📚 Cross-curricular Integrated Problem-Solving Practice | 跨学科综合题型训练
In today's world, mathematics is not an isolated subject; it provides essential tools for science, geography, music, sport and many other fields. This article presents a complete set of cross-curricular problem-solving tasks designed for Year 8 WJEC Advanced Mathematics. You will apply skills such as percentages, ratios, formulae, data handling and geometry to real-life contexts, strengthening both your mathematical fluency and your ability to connect ideas across disciplines.
当今世界，数学不是一门孤立的学科；它给科学、地理、音乐、体育等许多领域提供了基本工具。本文呈献一套完整的跨学科解题练习，针对 WJEC 八年级进阶数学设计。你将把百分比、比例、公式、数据处理和几何等技能应用到真实情境中，从而同时提升数学流畅度与跨学科联想能力。
1. What Cross-curricular Problem Solving Really Means | 跨学科解题的真正含义
Cross-curricular problem solving means that one realistic situation is described once, but the mathematics inside it belongs to several subjects at the same time. A single journey, for example, involves distance (geography), speed (science), averages (statistics) and money (business). You are not learning four separate topics; you are learning to select the right tool at the right moment.
跨学科解题的意思是：一个真实情境只描述一次，但其中包含的数学同时属于好几个学科。比如一次出行，就同时涉及距离（地理）、速度（科学）、平均数（统计）和金钱（商业）。你并不是在学四个互不相关的主题，而是在学习如何在恰当的时刻选出恰当的工具。
The examiner is looking for four things in an integrated question:
考官在综合题中主要看四点：
- Correct extraction: pulling the right numbers out of a paragraph, table or diagram.
- 正确提取：从文字、表格或图形中准确取出所需数据。
- Correct choice of method: knowing whether the question needs a ratio, a formula or an average.
- 正确选择方法：判断题目需要的是比例、公式还是平均数。
- Consistent units: never mixing minutes with hours, or grams with kilograms.
- 单位一致：绝不把分钟和小时、克和千克混在一起用。
- Clear communication: writing the formula, the substitution and the answer with its unit.
- 清晰表达：写出公式、代入过程以及带单位的答案。
A useful habit is to underline every quantity in the question and write its unit beside it before you start calculating. This single step prevents the majority of lost marks.
一个很实用的习惯是：动笔计算之前，先把题目中每一个数量划出来，并在旁边写上它的单位。仅仅这一步，就能避免大部分失分。
2. Science: Speed, Density and Compound Measures | 科学：速度、密度与复合单位
and then rearrange it when a different variable is required. Remember: speed = distance ÷ time, and density = mass ÷ volume. Always check that units are consistent before calculating.
在 KS3 科学中你会遇到速度、密度和流量等复合单位。核心技能是将数值代入三变量公式，并在需要求另一个变量时进行变换。记住：速度 = 距离 ÷ 时间；密度 = 质量 ÷ 体积。计算前务必检查单位是否一致。
distance ÷ time density = mass ÷ volume pressure = force ÷ area
速度 = 距离 ÷ 时间 密度 = 质量 ÷ 体积 压强 = 力 ÷ 面积
Example: A cyclist covers 45 km in 1 h 15 min. Find her average speed in km/h and in m/s.
示例：一位自行车手在 1 小时 15 分钟内骑行 45 公里。求她的平均速度，单位分别为 km/h 和 m/s。
Solution: s: 45 km = 45 000 m and 1.25 h = 1.25 × 3 600 = 4 500 s. So speed = 45 000 ÷ 4 500 = 10 m/s. Check with the shortcut: 36 × 1 000 ÷ 3 600 = 10 m/s. ✓
解答：先把时间换成小时：15 分钟 = 0.25 小时，所以总时间 = 1.25 小时。平均速度 = 45 ÷ 1.25 = 36 km/h。若要求米每秒，两个量都要换算：45 km = 45 000 m，1.25 h = 1.25 × 3 600 = 4 500 s。因此速度 = 45 000 ÷ 4 500 = 10 m/s。用快捷方法验证：36 × 1 000 ÷ 3 600 = 10 m/s。✓
Example: A metal block has a mass of 2.4 kg and a volume of 300 cm³. Find its density in g/cm³ and in kg/m³.
示例：一块金属质量为 2.4 kg，体积为 300 cm³。求其密度，单位分别为 g/cm³ 和 kg/m³。
Solution: = 8 000 kg/m³. Alternatively 300 cm³ = 0.0003 m³ and density = 2.4 ÷ 0.0003 = 8 000 kg/m³. ✓
解答：2.4 kg = 2 400 g，所以密度 = 2 400 ÷ 300 = 8 g/cm³。换算时记住 1 g/cm³ = 1 000 kg/m³，因此 8 g/cm³ = 8 000 kg/m³。也可以写成 300 cm³ = 0.0003 m³，密度 = 2.4 ÷ 0.0003 = 8 000 kg/m³。✓
| Measure | Formula | Typical units |
| Speed 速度 | distance ÷ time | km/h, m/s |
| Density 密度 | mass ÷ volume | g/cm³, kg/m³ |
| Pressure 压强 | force ÷ area | N/m², Pa |
| Flow rate 流量 | volume ÷ time | cm³/s, litres/min |
3. Geography: Map Scale, Distance and Unit Conversion | 地理：比例尺、距离与单位换算
Maps turn the real world into a scale drawing. A scale of 1 : 50 000 means that 1 cm on the map represents 50 000 cm in reality, which is 500 m or 0.5 km. Distances on a map are measured in centimetres, so the calculation always ends with a conversion.
地图把真实世界变成一幅按比例缩小的图。比例尺 1 : 50 000 表示图上 1 cm 代表现实中 50 000 cm，即 500 m 或 0.5 km。图上的距离都以厘米量出，所以计算最后总要做一次单位换算。
Example: On a map with scale 1 : 50 000, a footpath measures 6 cm. How long is it in kilometres?
示例：在比例尺为 1 : 50 000 的地图上，一条小路长 6 cm。它实际有多少公里？
Solution: Real distance = 6 × 50 000 = 300 000 cm = 3 000 m = 3 km.
解答：实际距离 = 6 × 50 000 = 300 000 cm = 3 000 m = 3 km。
Area works differently, and this is a classic trap. Because area is two-dimensional, the area scale factor is the square of the length scale factor.
面积的处理方式不同，这是经典陷阱。因为面积是二维的，面积缩放比是长度缩放比的平方。
length scale factor = k → area scale factor = k²
长度缩放比 = k → 面积缩放比 = k²
0.25 km². A wood covering 8 cm² on the map therefore has a real area of 8 × 0.25 = 2 km².
在 1 : 50 000 的地图上，1 cm 代表 0.5 km，所以 1 cm² 代表 0.5 × 0.5 = 0.25 km²。图上面积为 8 cm² 的林地，实际面积就是 8 × 0.25 = 2 km²。
- Always write the scale as a ratio, then multiply before converting.
- 始终把比例尺写成比的形式，先相乘，再换算单位。
- 1 km = 1 000 m = 100 000 cm.
- 1 km = 1 000 m = 100 000 cm。
- For area, square the length scale factor once, and only once.
- 求面积时，把长度缩放比平方，且只平方一次。
4. Music: Ratios, Frequency and String Length | 音乐：比例、频率与弦长
Music is mathematics you can hear. The pitch of a note is determined by frequency, measured in hertz (Hz), and the pleasing intervals of Western music correspond to simple whole-number ratios.
音乐是你能听见的数学。一个音的音高由频率决定，单位是赫兹（Hz），而西方音乐中悦耳的音程对应着简单的整数比。
| Interval 音程 | Frequency ratio 频率比 | Example 示例 |
| Octave 八度 | 2 : 1 | 440 Hz → 880 Hz |
| Perfect fifth 纯五度 | 3 : 2 | 440 Hz → 660 Hz |
| Perfect fourth 纯四度 | 4 : 3 | 440 Hz → 586.7 Hz |
| Major third 大三度 | 5 : 4 | 440 Hz → 550 Hz |
On a stringed instrument, frequency and string length are inversely proportional: halving the vibrating length doubles the frequency, which raises the note by one octave.
在弦乐器上，频率与弦长成反比：振动长度减半，频率加倍，音高就升高一个八度。
frequency × length = constant (f₁L₁ = f₂L₂)
频率 × 弦长 = 常数 (f₁L₁ = f₂L₂)
Example: A guitar string of length 650 mm sounds at 220 Hz. Where must you press to produce 440 Hz?
示例：一根长 650 mm 的吉他弦发出 220 Hz 的音。要发出 440 Hz，应按在什么位置？
Solution: ₂ = 143 000 ÷ 440 = 325 mm. This is exactly half the original length, at the twelfth fret. ✓
解答：由 f₁L₁ = f₂L₂，得 220 × 650 = 440 × L₂，所以 L₂ = 143 000 ÷ 440 = 325 mm。这正是原长的一半，也就是第十二品的位置。✓
5. Sport: Percentages, Averages and Performance Data | 体育：百分比、平均数与运动表现数据
Sport produces numbers constantly, and coaches use percentages and averages to compare athletes fairly. Percentage means "out of one hundred", so every percentage question begins with a fraction.
体育运动不断产生数据，教练用百分比和平均数来公平地比较运动员。百分数的意思是"每一百当中"，所以每道百分数题都从一个分数开始。
percentage = (part ÷ whole) × 100 %
百分数 = (部分 ÷ 整体) × 100 %
Example: A basketball player scores 17 free throws out of 20 attempts. What is her success rate?
示例：一名篮球运动员 20 次罚球中命中 17 次。她的命中率是多少？
Solution: 17 ÷ 20 = 0.85, so the rate is 0.85 × 100 % = 85 %.
解答：17 ÷ 20 = 0.85，所以命中率为 0.85 × 100 % = 85 %。
e change. A runner whose time drops from 50 s to 45 s has improved by (50 − 45) ÷ 50 × 100 % = 10 %. Note carefully that the original time is the denominator.
要比较两次表现，就计算百分比变化。一位跑者成绩从 50 s 降到 45 s，进步幅度为 (50 − 45) ÷ 50 × 100 % = 10 %。特别注意：分母是原来的成绩。
- Mean: add all values, divide by how many there are.
- 平均数：把所有数值相加，再除以个数。
- Median: put the values in order and take the middle one.
- 中位数：把数值排序后取中间的那个。
- Range: largest value minus smallest value, a simple measure of spread.
- 极差：最大值减最小值，是衡量分散程度最简单的量。
- Outlier: an unusual value that can pull the mean away from the typical result.
- 异常值：一个不寻常的数值，会把平均数拉离典型结果。
For example, the five times 12.1 s, 12.3 s, 12.2 s, 14.9 s and 12.0 s have a mean of 12.7 s, but the median is only 12.2 s. The slowest run is an outlier, and reporting the median gives a fairer picture.
例如，五次成绩 12.1 s、12.3 s、12.2 s、14.9 s、12.0 s 的平均数是 12.7 s，但中位数只有 12.2 s。最慢的那一次是异常值，此时报告中位数更公平。
6. Design and Technology: Area, Volume and Cost | 设计与技术：面积、体积与成本
Design and technology questions ask you to find the amount of material needed and then the cost. Surface area governs painting, wrapping and packaging; volume governs capacity and shipping weight.
设计与技术类题目要求你求出所需材料的数量，然后算出成本。表面积决定涂漆、包装和外包用料；体积决定容量和运输重量。
rectangle area = length × width cuboid volume = length × width × height
长方形面积 = 长 × 宽 长方体体积 = 长 × 宽 × 高
Example: A closed box is 40 cm long, 25 cm wide and 20 cm high. Find its total surface area and its volume.
示例：一个封闭的盒子长 40 cm、宽 25 cm、高 20 cm。求它的总表面积和体积。
Solution: 000 + 800 + 500) = 2 × 2 300 = 4 600 cm². Volume = 40 × 25 × 20 = 20 000 cm³ = 20 litres. ✓
解答：六个面分成三组：40 × 25 = 1 000 cm²，40 × 20 = 800 cm²，25 × 20 = 500 cm²。总表面积 = 2 × (1 000 + 800 + 500) = 2 × 2 300 = 4 600 cm²。体积 = 40 × 25 × 20 = 20 000 cm³ = 20 升。✓
Costing: If paint costs £0 per m² and 1 m² = 10 000 cm², then 4 600 cm² = 0.46 m² and the paint costs 0.46 × 3.20 = £1.472, which rounds to £1.47. Always round money to two decimal places at the very end.
成本核算：若油漆每平方米 £3.20，且 1 m² = 10 000 cm²，则 4 600 cm² = 0.46 m²，油漆费用为 0.46 × 3.20 = £1.472，四舍五入为 £1.47。金额一律在最后一步保留两位小数。
- Write down every dimension with its unit before substituting.
- 代入之前先把每个尺寸连同单位写下来。
- Convert cm² to m² by dividing by 10 000, and cm³ to m³ by dividing by 1 000 000.
- cm² 换算成 m² 要除以 10 000；cm³ 换算成 m³ 要除以 1 000 000。
- A rate such as "per m²" tells you exactly which quantity to multiply by the cost.
- "每平方米"这样的单价，直接提示你要用哪个量去乘费用。
7. Data Handling: Tables, Charts and Interpreting Evidence | 数据处理：表格、图表与证据解读
Data handling is where mathematics meets every other subject. A scientist, a geographer and a coach all read tables and graphs, and all of them can be misled by a badly drawn chart.
数据处理是数学与其他所有学科交汇的地方。科学家、地理学家和教练都要读表格和图表，而他们也都可能被一张画得不好的图误导。
Consider the table below, recording the number of cyclists passing a school gate each day.
请看下表，记录了每天经过校门的自行车数量。
| Day 日期 | Mon | Tue | Wed | Thu | Fri |
| Cyclists 人数 | 42 | 38 | 45 |
