单位，那么速度的单位就是千米每小时（km/h）。

Example 1: A train covers 210 km in 3 hours. Calculate its average speed.

例题 1：一列火车在 3 小时内行驶了 210 km。计算它的平均速度。

Write the formula: Speed = 210 ÷ 3 = 70 km/h.

写出公式：速度 = 210 ÷ 3 = 70 km/h。

Example 2: A cyclist rides at a steady speed of 15 km/h for 2 hours and 30 minutes. How far does she travel?

例题 2：一名自行车手以 15 km/h 的恒定速度骑行 2 小时 30 分钟。她总共骑行了多远？

First, convert 2 hours 30 minutes to 2.5 hours. Then use Distance = Speed × Time = 15 × 2.5 = 37.5 km.

首先，将 2 小时 30 分钟转换为 2.5 小时。然后使用距离 = 速度 × 时间 = 15 × 2.5 = 37.5 km。

When solving such problems, always check that your time is in the same unit as the speed’s time unit before multiplying.

解决这类问题时，务必在乘法之前检查时间单位是否与速度的时间单位一致。

2. Concentration and Mixing (Chemistry) | 浓度与混合（化学）

In chemistry, concentration describes how much solute is dissolved in a given volume of solvent. The basic concentration formula is:

在化学中，浓度描述的是在一定体积的溶剂中溶解了多少溶质。基本的浓度公式为：

Concentration = Mass of solute ÷ Volume of solution

浓度 = 溶质质量 ÷ 溶液体积

Common units are grams per litre (g/L).

常用单位是克每升（g/L）。

Example: A scientist dissolves 12 g of salt in water to make 400 mL of solution. Find the concentration in g/L.

例题：一位科学家将 12 g 盐溶于水，配制成 400 mL 溶液。计算浓度（单位 g/L）。

First, convert 400 mL to litres: 400 mL = 0.4 L. Then concentration = 12 ÷ 0.4 = 30 g/L.

首先，将 400 mL 转换为升：400 mL = 0.4 L。然后浓度 = 12 ÷ 0.4 = 30 g/L。

If you were told to prepare 250 mL of a 8 g/L sugar solution, you would calculate the mass of sugar needed as: Mass = Concentration × Volume = 8 × 0.25 = 2 g.

如果要求你配制 250 mL 的 8 g/L 糖溶液，你需要计算糖的质量：质量 = 浓度 × 体积 = 8 × 0.25 = 2 g。

3. Map Scales and Bearings (Geography) | 地图比例尺与方位角（地理）

Maps use ratios to shrink real-world distances. A scale of 1 : 50 000 means that 1 cm on the map represents 50 000 cm (or 0.5 km) on the ground.

地图利用比例来缩小真实世界的距离。比例尺 1 : 50 000 表示地图上的 1 cm 代表地面上的 50 000 cm（或 0.5 km）。

Example: On a 1 : 25 000 map, two villages are 8 cm apart. What is the actual distance in kilometres?

例题：在一张 1 : 25 000 的地图上，两个村庄相距 8 cm。实际距离是多少千米？

Map distance in cm: 8 cm. Actual distance in cm = 8 × 25 000 = 200 000 cm. Convert to km: 200 000 ÷ 100 000 = 2 km.

地图距离：8 cm。实际距离（cm）= 8 × 25 000 = 200 000 cm。换算成 km：200 000 ÷ 100 000 = 2 km。

You can also use the formula: Actual distance = Map distance × Scale factor, taking care to convert units.

你也可以使用公式：实际距离 = 地图距离 × 比例因子，注意单位换算。

Bearings are measured clockwise from north and expressed as three-digit angles. If a bearing is 045°, it points north‑east. Distances and bearings combine geometry with ratio; use a ruler and a protractor to measure, then apply the scale.

方位角是从正北方向顺时针量度，并用三位数表示的角度。如果方位角为 045°，则指向东北方向。距离和方位角将几何与比例结合起来；可用尺子和量角器测量，再依据比例尺计算。

4. Population Density and Area (Geography) | 人口密度与面积（地理）

Population density tells us how crowded a place is. It is calculated as:

人口密度告诉我们一个地方的拥挤程度。它的计算公式为：

Population Density = Total Population ÷ Land Area

人口密度 = 总人口 ÷ 土地面积

If the area is in square kilometres (km²), density is expressed as people per km².

如果面积以平方千米（km²）为单位，密度就用每平方千米人数表示。

Example: A city has 3.6 million people living in a region of area 1200 km². Find the population density.

例题：一个城市有 360 万人口，居住在一片 1200 km² 的区域。计算人口密度。

3.6 million = 3 600 000. Density = 3 600 000 ÷ 1200 = 3000 people/km².

360 万 = 3 600 000。密度 = 3 600 000 ÷ 1200 = 3000 人/km²。

If you know the density and area, you can find the total population: Population = Density × Area. Reverse problems help you check your understanding.

如果已知密度和面积，就可以求总人口：人口 = 密度 × 面积。反向问题有助于检验你的理解。

5. Simple Interest and Savings (Economics) | 单利与储蓄（经济学）

When you put money in a savings account, it earns interest. Simple interest is calculated only on the original principal. The formula is:

当你把钱存入储蓄账户时，它会赚取利息。单利只按原始本金计算。公式为：

Interest (I) = Principal (P) × Rate (r) × Time (t)

利息 (I) = 本金 (P) × 利率 (r) × 时间 (t)

Rate is usually given as a percentage per year, and time is in years.

利率通常以年百分数表示，时间以年为单位。

Example: You deposit £400 in a bank offering 3% simple interest per year. How much interest will you earn after 5 years?

例题：你将 400 英镑存入一家银行，年单利为 3%。5 年后你将获得多少利息？

Convert the rate to a decimal: 3% = 0.03. I = 400 × 0.03 × 5 = £60. The total amount after 5 years is £400 + £60 = £460.

将利率转换为小数：3% = 0.03。I = 400 × 0.03 × 5 = 60 英镑。5 年后的总金额为 400 + 60 = 460 英镑。

If you need to find the rate when you know the interest, principal and time, rearrange the formula: r = I ÷ (P × t).

如果已知利息、本金和时间，需要求利率，可以重新整理公式：r = I ÷ (P × t)。

6. Data Handling in Science Experiments | 科学实验中的数据处理

After carrying out an experiment, you often need to organise results in a table, find the mean, and draw a graph. This uses your statistics skills.

完成实验后，你常常需要将结果整理成一个表格，计算平均值，并绘制图表。这需要运用你的统计技能。

Example: A student measures the extension of a spring as she adds weights. The loads (in N) are 0, 1, 2, 3, 4, 5 and the extensions (in cm) are 0, 2, 4, 5.8, 8.2, 10. Find the mean extension and plot a scatter graph.

例题：一名学生测量弹簧在增加重物时的伸长量。负载（单位 N）为 0, 1, 2, 3, 4, 5，伸长量（单位 cm）为 0, 2, 4, 5.8, 8.2, 10。计算平均伸长量并绘制散点图。

Mean extension = (0 + 2 + 4 + 5.8 + 8.2 + 10) ÷ 6 = 30 ÷ 6 = 5 cm. The line of best fit shows that extension is roughly proportional to load, which is Hooke’s Law.

平均伸长量 = (0 + 2 + 4 + 5.8 + 8.2 + 10) ÷ 6 = 30 ÷ 6 = 5 cm。最佳拟合线表明伸长量大致与负载成正比，这就是胡克定律。

You must also identify any anomalous results. For example, if one point is far from the line of best fit, you might have made a measurement error and should consider repeating that part of the experiment.

你还必须识别任何异常数据点。例如，如果某个点远离最佳拟合线，可能是测量出现了错误，应考虑重做该部分的实验。

7. Unit Conversions in Measurement | 测量中的单位换算

Science and engineering problems often require switching between units. You should be confident with metric conversions:

科学和工程问题往往需要单位之间的转换。你应该熟练掌握公制单位的换算：

1 km = 1000 m
1 m = 100 cm
1 cm = 10 mm
1 kg = 1000 g
1 litre = 1000 mL
1 tonne = 1000 kg

1 km = 1000 m
1 m = 100 cm
1 cm = 10 mm
1 kg = 1000 g
1 升 = 1000 mL
1 吨 = 1000 kg

Example: A rectangular garden measures 0.05 km by 30 m. Find its area in m².

例题：一个长方形花园的长为 0.05 km，宽为 30 m。求它的面积，单位用 m²。

Convert 0.05 km to m: 0.05 × 1000 = 50 m. Area = length × width = 50 × 30 = 1500 m².

将 0.05 km 转换为 m：0.05 × 1000 = 50 m。面积 = 长 × 宽 = 50 × 30 = 1500 m²。

When converting square units, remember that 1 m² = 10 000 cm², because you must square the linear conversion factor (100 × 100).

转换平方单位时，记住 1 m² = 10 000 cm²，因为你需要将线性换算系数进行平方 (100 × 100)。

8. Geometric Shapes in Architecture | 建筑中的几何形状

Architects use geometry to design buildings. Understanding area and perimeter helps in calculating materials and costs.

建筑师运用几何学来设计建筑。理解面积和周长有助于计算材料和成本。

Example: A floor is made up of a square of side 6 m and a semicircle of diameter 6 m attached to one side. Find the total floor area.

例题：一个地板由一个边长为 6 m 的正方形和一个直径 6 m 的半圆形（附着在正方形的一条边上）组成。求地板的总面积。

Square area = 6 × 6 = 36 m². Radius of semicircle = 3 m. Area of full circle = π × 3² = π × 9. Use π ≈ 3.14: circle area ≈ 28.26 m². Semicircle area = 28.26 ÷ 2 = 14.13 m². Total area ≈ 36 + 14.13 = 50.13 m².

正方形面积 = 6 × 6 = 36 m²。半圆的半径 = 3 m。整个圆的面积 = π × 3² = π × 9。取 π ≈ 3.14：圆的面积 ≈ 28.26 m²。半圆面积 = 28.26 ÷ 2 = 14.13 m²。总面积 ≈ 36 + 14.13 = 50.13 m²。

If tiles cost £20 per m², the approximate cost is 50.13 × 20 ≈ £1002.60. Always show your working when rounding.

如果瓷砖每平方米 £20，那么大致成本为 50.13 × 20 ≈ £1002.60。在取近似值时，务必展示你的计算步骤。

9. Creating Graphs with Technology | 运用技术创建图表

Spreadsheets are powerful tools for displaying data. In Year 7 Further Mathematics, you may be asked to interpret graphs produced by software.

电子表格是展示数据的强大工具。在七年级进阶数学中，你可能会被要求解读由软件生成的图表。

Example: A spreadsheet line graph shows the temperature of a chemical mixture every minute. The temperatures recorded are: 20°C, 22°C, 25°C, 29°C, 34°C, 40°C. Describe the trend and estimate when the temperature reached 30°C.

例题：电子表格中的折线图显示了每分钟化学混合物的温度。记录的温度为：20°C, 22°C, 25°C, 29°C, 34°C, 40°C。描述变化趋势并估计温度达到 30°C 的时间。

The temperature rises; the rise is not constant but increasing. Between minute 3 (29°C) and minute 4 (34°C), the temperature passes 30°C. Interpolating: an increase of 5°C in one minute means roughly 1°C every 0.2 minutes. From 29°C to 30°C is a 1°C rise, so about 0.2 minutes after minute 3, i.e. at 3.2 minutes.

温度呈上升趋势，且上升幅度在逐渐增大。在第 3 分钟（29°C）和第 4 分钟（34°C）之间，温度超过了 30°C。进行内插：一分钟内上升 5°C 意味着每 0.2 分钟上升约 1°C。从 29°C 升到 30°C 需要 1°C 的升幅，因此大约在第 3 分钟后的 0.2 分钟，即 3.2 分钟时达到 30°C。

Reading values between data points is called interpolation; extending beyond the data is extrapolation, which is less reliable.

在已有数据点之间读取数值称为内插法；将趋势延伸到数据范围之外称为外推法，外推法的可靠性较低。

10. Solving Word Problems with Equations | 运用方程解应用题

Many real-life situations can be modelled with linear equations. The key is to define a variable, translate the description into an equation, solve it and then answer the question.

许多现实生活情境都可以用线性方程来建模。关键在于定义一个变量，将文字描述转化为方程，解方程，然后回答问题。

Example: Three identical boxes and a 5 kg weight have a total mass of 23 kg. Find the mass of one box.

例题：三个相同的箱子和一个 5 kg 的砝码总质量为 23 kg。求一个箱子的质量。

Let the mass of one box be b kg. Equation: 3b + 5 = 23. Subtract 5: 3b = 18. Divide by 3: b = 6 kg.

设一个箱子的质量为 b kg。方程：3b + 5 = 23。两边减 5：3b = 18。两边除以 3：b = 6 kg。

Check: 3 × 6 + 5 = 18 + 5 = 23. Correct. Always verify your answer in the context of the problem.

检验：3 × 6 + 5 = 18 + 5 = 23。正确。务必在题目情境下检验你的答案。

11. Percentages in Discounts and Profit | 折扣与利润中的百分数

Percentage calculations are used constantly in shops and business. Whether calculating sale prices or profit margins, understanding percent increase and decrease is vital.

百分数计算在商店和商业中经常用到。无论是计算打折价格还是利润率，理解百分数的增加和减少都至关重要。

Example: A jacket originally costs £80. In a sale, it is reduced by 15%. What is the sale price?

例题：一件夹克原价 80 英镑，打折 15%。打折后的价格是多少？

15% of £80 = 0.15 × 80 = £12. Sale price = £80 − £12 = £68. Alternatively, you keep 85% of the price: 0.85 × 80 = £68.

80 英镑的 15% = 0.15 × 80 = 12 英镑。打折后的价格 = 80 − 12 = 68 英镑。另一种方法：保持原价的 85%：0.85 × 80 = 68 英镑。

If a shopkeeper buys a toy for £15 and sells it for £21, the percentage profit is ((21 − 15) ÷ 15) × 100% = (6 ÷ 15) × 100% = 40%.

如果店主以 15 英镑买进一件玩具，以 21 英镑卖出，那么利润百分比为 ((21 − 15) ÷ 15) × 100% = (6 ÷ 15) × 100% = 40%。

12. Combining Skills: A Mixed Practice | 综合练习

In examinations, you might face problems that pull together several mathematical skills. Below is a multi‑step question that integrates ratio, area, speed and unit conversion.

在考试中，你可能会遇到需要综合运用多种数学技能的问题。下面是一个多步骤问题，整合了比例、面积、速度和单位换算。

Problem: A farmer’s field is in the shape of a rectangle 400 m by 250 m. The farmer spreads fertiliser at a rate of 3 kg per 100 m². The fertiliser costs £1.20 per kg. A tractor spreads fertiliser at a steady speed of 5 km/h and the spreader has a width of 2 m. How long will it take in hours to cover the whole field?

问题：一位农民的长方形田地尺寸为 400 m × 250 m。他按每 100 m² 3 kg 的比率撒肥料。肥料价格为每 kg £1.20。一台拖拉机以 5 km/h 的恒定速度撒肥料，撒肥机的宽度为 2 m。覆盖整块田地需要多少小时？

Field area = 400 × 250 = 100 000 m². Fertiliser needed: (100 000 ÷ 100) × 3 = 1000 × 3 = 3000 kg. Cost = 3000 × 1.20 = £3600. For time, the spreader covers a strip every metre the tractor moves forward. The area to cover is 100 000 m², and the strip area per metre travelled is 2 m². Total distance the tractor must travel = 100 000 ÷ 2 = 50 000 m. Convert to km: 50 000 ÷ 1000 = 50 km. Time = Distance ÷ Speed = 50 ÷ 5 = 10 hours.

田地面积 = 400 × 250 = 100 000 m²。所需肥料：(100 000 ÷ 100) × 3 = 1000 × 3 = 3000 kg。成本 = 3000 × 1.20 = £3600。对于时间：拖拉机每前进 1 m，撒肥机覆盖的条状面积为 2 m²。需覆盖的总面积为 100 000 m²，每前进 1 m 覆盖 2 m²，所以拖拉机必须行驶的总距离 = 100 000 ÷ 2 = 50 000 m。转换为 km：50 000 ÷ 1000 = 50 km。时间 = 距离 ÷ 速度 = 50 ÷ 5 = 10 小时。

This problem demonstrates how different mathematics topics connect; practising such challenges will strengthen your skills and prepare you for advanced study.

这个问题展示了不同数学主题如何相互关联；练习这类挑战性的题目将强化你的技能，为更高层次的学习做好准备。

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Interdisciplinary Problem-Solving Training | 跨学科综合题型训练