📚 Interdisciplinary Maths Problem-Solving Training | 跨学科综合题型训练
Mathematics is not an isolated subject; it is deeply woven into science, geography, history, business, art, sports and many other fields. This article presents a series of cross-curricular problem-solving exercises designed to strengthen your mathematical skills while showing how numbers, shapes, data and logic are used in real-world contexts. You will practise unit conversions, map scales, timelines, percentages, symmetry, statistics and volume calculations through scenarios drawn from daily life and other subjects. By the end, you will be better prepared for the kinds of integrated questions that appear in the CAIE Year 7 Mathematics curriculum and beyond.
数学并非一门孤立的学科，它深深融入科学、地理、历史、商业、艺术、体育等诸多领域。本文通过一系列跨学科综合题型训练，帮助你在真实情境中巩固数字、图形、数据和逻辑能力。你将练习单位换算、地图比例尺、时间线、百分数、对称、统计和体积计算。通过这些练习，你能更从容地应对CAIE Year 7数学课程中越来越常见的综合类问题。
1. Understanding Interdisciplinary Questions | 理解跨学科问题
Interdisciplinary questions require you to take a mathematical concept and apply it in a subject you might not expect, such as calculating the age of a fossil using negative numbers or finding the average rainfall from a climate graph. These challenges build higher-order thinking and reveal the true power of mathematics as a universal language.
跨学科问题要求你将数学概念应用到意想不到的学科中，例如利用负数计算化石的年代，或从气候图中求出平均降雨量。这类训练能培养高阶思维，并让你看到数学作为通用语言的真正力量。
When you solve an interdisciplinary problem, first identify the mathematical tools you will need—addition, subtraction, fractions, ratio, data handling, or measurement. Then carefully read the context, extract the numerical information, and perform the calculations step by step. Finally, check whether your answer makes sense within the original scenario.
解答跨学科问题时，先确定需要用到的数学工具——加减法、分数、比和比例、数据处理或测量。然后仔细阅读情境，提取数值信息，再一步步计算。最后，检验答案在原情境中是否合理。
The examples that follow are grouped by theme. Each section begins with a short real-world scenario, followed by a clear demonstration of the mathematics behind it. You are encouraged to attempt each question on your own before reading the solution.
下文按主题分组，每节先给出一个简短的现实情境，再清晰展示背后的数学。建议你在阅读解答之前，先试着自己完成每个问题。
2. Maths and Science: Measurements and Conversions | 数学与科学：测量与换算
In science lessons you often need to record lengths, masses, volumes, and temperatures, then convert between different metric units. For instance, a plant grows 15 mm in a week; express this growth in centimetres, and then in metres. The key is remembering the standard prefixes: kilo- (×1000), centi- (÷100), and milli- (÷1000).
科学课上经常需要记录长度、质量、体积和温度，并在不同公制单位之间进行换算。例如，一株植物一周长高15 mm；用厘米表示，再用米表示。关键是要记住标准词头：kilo-（×1000）、centi-（÷100）、milli-（÷1000）。
| Length unit | Equivalent |
|---|---|
| 1 kilometre (km) | 1000 metres (m) |
| 1 metre (m) | 100 centimetres (cm) |
| 1 centimetre (cm) | 10 millimetres (mm) |
Using this table, 15 mm = 15 ÷ 10 = 1.5 cm, and 1.5 cm = 1.5 ÷ 100 = 0.015 m. Conversions are simply a matter of multiplying or dividing by powers of ten.
利用上表，15 mm = 15 ÷ 10 = 1.5 cm，而 1.5 cm = 1.5 ÷ 100 = 0.015 m。换算其实就是乘以或除以10的幂。
Speed is another common link between maths and science. If a cyclist travels 9 kilometres in 30 minutes, what is her average speed in kilometres per hour? First convert 30 minutes to hours: 30 min = 0.5 h. Then average speed = distance ÷ time = 9 km ÷ 0.5 h = 18 km/h.
速度是数学与科学的另一常见联系。一位自行车骑手30分钟骑行9公里，她的平均速度是多少公里/小时？首先把30分钟化为小时：30 min = 0.5 h。然后平均速度 = 距离 ÷ 时间 = 9 km ÷ 0.5 h = 18 km/h。
Now try this: a beaker contains 0.25 litres of water. How many millilitres is this? (Remember 1 L = 1000 mL.)
现在请你试做：一个烧杯中有0.25升水。这是多少毫升？（记住1 L = 1000 mL。）
Solution: 0.25 × 1000 = 250 mL.
解答：0.25 × 1000 = 250 mL。
3. Geography and Maths: Map Scales and Coordinates | 地理与数学：地图比例尺与坐标
Map scales allow you to translate a small map distance into the real distance on the ground. A typical scale might be 1 : 50 000, meaning 1 cm on the map represents 50 000 cm in reality. Because 50 000 cm = 500 m = 0.5 km, you can quickly work out that 4 cm on the map equals 4 × 0.5 km = 2 km.
地图比例尺能将图上距离转化为实际地面距离。一个典型比例尺是 1 : 50 000，即图上1 cm代表实际50 000 cm。因为50 000 cm = 500 m = 0.5 km，你可以迅速算出图上4 cm等于4 × 0.5 km = 2 km。
Always write the full working step by step: map distance in cm × scale number gives real distance in cm; then divide by 100 to get metres, and divide by 1000 if you need kilometres. This process avoids mistakes with zeros.
务必一步步写出完整过程：图上厘米数 × 比例尺分母得到实际厘米数；再除以100得到米，如果需要再除以1000得到千米。这样可以避免数零出错。
Coordinates and grid references are often used together with maps. A point located at (3, 5) on a coordinate grid might represent the position of a mountain peak on a simplified topographic map. If the grid spacing represents 1 km, the distance between (3, 5) and (3, 9) is found by the difference in y-coordinates: 9 − 5 = 4 km.
坐标与网格参考常和地图一起使用。在坐标网格上，（3, 5）可能代表简化地形图上的一个山峰位置。如果网格间距代表1 km，点（3, 5）与（3, 9）之间的距离由y坐标差求得：9 − 5 = 4 km。
Question: On a map with scale 1 : 250 000, two towns are 6 cm apart. What is the actual straight-line distance in kilometres?
问题：在一幅比例尺为 1 : 250 000 的地图上，两个城镇相距6 cm。实际的直线距离是多少千米？
Solution: 6 × 250 000 = 1 500 000 cm → 15 000 m → 15 km.
解答：6 × 250 000 = 1 500 000 cm → 15 000 m → 15 km。
4. History and the Number Line: Timelines and BC/AD | 历史与数轴：时间线与公元前/公元后
In history, you encounter dates before and after the year 0. The BC (Before Christ) and AD (Anno Domini) system can be modelled on a number line, where BC years are negative numbers and AD years are positive. For example, 300 BC is represented as −300, and AD 150 is +150. The time gap between them is 150 − (−300) = 450 years.
历史中你会遇到公元前后年份。BC（公元前）与AD（公元）体系可以用数轴表示，其中BC年份为负数，AD年份为正数。例如，公元前300年表示为−300，公元150年表示为+150。二者之间的时间间隔为150 − (−300) = 450 年。
This idea is extremely useful when comparing lifespans of historical figures or the duration of empires. A mathematician born in 287 BC and died in 212 BC was alive for (−212) − (−287) = 75 years.
这种思路在比较历史人物寿命或帝国存续时间时非常有用。一位数学家生于公元前287年，卒于公元前212年，他活了 (−212) − (−287) = 75 岁。
Using a number line also helps solve problems like: the construction of a temple started in 450 BC and finished 38 years later. In which year was it completed? Working: starting at −450, adding 38 gives −450 + 38 = −412, i.e. 412 BC.
用数轴也可以解决此类问题：某神庙于公元前450年开始建造，38年后完工，是哪一年？计算：从−450开始，加上38得到−450 + 38 = −412，即公元前412年。
Practice: Alexander the Great died in 323 BC. If he lived for 33 years, in which year was he born?
练习：亚历山大大帝死于公元前323年，如果他活了33岁，他是哪一年出生的？
Solution: Died −323. Birth year = −323 − 33 = −356, i.e. 356 BC.
解答：逝世于−323。出生年份 = −323 − 33 = −356，即公元前356年。
5. Business Maths: Percentages, Profit and Discount | 商业数学：百分数、利润与折扣
Whether you are running a school tuck shop or analysing a shopping bill, percentages are everywhere. A discount of 20% on a cricket bat that originally costs £60 reduces the price by 0.20 × 60 = £12, so the sale price is £60 − £12 = £48. Alternatively, you can multiply the original price by (100% − 20%) = 80% = 0.8 to get £60 × 0.8 = £48 directly.
无论你是在经营校园小卖部还是在分析购物账单，百分数无处不在。一个原价£60的板球棒打八折（20% 折扣），降价0.20 × 60 = £12，因此售价为£60 − £12 = £48。你也可以直接用原价乘以（100% − 20%）= 80% = 0.8，立即得到£60 × 0.8 = £48。
Profit and loss are also expressed as percentages of the cost price. If a bookseller buys a set of revision guides for £25 and sells them for £35, the profit is £10. The percentage profit based on the cost price is (10 ÷ 25) × 100% = 40%.
利润和亏损也用成本价的百分数表示。如果一位书商用£25购入一套复习指南并以£35售出，利润为£10。基于成本价的利润率是(10 ÷ 25) × 100% = 40%。
You can also find the original price before a discount. If a backpack is sold for £36 after a 25% reduction, what was the original price? The sale price represents 75% of the original, so original × 0.75 = 36, thus original price = 36 ÷ 0.75 = £48.
你也可以反过来求折扣前的原价。一个背包降价25%后售价£36，原价是多少？售价相当于原价的75%，因此原价 × 0.75 = 36，原价 = 36 ÷ 0.75 = £48。
Now try: A video game console is bought for £180 and sold at a 15% profit. What is the selling price?
再试一题：一台游戏机以£180购入，加价15%出售。售价是多少？
Solution: Profit = 0.15 × 180 = £27, selling price = 180 + 27 = £207. (Or 180 × 1.15 = 207.)
解答：利润 = 0.15 × 180 = £27，售价 = 180 + 27 = £207。（或 180 × 1.15 = 207。）
6. Art and Symmetry: Reflection, Rotation and Tessellation | 艺术与对称：反射、旋转与密铺
Symmetry is a mathematical concept that artists, architects and designers use every day. A shape has reflection symmetry if you can draw a mirror line (line of symmetry) so that one half is the exact reflection of the other. An equilateral triangle has 3 lines of symmetry, a square has 4, and a regular pentagon has 5.
对称是一个数学概念，艺术家、建筑师和设计师每天都在使用。如果你能画出一条镜像线（对称轴），使得图形的一半恰好是另一半的反射，则该图形具有反射对称。等边三角形有3条对称轴，正方形有4条，正五边形有5条。
Rotational symmetry means the shape looks exactly the same after a turn of less than 360°. The order of rotational symmetry tells you how many times it matches itself in a full turn. For example, a square has rotational symmetry of order 4 (rotations of 90°, 180°, 270° and 360°).
旋转对称意味着图形旋转小于360°后与自身完全重合。旋转对称的阶数告诉你在一整圈旋转中它能与自身重合多少次。例如，正方形具有4阶旋转对称（旋转90°、180°、270°和360°）。
Tessellation is covering a plane with repeated shapes without any gaps or overlaps. Regular hexagons tessellate, as do equilateral triangles and squares. Artists like M.C. Escher created stunning tessellations using animal shapes, all built from underlying mathematical grids.
密铺指的是用重复图形无缝隙、不重叠地铺满平面。正六边形可以密铺，等边三角形和正方形也可以。像M.C.埃舍尔这样的艺术家利用动物图形创作了令人惊叹的密铺画，它们全部建立在数学网格之上。
Activity: Draw a capital letter “H”. How many lines of symmetry does it have? (Answer: 2 – one horizontal and one vertical.) Does it have rotational symmetry? (Yes, order 2.)
活动：写出大写字母“H”。它有多少条对称轴？（答案：2条——一条水平和一条竖直。）它具有旋转对称吗？（有，2阶。）
7. Sports Statistics: Mean, Median and Mode | 体育统计：平均数、中位数与众数
Statistics helps coaches and fans analyse player performance. Three common averages are the mean (the sum of all values divided by the number of values), the median (the middle value when ordered), and the mode (the value that appears most often).
统计学帮助教练和球迷分析球员表现。三种常见的平均数是：平均数（所有数值之和除以数值个数）、中位数（排序后正中间的数值）和众数（出现次数最多的值）。
Suppose a basketball player scores the following points in five matches: 18, 22, 18, 24, 18. The mean is (18+22+18+24+18) ÷ 5 = 100 ÷ 5 = 20 points. The median, after sorting: 18, 18, 18, 22, 24, is 18. The mode is 18 (it appears three times).
假设一位篮球运动员五场比赛分别得分为：18, 22, 18, 24, 18。平均数 = (18+22+18+24+18) ÷ 5 = 100 ÷ 5 = 20 分。排序后：18, 18, 18, 22, 24，中位数是18。众数也是18（出现三次）。
Different averages give different insights. The mean takes all scores into account, the median is not affected by an extremely high or low score, and the mode shows the most consistent performance.
不同的平均数各有特色。平均数考虑了所有得分；中位数不受极端高分或低分的影响；众数则体现出最稳定的表现。
Try this: A goalkeeper makes these saves in 8 matches: 3, 5, 2, 5, 4, 5, 3, 5. Find the mean, median and mode.
试一试：一位守门员在8场比赛中的扑救次数为：3, 5, 2, 5, 4, 5, 3, 5。求平均数、中位数和众数。
Solution: Sum = 3+5+2+5+4+5+3+5 = 32. Mean = 32 ÷ 8 = 4. Sorted: 2,3,3,4,5,5,5,5. Median = (4+5) ÷ 2 = 4.5. Mode = 5.
解答：和 = 3+5+2+5+4+5+3+5 = 32。平均数 = 32 ÷ 8 = 4。排序：2,3,3,4,5,5,5,5。中位数 = (4+5) ÷ 2 = 4.5。众数 = 5。
8. Environmental Data: Graphs and Charts | 环境数据：图表与统计图
Environmental scientists collect data on temperature, rainfall, air quality, and species populations. They then present the data using bar charts, line graphs, pictograms and pie charts. In Year 7, you are expected to interpret such diagrams and extract numerical information.
环境科学家会收集温度、降雨量、空气质量和物种数量等数据，然后用条形图、折线图、象形图和饼图来呈现。Year 7要求你能够解读这些图表并提取数值信息。
Here is a table showing the average maximum temperature (in °C) in a city over one week:
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Temp (°C) | 12 | 14 | 11 | 13 | 15 | 18 | 17 |
You could plot these points on a line graph with days on the horizontal axis and temperature on the vertical axis. Then calculate the mean temperature for the week: (12+14+11+13+15+18+17) ÷ 7 = 100 ÷ 7 ≈ 14.3 °C.
你可以把这些点画在折线图上，横轴为星期，纵轴为温度。然后计算本周平均最高气温：(12+14+11+13+15+18+17) ÷ 7 = 100 ÷ 7 ≈ 14.3 °C。
If you then collected rainfall data, you could draw a dual bar chart or a combined graph to compare temperature and rainfall. Analysing data graphically helps spot trends, such as rising temperatures towards the weekend.
如果再收集降雨量数据，你可以绘制双条形图或组合图来比较气温和降水。以图形方式分析数据有助于发现趋势，例如气温在周末前逐渐升高。
9. Engineering and Volume: Cubes and Cuboids | 工程与体积：立方体与长方体
Engineers use volume to design containers, water tanks and packaging. The volume of a cuboid (rectangular box) is found by multiplying length × width × height. All dimensions must be in the same unit, and the result is in cubic units, such as cm³ or m³.
工程师利用体积来设计容器、水箱和包装。长方体的体积 = 长 × 宽 × 高。所有尺寸必须使用同一单位，结果以立方单位表示，如 cm³ 或 m³。
For example, a box measuring 20 cm by 10 cm by 5 cm has a volume of 20 × 10 × 5 = 1000 cm³. Since 1 litre = 1000 cm³, this box can hold exactly 1 litre of liquid.
例如，一个尺寸为20 cm × 10 cm × 5 cm的盒子体积为 20 × 10 × 5 = 1000 cm³。因为1升 = 1000 cm³，这个盒子刚好可以容纳1升液体。
The relationship between volume and capacity is crucial: 1000 cm³ = 1 litre, and 1 m³ = 1 000 000 cm³ = 1000 litres. You can use these conversions in flood-control projects or aquarium design.
体积和容量之间的关系至关重要：1000 cm³ = 1升，1 m³ = 1 000 000 cm³ = 1000 升。你可以在防洪项目或水族箱设计中运用这些换算。
Problem: A fish tank is a cuboid of length 40 cm, width 25 cm and height 30 cm. What is its volume in cm³? How many litres of water can it hold when full?
问题：一个鱼缸是长方体，长40 cm，宽25 cm，高30 cm。它的体积是多少 cm³？装满水时能容纳多少升？
Solution: Volume = 40 × 25 × 30 = 30 000 cm³. Litres = 30 000 ÷ 1000 = 30 L.
解答：体积 = 40 × 25 × 30 = 30 000 cm³。升数 = 30 000 ÷ 1000 = 30 L。
10. Mixed Interdisciplinary Challenge | 综合跨学科挑战
Now combine your skills. Read the scenario below and answer all parts.
现在，综合运用你的技能。阅读以下情境并回答所有问题。
Scenario: An explorer plans a trek between two camps. The map scale is 1 : 25 000, and the camps are 8 cm apart on the map. She walks at an average speed of 4 km/h. She carries a rectangular water container with internal dimensions 30 cm × 20 cm × 15 cm. She drinks about 0.5 litres of water per hour. Before the trip, she lists her last
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