📚 Year 7 CIE Maths: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练
In Year 7 CIE Maths, you are often expected to step outside pure number work and apply your skills to real-world challenges that cross subject borders. This article brings together a rich set of interdisciplinary problems, blending science, geography, economics, and design. Working through these examples will sharpen your ability to spot the maths hidden in everyday situations and build lasting confidence.
在 Year 7 CIE 数学中，你常常需要跳出纯数字运算，将所学技能应用到跨越学科边界的实际挑战中。本文汇集了一系列丰富的跨学科问题，融合了科学、地理、经济学和设计。通过这些练习，你将练就一双从日常生活情境中发现数学的眼睛，并建立持久的解题信心。
1. Understanding Interdisciplinary Maths | 理解跨学科数学
Interdisciplinary problems ask you to use maths in contexts such as mixing ingredients for a science experiment, reading a map scale, or analysing data from a survey. The key skill is recognising which mathematical operation or concept fits the situation and then applying it correctly. These questions often appear in CIE assessments because they mirror how we use maths in adult life.
跨学科问题要求你在科学实验配制混合物、阅读地图比例尺或分析调查数据等情境中运用数学。关键技能在于识别适合该情境的数学运算或概念，然后正确应用。这类题目经常出现在 CIE 评估中，因为它们反映了我们成年后在现实生活中使用数学的方式。
When you encounter a mix of words, numbers and units from another subject, stay calm. Underline the digits and units, work out what the question is asking you to find, and then choose your tools: fractions, percentages, averages, conversion factors, or geometry. Every interdisciplinary problem is just a story with maths at its heart.
当你看到文字、数字和其他学科的单位混在一起时，请保持冷静。圈出数字和单位，想清楚题目要你求出什么，然后选择工具：分数、百分比、平均数、换算系数或几何。每一个跨学科问题其实都是一个以数学为核心的故事。
2. Maths Meets Science: Unit Conversions and Measurements | 数学遇见科学：单位换算与测量
In science lessons, you measure length, mass, volume and temperature. Converting between units is a fundamental maths skill. The metric system is based on powers of ten, so you multiply or divide by 10, 100 or 1000. For example, 1 cm = 10 mm, so to go from mm to cm you divide by 10. To change m to cm, you multiply by 100.
在科学课上，你会测量长度、质量、体积和温度。单位换算是基本的数学技能。公制系统以十的幂次为基础，因此你需要乘或除以 10、100 或 1000。例如，1 cm = 10 mm，要从毫米换算到厘米就除以 10；要把米换算为厘米则乘以 100。
Remembering a few key facts saves time: 1 kg = 1000 g, 1 litre = 1000 ml, 1 km = 1000 m, and 1 hour = 60 minutes. These appear in questions about plant growth, chemical solutions and human body measurements. Always write down the conversion factor before calculating.
记住几个关键数据能节省时间：1 kg = 1000 g，1 升 = 1000 ml，1 km = 1000 m，1 小时 = 60 分钟。这些会出现在植物生长、化学溶液和人体测量题目中。计算前务必将换算系数写下来。
Example: A beaker contains 0.45 litres of salt solution. The teacher asks for the volume in millilitres. Solution: 0.45 × 1000 = 450 ml. If the beaker originally held 500 ml, how much has been used? 500 – 450 = 50 ml.
示例：一个烧杯装有 0.45 升盐溶液。老师要求用毫升表示体积。解答：0.45 × 1000 = 450 ml。如果烧杯原先有 500 ml，用掉了多少？500 – 450 = 50 ml。
3. Geography and Scale Drawings | 地理与比例绘图
Maps use a scale to shrink real distances so they fit on paper. A scale of 1 : 50 000 means 1 cm on the map represents 50 000 cm on the ground. Since 100 cm = 1 m and 1000 m = 1 km, 50 000 cm = 500 m = 0.5 km. To find the actual distance, you multiply the map measurement by the scale factor and then convert to a sensible unit.
地图使用比例尺将实际距离缩小以便画在纸上。比例尺 1 : 50 000 表示图上 1 cm 代表实地 50 000 cm。因为 100 cm = 1 m，1000 m = 1 km，所以 50 000 cm = 500 m = 0.5 km。求实地距离时，用图上长度乘以比例因子，然后再换算成合适的单位。
You may also need to work backwards: given an actual distance, find the map distance. Divide the real distance by the scale factor. For instance, two landmarks are 3 km apart. On a 1 : 25 000 map, the distance in cm is (3 × 100 000) ÷ 25 000 = 12 cm. Always keep units consistent.
有时需要反向计算：已知实地距离，求图上距离。用实地距离除以比例因子。例如，两地标相距 3 km，在 1 : 25 000 的地图上，图上距离为 (3 × 100 000) ÷ 25 000 = 12 cm。始终要保持单位一致。
Practice: A hiking trail measures 14 cm on a 1 : 50 000 map. What is its true length in kilometres? Answer: 14 × 50 000 = 700 000 cm = 7 km.
练习：一条徒步路线在 1 : 50 000 地图上长 14 cm，实际距离是多少公里？答案：14 × 50 000 = 700 000 cm = 7 km。
4. Biology and Data Handling: Averages and Graphs | 生物与数据处理：平均数和图表
Biology experiments often produce lists of numbers – the heights of seedlings, the number of leaves per plant, or the heart rates of students after exercise. Maths helps you summarise this data. The mean is found by adding all values and dividing by the count. The mode is the value that appears most often, and the median is the middle value when the data is ordered.
生物实验常常产生一串数字——幼苗的高度、每株植物的叶片数量、或学生运动后的心率。数学能帮你概括这些数据。平均数（均值）由所有数值相加后除以个数得到。众数是最常出现的数值，中位数则是排序后位于中间的数值。
Displaying data on a bar chart or line graph makes trends visible. For example, a line graph showing the growth of a bean plant over seven days can tell you when growth speeded up. Always label axes and choose a sensible scale. In CIE papers, you may need to calculate the mean from a frequency table, so check carefully how many items are in each group.
将数据展示在条形图或折线图上能让趋势一目了然。例如，一张展示豆苗在七天内生长情况的折线图可以告诉你何时生长加速。务必给坐标轴添加标签并选择合适的刻度。在 CIE 试卷中，你可能需要根据频数表计算均值，因此要仔细检查每个组有多少数据。
Data set: Heights of 6 seedlings (cm): 4.5, 5.0, 4.5, 6.0, 5.5, 4.5. Mean = (4.5+5.0+4.5+6.0+5.5+4.5) ÷ 6 = 30 ÷ 6 = 5.0 cm. Mode = 4.5 cm. Median = 4.75 cm.
数据集：6 株幼苗的高度（cm）：4.5, 5.0, 4.5, 6.0, 5.5, 4.5。均值 = (4.5+5.0+4.5+6.0+5.5+4.5) ÷ 6 = 30 ÷ 6 = 5.0 cm。众数 = 4.5 cm。中位数 = 4.75 cm。
5. Physics: Speed, Distance, and Time | 物理：速度、距离和时间
The formula triangle for speed, distance and time is a classic cross-subject tool. Speed = Distance ÷ Time. If you know any two, you can calculate the third. Often the challenge is making sure the units match: if distance is in metres and time in seconds, speed will be in m/s. If distance is in km and time in hours, speed is in km/h.
速度、距离和时间的关系三角是经典的跨学科工具。速度 = 距离 ÷ 时间。只要知道其中两个量，就能求出第三个。常见挑战在于确保单位匹配：若距离以米为单位、时间以秒为单位，速度就是 m/s；若距离以 km 为单位、时间以小时为单位，速度就是 km/h。
When time is given in minutes, convert to hours by dividing by 60. For example, 30 minutes = 0.5 hours. You can also convert m/s to km/h by multiplying by 3.6, but at Year 7 level most problems will stick to simple conversions.
当时间以分钟给出时，除以 60 转为小时。例如，30 分钟 = 0.5 小时。你也可以将 m/s 乘以 3.6 转换成 km/h，不过在 Year 7 阶段，大部分题目只需简单换算。
Problem: A cyclist rides at 18 km/h for 40 minutes. How far does she travel? Convert 40 min to hours: 40 ÷ 60 = ⅔ hour (approximately 0.667 hours). Distance = 18 × ⅔ = 12 km.
问题：一位自行车手以 18 km/h 的速度骑行 40 分钟，她骑了多远？40 分钟换算：40 ÷ 60 = ⅔ 小时（约 0.667 小时）。距离 = 18 × ⅔ = 12 km。
Distance = Speed × Time
6. Economics: Profit, Loss, and Simple Interest | 经济：利润、亏损与简单利息
Business maths uses number skills in practical money situations. Profit occurs when the selling price is higher than the cost price; loss happens when it is lower. To express profit or loss as a percentage, use (Profit ÷ Cost Price) × 100%. This tells you how much return you made compared to your original outlay.
商业数学将数字技能运用到实际金钱情境中。当售价高于成本时产生利润，低于成本时则出现亏损。要以百分数表示利润或亏损，使用公式 (利润 ÷ 成本) × 100%。它能告诉你与原始投入相比获得了多少回报。
Simple interest is another real-world application: Interest = Principal × Rate × Time ÷ 100. Here the principal is the starting amount of money, rate is the annual interest percentage, and time is in years. Many CIE questions ask you to find the total amount after interest is added.
简单利息是另一个现实世界应用：利息 = 本金 × 利率 × 时间 ÷ 100。本金是初始金额，利率是年利率百分比，时间以年为单位。许多 CIE 题目会要求你求出加入利息后的总金额。
Example: A shopkeeper buys a basketball for £16 and sells it for £20. Profit = £4. Profit % = (4 ÷ 16) × 100 = 25%. If she deposits £300 in a bank at 4% simple interest for 5 years, interest = 300 × 4 × 5 ÷ 100 = £60. Total = £360.
示例：一位店主以 £16 购入一个篮球，以 £20 卖出。利润 = £4。利润百分比 = (4 ÷ 16) × 100 = 25%。如果她把 £300 存入银行，年利率 4%，存 5 年，利息 = 300 × 4 × 5 ÷ 100 = £60。总金额 = £360。
7. Art and Design: Symmetry, Tessellations, and Area | 艺术与设计：对称、镶嵌与面积
Geometry blends seamlessly with art and design. A tessellation is a pattern of shapes that fit together without gaps or overlaps. To check if a regular polygon tessellates, you look at its interior angle: it must divide exactly into 360°. For instance, a square (90°) and an equilateral triangle (60°) both tessellate because 360 ÷ 90 = 4 and 360 ÷ 60 = 6.
几何学与艺术设计无缝交融。镶嵌（密铺）是指形状之间无空隙、无重叠地拼合在一起的图案。要检验一个正多边形是否能镶嵌，只需看它的内角能否整除 360°。例如，正方形（90°）和等边三角形（60°）都可以，因为 360 ÷ 90 = 4，360 ÷ 60 = 6。
Line symmetry and rotational symmetry also appear in logos, wallpaper and architectural decorations. In maths, you can count lines of symmetry or figure out the order of rotational symmetry. Combined with area and perimeter calculations, these skills help designers estimate materials and costs.
线对称和旋转对称也出现在标志、壁纸和建筑装饰中。在数学里，你可以数出对称轴的条数或找出旋转对称的阶数。结合面积和周长计算，这些技能能帮助设计师估算材料和成本。
Task: A rectangular artboard is 60 cm by 40 cm. It will be covered with square ceramic tiles of side 5 cm. Area of board = 60 × 40 = 2400 cm². Area of one tile = 5² = 25 cm². Tiles needed = 2400 ÷ 25 = 96 tiles.
任务：一块矩形画板长 60 cm，宽 40 cm，将用边长 5 cm 的正方形瓷砖覆盖。画板面积 = 60 × 40 = 2400 cm²。一块瓷砖面积 = 5² = 25 cm²。所需瓷砖 = 2400 ÷ 25 = 96 块。
8. Real-World Word Problems: Combining Skills | 现实世界应用：综合技能训练
Now let’s look at problems that pull together several mathematical threads at once. Read each scenario carefully, extract the numbers and units, and then decide which operations you need.
现在来看看同时融合多个数学线索的综合问题。仔细阅读每个场景，提取数字和单位，然后决定需要哪些运算。
Problem A – Garden makeover: A rectangular garden is 12 m long and 9 m wide. The owner wants to build a fence around it and cover the ground with compost at a rate of 3 kg per m². a) Find the perimeter for fencing. b) Calculate the amount of compost needed.
问题 A – 花园改造：一个矩形花园长 12 m，宽 9 m。主人想建围栏并在地面铺堆肥，用量为每平方米 3 kg。a) 求围栏所需的周长。b) 计算需要的堆肥量。
Solution: Perimeter = 2 × (12 + 9) = 2 × 21 = 42 m. Area = 12 × 9 = 108 m². Compost = 108 × 3 = 324 kg.
解答：周长 = 2 × (12 + 9) = 2 × 21 = 42 m。面积 = 12 × 9 = 108 m²。堆肥 = 108 × 3 = 324 kg。
Problem B – School trip: A school hires a coach for a day trip to a castle 180 km away. The coach travels at an average speed of 60 km/h. The group spends 4 hours at the castle. If they leave at 8:00 am, at what time do they return? Assume no other stops.
问题 B – 学校旅行：一所学校租用大巴去 180 km 外的城堡一日游。大巴平均速度为 60 km/h。团队在城堡停留 4 小时。如果他们早上 8:00 出发，什么时候返回？假设无其他停留。
Solution: Travel time one way = 180 ÷ 60 = 3 hours. Total travel both ways = 6 hours. Total time = 6 + 4 = 10 hours. Return time = 8:00 am + 10 h = 6:00 pm.
解答：单程时间 = 180 ÷ 60 = 3 小时。往返 = 6 小时。总时间 = 6 + 4 = 10 小时。返回时间 = 早上 8:00 + 10 小时 = 傍晚 6:00。
Problem C – Pocket
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