📚 Year 7 WJEC Maths: Interdisciplinary Problem-Solving Training | Year 7 WJEC 数学：跨学科综合题型训练
In today's world, mathematics is not an isolated subject but a powerful tool used across all disciplines. From analysing data in geography to calculating proportions in science experiments, the skills you develop in Year 7 WJEC maths can help you solve real problems in a wide variety of contexts. This article presents a collection of interdisciplinary problem-solving questions that mirror the style of the WJEC curriculum, encouraging you to connect mathematical concepts with science, history, art, sport, and everyday life. By working through these examples, you will learn to transfer your knowledge, interpret information from different sources, and build confidence for both class assessments and future challenges.
在今天的世界里，数学不是一门孤立的学科，而是一种跨越所有领域的强大工具。从分析地理数据到计算科学实验中的比例，你在 Year 7 WJEC 数学中学到的技能可以帮助你在各种实际情境中解决问题。本文汇集了一系列跨学科综合题型，这些题目反映了 WJEC 课程风格，鼓励你将数学概念与科学、历史、艺术、体育和日常生活联系起来。通过这些例子，你将学会迁移知识、解释不同来源的信息，并为课堂评估和未来的挑战树立信心。
1. Proportion in Science: Mixing a Chemical Solution | 科学中的比例：配制化学溶液
A student is preparing a salt solution for a biology experiment. The instructions say to dissolve 15 g of salt in 120 ml of water. The student only needs 80 ml of the solution. How many grams of salt should they use so that the concentration stays the same?
一名学生正在为生物实验配制盐溶液。说明书上说要将 15 g 食盐溶解在 120 ml 水中。该学生只需要 80 ml 溶液。为了保持相同浓度，应该使用多少克食盐？
15 g ÷ 120 ml = x g ÷ 80 ml. Solving for x gives x = (15 × 80) ÷ 120 = 1200 ÷ 120 = 10 g.
为了保持浓度不变，盐的质量必须与水的体积成正比。我们可以列出比例式：15 g ÷ 120 ml = x g ÷ 80 ml。求解 x 得 x = (15 × 80) ÷ 120 = 1200 ÷ 120 = 10 g。
Understanding proportions helps scientists mix identical solutions at different scales without altering the outcome of experiments.
理解比例有助于科学家在不同规模下配制出完全相同的溶液，而不改变实验结果。
2. Data Analysis in Geography: Comparing Climate Graphs | 地理中的统计数据分析：比较气候图表
The table below shows average monthly rainfall (mm) for two cities, A and B, over three months. Use it to answer the questions that follow.
下表显示了两座城市 A 和 B 在三个月内的月平均降水量（毫米）。请使用它回答问题。
| Month | City A (mm) | City B (mm) |
|---|---|---|
| October | 58 | 72 |
| November | 64 | 85 |
| December | 71 | 94 |
Calculate the mean monthly rainfall for City A over this period. Which city experienced a larger total rainfall, and by how much?
计算城市 A 在这段时期的月平均降水量。哪座城市的总降水量更大，相差多少？
l place). City B total = 72 + 85 + 94 = 251 mm. City B had more rain by 251 - 193 = 58 mm.
城市 A 总量 = 58 + 64 + 71 = 193 mm。城市 A 的平均值 = 193 ÷ 3 = 64.3 mm（保留一位小数）。城市 B 总量 = 72 + 85 + 94 = 251 mm。城市 B 的降水多出 251 - 193 = 58 mm。
Being able to compare and interpret data like this helps geographers identify climate patterns and support decisions about farming or water supply.
能够像这样比较和解读数据，可以帮助地理学家识别气候模式，并支持农业或供水决策。
3. Symmetry and Tessellation in Art & Design | 艺术与设计中的对称与镶嵌
An artist is designing a floor pattern using regular hexagons. Each hexagon has interior angles of 120°. She wants to check if hexagons can tessellate on their own without gaps. Using the fact that the sum of angles around a point must equal 360°, prove whether regular hexagons tessellate. How many hexagons meet at a single point in this tiling?
一位艺术家正在使用正六边形设计地板图案。每个正六边形的内角为 120°。她想检查正六边形是否能单独无缝隙地密铺平面。利用围绕一个点的角度和必须等于 360° 的事实，证明正六边形是否可以密铺。在这个密铺中，有多少个六边形交于一点？
3 × 120° = 360°
Three regular hexagons can meet at a point because their interior angles add exactly to 360°, leaving no gap. Therefore, hexagons tessellate perfectly, with three meeting at each vertex.
三个正六边形可以交于一点，因为它们的总内角恰好为 360°，没有空隙。因此，六边形可以完美密铺，每个顶点处相交三个六边形。
This geometric property is why we often see hexagonal tiling in art, architecture, and even honeycombs in nature.
这种几何属性解释了为什么我们经常在艺术、建筑甚至天然的蜂巢中看到六边形密铺。
4. Timeline Calculations in History: Eras and Intervals | 历史中的时间线计算：纪元与时间间隔
An ancient coin was minted in 132 BCE. A museum acquires a collection that contains this coin and another minted in 218 CE. How many years separate the two coins? Remember that there is no year 0; the year after 1 BCE is 1 CE.
一枚古币铸造于公元前 132 年。一家博物馆获得一组藏品，包含这枚钱币和另一枚铸造于公元 218 年的钱币。这两枚钱币相隔多少年？请记住，没有公元 0 年；公元前 1 年之后就是公元 1 年。
years. (Think: from 132 BCE to 1 BCE is 131 years, then from 1 CE to 218 CE is 217 years; total 131 + 217 = 348, plus one more year to move from 1 BCE to 1 CE? Actually careful: 132 BCE to 1 BCE inclusive is 132 years? Let' 1 BCE is 131 years; from Jan 1, 1 BCE to Jan 1, 1 CE is 1 year; from Jan 1, 1 CE to Jan 1, 218 CE is 217 years. Total = 131 + 1 + 217 = 349 years.
从公元前某年到公元某年的间隔可以通过将两个数字相加再减一得到。132 + 218 - 1 = 349 年。（设想：从公元前 132 年到公元前 1 年是 131 年，从公元前 1 年到公元 1 年是 1 年，从公元 1 年到公元 218 年是 217 年，总计 349 年。）
Historians and archaeologists rely on such calculations to build accurate chronologies of past events.
历史学家和考古学家依靠这类计算来建立过去事件的精确年表。
5. Measurement in PE: Analysing Athletic Performance | 体育中的成绩测量：分析运动表现
In a PE lesson, four pupils ran 100 metres. Their times in seconds were: Alice 14.2, Ben 12.8, Chloe 15.5, and Dan 13.6. Calculate the range of these times. Then, if the average speed is calculated as distance ÷ time, find Dan's average speed in metres per second to one decimal place.
在一节体育课上，四名学生进行了 100 米赛跑。他们的成绩（秒）为：Alice 14.2, Ben 12.8, Chloe 15.5, Dan 13.6。计算这些时间的范围。然后，如果平均速度 = 距离 ÷ 时间，求 Dan 的平均速度，以米/秒为单位，保留一位小数。
Range = highest time - lowest time = 15.5 - 12.8 = 2.7 seconds. Dan's average speed = 100 m ÷ 13.6 s = 7.35294... m/s ≈ 7.4 m/s (to 1 d.p.).
范围 = 最高时间 - 最低时间 = 15.5 - 12.8 = 2.7 秒。Dan 的平均速度 = 100 m ÷ 13.6 s = 7.35294... m/s ≈ 7.4 m/s（保留一位小数）。
PE teachers use these stats to monitor improvement and compare performance across different sports.
体育老师利用这些统计数据来监测进步情况，并比较不同运动项目的表现。
6. Business Scenario: Profit and Loss in a School Fair | 商业中的盈亏计算：学校义卖会
A group of Year 7 pupils runs a stall at a school fair. They buy 50 cupcakes for £0.45 each and sell them all at £1.20 each. They also spend £6.50 on decorations. Calculate: the total revenue, the total costs, and the profit.
一组七年级学生在学校义卖会摆了一个摊位。他们以每个 0.45 英镑的价格购入 50 个纸杯蛋糕，并以每个 1.20 英镑全部售出。他们还花了 6.50 英镑购买装饰品。计算：总收入、总成本和利润。
Total revenue = 50 × £1.20 = £60.00. Total cost of cakes = 50 × £0.45 = £22.50. Total costs = £22.50 + £6.50 = £29.00. Profit = revenue - costs = £60.00 - £29.00 = £31.00.
总收入 = 50 × 1.20 英镑 = 60.00 英镑。蛋糕成本 = 50 × 0.45 英镑 = 22.50 英镑。总成本 = 22.50 英镑 + 6.50 英镑 = 29.00 英镑。利润 = 收入 - 成本 = 60.00 英镑 - 29.00 英镑 = 31.00 英镑。
Simple business calculations like these form the foundation of financial literacy and enterprise skills.
诸如此类的简单商业计算构成了财商和创业技能的基础。
7. Fractions and Rhythm in Music | 音乐中的分数与节拍
consisting of one minim, one crotchet, and two quavers. Has the bar been filled correctly? (Remember: a minim = 2 beats, a crotchet = 1 beat, a quaver = ½ beat)
一段音乐采用 4/4 拍，意味着每小节有四个四分音符的时值。一位作曲家写了一个小节，包括一个二分音符、一个四分音符和两个八分音符。这个小节的时值正确填满了吗？（记住：二分音符 = 2 拍，四分音符 = 1 拍，八分音符 = ½ 拍）
1 minim (2) + 1 crotchet (1) + 2 quavers (2 × ½ = 1) = 2 + 1 + 1 = 4 beats
The total is exactly 4 beats, so the bar is perfectly filled without any overrun or shortage.
总和恰好为 4 拍，因此这个小节完美填满，没有超出或不足。
Musicians use fraction addition constantly to read and write rhythms correctly.
音乐家不断使用分数加法来正确地读谱和记谱。
8. Scale Drawing in D&T: Designing a Phone Stand | 技术中的测量与缩放：设计手机支架
In Design and Technology, a pupil draws a plan of a wooden phone stand. The actual length of the base is 18 cm. On the plan, the length is drawn as 6 cm. What is the scale factor of the drawing? If the actual height of the stand is 12 cm, what will be the height on the drawing?
在设计与技术课上，一名学生画了一个木制手机支架的平面图。底座实际长度为 18 cm。在平面图上，长度画为 6 cm。图纸的比例尺是多少？如果支架的实际高度是 12 cm，图纸上的高度是多少？
e drawing is a reduction, scale 1:3). To find the height on drawing: 12 cm × (1/3) = 4 cm.
比例尺 = 图纸长度 ÷ 实际长度 = 6 cm ÷ 18 cm = 1/3（图纸为缩小，比例 1:3）。求图纸上的高度：12 cm × (1/3) = 4 cm。
Using scale drawings allows designers and engineers to plan objects accurately before manufacturing them.
使用比例图纸使设计师和工程师能够在制造之前精确地规划物体。
9. Interpreting Graphs in Environmental Science | 环境科学中的图表解释
A line graph shows the concentration of carbon dioxide (CO₂) in the air from 2000 to 2020. The values read from the graph are: 2000: 370 ppm, 2005: 380 ppm, 2010: 390 ppm, 2015: 400 ppm, 2020: 415 ppm. By how many parts per million did CO₂ increase between 2000 and 2020? Estimate the average annual increase over this 20-year period.
一张折线图显示了 2000 年至 2020 年空气中二氧化碳的浓度。从图中读取的数值为：2000 年：370 ppm，2005 年：380 ppm，2010 年：390 ppm，2015 年：400 ppm，2020 年：415 ppm。从 2000 年到 2020 年，二氧化碳浓度增加了多少 ppm？估算这 20 年间的年平均增长量。
Total increase = 415 - 370 = 45 ppm. Average per year = 45 ÷ 20 = 2.25 ppm/year.
总增长 = 415 - 370 = 45 ppm。年平均增长 = 45 ÷ 20 = 2.25 ppm/年。
Environmental scientists interpret such trends to understand climate change and make projections for the future.
环境科学家解读此类趋势，以理解气候变化并预测未来。
10. Everyday Money: Budgeting for a Trip | 日常金钱问题：为一次旅行做预算
A family is planning a day trip. The train tickets cost £34.80 for one adult and two children. They budget £55 for meals and £22.50 for entry fees. If they have saved £120, how much money will be left after the trip? If there are no other expenses, express the remaining amount as a fraction of the total savings, simplified to its lowest terms.
一个家庭正在计划一日游。火车票花费为一名成人和两名儿童共 34.80 英镑。他们为餐费预算 55 英镑，门票费用 22.50 英镑。如果他们存了 120 英镑，旅行结束后还剩多少钱？如果没有其他开支，将余额表示为总储蓄的分数，并化简为最简形式。
Total spending = £34.80 + £55 + £22.50 = £112.30. Amount left = £120 - £112.30 = £ no common factors (77 = 7 × 11, 1200 is 2⁴×3×5²), so the fraction in simplest form is 77/1200.
总支出 = 34.80 英镑 + 55 英镑 + 22.50 英镑 = 112.30 英镑。剩余金额 = 120 英镑 - 112.30 英镑 = 7.70 英镑。余额占总储蓄的 7.70/120。化简：分子分母同乘 100 得 770/12000，再除以 10 得 77/1200。77 和 1200 没有公因数（77 = 7 × 11，1200 = 2⁴×3×5²），所以最简分数为 77/1200。
Managing a budget involves adding, subtracting, and representing numbers in different forms – skills practised throughout Year 7.
管理预算涉及加法、减法和用不同形式表示数字——这些都是在 Year 7 全程需要练习的技能。
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