📚 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
late speed using the formula: Speed = Distance ÷ Time. The units must be consistent – ange 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: to 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: s = 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
0000 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: 1:25000. Two villages are 8 cm apart on the map. What is the actual distance in kilometres? y 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
ver 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? = 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
unt 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% ded 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: games console. How many weeks must you save if you also spend £3 per week on sweets? minus; 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
us; 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' over six matches are: −2, +5, 0, −1, +4, +3. What is the mean goal difference? d 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
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: 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 ; 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: ny metres of skirting are required if there is a door 2 m wide where no skirting is placed? 8 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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