📚 Year 7 CIE Advanced Mathematics: Interdisciplinary Integrated Problem-Solving Practice | 7年级CIE进阶数学：跨学科综合题型训练
Welcome to the Year 7 CIE Advanced Mathematics revision guide, focused on interdisciplinary integrated problem-solving. This special topic trains you to apply mathematical thinking to real-world contexts across science, geography, economics, and design. By blending numerical, algebraic, and geometric skills with other subjects, you will develop a deeper understanding of how mathematics describes the world. Each section presents paired English and Chinese explanations, followed by practical examples to build confidence for CIE-style assessments.
欢迎来到7年级CIE进阶数学的跨学科综合题型训练专题。本指南帮助你运用数学思维解决科学、地理、经济和设计等领域的实际问题。通过将算术、代数与几何技能融入其他学科，你将对数学如何描述世界产生更深刻的理解。每个部分都提供中英对照的讲解和实例，助你轻松应对CIE风格的评估。
1. Why Interdisciplinary Problems Matter | 为什么跨学科问题很重要
Mathematics does not exist in isolation; it is the language of science, technology, and everyday decision-making. In CIE Advanced Mathematics for Year 7, you will often encounter questions that mix pure math with contexts like speed, mixtures, map scales, or artistic patterns. Training in interdisciplinary problem-solving sharpens your ability to extract relevant data, choose appropriate operations, and interpret results meaningfully. It also prepares you for higher-level IGCSE and beyond.
数学并非孤立存在，它是科学、技术和日常决策的语言。在7年级CIE进阶数学中，你常会遇到混合了速度、混合物、地图比例尺或艺术图案等情境的题目。跨学科解题训练能提升你提取数据、选择合适的运算并合理诠释结果的能力，同时也为更高级别的IGCSE及未来学习做好准备。
2. Physics in Action: Speed, Distance, and Time | 物理应用：速度、距离与时间
A cyclist travels at a constant speed of 12 km/h for 45 minutes. How far does the cyclist go? Here you must convert time into hours (45 min = 0.75 h), then use the formula: distance = speed × time. So, distance = 12 × 0.75 = 9 km. Another common variation asks for average speed when distances or times vary. You need to calculate total distance and total time before dividing.
一名骑自行车的人以12 km/h的恒定速度骑行45分钟，他骑了多远？你需要先把时间换算为小时（45 分钟 = 0.75 小时），再用公式：距离 = 速度 × 时间，得出 12 × 0.75 = 9 km。另一种常见变化是求不同路段或时间的平均速度，必须先算出总距离和总时间再相除。
Practice: A car travels 30 km in 0.5 h, then another 50 km in 1.5 h. Find the average speed for the whole journey. Solution: total distance = 80 km, total time = 2 h, so average speed = 80 ÷ 2 = 40 km/h. Notice the units always remain consistent.
练习：一辆汽车先以0.5小时行驶30 km，又以1.5小时行驶50 km。求全程的平均速度。解答：总距离 = 80 km，总时间 = 2 h，平均速度 = 80 ÷ 2 = 40 km/h。注意单位始终保持一致。
3. Chemical Mixtures: Ratios and Concentration | 化学混合物：比例与浓度
Chemists often prepare solutions by mixing a solute with water in a specific ratio. Suppose a saline solution requires salt and water in the ratio 1 : 20 by mass. To make 420 grams of solution, find the mass of salt needed. The total parts = 1 + 20 = 21 parts. One part mass = 420 ÷ 21 = 20 g, so salt = 20 g. Such proportional reasoning is a core Year 7 skill.
化学家常常按特定比例将溶质与水混合制成溶液。假设一种盐水溶液的质量比为盐:水 = 1:20。要配制420克溶液，需要多少克盐？总份数 = 1+20=21份，每份质量 = 420÷21=20 g，因此盐为20 g。这类比例推理是7年级的核心技能。
Another scenario: You have a 30% sugar solution and you want to increase the concentration to 40% without changing the amount of sugar. You must evaporate some water. Let the original mass be 200 g. Sugar mass = 30% × 200 = 60 g. For a 40% solution, 60 g should be 40% of the new total mass, so new total = 60 ÷ 0.40 = 150 g. Water to remove = 200 – 150 = 50 g.
另一种情境：你有200克30%的糖溶液，想在不改变糖量的情况下将浓度提升到40%，需要蒸发掉一些水。糖的质量 = 200 × 30% = 60 g。对于40%溶液，60 g应为新总质量的40%，新总质量 = 60 ÷ 0.40 = 150 g，因此需蒸发掉 200 – 150 = 50 g 水。
4. Biological Growth: Exponential Thinking | 生物应用：指数式生长思维
Bacteria often double in number at regular intervals. If a colony starts with 5 bacteria and doubles every hour, the population after 3 hours is calculated as: 5 × 2³ = 5 × 8 = 40. Although Year 7 students may not have formally learned exponents, repeated multiplication is accessible. The pattern uses powers of 2: after n hours, population = 5 × 2ⁿ.
细菌常常每隔固定时间数量翻倍。如果一个菌落从5个细菌开始，每小时翻倍一次，3小时后的数量为：5 × 2³ = 5 × 8 = 40。虽然7年级学生可能尚未正式学习指数，但重复乘法完全可以理解。其模式使用的是2的幂：n 小时后，数量 = 5 × 2ⁿ。
A practical tree-planting problem: In a rewilding project, the number of trees planted increases by a factor of 1.5 each year. In Year 1, 200 trees are planted. Estimate the number planted in Year 3. Multiply 200 × 1.5 = 300 (Year 2), then 300 × 1.5 = 450 (Year 3). This introduces the idea of geometric sequences.
一道植树问题：在再野化项目中，每年种植的树木数量以上一年的1.5倍增长。第一年种植200棵，估算第三年的种植量。200 × 1.5 = 300（第二年），300 × 1.5 = 450（第三年）。这里引入了等比数列的初步概念。
5. Geographical Scales: Map Reading and Area | 地理应用：比例尺与面积
A map uses a scale of 1 : 50 000. If two towns are 4 cm apart on the map, what is the real distance in kilometres? Real distance = 4 cm × 50 000 = 200 000 cm = 2 km. You must be comfortable converting between cm, m, and km: 100 cm = 1 m, 1000 m = 1 km.
一张地图的比例尺为 1:50 000。如果两镇在地图上相距4 cm，实际距离是多少千米？实际距离 = 4 cm × 50 000 = 200 000 cm = 2 km。你必须熟练地在cm、m和km之间换算：100 cm = 1 m，1000 m = 1 km。
Area calculations on maps require squaring the scale factor. A nature reserve appears as a rectangle of 3 cm by 5 cm on a 1:25 000 map. Actual length = 5 × 25 000 cm = 125 000 cm = 1250 m = 1.25 km; width = 3 × 25 000 = 75 000 cm = 0.75 km. Area = 1.25 × 0.75 = 0.9375 km². Alternatively, find map area 15 cm², then multiply by 25 000² (625 000 000) and convert to km². Both methods test careful unit handling.
地图上的面积计算需要将比例尺平方。一个自然保护区在1:25 000地图上呈3 cm × 5 cm的矩形。实际长度 = 5 × 25 000 = 125 000 cm = 1.25 km；宽度 = 3 × 25 000 = 75 000 cm = 0.75 km，面积 = 1.25 × 0.75 = 0.9375 km²。你也可以先算地图面积15 cm²，再乘以25 000²（625 000 000）并换算为km²。两种方法都考验单位换算的细致度。
6. Economics and Finance: Profit, Loss, and Discount | 经济与金融：利润、亏损与折扣
Mathematics for business contexts appears frequently. A shop buys a scarf for £15 and sells it for £24. The profit is £24 – £15 = £9. The percentage profit based on cost price = (9 ÷ 15) × 100% = 60%. If a discount of 20% is offered on the selling price, the new price becomes £24 × 0.80 = £19.20. Use multipliers to work efficiently with percentages.
商业情境中的数学问题很常见。一家商店以£15购入围巾，以£24卖出。利润 = £24 – £15 = £9。基于成本价的利润率 = (9 ÷ 15) × 100% = 60%。如果在售价基础上打八折，新售价 = £24 × 0.80 = £19.20。使用乘数能高效地处理百分比。
A more integrated problem: A mobile phone originally costs £320. It is increased by 15% and then later decreased by 10%. Is the final price higher or lower than the original? After increase: £320 × 1.15 = £368; after decrease: £368 × 0.90 = £331.20. The final price is £11.20 higher. This shows that a successive percentage increase and decrease do not cancel out.
一道综合题：一部手机原价£320，先涨价15%，随后又降价10%。最终价格比原价高了还是低了？涨价后：£320 × 1.15 = £368；降价后：£368 × 0.90 = £331.20，比原价高£11.20。这说明连续的百分比增减并不会相互抵消。
7. Art and Design: Geometry, Symmetry, and Tessellations | 艺术与设计：几何、对称与密铺
Artists use transformations to create patterns. Reflecting a shape across the y-axis while keeping its distance equal demonstrates symmetry. Rotating an equilateral triangle by 120° about its centre maps it onto itself. These concepts link algebra (coordinates) with visual design. Practice plotting points like (2,3), (-2,3) after reflection in the y-axis.
艺术家利用变换来创造图案。将图形关于y轴反射并保持距离相等，就展现了对称。将等边三角形绕其中心旋转120°会让它自身重合。这些概念将代数（坐标）与视觉设计联系起来。试着绘制点(2,3)以及反射后的(-2,3)。
Tessellations cover a plane without gaps using regular or irregular tiles. Only equilateral triangles, squares, and regular hexagons can tessellate by themselves. However, students can design their own tessellating shape by modifying a square. This requires calculating interior angles: for a regular polygon with n sides, interior angle = (n-2)×180°/n. For a hexagon, (6-2)×180°/6 = 120°.
密铺是指用规则或不规则的图形无缝隙地覆盖平面。只有等边三角形、正方形和正六边形可以单独密铺。但学生可以通过改造正方形来设计自己的密铺形状。这需要计算内角：n 边形内角和 = (n-2)×180°，每个内角 = (n-2)×180°/n。对正六边形，(6-2)×180°/6 = 120°。
8. Environmental Science: Data Handling and Graphs | 环境科学：数据处理与图表
Interpreting line graphs, bar charts, and scatter plots is crucial when studying climate or pollution. Given a table of monthly rainfall in millimetres, you might be asked to calculate the total rainfall for the year, identify the mode month, or draw a line graph. You might also compute the mean monthly rainfall: sum of 12 values divided by 12.
在研究气候或污染时，解读折线图、条形图和散点图至关重要。给定一张月降水量（毫米）表格，你可能需要计算年总降水量、找出降水最多的月份，或者绘制折线图。你还可能需要计算月平均降水量：12个数值的总和除以12。
Consider a data set of carbon dioxide concentration over 10 years. After plotting, a general upward trend is clear. A rate of change can be estimated: if concentration rises from 380 ppm to 420 ppm over 10 years, the average annual increase is (420-380)÷10 = 4 ppm per year. This uses the gradient formula in a real context.
考虑一组10年内二氧化碳浓度的数据。作图后可发现明显的上升趋势。可以估算变化率：如果浓度从380 ppm升至420 ppm跨越10年，平均年增量为(420-380)÷10 = 4 ppm/年。这里在实际情境中运用了斜率公式。
9. Engineering and Technology: Scale Drawings and 3D Volume | 工程与技术：比例图和立体体积
Engineers use scale drawings to represent objects. A model car is built to a scale of 1:18. If the real car is 4.5 m long, the model length = 4.5 ÷ 18 = 0.25 m or 25 cm. Volume scales by the cube of the linear scale. The real car’s interior volume is 10 m³; the model’s volume will be 10 ÷ (18³) = 10 ÷ 5832 ≈ 0.00171 m³, which converts to 1710 cm³. This vividly shows why models feel so light.
工程师用比例图表示物体。一辆汽车模型按1:18制作。如果实车长4.5 m，模型长度 = 4.5 ÷ 18 = 0.25 m，即25 cm。体积按线性比例的立方缩放。若实车内部空间为10 m³，模型体积为 10 ÷ (18³) = 10 ÷ 5832 ≈ 0.00171 m³，换算为1710 cm³。这生动地说明了模型为何如此轻巧。
For composite 3D shapes, like a cylinder topped with a cone, you may need to find total volume. The formula for cylinder volume is V = πr²h, and for a cone V = (1/3)πr²h. All measurements must be in the same units. Use π ≈ 3.14 or leave in terms of π if asked.
对于复合立体图形，如圆柱上方加一个圆锥，你可能需要计算总体积。圆柱体积公式 V = πr²h，圆锥 V = (1/3)πr²h。所有测量单位必须统一。根据要求可令π ≈ 3.14或保留π。
10. Sports Analytics: Averages and Probability | 体育分析：平均数和概率
A basketball player’s points over 5 games: 18, 22, 15, 30, 20. Mean = (18+22+15+30+20)÷5 = 105÷5 = 21 points. Median: order the list (15,18,20,22,30), middle value is 20. Mode: no repeat, so no mode. Range = 30 – 15 = 15. These statistics can be compared between players.
一名篮球运动员在5场比赛中的得分：18、22、15、30、20。平均数 = (18+22+15+30+20)÷5 = 105÷5 = 21分。中位数：排序后为15,18,20,22,30，中间值为20。众数：无重复，故无众数。极差 = 30-15=15。可以在运动员之间比较这些统计量。
Probability helps predict outcomes. If the player’s free-throw success rate is 70%, the probability of making both of two free throws is 0.7 × 0.7 = 0.49 or 49%. A tree diagram can illustrate all possibilities: hit-hit, hit-miss, miss-hit, miss-miss. The probability of missing both is 0.3 × 0.3 = 0.09.
概率有助于预测结果。如果该球员罚球成功率为70%，那么连续两次都命中的概率为0.7 × 0.7 = 0.49即49%。树状图可以展示全部可能性：中-中、中-不中、不中-中、不中-不中。两次皆不中的概率为0.3 × 0.3 = 0.09。
11. Food Technology: Recipe Scaling and Proportions | 食品工艺：配方比例缩放
A pancake recipe for 4 people uses 200 g flour, 2 eggs, and 300 ml milk. To adapt for 10 people, find the scaling factor: 10 ÷ 4 = 2.5. Then multiply each quantity: flour = 200 × 2.5 = 500 g; eggs = 2 × 2.5 = 5; milk = 300 × 2.5 = 750 ml. This is direct proportion, a fundamental concept in applied mathematics.
一份供4人食用的煎饼配方使用200 g面粉、2个鸡蛋和300 ml牛奶。调整到供10人食用，比例因子 = 10÷4 = 2.5。然后各项材料乘以2.5：面粉 = 200×2.5=500 g；鸡蛋 = 2×2.5=5个；牛奶 = 300×2.5=750 ml。这体现了正比例关系，是应用数学的基本概念。
Sometimes ingredients are given as percentages of total mass. If the total dough mass is 1.2 kg and butter constitutes 15%, butter mass = 0.15 × 1200 g = 180 g. Students must be able to convert between grams and kilograms seamlessly.
有时配方中的成分以总质量的百分比给出。如果面团总质量为1.2 kg，黄油占15%，则黄油质量 = 0.15 × 1200 g = 180 g。学生必须能熟练地在克与千克之间换算。
12. Integrated Challenge: Designing a School Garden | 综合挑战：设计学校花园
Now combine multiple cross-curricular skills. Your task is to design a rectangular school garden measuring 12 m by 8 m. Paths 1 m wide run along the perimeter inside the garden. Calculate the area available for planting. The inner rectangle dimensions will be reduced by 2 m in both length and width (paths on both sides), so planting length = 12 – 2 = 10 m, width = 8 – 2 = 6 m, area = 60 m².
现在综合运用多种跨学科技能。任务：设计一个长12 m、宽8 m的矩形学校花园。花园内部沿周边有1 m宽的小路。计算可种植区域的面积。内部矩形长宽均因两侧小路各减2 m，因此种植区长 = 12-2=10 m，宽 = 8-2=6 m，面积 = 60 m²。
You plan to fill the planting area with soil to a depth of 0.3 m. Volume of soil needed = 60 × 0.3 = 18 m³. Soil is sold in bags of 0.75 m³. Number of bags = 18 ÷ 0.75 = 24 bags. If each bag costs £4.50, total cost = 24 × 4.50 = £108. If the budget was £120, what percentage remains? Remaining = £12, as a percentage = (12 ÷ 120) × 100% = 10%.
你计划将种植区域填土至0.3 m深。所需土壤体积 = 60 × 0.3 = 18 m³。土壤以每袋0.75 m³出售，袋数 = 18 ÷ 0.75 = 24袋。每袋£4.50，总花费 = 24 × 4.50 = £108。如果预算为£120，剩余金额占预算的比例是多少？剩余£12，百分比 = (12 ÷ 120) × 100% = 10%。
Finally, record the perimeter for fencing: outer rectangle perimeter = 2×(12+8)=40 m, but you only need to fence the outer edge. Interdisciplinary mathematics helps you plan real projects precisely.
最后，计算围栏所需的周长：外部矩形周长 = 2×(12+8)=40 m，你只需围起外边缘。跨学科数学让你精确规划真实项目。
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