📚 Year 8 CIE Maths: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练
Mathematics is not an isolated subject – it appears in physics, biology, geography, economics, design, music and sport. In Year 8 CIE Maths, you will often meet problems that mix pure mathematical skills with real-world contexts from other disciplines. This article presents a collection of interdisciplinary practice questions to help you build confidence in applying fractions, percentages, ratios, algebra, geometry and statistics to situations borrowed from science, humanities and the arts. Work through each section and notice how the same number skills keep cropping up in different disguises.
数学并不是一门孤立的学科——它出现在物理、生物、地理、经济、设计、音乐和体育中。在八年级 CIE 数学中，你经常会遇到将纯数学技能与其他学科的真实情境相结合的题目。本文提供了一系列跨学科综合练习题，帮助你自信地把分数、百分数、比和比例、代数、几何以及统计等知识运用到从科学、人文和艺术中借来的情境里。请逐一完成每个小节，你会发现同样的数字技能会以不同的面貌反复出现。
1. Physics: Speed, Distance and Time | 物理：速度、距离与时间
In physics, the formula speed = distance ÷ time n metres per second and kilometres per hour. For example, if a cyclist travels 24 km in 1.5 hours, the average speed is 24 ÷ 1.5 = 16 km/h. To change 16 km/h to m/s, multiply by 1000 and divide by 3600, giving approximately 4.44 m/s. Always check that units match before calculating.
在物理学中，我们用公式 速度＝距离÷时间 来描述运动。八年级的题目可能会给出三个量中的两个，让你求第三个。你可能还需要在米每秒和千米每小时之间进行单位换算。例如，如果一名自行车骑手 1.5 小时骑了 24 km，平均速度就是 24÷1.5＝16 km/h。要把 16 km/h 化成 m/s，需要乘以 1000 再除以 3600，得到约 4.44 m/s。计算前一定要检查单位是否统一。
Practice question 1: A runner completes 400 metres in 50 seconds. Calculate her speed in m/s and in km/h.
练习题 1：一名跑步者用 50 秒跑完 400 米。请计算她的速度，以 m/s 和 km/h 表示。
Practice question 2: A car travels at a constant speed of 54 km/h. How far does it go in 40 minutes?
练习题 2：一辆汽车以 54 km/h 的恒定速度行驶。40 分钟内它行驶了多远？
2. Biology: Populations and Sampling | 生物：种群与取样
Ecologists estimate animal populations using the capture‑ e marked. The estimated population N satisfies the proportion 30/N = 6/40. By cross-multiplying, 6N = 1200, so N = 200. This technique uses equivalent ratios and simple algebra, typical of Year 8 work. You might also be asked to express results as percentages or draw a conclusion from data given in a table.
生态学家使用捕捉‑重捕法来估算动物种群数量，这种方法依赖比例。假设你捕捉了 30 只甲虫并做标记，然后放回。一周后你捕捉了 40 只甲虫，其中 6 只有标记。估算的种群数量 N 满足比例式 30/N＝6/40。通过交叉相乘，6N＝1200，所以 N＝200。这一方法用到了等比和简单代数，正是八年级的典型内容。你也可能被要求将结果用百分数表示，或根据表格给出的数据得出结论。
Practice question: In a pond, 50 fish are caught, tagged and released. Later, 60 fish are caught, and 10 are tagged. Estimate the total number of fish in the pond. Then express the proportion of tagged fish in the second catch as a percentage.
练习题：一个池塘中捕获了 50 条鱼，做标记后放回。后来再次捕获 60 条鱼，其中 10 条有标记。请估算池塘中鱼的总数，并把第二次捕获中有标记鱼的比例用百分数表示。
3. Geography: Scale Drawings and Map Reading | 地理：比例图与地图判读
may ask you to convert a length measured on a map into a real distance, or to calculate an area from a grid. For example, if a nature reserve appears as a rectangle 3 cm by 5 cm on a 1:50 000 map, the actual sides are 3×500 = 1500 m and 5×500 = 2500 m, so the real area is 1500×2500 = 3 750 000 m², or 3.75 km². This uses multiplication, unit conversion and area formulas.
地图用比例尺来表示实际距离。比例尺 1:25 000 表示地图上的 1 cm 代表实际 25 000 cm（即 250 m）。八年级的题目可能要求你把地图上量出的长度换算成实际距离，或者根据网格计算面积。比如，一个自然保护区在 1:50 000 的地图上呈长方形，边长 3 cm 和 5 cm，那么实际边长为 3×500＝1500 m 和 5×500＝2500 m，实际面积就是 1500×2500＝3 750 000 m²，即 3.75 km²。这里用到了乘法、单位换算和面积公式。
Practice question: ind the real perimeter of the field in metres and its area in hectares (1 ha = 10 000 m²).
练习题：在一张比例尺为 1:10 000 的地图上，学校操场是一个边长为 8 cm 的正方形。求操场的实际周长（以米为单位）及其面积（以公顷为单位，1 ha＝10 000 m²）。
4. Economics: Profit, Loss and Discounts | 经济：利润、亏损与折扣
sell goods, percentages help you work out discounts, profit and loss. A shop may offer 15% off an item originally priced at £80. The discount is 0.15×80 = £12, so the sale price is £80 − £12 = £68. If the shopkeeper bought the item for £50 and sold it for £68, the profit is £18, and the percentage profit based on the cost price is (18/50)×100% = 36% age of an amount, percentage change and simple business contexts that appear in economics.
在购买或出售商品时，百分数可以帮助我们计算折扣、利润和亏损。一家商店可能对原价 80 英镑的商品打 85 折（15% 折扣）。折扣额为 0.15×80＝12 英镑，因此售价为 80−12＝68 英镑。如果店主以 50 英镑购入该商品，以 68 英镑售出，利润为 18 英镑，基于成本价的利润率为 (18/50)×100%＝36%。这些问题综合了求一个数的百分之几、百分比变化等知识，以及出现在经济学中的简单商业情境。
Practice question 1: A music store buys headphones for $30 and sells them with a 40% mark-up. During a sale, the store offers an extra 10% off the marked price. Find the final sale price and the store’s profit per headphone.
练习题 1：一家音乐店以 30 美元购入耳机，并以 40% 的加成出售。在促销期间，商店再打九折。求最终售价和商店每副耳机的利润。
Practice question 2: A bicycle is reduced by 25% to £180. What was the original price?
练习题 2：一辆自行车降价 25% 后售价为 180 英镑。原价是多少？
5. Chemistry: Mixtures and Ratios | 化学：混合物与比
pe up or down. If a laboratory needs 250 ml of a solution where water and acid are in the ratio 5:2, the teacher first finds the total number of parts (5+2=7). Each part is 250÷7 ≈ 35.714 ml, so water is 5×35.714 ≈ 178.6 ml and acid is 2×35.714 ≈ 71.4 ml. These calculations involve dividing in a given ratio and rounding to a suitable degree of accuracy.
化学家通过按固定比例混合物质来配制溶液。当需要按比例放大或缩小时，必须保持两种化学物质之间的比例，比如 2:3。如果实验室需要 250 ml 的溶液，其中水与酸的比为 5:2，老师首先要计算总份数（5＋2＝7）。每一份是 250÷7≈35.714 ml，所以水需要 5×35.714≈178.6 ml，酸需要 2×35.714≈71.4 ml。这些计算涉及按给定比例分配和四舍五入到适当的精确度。
Practice question: A weed-killer is made by mixing concentrate and water in the ratio 1:20. How much concentrate and how much water are needed to make 3.15 litres of mixture?
练习题：一种除草剂由浓缩液和水按 1:20 的比例混合制成。要配制 3.15 升混合物，需要多少浓缩液和多少水？
6. Design & Technology: Geometric Constructions and Nets | 设计技术：几何作图与展开图
m with given measurements, calculating the surface area, or constructing triangles using a compass and ruler. For instance, to draw a triangle with sides 6 cm, 8 cm and 10 cm, you must set your compass to 6 cm, draw one side, and then use arcs of radii 8 cm and 10 cm to locate the third vertex. After constructing the triangle, you can check it is right-angled using Pythagoras’ theorem, since 6² + 8² = 10². These skills link geometry, measurement and basic trigonometry concepts.
设计一件产品往往需要准确地绘制形状和展开图。八年级的任务可能包括绘制给定尺寸的圆柱或三棱柱的展开图，计算表面积，或者用圆规和直尺作三角形。例如，要画一个边长分别为 6 cm、8 cm 和 10 cm 的三角形，你需要把圆规张开到 6 cm，画出第一条边，然后用半径 8 cm 和 10 cm 的弧线来确定第三个顶点。作完三角形后，你可以用勾股定理检验它是否为直角三角形，因为 6²＋8²＝10²。这些技能将几何、测量和基本的三角学概念联系在一起。
Practice question: ngle with a protractor. Use the cosine rule (or careful measurement) to verify your result – th the formula a² = b² + c² − 2bc cos A if you have met it, or simply estimate by drawing.
练习题：准确地作出一个边长分别为 5 cm、7 cm 和 9 cm 的三角形。用量角器测量最大的内角。如果已经学过，可尝试用余弦公式 a²＝b²＋c²−2bc cos A 进行验证，或者通过画图进行估算。
7. Music: Fractions and Beats | 音乐：分数与节拍
Rhythm in music is built from fractions of a whole note. A crotchet is ¼ of a semibreve, a quaver is ⅛, and a triplet of three quavers in the time of two makes each one 1/12 of a whole note. Time signatures such as 4/4, 3/4 or 6/8 tell you how many beats are in a bar and which note value gets one beat. Year 8 problems might ask: “In a piece with time signature 6/8, a bar contains a dotted crotchet (3/8), a quaver (1/8), and two semiquavers (1/16 each). How many more semiquavers are needed to fill the bar?” You would add the existing durations, find the missing fraction, and convert to the required note values. This is pure fraction addition and subtraction in a musical context.
音乐中的节奏是由全音符的分数构成的。四分音符是全音符的 ¼，八分音符是 ⅛，而在两拍时间内均匀奏出三个八分音符的三连音中，每个音是全音符的 1/12。拍号如 4/4、3/4 或 6/8 告诉你每小节有几拍，以及以哪个音符为一拍。八年级的题目可能会问：“一首乐曲用 6/8 拍，一个小节包含一个附点四分音符（3/8 拍）、一个八分音符（1/8 拍）和两个十六分音符（每个 1/16 拍）。还需要多少个十六分音符才能填满这一小节？”你需要把已有的时值相加，求出缺失的分数，再换算成指定的音符时值。这就纯粹是在音乐情境中的分数加法与减法。
Practice question: A passage in 4/4 time begins with a minim (½), a dotted crotchet (3/8), and a quaver (1/8). What fraction of the bar is left? Express your answer as a fraction of a semibreve and suggest one possible rhythm to fill it using notes you know.
练习题：一段 4/4 拍的乐谱，开始是一个二分音符（½ 拍）、一个附点四分音符（3/8 拍）和一个八分音符（1/8 拍）。该小节还剩下几分之几的全音符？用全音符的分数表示你的答案，并建议一种用你认识的音符组合来填满它的可能节奏。
8. Sports: Statistics and Averages | 体育：统计与平均数
Coaches and analysts use the three measures of average – mean, median and mode – add the scores (140) and divide by 10, giving 14. The mode is 12 because it occurs most often. To find the median, order the data: 10, 12, 12, 12, 12, 14, 15, 15, 18, 20. With ten numbers, the median is the average of the 5th and 6th values: (12+14)/2 = 13. Discussing which average best represents the player’s typical performance develops your statistical reasoning. Year 8 problems may also include range and simple box plots.
教练和分析人员用三种平均数——平均数（均值）、中位数和众数——来汇总运动表现数据。一名篮球运动员在 10 场比赛中的得分可能是：12、15、10、18、12、14、12、20、15、12。要算均值，把得分相加得到 140，再除以 10，等于 14。众数是 12，因为它出现得最频繁。要算中位数，先把数据排序：10、12、12、12、12、14、15、15、18、20。因为有十个数，中位数是第 5 和第 6 个值的平均数：(12＋14)/2＝13。讨论哪一个平均数最能代表该运动员的典型表现，可以锻炼你的统计推理能力。八年级的题目还可能包括极差和简单的箱线图。
Practice question: A cricketer’s scores in 8 innings are: 8, 24, 0, 36, 14, 14, 52, 14. Calculate the mean, median, mode and range. Which measure would you use to persuade the sports club to keep the player? Justify your choice.
练习题：一名板球运动员在 8 局比赛中的得分是：8、24、0、36、14、14、52、14。计算其均值、中位数、众数和极差。你会用哪一个平均数来说服体育俱乐部留下这名运动员？解释你的选择。
9. Environmental Science: Interpreting Graphs and Data | 环境科学：图表与数据解读
d to find the increase between two dates, express it as a percentage of the original, or predict a future value assuming a steady rate. For example, if CO₂ was 315 ppm in 1960 and 420 ppm in 2020, the increase is 105 ppm. As a percentage of the 1960 level, this is (105/315)×100% ≈ 33.3%. If the trend continues at the same rate, what would the level be in 2040? ch 455 ppm. This sequence uses subtraction, division, multiplication and percentage change – all core Year 8 skills.
环境数据通常用折线图、条形图或饼图来呈现。题目可能会给你一张二氧化碳浓度随时间变化的图，要求你找出两个日期之间的增加值，将其表示为原始水平的百分数，或按恒定速率预测未来的数值。例如，如果二氧化碳浓度在 1960 年是 315 ppm，2020 年是 420 ppm，那么增加量就是 105 ppm。与 1960 年水平相比的百分数为 (105/315)×100% ≈ 33.3%。如果按同样速率持续，2040 年的浓度会是多少？60 年的增加速率为 105 ppm，即每年 1.75 ppm。20 年后需再增加 20×1.75＝35 ppm，达到 455 ppm。这个过程用到了减法、除法、乘法和百分数变化——全部是八年级的核心技能。
Practice question: The table below shows the area of Arctic sea ice each September from 2015 to 2024 (in million km²). Use the data to calculate the percentage decrease from 2015 to 2024. Then work out the average annual decrease and predict the value for 2030 if the trend stays constant.
| Year | 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 | 2024 |
| Area | 4.63 | 4.53 | 4.80 | 4.68 | 4.25 | 3.82 | 4.67 | 4.81 | 4.23 | 3.97 |
练习题：下表显示了 2015 至 2024 年每年 9 月的北极海冰面积（单位：百万 km²）。用数据计算 2015 年到 2024 年的百分比减少量。然后计算出年均减少量，并预测如果趋势保持不变，2030 年的值会是多少。
10. Art: Symmetry, Patterns and Transformations | 艺术：对称、模式与变换
precisely. A shape may be reflected in the line y = x, rotated 90° clockwise about the point (1,2), or enlarged by a scale factor of 2 from the origin. Year 8 tasks often ask you to carry out such transformation on a coordinate grid and compare the original and image to identify which transformation has been used. This brings together coordinates, vectors (for translations) and scale factors. Questions might say: “ 1), (2,4). Rotate triangle A 90° anticlockwise about the origin and give the new vertices.” Applying the rotation rule (−y, x) to each vertex gives (−1,1), (−1,3), (−4,2).
艺术家和设计师利用反射、旋转、平移和缩放来创作图案。在数学中，你需要精确描述一种变换。一个图形可能关于直线 y＝x 反射，绕点 (1,2) 顺时针旋转 90°，或者以原点为中心放大 2 倍。八年级的题目经常要求你在坐标网格上完成这样的变换，并比较原图与像图，判断使用了哪种变换。这综合运用了坐标、向量（用于平移）和比例因子。题目可能会说：“三角形 A 的顶点为 (1,1)、(3,1)、(2,4)。将三角形 A 绕原点逆时针旋转 90°，写出新顶点的坐标。”将旋转规律 (−y, x) 应用于每个顶点，得到 (−1,1)、(−1,3)、(−4,2)。
Practice question: Shape P has vertices (−2,0), (0,2), (2,0), (0,−2). Reflect P in the x-axis, then translate the result by the vector (3, −1). Write the coordinates of the final image. Can you combine the two transformations into a single instruction?
练习题：图形 P 的顶点为 (−2,0)、(0,2)、(2,0)、(0,−2)。先将 P 关于 x 轴对称，然后将得到的结果按向量 (3, −1) 平移。写出最终图像的坐标。你能否将这两步变换合并为一条指令？
11. Mixed Interdisciplinary Challenge | 跨学科综合挑战
Now try a multi-step question that blends skills from several of the topics above. A young scientist records how a plant grows under different light conditions. She measures the height of a bean plant every day for 10 days. The initial height is 2.3 cm. Each day the plant grows by a rate that is 15% of its height the previous day. (a) Calculate the height after 10 days, rounding to the nearest mm. (b) The scientist wants to present the daily heights in a line graph; suggest suitable scales for the axes. (c) The plant pot sits on a shelf that casts a triangular shadow. The shadow area on day 1 is measured using a coordinate grid: vertices (0,0), (5,0), (0,3) in cm. Under a different lamp, the shadow is an enlargement of this triangle with scale factor 2.5. Find the area of the enlarged shadow. (d) If the cost of electricity to run the lamp is £0.18 per hour and the lamp runs for 12 hours a day, what is the total electricity cost over the 10 days? Express your answer as a fraction of £10.
现在来尝试一道综合了上述多个主题的多步应用题。一位小科学家记录了植物在不同光照条件下的生长情况。她每天测量一棵豆苗的高度，连续 10 天。初始高度为 2.3 cm。每天的高度都比前一天增加 15%。（a）计算 10 天后的高度，精确到毫米。（b）她想用折线图展示每天的高度；请你为横轴和纵轴建议合适的刻度。（c）花盆放在一个架子上，投下一个三角形的影子。第一天的影子在坐标网格上测得顶点为 (0,0)、(5,0)、(0,3)，单位为 cm。换了一盏灯后，影子变成该三角形的放大，放大因子为 2.5。求放大后影子的面积。（d）如果电费为每小时 0.18 英镑，且灯每天开 12 小时，10 天的总电费是多少？用 10 英镑的分数表示你的答案。
12. Summary and Revision Tips | 小结与复习建议
Interdisciplinary problems are simply real-life scenarios dressed in the language of science, geography, art or business. The mathematical core remains the same: ratio, percentage, algebra, geometry and statistics. To prepare for these questions, practise extracting numerical information from written descriptions, draw diagrams where possible, and always check that your answer makes sense in the context. Revise by converting everyday observations into quick maths questions – at the supermarket, calculate discounts; on a journey, work out average speed; in the kitchen, adjust a recipe's ratio. This will train your brain to spot the mathematics hiding in all corners of the curriculum.
跨学科问题不过是用科学、地理、艺术或商业的语言包装起来的真实生活场景。其数学核心仍然是：比和比例、百分数、代数、几何和统计。为准备这类题目，要多练习从文字描述中提取数字信息，尽可能画图辅助思考，并始终检查答案在给定情境中是否合理。复习时，可以把日常观察变成一道道速算题——在超市，计算折扣；旅途中，算一算平均速度；在厨房里，按比例调整食谱。这样训练，你的大脑就能敏锐地发现隐藏在课程各个角落的数学。
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