📚 Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练
Mathematics is not only about numbers and equations in a textbook; it is a powerful tool used across many subjects. In Year 7 CCEA Further Mathematics, you will encounter interdisciplinary problem-solving questions that combine maths with science, geography, art, finance, sport, and more. Practising these helps you see how maths applies to real-world situations and strengthens your analytical thinking. This article presents a series of cross-curricular practice questions with step-by-step solutions to build your confidence.
数学不仅仅是课本中的数字和方程式；它也是许多学科中使用的强大工具。在七年级 CCEA 进阶数学中，你会遇到跨学科的应用题，将数学与科学、地理、艺术、金融、体育等结合起来。练习这些题目能帮助你理解数学在现实世界中的应用，并增强你的分析思维能力。本文提供一系列跨学科练习题并配有逐步解答，以建立你的信心。
1. Maths and Science: Speed, Distance and Time | 数学与科学：速度、距离和时间
In science, you often calculate speed using the formula: Speed = Distance ÷ Time. The units must be consistent – for example, if distance is in kilometres and time is in hours, the speed will be in kilometres per hour (km/h). Knowing how to rearrange the formula is useful: Distance = Speed × Time, Time = Distance ÷ Speed.
在科学中，经常使用公式：速度 = 距离 ÷ 时间。单位必须一致——例如，如果距离以公里为单位，时间以小时为单位，速度就是公里/小时。了解如何变换公式也很有用：距离 = 速度 × 时间，时间 = 距离 ÷ 速度。
Example problem: A car travels 165 km in 2 hours and 30 minutes. Calculate its average speed in km/h. First, convert 30 minutes into hours: 30 min = 0.5 h, so total time = 2.5 h. Then, Speed = 165 ÷ 2.5 = 66 km/h.
例题：一辆汽车在2小时30分钟内行驶了165公里。计算其平均速度（公里/小时）。首先，将30分钟转换为小时：30分钟 = 0.5小时，因此总时间 = 2.5小时。然后，速度 = 165 ÷ 2.5 = 66公里/小时。
Try this: A cyclist rides 36 km in 1 hour 20 minutes. Find the average speed. (Hint: 20 minutes = 1/3 hour, total time = 4/3 hours; Speed = 36 ÷ 4/3 = 36 × 3/4 = 27 km/h.)
试一试：一名骑车人在1小时20分钟内骑行36公里。求平均速度。（提示：20分钟 = 1/3小时，总时间 = 4/3小时；速度 = 36 ÷ 4/3 = 36 × 3/4 = 27公里/小时。）
2. Maths and Geography: Scale and Map Reading | 数学与地理：比例尺与地图阅读
Map scales allow you to convert a distance measured on a map to the real distance on the ground. A scale of 1:50000 means that 1 cm on the map represents 50000 cm in reality. Since 50000 cm = 500 m = 0.5 km, you can quickly work out real distances. Always convert to a sensible unit, such as kilometres, for long distances.
地图比例尺可以将地图上测量的距离转换为地面上的实际距离。比例尺 1:50000 表示地图上的1厘米代表现实中的50000厘米。因为50000厘米 = 500米 = 0.5公里，你可以很快计算出实际距离。对于长距离，始终转换为合适的单位，例如公里。
Problem: A map has a scale of 1:25000. Two villages are 8 cm apart on the map. What is the actual distance in kilometres? Real distance = 8 × 25000 = 200000 cm. Divide by 100000 to get kilometres: 200000 ÷ 100000 = 2 km. So the villages are 2 km apart.
问题：一张地图的比例尺是1:25000。两个村庄在地图上相距8厘米。实际距离是多少公里？实际距离 = 8 × 25000 = 200000厘米。除以100000得到公里：200000 ÷ 100000 = 2公里。因此村庄相距2公里。
Challenge: On a 1:10000 map, a park has an area of 5 cm2. What is the real area in square metres? (Hint: linear scale factor is 10000, so area scale factor is 100002 = 100 million. Real area = 5 cm2 × 100000000 = 500000000 cm2. Convert: 1 m2 = 10000 cm2, so 500000000 ÷ 10000 = 50000 m2.)
挑战：在1:10000的地图上，一个公园的面积为5平方厘米。实际面积是多少平方米？（提示：线性比例因子为10000，因此面积比例因子为100002=1亿。实际面积=5 × 1亿 = 5亿平方厘米。换算：1平方米=10000平方厘米，因此5亿÷10000=50000平方米。）
3. Maths and Art: Symmetry and Tessellation | 数学与艺术：对称与镶嵌
Artists and designers use mathematics to create patterns. Symmetry can be reflective (mirror line) or rotational. A shape tessellates if it can cover a surface without gaps or overlaps. Regular polygons tessellate only when the interior angle divides exactly into 360°. For example, equilateral triangles (60°), squares (90°) and regular hexagons (120°) tessellate, but regular pentagons (108°) do not because 360 ÷ 108 = 3.33, which is not a whole number.
艺术家和设计师使用数学来创作图案。对称可以是反射（镜像线）或旋转对称。如果一个图形可以无间隙、无重叠地覆盖一个表面，则它能够镶嵌。只有当内角能整除360°时，正多边形才能镶嵌。例如，等边三角形（60°）、正方形（90°）和正六边形（120°）可以镶嵌，但正五边形（108°）不能，因为360÷108≈3.33，不是整数。
Example: How many regular hexagons meet at a single vertex in a tessellation? Hexagon interior angle = 120°. 360° ÷ 120° = 3. So exactly three hexagons meet at each point.
例子：在一个镶嵌中，多少个正六边形会聚在一个顶点？六边形内角=120°。360°÷120°=3。因此每个顶点恰好有三个六边形。
Creative task: Draw a shape with exactly two lines of reflective symmetry and rotational symmetry of order 2. Write down its name (e.g. a rectangle). Explain how knowing the interior angles helps you decide if it will tessellate.
创意任务：画一个恰好有两条反射对称线且旋转对称的阶数为2的图形。写出它的名称（例如矩形）。解释如何通过了解内角来判断它是否能镶嵌。
4. Maths and Finance: Discounts, Tax and Budgeting | 数学与金融：折扣、税费与预算
When shopping, you often see percentage discounts. To find the sale price, first calculate the discount amount and subtract it from the original price, or multiply the original price by (100% − discount percentage). For example, a 20% discount on a £50 item means you pay 80% of £50 = 0.8 × 50 = £40. In the UK, VAT (Value Added Tax) at 20% is often added to goods. A £36 item before VAT becomes 1.2 × 36 = £43.20 after tax.
购物时，你常常看到百分比折扣。要找到折扣价，先计算折扣金额然后从原价中减去，或将原价乘以（100% − 折扣百分比）。例如，一件£50的商品打八折意味着你支付£50的80% = 0.8 × 50 = £40。在英国，商品经常加收20%的增值税。一件不含税£36的商品，加税后变为1.2 × 36 = £43.20。
Problem: A jacket has a price tag of £80 excluding VAT. During a sale, the shop offers 15% off the price including VAT. How much do you pay? First add VAT: £80 × 1.2 = £96. Then 15% discount means you pay 85%: 0.85 × 96 = £81.60. So the final cost is £81.60.
问题：一件夹克标价£80（不含增值税）。促销期间，商店对含税价格打八五折。你应付多少钱？首先加增值税：£80 × 1.2 = £96。然后八五折意味着支付85%：0.85 × 96 = £81.60。最终价格为£81.60。
Budgeting: You earn £10 pocket money per week and want to buy a £120 games console. How many weeks must you save if you also spend £3 per week on sweets? Savings per week = 10 − 3 = £7. Number of weeks = 120 ÷ 7 ≈ 17.14, so you need 18 weeks.
预算：你每周零花钱为£10，想买一个£120的游戏机。如果每周你还花£3买糖果，需要储蓄多少周？每周储蓄 = 10 − 3 = £7。所需周数 = 120 ÷ 7 ≈ 17.14，因此需要18周。
5. Maths and PE: Statistics and Averages | 数学与体育：统计数据与平均值
In sport, coaches use statistics to analyse performance. The mean (average), median (middle value), mode (most frequent) and range (maximum − minimum) help summarise data. For a basketball player, the points scored over five games were: 18, 24, 15, 30, 13. Mean = (18+24+15+30+13) ÷ 5 = 100 ÷ 5 = 20 points. To find the median, order the numbers: 13, 15, 18, 24, 30; median is 18. The range is 30 − 13 = 17, showing consistency or variation.
在体育中，教练使用统计数据来分析表现。平均数（均值）、中位数（中间值）、众数（最常见值）和极差（最大值 − 最小值）有助于总结数据。一名篮球运动员在五场比赛中的得分分别为：18, 24, 15, 30, 13。均值 = (18+24+15+30+13) ÷ 5 = 100 ÷ 5 = 20分。要找中位数，将数据排序：13, 15, 18, 24, 30；中位数是18。极差为30 − 13 = 17，显示稳定性或变化幅度。
Question: A netball team’s goal differences over six matches are: −2, +5, 0, −1, +4, +3. What is the mean goal difference? Sum = −2+5+0−1+4+3 = 9. Mean = 9 ÷ 6 = 1.5. The median of ordered data −2, −1, 0, 3, 4, 5 is (0+3)/2 = 1.5. What does a positive mean tell you?
问题：一支无挡板篮球队在六场比赛中的净胜球为：−2, +5, 0, −1, +4, +3。平均净胜球是多少？总和 = −2+5+0−1+4+3 = 9。均值 = 9 ÷ 6 = 1.5。排序后的数据为 −2, −1, 0, 3, 4, 5，中位数 = (0+3)/2 = 1.5。均值为正说明了什么？
6. Maths and Design: Area, Perimeter and Costing | 数学与设计：面积、周长与成本
Designing rooms, gardens or objects requires calculating area and perimeter. For a rectangular space, Area = length × width, Perimeter = 2(length + width). When planning materials, you often need to find how many tiles, rolls of turf, or litres of paint are needed, then compute the total cost.
设计房间、花园或物品需要计算面积和周长。对于一个矩形空间，面积 = 长 × 宽，周长 = 2（长 + 宽）。在规划材料时，你通常需要计算需要多少块瓷砖、卷草皮或多少升油漆，然后计算总成本。
Scenario: A rectangular classroom measures 12 m by 8 m. The floor is to be covered with square carpet tiles of side 50 cm (0.5 m). How many tiles are needed? Area of room = 12 × 8 = 96 m2. Area of one tile = 0.5 × 0.5 = 0.25 m2. Number of tiles = 96 ÷ 0.25 = 384. If each tile costs £3.50, total cost = 384 × 3.50 = £1344.
情境：一间矩形教室长12米，宽8米。地面要铺边长为50厘米（0.5米）的正方形地毯砖。需要多少块瓷砖？教室面积 = 12 × 8 = 96平方米。一块瓷砖面积 = 0.5 × 0.5 = 0.25平方米。瓷砖数量 = 96 ÷ 0.25 = 384块。如果每块瓷砖£3.50，总成本 = 384 × 3.50 = £1344。
Extension: The room also needs a border skirting board around its perimeter. How many metres of skirting are required if there is a door 2 m wide where no skirting is placed? Perimeter = 2(12+8) = 40 m. Subtract door width: 40 − 2 = 38 m. If skirting costs £4.20 per metre, the cost is 38 × 4.20 = £159.60.
拓展：这间教室的周边还需要贴踢脚线。如果有一扇2米宽的门不贴踢脚线，需要多少米踢脚线？周长 = 2(12+8) = 40米。减去门宽：40 − 2 = 38米。如果踢脚线每
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