📚 Cross-curricular Integrated Problem-solving Practice | 跨学科综合题型训练
In Year 9 CCEA Mathematics, students begin to see how number, algebra, geometry, and statistics appear naturally in other subjects. Solving problems that combine maths with science, geography, business, health, and technology not only reinforces mathematical skills but also prepares learners for real-world applications. This article presents a variety of integrated tasks with full explanations, giving you the chance to practise reading, interpreting, and calculating in a cross-curricular context.
在 CCEA 九年级数学中，学生会开始发现数字、代数、几何和统计如何自然地出现在其他科目中。解决将数学与科学、地理、商业、健康和技术结合起来的实际问题，不仅能强化数学技能，还能为现实生活中的应用做好准备。本文呈现了多种综合题型并附有详细解析，让你有机会在跨学科情境中练习阅读、解读和计算。
1. Science and Maths: Speed, Density, and Ratios | 科学与数学：速度、密度与比例
Science experiments often require you to calculate speed, density, or concentration. These quantities rely on ratios and formula rearrangement, which are key algebra skills in Year 9.
科学实验经常需要计算速度、密度或浓度。这些量依赖于比例和公式变形，这也是九年级代数的重要技能。
Example 1: Speed. A cyclist travels 45 km in 2 hours and 30 minutes. Find the average speed in km/h. First, convert the time: 2 hours 30 minutes = 2.5 hours. Then use the formula:
例1：速度。一名自行车手在2小时30分钟内行驶了45公里。求平均速度（公里/小时）。首先转换时间：2小时30分钟 = 2.5小时。然后使用公式：
Average speed = Total distance ÷ Total time
平均速度 = 总距离 ÷ 总时间
Average speed = 45 ÷ 2.5 = 18 km/h. The cyclist’s average speed is 18 km/h.
平均速度 = 45 ÷ 2.5 = 18 公里/小时。这位自行车手的平均速度是18公里/小时。
Example 2: Density. A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³.
例2：密度。一块金属的质量为540克，体积为200立方厘米。计算其密度（克/立方厘米）。
Density = Mass ÷ Volume
密度 = 质量 ÷ 体积
Density = 540 ÷ 200 = 2.7 g/cm³. This value is typical for aluminium.
密度 = 540 ÷ 200 = 2.7 克/立方厘米。这个数值是铝的典型密度。
Practice by rearranging: if you know density and mass, volume = mass ÷ density. A 135 g sample of the same metal would have a volume of 135 ÷ 2.7 = 50 cm³.
试着变形公式练习：如果已知密度和质量，体积 = 质量 ÷ 密度。同种金属135克样品的体积为 135 ÷ 2.7 = 50 立方厘米。
2. Geography and Maths: Population Density and Scale | 地理与数学：人口密度与比例尺
Geographers use population density to compare how crowded different regions are. Map scales also rely on ratio and proportion.
地理学家用人口密度来比较不同地区的拥挤程度。地图比例尺也依赖于比例和比率。
A city has a population of 240,000 and an area of 120 km². Population density is the number of people per square kilometre:
某城市有240,000人口，面积为120平方公里。人口密度即每平方公里的人数：
Population density = Population ÷ Area
人口密度 = 人口数 ÷ 面积
Population density = 240,000 ÷ 120 = 2000 people/km². If a second city has an area of 80 km² and the same density, its population would be 2000 × 80 = 160,000.
人口密度 = 240,000 ÷ 120 = 2000 人/平方公里。如果第二个城市面积为80平方公里且密度相同，其人口将是 2000 × 80 = 160,000。
Map scales are often written as a ratio, e.g., 1 : 50,000. If two towns are 8 cm apart on the map, the real straight-line distance is 8 × 50,000 cm = 400,000 cm = 4000 m = 4 km. Always convert to the required unit.
地图比例尺通常写为比值形式，如1 : 50,000。如果地图上两镇相距8厘米，则实际直线距离为 8 × 50,000 厘米 = 400,000 厘米 = 4000 米 = 4 公里。注意转换到所需单位。
3. Business Studies and Maths: Profit, Discount, and Simple Interest | 商业与数学：利润、折扣与单利
Business applications involve percentages, profit margins, and simple interest calculations – all core Year 9 topics.
商业应用涉及百分比、利润率和单利计算——这些全是九年级的核心内容。
A shop buys a jacket for £40 and wants to make a 35% profit on the cost price. Profit amount = 35% of £40 = 0.35 × 40 = £14. Selling price = cost + profit = £40 + £14 = £54.
一家商店以40英镑购进一件夹克，想按成本价获得35%的利润。利润额 = 40英镑的35% = 0.35 × 40 = 14英镑。售价 = 成本 + 利润 = 40 + 14 = 54英镑。
During a sale, the jacket is reduced by 20%. Sale price = £54 – 20% of £54 = 54 – (0.20 × 54) = 54 – 10.80 = £43.20. Notice that reducing the selling price by 20% still leaves a small profit compared to the cost price.
促销期间，夹克降价20%。促销价 = 54 – 54的20% = 54 – (0.20 × 54) = 54 – 10.80 = 43.20英镑。可以看到，即使售价降低20%，与成本相比仍有一定利润。
Simple interest (I) is found using I = PRT ÷ 100, where P is the principal amount, R is the annual interest rate, and T is the time in years. For a savings of £300 at 5% per year for 3 years, I = (300 × 5 × 3) ÷ 100 = 4500 ÷ 100 = £45. Total amount = £300 + £45 = £345.
单利 (I) 使用公式 I = PRT ÷ 100，其中P为本金，R为年利率，T为时间（年）。若300英镑存款年利率5%，存期3年，利息 I = (300 × 5 × 3) ÷ 100 = 4500 ÷ 100 = 45英镑。总金额 = 300 + 45 = 345英镑。
4. Health and Nutrition: BMI and Calorie Calculations | 健康与营养：BMI与卡路里计算
Body Mass Index (BMI) combines weight and height to classify whether a person is underweight, normal, overweight, or obese. It uses metric units and squares.
身体质量指数 (BMI) 结合体重和身高来判断一个人是偏瘦、正常、超重还是肥胖。它使用公制单位并包含平方运算。
BMI = Weight (kg) ÷ [Height (m)]²
BMI = 体重（公斤） ÷ 身高（米）²
For a person weighing 65 kg and 1.72 m tall, height squared = 1.72² ≈ 2.9584. BMI = 65 ÷ 2.9584 ≈ 22.0 kg/m². This falls within the healthy range (18.5 – 24.9).
一位体重65公斤、身高1.72米的人，身高平方 = 1.72² ≈ 2.9584。BMI = 65 ÷ 2.9584 ≈ 22.0 公斤/米²。这属于健康范围 (18.5 – 24.9)。
Calorie calculations often use direct proportion. If 100 g of a snack contains 520 calories, how many calories are in 35 g? Calories in 35 g = (520 ÷ 100) × 35 = 5.2 × 35 = 182 calories.
卡路里计算常用正比例。如果100克零食含520卡路里，那么35克含有多少卡路里？35克的卡路里 = (520 ÷ 100) × 35 = 5.2 × 35 = 182 卡路里。
Understanding percentages from nutrition labels also helps: if a serving provides 18% of the recommended daily allowance of a nutrient, you can work out the total RDA. 18% of RDA = the given amount. For example, if 18% equals 9 g, then 1% = 9 ÷ 18 = 0.5 g, so 100% RDA = 0.5 × 100 = 50 g.
读懂营养成分表中的百分比同样有帮助：如果一份食物提供某种营养素日推荐摄入量的18%，你可以计算出总的日推荐摄入量。18%对应量已知。例如，若18%等于9克，则1% = 9 ÷ 18 = 0.5克，因此100%日推荐摄入量 = 0.5 × 100 = 50克。
5. Environmental Science: Carbon Footprint and Recycling | 环境科学：碳足迹与回收
Environmental data often involve reading tables, calculating totals, and comparing percentages. A carbon footprint problem: a household uses electricity that produces 2.3 kg of CO₂ per kWh and uses 180 kWh in a month. Total CO₂ = 2.3 × 180 = 414 kg.
环境数据常需要读表、计算总量和比较百分比。碳足迹问题：某家庭用电每千瓦时产生2.3公斤二氧化碳，一个月用电180千瓦时。总共排放的二氧化碳 = 2.3 × 180 = 414 公斤。
If the household reduces its usage by 15%, new usage = 180 × 0.85 = 153 kWh. New CO₂ = 2.3 × 153 = 351.9 kg. The reduction is 414 – 351.9 = 62.1 kg, about a 15% drop.
如果该家庭用电量减少15%，新用电量 = 180 × 0.85 = 153 千瓦时。新的二氧化碳排放 = 2.3 × 153 = 351.9 公斤。减少量为 414 – 351.9 = 62.1 公斤，约降低15%。
Recycling rates: a school recycles 65% of its 800 kg of waste each week. Recycled amount = 0.65 × 800 = 520 kg. The remaining waste sent to landfill = 800 – 520 = 280 kg. If the school aims to recycle 80% eventually, the target recycled amount = 0.80 × 800 = 640 kg, so an extra 640 – 520 = 120 kg must be recycled.
回收率：一所学校每周产生800公斤垃圾，其中65%被回收。回收量 = 0.65 × 800 = 520 公斤。送往填埋场的垃圾 = 800 – 520 = 280 公斤。如果学校目标是将回收率提高到80%，目标回收量 = 0.80 × 800 = 640 公斤，因此还需要额外回收 640 – 520 = 120 公斤。
6. Art and Design: Golden Ratio and Symmetry | 艺术与设计：黄金比例与对称
The golden ratio (approximately 1.618 : 1) appears in art, architecture, and nature. It is often denoted by the Greek letter φ (phi). If a rectangle’s length-to-width ratio equals φ, it is called a golden rectangle. You can test whether a shape follows the golden ratio by dividing the longer side by the shorter side.
黄金比例（大约为1.618 : 1）出现在艺术、建筑和自然界中，通常用希腊字母 φ (phi) 表示。如果一个矩形的长宽比等于φ，则称为黄金矩形。你可以用长边除以短边来检验一个形状是否符合黄金比例。
For example, a painting is 80 cm long and 49.5 cm wide. Ratio = 80 ÷ 49.5 ≈ 1.616. This is very close to the golden ratio, showing the artist may have used it intentionally.
例如，一幅画长80厘米，宽49.5厘米。比值 = 80 ÷ 49.5 ≈ 1.616。非常接近黄金比例，显示画家可能有意采用了这一比例。
Symmetry is another link: a design with line symmetry can be analysed using coordinates. If a shape on a grid has coordinates A(2,3), B(5,3), C(5,1), and its mirror line is x = 6, the reflected points are found by subtracting the x-distance from 6 twice or by using the formula: reflected x = 2 × mirror line – original x. So A’ x = 12 – 2 = 10, giving A’(10,3).
对称性是另一种联系：可以用坐标分析具有线对称的设计。如果一个网格上的图形坐标为 A(2,3), B(5,3), C(5,1)，其对称轴为 x = 6，反射点的求法是从对称轴减去原x距离的两倍，或使用公式：反射后的 x = 2 × 对称轴 – 原 x。因此 A’ 的 x = 12 – 2 = 10，得到 A’(10,3)。
7. Technology: Coding Coordinates and Simple Algorithms | 技术：坐标编码与简单算法
Programming and game design use coordinates and sequences. Moving a character on a screen involves adding or subtracting from (x, y) positions. If a sprite starts at (10, 20) and moves right 5 units and down 3 units, its new position is (10+5, 20–3) = (15, 17). This uses integer arithmetic and an understanding of the coordinate plane.
编程和游戏设计使用坐标和序列。在屏幕上移动角色涉及在 (x, y) 位置上加减。如果一个精灵起始于 (10, 20)，向右移动5个单位，向下移动3个单位，其新位置为 (10+5, 20–3) = (15, 17)。这用到整数运算以及对坐标平面的理解。
Simple algorithms can be expressed as number patterns. For example, a robot is programmed to draw a spiral by repeating: move forward n units, turn 90°, increase n by 2. Starting with n = 2, the distance sequence is 2, 4, 6, 8, … This is the 2 times table. The total distance after 6 moves is the sum of the first six even numbers: 2 + 4 + 6 + 8 + 10 + 12 = 42 units.
简单算法可以表达为数字规律。例如，一个机器人被编程绘制螺旋图案，重复执行：向前移动n个单位，转90度，将n增加2。从 n = 2 开始，距离序列为 2, 4, 6, 8, … 这正好是2的倍数。6次移动后的总距离为前六个偶数之和：2 + 4 + 6 + 8 + 10 + 12 = 42 个单位。
Binary and data: computers store information in binary (base-2). Converting a decimal number like 29 to binary: 29 = 16 + 8 + 4 + 1, so in binary (using 8 bits) it is 00011101. Year 9 students may be asked to convert small numbers or to understand that an 8-bit binary number can represent values from 0 to 255.
二进制与数据：计算机以二进制（基数为2）存储信息。将十进制数29转换为二进制：29 = 16 + 8 + 4 + 1，所以8位二进制形式为 00011101。九年级学生可能需要转换小数字，或理解8位二进制数可以表示从0到255的值。
8. Physical Education: Statistics and Average Speed | 体育：统计与平均速度
Sports provide rich data for statistical analysis. A basketball player’s points over five games: 12, 15, 8, 20, 10. Calculate the mean (average): (12+15+8+20+10) ÷ 5 = 65 ÷ 5 = 13 points per game. The median is found by ordering: 8, 10, 12, 15, 20 → median = 12. The range = 20 – 8 = 12 points, showing variability.
体育为统计分析提供了丰富的数据。某篮球运动员五场比赛的得分：12, 15, 8, 20, 10。计算平均数：(12+15+8+20+10) ÷ 5 = 65 ÷ 5 = 13 分/场。将数据排序求中位数：8, 10, 12, 15, 20 → 中位数 = 12。极差 = 20 – 8 = 12分，显示出波动性。
Athletes often use average speed. A runner completes 400 m in 50 seconds. Average speed = distance ÷ time = 400 m ÷ 50 s = 8 m/s. To convert to km/h, multiply by 3.6: 8 × 3.6 = 28.8 km/h. This conversion factor comes from 1 m/s = 3.6 km/h (since 1 hour = 3600 s and 1 km = 1000 m, 3600/1000 = 3.6).
运动员常用平均速度。一名跑步者用50秒跑完400米。平均速度 = 距离 ÷ 时间 = 400米 ÷ 50秒 = 8 米/秒。换算为公里/小时，乘以3.6：8 × 3.6 = 28.8 公里/小时。这个换算系数是因为 1 m/s = 3.6 km/h（1小时 = 3600秒，1公里 = 1000米，3600/1000 = 3.6）。
Heart rate zones: a 14-year-old’s maximum heart rate is roughly 220 – age = 206 beats per minute (bpm). Moderate exercise is 50–70% of max, so lower limit = 206 × 0.5 = 103 bpm, upper limit = 206 × 0.7 ≈ 144 bpm. These calculations combine algebra and percentages.
心率区间：14岁青少年的最大心率大约是 220 – 年龄 = 206 次/分钟 (bpm)。中等强度运动为最大值的50%–70%，因此下限 = 206 × 0.5 = 103 bpm，上限 = 206 × 0.7 ≈ 144 bpm。这些计算结合了代数与百分比。
9. Everyday Finance: Budgeting and Best Buys | 日常金融：预算与最佳购买
Making sensible spending decisions requires comparing unit prices. For example, a 750 g box of cereal costs £2.70, while a 1.2 kg box costs £3.96. Which is better value? Calculate price per 100 g: first box → £2.70 ÷ 7.5 = £0.36 per 100 g; second box → £3.96 ÷ 12 = £0.33 per 100 g. The larger box is cheaper per 100 g.
做出明智的消费决策需要比较单位价格。例如，一盒750克麦片售价2.70英镑，一盒1.2公斤售价3.96英镑。哪个更划算？计算每100克价格：第一盒 → 2.70 ÷ 7.5 = 0.36英镑/100克；第二盒 → 3.96 ÷ 12 = 0.33英镑/100克。大盒的每100克价格更低。
A monthly budget: a family’s income is £2400. They plan to spend 30% on housing, 20% on food, 15% on transport, and save 10%. Housing amount = 0.30 × 2400 = £720; food = 0.20 × 2400 = £480; transport = 0.15 × 2400 = £360; savings = 0.10 × 2400 = £240. The remainder = 2400 – (720+480+360+240) = 2400 – 1800 = £600 for other expenses.
月度预算：一个家庭收入为2400英镑。他们计划30%用于住房，20%用于食物，15%用于交通，并储蓄10%。住房金额 = 0.30 × 2400 = 720英镑；食物 = 0.20 × 2400 = 480英镑；交通 = 0.15 × 2400 = 360英镑；储蓄 = 0.10 × 2400 = 240英镑。剩余 = 2400 – (720+480+360+240) = 2400 – 1800 = 600英镑用于其他支出。
Exchange rates: when travelling, you may need to convert money. If £1 = €1.15, how many euros do you get for £250? € = 250 × 1.15 = €287.50. Returning, if you have €50 left and the buy-back rate is €1 = £0.85, you receive 50 × 0.85 = £42.50. Exchange rates involve proportional reasoning.
汇率：旅行时你可能需要兑换货币。如果 £1 = €1.15，那么250英镑可以兑换多少欧元？欧元 = 250 × 1.15 = 287.50欧元。返回时，如果你还剩50欧元，回购汇率为 €1 = £0.85，那么你得到 50 × 0.85 = 42.50英镑。汇率问题涉及比例推理。
10. Conclusion: Strategies for Cross-curricular Problems | 总结：跨学科问题的解题策略
To succeed with cross-curricular problems, first identify the mathematical topic hidden in the context. Is it ratio, percentage, formula, or statistics? Then extract the relevant numbers and units. Write down the formula or relationship, substitute carefully, and check that your answer makes sense in the real-world scenario. Always show your working step by step.
要成功解决跨学科问题，首先要识别隐藏在情境中的数学主题。它是比例、百分比、公式还是统计？然后提取相关的数字和单位。写下公式或关系式，仔细代入，并检查你的答案在现实情境中是否合理。务必逐步展示你的计算过程。
Practice with a wide range of contexts—science experiments, maps, business offers, health data, sports statistics—builds confidence and fluency. Remember, mathematics is a tool for understanding the world, and Year 9 CCEA Maths provides the perfect foundation to use it across all subjects.
在多种情境中练习——科学实验、地图、商业报价、健康数据、体育统计——可以建立信心和流畅度。请记住，数学是理解世界的工具，而 CCEA 九年级数学正为你提供了在所有科目中使用这一工具的完美基础。
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