📚 Year 7 OCR Advanced Maths: Cross-Curricular Problem Solving | 7年级OCR进阶数学：跨学科综合题型训练
Mathematics is not just an abstract subject; it is a powerful tool used across all disciplines. In Year 7 OCR Advanced Maths, cross-curricular problem-solving helps you connect mathematical concepts to real-world scenarios in science, geography, design, finance, and more. This article provides engaging exercises that blend maths with other subjects, giving you the skills to think critically and apply your knowledge.
数学不仅仅是一门抽象的学科，它还是跨学科的强大工具。在7年级OCR进阶数学中，跨学科问题解决帮助你将数学概念与科学、地理、设计、金融等现实场景联系起来。本文提供融合数学与其他学科的趣味练习，让你学会批判性思考并应用知识。
1. Ratios in Chemistry and Physics | 化学与物理中的比例
stance, the reaction between hydrogen and oxygen to form water has a mass ratio of H:O = 1:8.
在科学中，比例描述量之间的关系。在化学反应中，反应物和生成物的质量比始终保持恒定。例如，氢气与氧气反应生成水的质量比为H:O = 1:8。
or 5 g of hydrogen. Since the ratio is 1:8, for 5 g hydrogen you need 5 × 8 = 40 g oxygen.
例子：如果2 g氢气与16 g氧气恰好反应，生成水的质量为2+16=18 g。利用这一比例计算5 g氢气需要多少氧气。因为比例是1:8，5 g氢气需要5×8=40 g氧气。
ver lifts a 300 N load with an effort of 100 N, the mechanical advantage is 300 ÷ 100 = 3.
在物理中，比例也出现在简单机械里。杠杆的机械效益是阻力臂与动力臂的比值。如果一根杠杆用100 N的力举起300 N的重物，机械效益为300÷100=3。
2. Scale and Map Reading | 地图比例尺与阅读
Map scales allow you to convert distances on a map to real distances on the ground. A scale of 1:25,000 means 1 cm on the map represents 25,000 cm in reality, which is 250 m.
地图比例尺让你把图上的距离转换为实际地面距离。1:25,000的比例尺意味着图上1 cm代表实地25,000 cm，即250 m。
00 = 320,000 cm = 3,200 m = 3.2 km. You can also use a scale line to measure curved paths.
如果两点在1:50,000地图上相距6.4 cm，实际距离=6.4×50,000=320,000 cm=3,200 m=3.2 km。你还可以使用比例尺线段来测量弯曲的路径。
| Map scale | 1 cm represents |
| 1:10,000 | 100 m |
| 1:25,000 | 250 m |
| 1:50,000 | 500 m |
| 1:100,000 | 1 km |
le 1:12,500. If the route measures 14.8 cm on the map, what is the actual walking distance? Answer: 14.8 × 12,500 cm = 185,000 cm = 1,850 m = 1.85 km.
练习：学校旅行地图的比例尺为1:12,500。如果路线在图上长14.8 cm，实际步行距离是多少？答案：14.8×12,500 cm=185,000 cm=1,850 m=1.85 km。
3. Geometry in Design and Technology | 设计与技术中的几何学
ctangular wooden board of length 120 cm and width 80 cm, the area is 120 × 80 = 9,600 cm².
设计师利用面积和周长计算所需材料。一块长120 cm、宽80 cm的长方形木板，面积为120×80=9,600 cm²。
a ≈ 3.14 × 15² = 3.14 × 225 ≈ 706.5 cm². The remaining area = 9,600 - 706.5 = 8,893.5 cm².
如果切出一个半径为15 cm的圆孔，去掉的面积=π×r²。取π≈3.14，面积≈3.14×15²=3.14×225≈706.5 cm²。剩余面积=9,600-706.5=8,893.5 cm²。
and 30 cm, the area = ½ × base × height = ½ × 40 × 30 = 600 cm². The hypotenuse length is √40² + 30² = √1600 + 900 = √2500 = 50 cm.
三角形支撑在桥梁中很常见。对于一个直角边为40 cm和30 cm的直角三角形，面积=½×底×高=½×40×30=600 cm²。斜边长度=√40²+30²=√1600+900=√2500=50 cm。
4. Data Handling in Biology | 生物中的数据管理
Biologists use graphs to show growth patterns. A bacteria population doubles every hour. Starting with 50 bacteria at 9 am, the numbers are recorded: 9am:50, 10am:100, 11am:200, 12pm:400, 1pm:800.
生物学家用图表展示生长模式。一个细菌种群每小时翻倍。从上午9点开始有50个细菌，记录数据：9:00 50, 10:00 100, 11:00 200, 12:00 400, 13:00 800。
Plot these on a line graph with time on the x-axis and population on the y-axis. The line curves upward, showing exponential growth. From the graph, you can estimate the population at 10:30 am (about 141) and find how long it takes to reach 1000 (around 4 hours 20 minutes).
在折线图上绘制这些数据，时间在x轴，数量在y轴。曲线向上弯曲，表明指数增长。从图中可以估算10:30的数量（约141），并找出达到1000所需的时间（大约4小时20分钟）。
, and 19 cm, the mean = (12+15+10+14+19) ÷ 5 = 70 ÷ 5 = 14 cm. The range = 19 - 10 = 9 cm.
平均数和极差可以概括植物的高度。五株幼苗高度分别为12 cm、15 cm、10 cm、14 cm和19 cm，平均数=(12+15+10+14+19)÷5=70÷5=14 cm。极差=19-10=9 cm。
5. Speed, Distance, and Time in Physics | 物理中的速度、距离与时间
÷ time (s = d/t). If a cyclist travels 36 km in 2 hours, average speed = 36 ÷ 2 = 18 km/h.
速度、距离和时间的关系由公式 速度=距离÷时间 (s=d/t) 给出。如果自行车手2小时骑行36 km，平均速度=36÷2=18 km/h。
600 s) = 5/18. So 18 km/h = 18 × (5/18) = 5 m/s. Conversely, m/s to km/h multiply by 18/5.
要将km/h转换为m/s，乘以(1000 m/3600 s)=5/18。所以18 km/h=18×(5/18)=5 m/s。反过来，m/s转km/h则乘以18/5。
÷ 8 = 25 m/s. In km/h, that’s 25 × 18/5 = 90 km/h. How long to travel 540 km at this speed? Time = d/s = 540 ÷ 90 = 6 hours.
一列火车8秒行驶200 m。速度=200÷8=25 m/s。以km/h计，25×18/5=90 km/h。以该速度行驶540 km需要多长时间？时间=距离/速度=540÷90=6小时。
6. Percentages in Finance | 金融中的百分比
Understanding percentages is vital for managing money. A shop offers 20% off a jacket priced at £45. Discount = 20% of £45 = 0.2 × 45 = £9. Sale price = £45 - £9 = £36.
理解百分比对理财至关重要。一件夹克标价£45，商店打八折（20% off）。折扣额=£45的20%=0.2×45=£9。售价=£45-£9=£36。
Simple interest earned in a savings account: if you deposit £200 at 3% per year, interest after 1 year = 3% × 200 = £6. After 3 years, total interest = £6 × 3 = £18, so balance = £218.
储蓄账户的单利：如果你存入£200，年利率3%，一年后利息=3%×200=£6。三年后总利息=£6×3=£18，余额=£218。
VAT (Value Added Tax) at 20% is added to many items. A laptop costs £400 before VAT. Price after VAT = £400 + 20% of £400 = £400 + £80 = £480. To find the pre-VAT price from a total of £480, divide by 1.2: £480 ÷ 1.2 = £400.
增值税(VAT)按20%加在多数商品上。一台笔记本电脑税前£400。税后价格=£400+£400的20%=£400+£80=£480。要从总价£480倒推税前价，除以1.2：£480÷1.2=£400。
7. Symmetry and Tessellation in Art | 艺术中的对称与镶嵌
Symmetry creates balanced designs. A regular hexagon has 6 lines of symmetry and rotational symmetry of order 6. The interior angle of a regular hexagon is 120°, so six hexagons can meet at a point without gaps, making a tessellation.
对称创造平衡的设计。正六边形有6条对称轴和6阶旋转对称。正六边形的内角为120°，因此六个六边形可以在一点无缝隙地拼接，形成镶嵌。
You can design a pattern by reflecting a shape across a line. If an original triangle has vertices (2,3), (4,3), (3,5) and you reflect it in the y-axis, the image vertices become (-2,3), (-4,3), (-3,5). Repeating reflections builds a frieze pattern.
你可以通过镜面反射设计图案。如果原三角形的顶点是(2,3), (4,3), (3,5)，关于y轴反射后，镜像顶点变为(-2,3), (-4,3), (-3,5)。重复反射可以构造出带状装饰图案。
each vertex. Combining squares and equilateral triangles is not possible at a single vertex because the angles 90° and 60° cannot sum to 360° exactly with integer counts.
仅使用等边三角形（内角60°）的镶嵌可以实现，因为360°÷60°=6个三角形围绕每个顶点。将正方形与等边三角形组合在一个顶点则不行，因为90°和60°无法用整数个恰好拼成360°。
8. Sequences and Logic in Computing | 计算机科学中的序列与逻辑
quences form the basis of algorithms. What is the 10th term of the sequence 5, 11, 17, 23, …? rence d=6. The nth term formula is a + (n-1)d. So the 10th term = 5 + (9×6) = 5 + 54 = 59.
序列是算法的基础。序列5, 11, 17, 23, ...的第10项是什么？这是个等差数列，首项a=5，公差d=6。第n项公式为a+(n-1)d。因此第10项=5+(9×6)=5+54=59。
Flowcharts use decision boxes: 'Is x > 10?' leads to 'Yes' or 'No' d) → 10, 10 → 5, 5 → 16, 16 → 8, 8 → 4, 4 → 2, 2 → 1. This is the famous Collatz sequence.
流程图使用判断框："x > 10?" 引出 "是" 或 "否" 分支。考虑以下逻辑：输入一个数，若是偶数，除以2；若是奇数，乘以3再加1。从6开始，序列：6 (偶)→3, 3 (奇)→10, 10→5, 5→16, 16→8, 8→4, 4→2, 2→1。这就是著名的科拉茨序列。
+ 4 + 0 + 1 = 13 in decimal. Working with powers of 2 is fundamental in computer science.
二进制模式有助于理解数据：二进制数1101代表(1×2³)+(1×2²)+(0×2¹)+(1×2⁰)=8+4+0+1=13（十进制）。运用2的幂是计算机科学的基础。
9. Statistics in Sports | 体育中的统计学
0 ÷ 5 = 18 points. The median after arranging data: 10, 14, 18, 22, 26 is the middle value 18.
教练利用统计数据分析表现。一名篮球运动员五场比赛得分分别为14, 22, 18, 10, 26。平均数=(14+22+18+10+26)÷5=90÷5=18分。将数据排序：10,14,18,22,26，中位数是中间值18。
s median of upper half (22,26) = 24. The interquartile range IQR = Q3 - Q1 = 24 - 12 = 12.
箱线图概括离散程度。上述数据的下四分位数Q1是下半部分(10,14)的中位数=12；上四分位数Q3是上半部分(22,26)的中位数=24。四分位距IQR=Q3-Q1=24-12=12。
Comparing two players: Player A has mean 18, range 16; Player B has mean 20, range 8. Player B is more consistent but scores higher on average. This helps select the right player for different match situations.
比较两名球员：球员A平均18分，极差16；球员B平均20分，极差8。球员B更稳定且平均得分更高。这有助于在不同比赛情形下选择合适球员。
10. Environmental Maths: Carbon Footprint | 环境数学：碳足迹
daily emission = 120 × 45 = 5,400 g = 5.4 kg. In a 5-day week, emission = 5.4 × 5 = 27 kg.
计算碳足迹需结合加法、乘法和单位换算。一辆汽车每公里排放120 g CO₂。对于每天45 km的通勤，日排放量=120×45=5,400 g=5.4 kg。每周5天，排放量=5.4×5=27 kg。
monthly emissions from electricity = 300 × 0.25 = 75 kg. Annual total = 75 × 12 = 900 kg.
如果一户家庭每月用电300 kWh，排放因子为每kWh 0.25 kg CO₂，每月用电碳排放=300×0.25=75 kg。年总量=75×12=900 kg。
Reducing by 15%: a target decrease of 15% k. New emission = 27 - 4.05 = 22.95 kg. This encourages using public transport or cycling.
减少15%：每周汽车排放27 kg的目标减少量为15%×27=4.05 kg。新排放量=27-4.05=22.95 kg。这鼓励使用公共交通或骑车。
Published by TutorHao | Advanced 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