📚 Year 7 CCEA Maths: Cross-curricular Problem-Solving Training | CCEA 七年级数学：跨学科综合题型训练
Cross-curricular problem solving helps you see how mathematics connects to science, geography, history, art and everyday life. In this article, you will work through typical Year 7 CCEA style questions that mix number skills, measures, data and geometry with other subjects. Each example shows the thinking steps you need to become a confident problem solver.
跨学科综合题能帮助你看到数学如何与科学、地理、历史、艺术和日常生活紧密相连。本文将带你练习典型的 CCEA 七年级题型，它们把数字运算、计量、数据处理和几何与其它学科结合在一起。每个例子都会展示解题步骤，帮你建立解题信心。
1. Science: Temperature Changes and Negative Numbers | 科学：温度变化与负数
In a science lesson, a liquid is cooled to −5 °C and then heated until it reaches 23 °C. Find the total increase in temperature.
在一节科学课上，一种液体先被冷却到 −5 °C，然后加热直至达到 23 °C。求温度总共上升了多少。
Draw a number line or use the idea that subtracting a negative is like adding. The change is 23 − (−5) = 23 + 5 = 28 °C. So the temperature rises by 28 °C.
画一条数轴，或者利用「减去负数等于加上正数」的思路。变化量是 23 − (−5) = 23 + 5 = 28 °C。因此温度升高了 28 °C。
2. Geography: Map Scale and Distance | 地理：地图比例尺与距离
A footpath is shown as 8 cm on a map with a scale of 1 : 50 000. Calculate the real distance in kilometres.
一条步行道在比例尺为 1 : 50 000 的地图上长 8 cm。计算真实距离，并以千米为单位表示。
Real distance in cm = 8 × 50 000 = 400 000 cm. Since 100 cm = 1 m and 1000 m = 1 km, we know 100 000 cm = 1 km. So 400 000 ÷ 100 000 = 4 km. The actual footpath is 4 km long.
实际距离（厘米）= 8 × 50 000 = 400 000 cm。因为 100 cm = 1 m，1000 m = 1 km，可得 100 000 cm = 1 km。因此 400 000 ÷ 100 000 = 4 km。这条步行道的实际长度是 4 千米。
3. History: Timeline Calculations | 历史：时间线计算
The Roman Empire began in 27 BC, and the Western Roman Empire fell in AD 476. How many years passed between these two events?
罗马帝国始于公元前 27 年，西罗马帝国灭亡于公元 476 年。这两个事件之间相隔了多少年？
A simple method for Year 7 is to add the BC and AD numbers: 27 + 476 = 503 years. (Historians note that there is no year zero, which can make the exact count 502 years, but for most maths problems we add them.)
七年级常用的简单方法是将公元前年数与公元年数相加：27 + 476 = 503 年。（历史学家指出没有公元 0 年，所以精确的计时可能是 502 年，但在大多数数学题中我们直接相加。）
4. Physical Education: Averages and Data Handling | 体育：平均数与数据处理
A basketball team records the points scored by one player in five matches: 12, 15, 8, 22, 10. Find the mean, median and mode.
一支篮球队记录了某名球员在五场比赛中的得分：12, 15, 8, 22, 10。求平均数、中位数和众数。
Mean = (12 + 15 + 8 + 22 + 10) ÷ 5 = 67 ÷ 5 = 13.4 points. To find the median, order the scores: 8, 10, 12, 15, 22. The middle number is 12. There is no mode because no score repeats.
平均数 = (12 + 15 + 8 + 22 + 10) ÷ 5 = 67 ÷ 5 = 13.4 分。找中位数时先将得分排序：8, 10, 12, 15, 22，中间的数字是 12。由于所有得分都没有重复，所以没有众数。
5. Art & Design: Symmetry and Tessellation | 艺术与设计：对称与镶嵌
You want to tile a wall using only one type of regular polygon. Explore which polygons tessellate by checking if the interior angle divides exactly into 360°.
你想只用一种正多边形铺满一面墙。通过检验内角是否能整除 360°，探讨哪些正多边形可以完成镶嵌。
Equilateral triangle: interior angle 60°, 360 ÷ 60 = 6 – tessellates. Square: 90°, 360 ÷ 90 = 4 – tessellates. Regular pentagon: 108°, 360 ÷ 108 = 3.333… – does not tessellate. Regular hexagon: 120°, 360 ÷ 120 = 3 – tessellates. So equilateral triangles, squares and regular hexagons work.
等边三角形：内角 60°，360 ÷ 60 = 6 —— 可以镶嵌。正方形：90°，360 ÷ 90 = 4 —— 可以镶嵌。正五边形：108°，360 ÷ 108 = 3.333… —— 不行。正六边形：120°，360 ÷ 120 = 3 —— 可以镶嵌。因此，等边三角形、正方形和正六边形能够实现单种正多边形镶嵌。
6. Music: Fractions and Rhythm | 音乐：分数与节奏
In 4/4 time, a bar holds 4 beats. A dotted minim lasts for 3 beats. What note or notes could you add to complete the bar?
在 4/4 拍中，一个小节有 4 拍。一个附点二分音符的时值为 3 拍。你可以加入什么音符使该小节完整？
Remaining beats = 4 − 3 = 1 beat. You could add a crotchet (quarter note, 1 beat) or two quavers (eighth notes, ½ beat each). For example, a dotted minim plus a crotchet makes 3 + 1 = 4 beats.
剩余拍数 = 4 − 3 = 1 拍。你可以加入一个四分音符（1 拍），或者两个八分音符（每个 ½ 拍）。例如，一个附点二分音符加一个四分音符，总时值为 3 + 1 = 4 拍。
7. Cooking and Nutrition: Ratio and Proportion | 烹饪与营养：比例与比率
A recipe for 4 people uses 200 g of flour, 100 g of sugar and 50 g of butter. Adapt the quantities for 6 people.
一份供 4 人食用的食谱需要 200 g 面粉、100 g 糖和 50 g 黄油。请调整为 6 人份的用量。
The scale factor is 6 ÷ 4 = 1.5. Multiply each ingredient by 1.5: flour = 200 × 1.5 = 300 g, sugar = 100 × 1.5 = 150 g, butter = 50 × 1.5 = 75 g. Always check that the ratios stay the same.
比例因子是 6 ÷ 4 = 1.5。将每种原料的量乘以 1.5：面粉 = 200 × 1.5 = 300 g，糖 = 100 × 1.5 = 150 g，黄油 = 50 × 1.5 = 75 g。记得检查调整后的比例是否与原食谱一致。
8. Design Technology: Perimeter and Area | 设计与技术：周长与面积
A rectangular garden must have an area of 24 m² and whole-number side lengths in metres. Which dimensions give the smallest perimeter?
一个矩形花园的面积必须是 24 m²，且边长（米）均为整数。哪种长宽组合能使周长最小？
| Length (m) | Width (m) | Perimeter (m) |
|---|---|---|
| 1 | 24 | 50 |
| 2 | 12 | 28 |
| 3 | 8 | 22 |
| 4 | 6 | 20 |
From the table, the combination 4 m by 6 m gives the smallest perimeter of 20 m. The closer a rectangle is to a square, the smaller its perimeter for a fixed area.
从表中可以看出，4 m × 6 m 的组合拥有最小的周长 20 m。对于固定的面积，矩形的长宽越接近正方形，周长就越小。
9. Financial Capability: Budgeting and Percentages | 金融能力：预算与百分比
You receive £40 pocket money. You decide to save 25% and spend £15 on a game. How much is left for snacks?
你收到 £40 零花钱。你决定把 25% 存起来，并花 £15 买游戏。还剩下多少钱买零食？
Savings = 25% of £40 = 0.25 × 40 = £10. Money remaining after saving = £40 − £10 = £30. After spending £15 on a game, the amount for snacks is £30 − £15 = £15.
存款 = £40 的 25% = 0.25 × 40 = £10。存款后剩余金额 = £40 − £10 = £30。花 £15 买游戏后，可用于零食的钱为 £30 − £15 = £15。
10. Science: Speed, Distance and Time | 科学：速度、距离与时间
A cyclist rides at a steady speed of 12 km/h. How far does she travel in 30 minutes?
一名自行车骑手以 12 km/h 的恒定速度骑行。她在 30 分钟内能骑多远？
First change 30 minutes to hours: 30 min = 0.5 h. Distance = speed × time = 12 × 0.5 = 6 km. Always check that time units match the speed units before multiplying.
首先将 30 分钟转换为小时：30 min = 0.5 h。距离 = 速度 × 时间 = 12 × 0.5 = 6 km。在做乘法之前，务必确保时间单位与速度单位相匹配。
11. Geography: Population Density | 地理：人口密度
A town covers an area of 50 km² and has a population of 200 000. Calculate its population density (people per km²).
一座小镇的面积为 50 km²，人口为 200 000。计算它的人口密度（每平方千米人数）。
Population density = population ÷ area = 200 000 ÷ 50 = 4000 people per km². This tells us how crowded the town is on average.
人口密度 = 人口 ÷ 面积 = 200 000 ÷ 50 = 4000 人/km²。这个数值反映了该镇的平均拥挤程度。
12. Environment: Recycling Data and Bar Charts | 环境：回收数据与条形图
In one week a school collects 25 kg of paper, 15 kg of plastic and 10 kg of glass for recycling. Draw a bar chart and find the total mass.
在一周内，某所学校收集了 25 kg 废纸、15 kg 塑料和 10 kg 玻璃用于回收。试画一张条形图，并计算总质量。
Total mass = 25 + 15 + 10 = 50 kg. On a bar chart, label the horizontal axis with material types and the vertical axis with mass in kg. Draw bars of heights 25, 15 and 10. This simple data set is perfect for practising scale and accurate plotting.
总质量 = 25 + 15 + 10 = 50 kg。绘制条形图时，横轴标出材料种类，纵轴标明质量（kg），然后分别画出高度为 25、15 和 10 的直条。这个简单的数据集非常适合练习比例和准确绘图。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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