📚 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
cyclist travels at a constant speed of 12 km/h for 45 minutes. How far does the cyclist go? ation 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。
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
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% t 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% 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
ulated 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。
e 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
: 50 000. If two towns are 4 cm apart on the map, what is the real distance in kilometres? 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。
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 £0 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)。
late 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。
: 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
= 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' 832 ≈ 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。
e 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' 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% t-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
antity: 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。
as percentages of total mass. If the total dough mass is 1.2 kg and butter constitutes 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
rimeter 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。
. 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%。
ence 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。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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