📚 Year 7 AQA Further Maths: Interdisciplinary Problem-Solving Training | Year 7 AQA 进阶数学：跨学科综合题型训练
Interdisciplinary problem‑solving puts your mathematical skills into real‑world contexts taken from science, geography, economics and beyond. In Year 7 Further Maths, you are expected to apply fractions, percentages, ratios, equations, area, volume and data handling to unfamiliar situations. This training set will guide you through typical AQA‑style exercises, showing how the same core mathematics appears again and again across different subjects.
跨学科解题训练是把数学技能放进科学、地理、经济学等真实情境中。Year 7 进阶数学要求你将分数、百分数、比、方程、面积、体积和数据处理应用到全新场景。以下训练围绕 AQA 常见题型展开，让你看到同一核心数学知识如何在各个学科中反复出现。
1. Physics: Speed, Distance and Proportion | 物理：速度、距离与比例
In physics, the relationship between speed, distance and time is governed by direct proportion. When speed stays the same, distance and time increase together. A typical question might ask: "A cyclist travels 24 km in 1.5 hours at a constant speed. How far can she travel in 2.5 hours?"
在物理中，速度、距离与时间的关系由正比例控制。速度不变时，距离与时间同步增加。典型题目如：“一位自行车手以恒定速度 1.5 小时行驶 24 公里，她 2.5 小时能骑多远？”
5 = 40 km. The same structure works when you are given two speeds and need to average them – ul: average speed is total distance divided by total time, not the mean of the two speeds.
先求速度：速度 = 距离 ÷ 时间 = 24 ÷ 1.5。把 1.5 写作 3/2，得到 24 ÷ 3/2 = 24 × 2/3 = 16 km/h。再算 2.5 小时：距离 = 16 × 2.5 = 40 km。当题目给出两段不同速度并要求平均速度时，务必注意：平均速度是总距离除以总时间，而不是两个速度的平均值。
or example changing m/s to km/h. Since 1 m/s = 3.6 km/h, a speed of 10 m/s becomes 36 km/h – a useful rule built on multiplying by 3.6.
单位换算也用正比例，例如 m/s 转换成 km/h。因为 1 m/s = 3.6 km/h，所以 10 m/s 即 36 km/h——记住这个乘以 3.6 的实用规则。
2. Chemistry: Mixtures, Ratios and Fractions | 化学：混合物、比与分数
Chemistry practicals often involve mixing solutions or substances in specific ratios. A cleaner is made by mixing concentrate and water in the ratio 1 : 9. If you need 2.5 litres of cleaner, how much concentrate do you use?
化学实验中常常按特定比混合溶液或物质。一种清洁剂由浓缩液与水按 1 : 9 混合而成。如果需要 2.5 升清洁剂，需要多少浓缩液？
ed = (1/10) × 2.5 = 0.25 litres, or 250 ml. This is a simple ratio problem, but it becomes "cross‑curricular" when linked to safety concentration levels.
总份数为 1+9=10。浓缩液占混合物的 1/10。2.5 升中需浓缩液 (1/10) × 2.5 = 0.25 升，即 250 毫升。这是一个简单的比的问题，但当与安全浓度联系起来时，便成为跨学科题目。
. If 40 g of reactant A produces 56 g of product, how much product is made from 100 g of A? 0) ÷ 40 = 140 g. This is the heart of stoichiometry, which is simply fraction equivalence.
另一常见题型用比例因子调整化学反应用量。若 40 g 反应物 A 生成 56 g 产物，100 g A 能生成多少？比值不变，因此建立等效分数：40/56 = 100/x，解得 x = (56 × 100) ÷ 40 = 140 g。这就是化学计量学的核心——本质上就是分数对等。
3. Biology: Populations, Sampling and Statistics | 生物：种群、取样与统计
Ecologists collect data about populations using sampling methods. Imagine a field where a student captures 50 woodlice, marks them and releases them. A day later she captures 60 woodlice, and 12 of them are marked. Estimate the total population.
生态学家通过取样收集种群数据。设想一块田地，一位学生捕获 50 只潮虫，做标记后放回。一天后她再捕获 60 只，其中 12 只有标记。估计总种群数量。
This is the Lincoln– × 60) ÷ 12 = 3000 ÷ 12 = 250. The estimated population is 250 woodlice. The calculation relies on the assumption that marked and unmarked individuals mix randomly – a key concept of proportionality.
这叫做 Lincoln–Petersen 指数：总种群 N = (M × C) ÷ R，其中 M 为首次标记数（50），C 为第二次捕获总数（60），R 为再次捕获到的标记数（12）。N = (50 × 60) ÷ 12 = 3000 ÷ 12 = 250。估计总数为 250 只潮虫。计算假设标记与未标记个体随机混合——这是比例概念的关键。
You may then be asked to compare two habitats using a bar chart or line graph of species counts. Practice reading scales, calculating the mean, and identifying the range. These data skills are directly transferable to geography and economics.
接下来可能要求你使用柱状图或折线图比较两个栖息地的物种数量。练习读刻度、计算平均值、找出极差。这些数据技能在地理和经济学科中同样适用。
4. Geography: Maps, Scale and Bearings | 地理：地图、比例尺与方位角
500 m or 0.5 km. Therefore, a distance of 3.5 cm on the map equals 3.5 × 0.5 km = 1.75 km.
地图离不开比例尺。若比例尺为 1 : 50 000，那么地图上 1 厘米代表实际 50 000 厘米，即 500 米或 0.5 千米。因此，图上 3.5 厘米的距离相当于实际 3.5 × 0.5 km = 1.75 km。
Bearings are measured clockwise from north, always given as three figures (e.g. 045°). In interdisciplinary exercises you might need to use a protractor to measure a bearing and then use the scale to find the distance between two points. This blends geometry (angles) with proportion (scale factor).
方位角从正北顺时针度量，始终用三位数表示（如 045°）。跨学科练习中，你可能需要用量角器测量方位角，再用比例尺计算两点距离。这融合了几何（角）与比例（比例因子）。
0 000 cm². Change to m² by dividing by 10 000 (since 1 m² = 10 000 cm²), so real area = 8 × 10 000 = 80 000 m². Always be careful to square the linear scale for area.
面积也可能出现：某农场在地图上依比例尺 1 : 10 000 画出，图上面积为 8 cm²。实际面积（以 m² 计）为 8 × (10 000)² cm² = 8 × 100 000 000 cm²。换算成 m² 需除以 10 000（1 m² = 10 000 cm²），故实际面积 = 8 × 10 000 = 80 000 m²。注意处理面积时必须将线性比例尺平方。
5. Economics: Interest, Percentages and Budgeting | 经济学：利息、百分数与预算
Simple interest problems are common in economics‑linked maths. A savings account pays 4% simple interest per year. If you invest £300, how much interest do you earn in 3 years?
单利问题在经济数学中很常见。一个储蓄账户年单利 4%。存入 £300，3 年后获得多少利息？
Interest per year = 4% of £300 = 0.04 × 300 = £12. Over 3 years: 3 × 12 = £36. The total amount is £300 + £36 = £ ompound interest appears in some Year 7 extension tasks, where you calculate 300 × (1.04)³ – a chance to use index notation.
年利息 = 300 的 4% = 0.04 × 300 = £12。3 年：3 × 12 = £36。总额为 £300 + £36 = £336。这结合了百分数增长与乘法。在 Year 7 拓展任务中，也可能出现复利，即计算 300 × (1.04)³——这是使用指数表示法的好机会。
s require you to manage money using all four operations. For instance, a school trip costs £18 per student and the school has a budget of £540. How many students can go? 540 ÷ 18 = 30. Then, if 20% of the trip cost is subsidised, what is the new cost per student? 18 × (1 – 0.2) = 18 × 0.8 = £14.40. This blends percentages, division and real‑life decision‑making.
预算练习需要你运用四则运算管理资金。例如，一次学校旅行每人花费 £18，学校预算为 £540，可供多少名学生参加？540 ÷ 18 = 30 人。若旅行费用有 20% 补贴，每人新成本是多少？18 × (1 – 0.2) = 18 × 0.8 = £14.40。这融合了百分数、除法与现实决策。
6. Design & Technology: Area, Volume and Nets | 设计技术：面积、体积与展开图
In DT, you often need to calculate the amount of material for a box or packaging. A rectangular box has length 20 cm, width 15 cm and height 10 cm. Find the surface area of the net (excluding overlaps).
在设计技术中，常需计算盒子或包装材料用量。一个长方体盒子长 20 cm、宽 15 cm、高 10 cm。求展开图的表面积（不含重叠部分）。
tal surface area = 600 + 400 + 300 = 1300 cm². This is directly linked to the algebra of area, which can be extended to using variables for dimensions: SA = 2(lw + lh + wh).
盒子有三对面：上下 2×(20×15)=600 cm²，前后 2×(20×10)=400 cm²，两侧 2×(15×10)=300 cm²。总表面积 = 600+400+300 = 1300 cm²。这直接关联到面积计算，也可扩展到用变量表示尺寸：SA = 2(lw + lh + wh)。
o hold exactly 3 litres (3000 cm³) and the base is 20 cm by 15 cm, what must its height be? ÷ 300 = 10 cm. Being able to rearrange simple equations is a crucial further maths skill.
体积计算同样常见。若盒子恰好装下 3 升（3000 cm³），底面为 20 cm × 15 cm，高度应为多少？体积 = 长×宽×高 → 3000 = 20×15×h → h = 3000 ÷ (20×15) = 3000 ÷ 300 = 10 cm。会变形简单方程是进阶数学的关键能力。
7. Art: Symmetry, Patterns and Fractions | 艺术：对称、图案与分数
esign repeats every 60° around a centre. How many identical sectors fit in the full circle? 360 ÷ 60 = 6 – e circle, so if you paint 2 sectors blue, the fraction of blue in the design is 2/6 = 1/3.
艺术项目常使用镶嵌与旋转对称。一个图案围绕中心每 60° 重复一次。整个圆能容纳多少个相同扇形？360 ÷ 60 = 6，即 6 阶旋转对称。每个扇形占圆的 1/6，所以若把 2 个扇形涂成蓝色，蓝色所占分数为 2/6 = 1/3。
ect ratio. The scale factor is 20/30 = 2/3. New height = 42 × 2/3 = 28 cm. This is directly taken from similar shapes – a geometry topic with deep cross‑curricular links.
放大与缩小就是比例任务。一张海报尺寸 30 cm × 42 cm。艺术家想按比例缩放到宽 20 cm。比例因子为 20/30 = 2/3。新高度 = 42 × 2/3 = 28 cm。这直接来自相似形——一个具有深厚跨学科联系的几何主题。
mixes blue and red in the ratio 3 : 2. If they need 60 ml of paint, how much blue is used? so blue = 3/5 × 60 = 36 ml. This merges art with the same ratio skills seen in chemistry.
调色同样用到比：为调出紫色，艺术家将蓝与红按 3 : 2 混合。若需要 60 ml 颜料，需用多少蓝色？总份数 5，蓝色 = 3/5 × 60 = 36 ml。这与化学中见到的比技能如出一辙。
8. Music: Rhythms, Fractions and Sequences | 音乐：节奏、分数与数列
bar. If a rhythm includes a dotted minim (1/2 + 1/4 = 3/4 of a bar) plus a quaver (1/8), what fraction of the bar remains?
音符时值基于全音符的分数：四分音符为 1/4，八分音符为 1/8。在 4/4 拍的小节中，四个四分音符填满一小节。若某节奏包含一个附点二分音符（1/2 + 1/4 = 3/4 小节）加一个八分音符（1/8），还剩几分之几小节？
Used fraction = 3/4 + 1/8 = 6/8 + 1/8 = 7/8. Remaining fraction = 1 – 1/8. So one quaver remains. This is pure fraction addition, wrapped in a musical context.
已用分数 = 3/4 + 1/8 = 6/8 + 1/8 = 7/8。剩余分数 = 1 – 7/8 = 1/8。所以还剩一个八分音符的时长。这纯粹是分数加法，套上音乐背景而已。
Sequences appear when repeating rhythmic patterns. A drum beat repeats every 3 beats, a bass line every 4 beats. After how many beats do they coincide for the first time? This is the LCM of 3 and 4: 12. Number patterns like this are fundamental to polyrhythms and also appear in science when two cycles align (e.g. predator‑prey cycles).
数列在重复的节奏型中出现。鼓点每 3 拍重复一次，贝斯线每 4 拍重复一次。多少拍后它们第一次同时开始？这是 3 和 4 的最小公倍数：12。这类数字规律是复合节奏的基础，也出现在两个周期对齐的科学情境中（如捕食者‑猎物周期）。
9. Environmental Science: Carbon Footprint and Simple Equations | 环境科学：碳足迹与简单方程
ut 120 g of CO₂ per km. If a family drives 250 km in a week, how many grams of CO₂ is that? Use direct proportion: 120 × 250 = 30 000 g, or 30 kg.
一项环境研究报告称汽车每公里排放约 120 g CO₂。若某家庭一周驾车 250 公里，排放多少克 CO₂？用正比例：120 × 250 = 30 000 g，即 30 kg。
of trees T needed to offset W kg of CO₂ per year: T = W ÷ 7. If W = 210 kg, how many trees? T = 210 ÷ 7 = 30 trees. This is a simple linear equation in a real‑world setting, encouraging algebraic thinking.
再考虑每年吸收 7 kg CO₂ 的树。写出抵消每年 W kg CO₂ 所需树木数量 T 的方程：T = W ÷ 7。若 W = 210 kg，需要多少树？T = 210 ÷ 7 = 30 棵树。这是一个真实情境下的简单线性方程，培养代数思维。
Comparative problems can also involve percentages: If a recycle centre reduces waste by 35% from 800 tonnes, the new waste amount is 800 × (1 – . Chain calculations like this test your ability to combine percentages and large numbers.
比较类问题也可涉及百分数：若回收中心将 800 吨垃圾减少 35%，新垃圾量为 800 × (1 – 0.35) = 800 × 0.65 = 520 吨。这种链式运算考验你结合百分数和大数的能力。
10. Sports: Statistics, Averages and Graphs | 体育：统计、平均数与图表
Sports analytics uses numbers to evaluate performance. A basketball player scores the following points over 6 games: 12, 18, 9, 24, 15, 6. Find the mean, median and range.
体育数据分析用数字评估表现。一名篮球运动员 6 场比赛得分如下：12, 18, 9, 24, 15, 6。求平均数、中位数和极差。
. Arrange in order: 6, 9, 12, 15, 18, 24. Median = (12 + 15) ÷ 2 = 13.5 points. Range = 24 – atistics are identical to those used in biology and geography, just in a sporting context.
平均数 = 总和 ÷ 数值个数 = (12+18+9+24+15+6) ÷ 6 = 84 ÷ 6 = 14 分。排序：6, 9, 12, 15, 18, 24。中位数 = (12+15) ÷ 2 = 13.5 分。极差 = 24 – 6 = 18 分。这些统计量与生物、地理中使用的完全相同，只是换上了体育背景。
Fitness tests often collect time data. A swimmer records 50 m lap times in seconds: 42, 38, 40, 37, 39. Draw a line graph and describe the trend. The improvement can be represented as a decreasing sequence, and you might calculate the mean improvement per lap. This bridges time series graphs with basic algebra of change.
体能测试常收集时间数据。一位游泳者记录 50 米每趟用时（秒）：42, 38, 40, 37, 39。画折线图并描述趋势。进步可用递减数列表示，并可计算每趟平均提升。这连接了时间序列图与基本的“变化”代数。
11. Integrated Challenge: Planning a School Fair | 综合挑战：策划校园义卖
This final challenge combines several skills in one task. Your school wants to run a fair: a stall sells cakes at £1.20 each, and the ingredients cost £0.45 per cake. There is a fixed hire charge of £15 for the stall. Write an expression for the profit if n cakes are sold.
最后这道挑战把多种技能合而为一。学校要办义卖：一个摊位出售蛋糕，每个 £1.20，单个原料成本 £0.45。摊位固定租赁费 £15。写出销售 n 个蛋糕的利润表达式。
Revenue = 1.20n, cost = 0.45n + 15. Profit = revenue – cost = 1.20n – (0.45n + 15) = 0.75n – 15. This is algebra built from a real‑life situation.
收入 = 1.20n，成本 = 0.45n + 15。利润 = 收入 – 成本 = 1.20n – (0.45n + 15) = 0.75n – 15。这正是由真实情境构建的代数式。
Now answer: how many cakes must be sold to break even? Set profit = 0: 0.75n – 15 = 0 → 0.75n = 15 → n = 15 ÷ 0.75 = 20 cakes. If the school wants a profit of at least £30, solve 0.75n – 30 → 0.75n ≥ 45 → n ≥ 60 cakes. Finally, represent this on a number line. All of these sub‑tasks involve equations, inequalities, fractions and decimals – the full set of Year 7 further maths tools.
再回答问题：必须售出多少蛋糕才能盈亏平衡？令利润=0：0.75n – 15 = 0 → 0.75n = 15 → n = 15 ÷ 0.75 = 20 个。若学校希望利润至少 £30，解 0.75n – 15 ≥ 30 → 0.75n ≥ 45 → n ≥ 60 个。最后在数轴上表示。这些子任务全面覆盖了方程、不等式、分数和小数——Year 7 进阶数学的全部工具。
12. Key Takeaways for Interdisciplinary Success | 跨学科成功的关键要点
Across all these subjects, the message is the same: identify the mathematical structure – is it a ratio, a fraction, a linear relationship, an average? – then apply the method you know. Do not be distracted by unfamiliar vocabulary; translate it into numbers and symbols. Practice by creating your own problems, swapping contexts between physics, music or geography. The more you connect concepts, the deeper your understanding becomes.
所有这些学科传递出同一个信息：认清数学结构——这是比例、分数、线性关系还是平均值？——然后运用已知方法。不要被不熟悉的词汇干扰，把它们转换为数字和符号。主动编题练习，把物理、音乐、地理的背景互换。概念联系得越多，理解就越深入。
Published by TutorHao | Further Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图