📚 Year 7 WJEC Further Mathematics: Interdisciplinary Problem-Solving Training | Year 7 WJEC 进阶数学：跨学科综合题型训练
Further Mathematics in Year 7 is not just about numbers and algebra – it is a toolkit for solving problems across science, geography, design and many other subjects. This article brings together key topics from the WJEC curriculum and shows how ratio, percentage, measurement and data skills are used in real cross-curricular contexts. Each section presents a typical problem style, with guided reasoning and numerical practice to build confidence for mixed assessments.
七年级的进阶数学不仅仅是数字和代数，它是一套解决科学、地理、设计等跨学科问题的工具。本文结合WJEC课程中的重点内容，展示如何在真实的跨学科情境中运用比和比例、百分数、测量以及数据处理技能。每个小节都提供了一种典型题型，并配有推导过程和数值练习，帮助你建立应对综合评估的信心。
1. Distance, Speed and Time in Science | 科学中的路程、速度和时间
When a cyclist travels 36 km in 2 hours, we can find her average speed. The formula that links distance, speed and time is written as:
骑自行车的人2小时行驶36公里，我们可以求出她的平均速度。路程、速度和时间的关系式为：
speed = distance ÷ time
we needed the speed in metres per second, we convert: 18 km/h = 18 000 m ÷ 3600 s = 5 m/s.
因此她速度为 36 公里 ÷ 2 小时 = 18 公里/小时。如果需要用米/秒表示，可换算：18 公里/小时 = 18 000 米 ÷ 3600 秒 = 5 米/秒。
= 4 × 25 = 100 m. Rearranging the formula gives time = distance ÷ speed, so swimming 200 m at 2 m/s takes 100 seconds.
综合题往往也会要求你求路程或时间。例如，一名跑步者以4米/秒的速度跑了25秒，经过的路程 = 速度 × 时间 = 4 × 25 = 100 米。将公式变形可得 时间 = 路程 ÷ 速度，因此以2米/秒的速度游200米需要100秒。
| Quantity | Formula | Units |
|---|---|---|
| Speed (s) | s = d ÷ t | km/h, m/s |
| Distance (d) | d = s × t | km, m, miles |
| Time (t) | t = d ÷ s | h, min, s |
2. Mixing Solutions and Ratios in Chemistry | 化学中的溶液混合与比
use 300 ml of concentrate, the volume of water needed is 300 ml × 4 = 1200 ml. The total mixture is 300 + 1200 = 1500 ml.
在化学中，学生经常会按给定比例混合液体。假设一种清洁液需要将浓缩液与水按1:4混合。如果用了300毫升浓缩液，所需水的体积为 300 毫升 × 4 = 1200 毫升。总的混合液为 300 + 1200 = 1500 毫升。
9 = 10. One part corresponds to 2000 ml ÷ 10 = 200 ml. So you need 200 ml of squash and 1800 ml of water. This method of dividing by total parts is essential for fair shares, recipes and chemical mixtures.
有时题目会给出总体积。一瓶2升的稀释果汁按浓缩汁与水1:9的比例配制。总份数为 1 + 9 = 10。每一份相当于 2000 毫升 ÷ 10 = 200 毫升。因此需要200毫升浓缩汁和1800毫升水。这种除以总份数的方法在公平分配、食谱和化学混合物中非常关键。
A real cross-curricular check: after mixing, the squash bottle says 'dilute 1 part squash to 4 parts water'. That is a ratio 1 : 4, meaning for every 1 ml of squash you add 4 ml water, total 5 parts. Practise converting between part-to-part and part-to-whole ratios.
一个真实的跨学科检查：稀释果汁瓶上写着“1份浓缩汁加4份水”，即比例为1:4，这意味着每1毫升浓缩汁加4毫升水，共5份。练习在部分与部分之比以及部分与整体之比之间进行转换。
3. Population Growth and Percentages in Biology | 生物学中的人口增长与百分数
Biologists study how populations change. A colony of 500 bacteria increases by 20% in one day. The increase is 20% so the new population is 500 + 100 = 600 bacteria. This is a percentage increase problem.
生物学家研究种群数量的变化。一个含有500个细菌的菌落在一天内增加了20%。增加的数量是500的20% = 0.20 × 500 = 100，因此新的种群数量为 500 + 100 = 600 个细菌。这是一个百分数增加的问题。
hen a population decreases, we use percentage decrease. If a pond has 1200 tadpoles and 25% are eaten by fish, the number lost is 0.25 × 1200 = 300. The remaining tadpoles: 1200 – 300 = 900. You can also calculate the final amount directly by multiplying by (100% – 25%) = 75% = 0.75, so 1200 × 0.75 = 900. Practise both methods.
当种群减少时，我们使用百分数减少。如果一个池塘里有1200只蝌蚪，25%被鱼吃掉，损失的数量为 0.25 × 1200 = 300。剩下的蝌蚪：1200 – 300 = 900。你也可以直接用 (100% – 25%) = 75% = 0.75 乘，即 1200 × 0.75 = 900。两种方法都要练习。
sciplinary questions may ask you to calculate original amounts before a change. After a 10% ease, a rabbit population is 550. The original population = 550 ÷ 1.10 = 500 rabbits. This 'reverse percentage' thinking is used in many science and geography contexts.
跨学科题目可能会要求你计算变化前的原始数量。经过10%的增长后，兔子数量为550只。原始数量 = 550 ÷ 1.10 = 500 只。这种“逆推百分数”的思路在许多科学和地理情境中都会用到。
4. Map Scales and Bearings in Geography | 地理中的地图比例尺与方位角
to metres: 200 000 cm ÷ 100 = 2000 m, or 2 km. Always remember to convert units sensibly.
在一张比例尺为1:25 000的英国地形测量图上，图上8厘米的距离代表实际距离为 8 厘米 × 25 000 = 200 000 厘米。换算成米：200 000 厘米 ÷ 100 = 2000 米，即2公里。要记住合理换算单位。
Bearings are measured clockwise from north, always given as three-digit angles. A ship is spotted on a bearing of 125° from a lighthouse. To draw this, place the protractor with 0° pointing north and measure clockwise to 125°. This skill links geometry and geography.
方位角是从正北方向顺时针测量的角度，通常用三位数表示。一艘船从灯塔看去的方位角为125°。绘制时，将量角器0°对准正北，顺时针对125°。这一技能将几何与地理联系起来。
iving an area of 150 000 m². You can express this in hectares (1 ha = 10 000 m²), so the field is 15 ha. This shows how scale factors apply to length, but area scales by the square of the linear factor.
结合比例尺与面积也很常见。如果一块矩形土地在1:10 000的地图上的尺寸为3厘米乘5厘米，实际尺寸则为300米乘500米，面积为 150 000 平方米。你可以用公顷表示（1 公顷 = 10 000 平方米），所以这块地为15公顷。这表明比例因子作用于长度，而面积则要按线性因子的平方来换算。
5. Currency Exchange and Budgeting in Economics | 经济学中的货币兑换与预算
When travelling, you exchange money using a given rate. If the exchange rate is £1 = $1.30, converting £200 to dollars gives 200 × 1.30 = $260. To change back, you divide by the same rate: $156 in pounds is 156 ÷ 1.30 = £120. Always state which currency you are converting to.
旅行时，你会按照给定的汇率兑换货币。如果汇率为 £1 = $1.30，将200英镑兑换成美元为 200 × 1.30 = 260 美元。换回来时，除以同一汇率：156美元等于 156 ÷ 1.30 = 120 英镑。要始终说明你兑换的目标货币。
ting tasks require adding costs and finding change. A family plans a trip with a budget of £400. They spend £68.50 on fuel, £115.30 on food and £192.00 on accommodation. Total spent = 68.50 + 115.30 + 192.00 = £375.80. The remaining balance is 400 – 375.80 = £24.20. These calculations use decimal addition and subtraction accurately.
预算任务需要累加成本并计算余额。一个家庭计划旅行，预算为400英镑。他们支付了燃油费68.50英镑、食物115.30英镑和住宿192.00英镑。总支出 = 68.50 + 115.30 + 192.00 = 375.80 英镑。剩余金额为 400 – 375.80 = 24.20 英镑。这些计算需精确运用小数加减法。
rices when items are measured in different units. For example, a 1.5 kg bag of flour costs £1.20, so the price per kg is £1.20 ÷ 1.5 = £0.80. A 2 kg bag costs £1.50, giving £ ter value. Using division to find unit price links maths with everyday economic decisions.
有时你需要比较不同单位量商品的价格。比如，一袋1.5千克面粉售价1.20英镑，每千克价格为 1.20 ÷ 1.5 = 0.80英镑。一袋2千克售价1.50英镑，每千克为0.75英镑。大袋更划算。用除法求单价把数学与日常经济决策联系起来。
6. Area and Perimeter of Composite Shapes in Design & Technology | 设计与技术中的复合图形面积和周长
. For an L-shape with side lengths 8 m, 5 m, 3 m, 2 m, the perimeter is 8 + 5 + 3 + 2 + (8–3) + (5–2) = 8+5+3+2+5+3 = 26 m. Be careful to include every outer segment exactly once.
在设计项目中，你经常要处理L形或T形图形。求周长时，只需将所有外部边长相加。对于边长分别为8米、5米、3米、2米的L形，其周长为 8 + 5 + 3 + 2 + (8–3) + (5–2) = 8+5+3+2+5+3 = 26 米。注意每条外部线段只加一次，不要遗漏。
; total area = 25 m². Alternatively, enclose the shape in a large rectangle 8 m by 5 m (40 m²) and subtract the missing 3 m by 5 m cut-out (15 m²), giving 25 m². Both methods work well.
求面积时，将图形分割成两个矩形。对于同一个L形，可分割为一个8米×2米的矩形（面积16平方米）和一个3米×3米的矩形（面积9平方米）；总面积为25平方米。或者，将图形放入一个8米×5米的大矩形中（面积40平方米），减去被切掉的部分3米×5米（15平方米），同样得到25平方米。两种方法都可以。
eeded. If a wooden frame is to be built along the perimeter, you must add all lengths, but subtracting doorways. A room 6 m by 4 m with a 1 m wide door needs (6+4+6+4 – ting board. Combining area and perimeter with real constraints teaches practical planning.
设计师还要计算所需材料。如果要沿周长制作木框，需要加总长度，但要减去门洞。一个6米×4米的房间，有一扇1米宽的门，所需的踢脚线长度为 (6+4+6+4 – 1) = 19 米。将面积和周长与现实约束结合起来可以培养实际规划能力。
7. Recipes and Proportional Reasoning in Food Technology | 食品科技中的食谱与比例推理
A biscuit recipe needs 200 g of flour, 100 g of butter, 50 g of sugar and 1 egg. This makes 10 biscuits. If you need 30 biscuits, multiply each ingredient by 3: 600 g flour, 300 g butter, 150 g sugar and 3 eggs. This is a direct proportion problem: the amount of each ingredient is directly proportional to the number of biscuits.
一份饼干配方需要200克面粉、100克黄油、50克糖和1个鸡蛋，可以做出10块饼干。如果你需要30块饼干，将每种原料乘以3：600克面粉、300克黄油、150克糖和3个鸡蛋。这是一个正比例问题：每种原料用量与饼干数量成正比。
cone first: 240 ml ÷ 4 = 60 ml per scone, then for 6 scones: 60 × 6 = 360 ml. The unitary method is safest for any scaling up or down.
有时配方给出的是特定份数，而你需要减少。做4个司康饼需要240毫升牛奶。要做6个，先求出单个司康饼的牛奶量：240 毫升 ÷ 4 = 60 毫升每个，然后6个需要 60 × 6 = 360 毫升。单位量法对于任何规模的调整都最为稳妥。
= 5/6 of the batch; adjust all other ingredients by 5/6. This introduces the concept of limiting reactant that appears later in chemistry.
比例问题可能会涉及限制性原料。如果黄油是250克一块，而你完整一批需要300克，那么你只能做 (250/300) = 5/6 批；所有其他原料也要乘以 5/6。这引入了限制反应物的概念，今后在化学中也会出现。
8. Data Interpretation and Line Graphs in Environmental Science | 环境科学中的数据解读与折线图
Environmental scientists monitor temperature changes. A table shows the temperature recorded at midday over one week in July: Mon 22°C, Tue 24°C, Wed 21°C, Thu 23°C, Fri 25°C, Sat 27°C, Sun 26°C. Plotting these as a line graph helps to spot trends, such as a steady rise from Wednesday to Saturday.
环境科学家监测温度变化。某年7月一周中午记录的温度表显示：周一22°C，周二24°C，周三21°C，周四23°C，周五25°C，周六27°C，周日26°C。将这些数据绘制成折线图有助于发现趋势，比如从周三到周六持续上升。
From a line graph you can read intermediate values (interpolation) or extend the line sensibly to predict (extrapolation). If the temperature continued rising by about 1°C per day, Monday might be 28°C, but scientists know extrapolation is only an estimate.
从折线图中可以读取中间值（内插），或合理延长线来进行预测（外推）。如果温度每天大约上升1°C，那么下周一可能为28°C，但科学家知道外推只是一种估算。
ity to match scales. For example, in June rainfall is 65 mm and temperature is 20°C. Questions may ask: 'How much more rain fell in October than in September?' 74 mm – 58 mm = 16 mm. Interpret carefully.
气候图常在同一坐标系中结合表示降雨量(mm)的条形图和表示温度(°C)的折线。阅读双轴图考验对刻度的匹配能力。例如，6月降雨量65毫米，气温20°C。问题可能问：“10月比9月多下了多少雨？” 74 毫米 – 58 毫米 = 16 毫米。需仔细解读。
9. Density, Mass and Volume in Physics | 物理学中的密度、质量和体积
Density tells us how much mass is packed into a given volume. The formula is:
密度告诉我们给定体积内包含多少质量。其公式为：
density = mass ÷ volume
looked up in a table to confirm the material. Different materials have different densities – for example, gold is 19.3 g/cm³ and pine wood is about 0.5 g/cm³.
一块铝的质量为540克，体积为200立方厘米。其密度为 540 ÷ 200 = 2.7 克/立方厘米。可以在表中查到这个数值来确认材料。不同材料有不同密度，例如金子为19.3克/立方厘米，松木约为0.5克/立方厘米。
2700 × 0.4 = 1080 kg. To find volume, rearrange: volume = mass ÷ density. If a 300 g liquid has density 1.2 g/cm³, its volume is 300 ÷ 1.2 = 250 cm³. Practise all three forms.
已知密度和体积求质量时，相乘即可：质量 = 密度 × 体积。一尊花岗岩雕像体积0.4立方米，密度2700千克/立方米，质量为 2700 × 0.4 = 1080 千克。要求体积时，公式变形：体积 = 质量 ÷ 密度。如果300克液体的密度为1.2克/立方厘米，其体积为 300 ÷ 1.2 = 250 立方厘米。三种形式都要练习。
ty = 36 g ÷ 8 cm³ = 4.5 g/cm³. Check the table: that matches titanium. This integrates geometry with physics.
跨学科题目可能会要求你鉴别物质。如果一个边长为2厘米的金属立方体的质量为36克，其体积为 2厘米 × 2厘米 × 2厘米 = 8 立方厘米。密度 = 36 克 ÷ 8 立方厘米 = 4.5 克/立方厘米。查表：这与钛相符。这结合了几何与物理。
10. Averages and Range in Sports Science | 体育科学中的平均数和极差
Coaches use statistics to analyse athletes' 40+147) ÷ 5 = 712 ÷ 5 = 142.4 s. The median is found by ordering the list: 138, 140, 142, 145, 147, the middle value is 142 s. The mode is the most frequent value, here there is none, but sometimes two results are the same.
教练用统计数据来分析运动员的表现。五次800米跑的成绩（单位秒）为：145，138，142，140，147。平均数（均值）为 (145+138+142+140+147) ÷ 5 = 712 ÷ 5 = 142.4 秒。中位数的求法是将数据排序：138, 140, 142, 145, 147，中间值为142秒。众数是最频繁出现的值，此处没有，但有时会有两个相同成绩。
The range (maximum – minimum) = 147 – ent and slower overall speed. Comparing data sets builds critical thinking in physical education.
极差（最大值 – 最小值）= 147 – 138 = 9 秒。极差较小表示表现稳定。如果一位新运动员跑出162秒，平均数和极差都会增加，表明整体速度较慢且一致性较差。比较数据集可以培养体育中的批判性思维。
You might also represent the data on a bar chart or a stem-and-leaf diagram. A stem-and-leaf for the times 138, 140, 142, 145, 147 could show stem 13 | 8, stem 14 | 0 2 5 7, making it easy to see the shape of the distribution. Choose appropriate graphs for different types of data – a skill tested across subjects.
你也可以用条形图或茎叶图来呈现数据。对于时间138, 140, 142, 145, 147，茎叶图可以表示为 茎13 | 叶8，茎14 | 叶0 2 5 7，这样很容易看出分布的形状。为不同类型的数据选择合适的图表是一项跨学科考试中会考查的技能。
11. Symmetry and Tessellations in Art & Design | 艺术与设计中的对称与镶嵌
Artists and designers use reflective and rotational symmetry. A regular pentagon has 5 lines of symmetry and rotational symmetry of order 5 (angle 72°). When tiling a floor, shapes that tessellate must fit together with no gaps. Squares, equilateral triangles and regular hexagons all tessellate, but regular pentagons alone do not because interior angle 108° does not divide 360° exactly. However, pentagons combined with other
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