📚 Interdisciplinary Problem-Solving for CCEA Year 10 Mathematics | CCEA 十年级数学跨学科综合题型训练
Mathematics is not confined to abstract exercises; it is the universal language that connects science, geography, economics, sport, and everyday life. In CCEA Year 10, you are expected to apply numerical, algebraic, and graphical skills to unfamiliar contexts taken from other subjects. This article presents a series of cross-curricular problem sets, each blending real-world scenarios with core maths techniques. By working through these examples, you will strengthen your ability to interpret, model, and solve problems that transcend traditional topic boundaries.
数学并不局限于抽象练习；它是连接科学、地理、经济、体育和日常生活的通用语言。在CCEA十年级课程中，你需要将数、代数与图表技能应用到来自其他学科的真实情境中。本文提供一系列跨学科问题集，每个问题都将现实场景与核心数学方法融合。通过演练这些例题，你将提升跨越传统主题边界进行解释、建模和解题的能力。
1. Science – Concentration and Dilution | 科学——浓度与稀释
In a biology lab, a technician has 200 mL of a 25% salt solution. She needs to dilute it with pure water to obtain a 10% solution for an osmosis experiment. The mass of salt remains constant during dilution. We can set up an equation using the relationship: (initial mass of solute) = (final mass of solute). Let the volume of water added be x mL.
在生物实验室中，技术员有200 mL的25%盐水。她需要用纯水稀释，得到适用于渗透实验的10%溶液。稀释过程中盐的质量不变。我们可以利用关系式建立方程：(溶质初始质量) = (溶质最终质量)。设加入的水为 x mL。
The initial mass of salt is 25% of 200 mL, which gives 0.25 × 200 = 50 grams (assuming 1 mL of solution has a mass of approximately 1 g). After adding x mL of water, the total volume becomes (200 + x) mL, and the final concentration is 10%. Thus, 0.10 × (200 + x) = 50. Solving 0.10(200 + x) = 50 leads to 20 + 0.10x = 50, so 0.10x = 30, and x = 300 mL. Always check: adding 300 mL gives a total of 500 mL, and 50 g in 500 mL is indeed 10%.
盐的初始质量是200 mL的25%，即0.25 × 200 = 50克（假设1 mL溶液质量约1 g）。加入 x mL 水后，总体积变为(200 + x) mL，最终浓度为10%。因此，0.10 × (200 + x) = 50。解方程 0.10(200 + x) = 50 得到 20 + 0.10x = 50，则 0.10x = 30，得 x = 300 mL。检验：加入300 mL水后总体积为500 mL，其中50 g溶质的确构成10%浓度。
Dilution formula used: C₁V₁ = C₂V₂, where C = concentration, V = volume
所用稀释公式：C₁V₁ = C₂V₂，其中 C 为浓度，V 为体积
2. Geography – Population Density and Growth | 地理——人口密度与增长
A region has a land area of 15,200 km² and a population of 2.4 million. Calculate the population density in people per square kilometre. If the population grows by 1.8% annually, what will the population be after three years, assuming no migration and continuous growth? Use the compound growth model.
某地区土地面积为15,200 km²，人口为240万。计算人口密度（人/km²）。如果人口每年增长1.8%，假设无迁移且连续增长，三年后人口是多少？请使用复合增长模型。
Density = total population ÷ area = 2,400,000 ÷ 15,200 ≈ 157.9 people/km². For growth, the multiplier per year is 1 + 1.8% = 1.018. After 3 years, population ≈ 2,400,000 × 1.018³. Calculate stepwise: 1.018² = 1.036324, times 1.018 again ≈ 1.0550. Then 2,400,000 × 1.0550 ≈ 2,532,000. So approximately 2.53 million. This model links multiplication and powers to demographic trends.
密度 = 总人口 ÷ 面积 = 2,400,000 ÷ 15,200 ≈ 157.9 人/km²。对于增长，每年乘数为 1 + 1.8% = 1.018。3年后，人口 ≈ 2,400,000 × 1.018³。逐步计算：1.018² = 1.036324，再乘以1.018 ≈ 1.0550。然后 2,400,000 × 1.0550 ≈ 2,532,000。即约253万人。此模型将乘法和幂运算与人口趋势联系起来。
When solving percentage growth problems, it is often helpful to express the rate as a decimal multiplier and apply indices for multiple periods.
在求解百分比增长问题时，将增长率表示为小数乘数并用指数处理多个周期，通常非常有效。
3. Economics – Currency Exchange and Budgeting | 经济——货币兑换与预算
You are planning a school trip to the eurozone. The exchange rate is £1 = €1.14. You have a budget of £450 for accommodation, meals, and souvenirs. Convert your total budget into euros. Then allocate 35% for accommodation, 45% for meals, and the rest for souvenirs. Display the allocation in a pie chart and find the exact amounts in euros.
你正计划一次去欧元区的学校旅行。汇率为 £1 = €1.14。你的住宿、餐饮和纪念品总预算为450英镑。将总预算换算为欧元。然后分配35%用于住宿，45%用于餐饮，剩余用于纪念品。用饼图展示分配，并求出各项欧元金额。
Total in euros = 450 × 1.14 = €513. Accommodation = 35% of 513 = 0.35 × 513 = €179.55. Meals = 0.45 × 513 = €230.85. Souvenirs = remainder = 20% = 0.20 × 513 = €102.60. In a pie chart, the angles are: accommodation 0.35 × 360° = 126°, meals 0.45 × 360° = 162°, souvenirs 0.20 × 360° = 72°. This integrates percentages, multiplication, and data representation.
欧元总额 = 450 × 1.14 = €513。住宿 = 513的35% = 0.35 × 513 = €179.55。餐饮 = 0.45 × 513 = €230.85。纪念品 = 剩余20% = 0.20 × 513 = €102.60。在饼图中，角度分别为：住宿 0.35 × 360° = 126°，餐饮 0.45 × 360° = 162°，纪念品 0.20 × 360° = 72°。这融合了百分比、乘法与数据表示方法。
| Category / 类别 | Amount (€) | Angle / 角度 |
|---|---|---|
| Accommodation / 住宿 | 179.55 | 126° |
| Meals / 餐饮 | 230.85 | 162° |
| Souvenirs / 纪念品 | 102.60 | 72° |
4. Sports – Speed, Time and Distance | 体育——速度、时间与距离
A runner completes a 10 km race in 42 minutes 30 seconds. Express this time in hours and calculate the average speed in km/h. Then adjust the pace if the runner wants to finish in 40 minutes, determining the required speed increase. This situation demands unit conversion and rearrangement of the formula v = d ÷ t.
一名跑步者用42分30秒跑完10公里比赛。将时间转换为小时，并计算平均速度（km/h）。然后调整配速，若他希望40分钟完赛，求所需速度增量。此情境需要单位换算以及公式 v = d ÷ t 的变形。
Time in hours: 42 minutes = 42/60 = 0.7 h, 30 seconds = 30/3600 = 0.00833 h, total = 0.70833 h. Average speed = 10 ÷ 0.70833 ≈ 14.12 km/h. Alternatively, using fractions: 42½ minutes = 42.5/60 = 0.7083 h. For a 40-minute finish, time = 40/60 = 0.6667 h. Required speed = 10 ÷ 0.6667 ≈ 15.0 km/h. The increase is about 0.88 km/h. This exercise strengthens fluency with time units and decimal operations.
以小时计的时间：42分钟 = 42/60 = 0.7小时，30秒 = 30/3600 = 0.00833小时，合计 ≈ 0.70833小时。平均速度 = 10 ÷ 0.70833 ≈ 14.12 km/h。或者用分数：42½分钟 = 42.5/60 = 0.7083小时。若用40分钟完赛，时间 = 40/60 = 0.6667小时。所需速度 = 10 ÷ 0.6667 ≈ 15.0 km/h。增速约0.88 km/h。此练习可强化时间单位和小数运算的熟练度。
v = d ÷ t, and t = d ÷ v when rearranged
公式变形：v = d ÷ t，以及 t = d ÷ v
5. Engineering – Volume and Surface Area | 工程——体积与表面积
An engineer designs a cylindrical water tank with a radius of 1.2 m and a height of 3.5 m. Calculate the volume of water it can hold in cubic metres, then convert to litres (1 m³ = 1000 L). Also, find the total surface area of the closed cylinder to estimate the amount of sheet metal required. Use π ≈ 3.14.
工程师设计了一个圆柱形水箱，半径1.2 m，高3.5 m。计算它能容纳的水体积（立方米），并换算成升（1 m³ = 1000 L）。同时求封闭圆柱体的总表面积，以估算所需金属板材。取 π ≈ 3.14。
Volume of a cylinder = πr²h = 3.14 × (1.2)² × 3.5 = 3.14 × 1.44 × 3.5. Multiply step by step: 3.14 × 1.44 = 4.5216, then 4.5216 × 3.5 = 15.8256 m³. In litres: 15.8256 × 1000 = 15,825.6 L. Surface area of a closed cylinder = 2πrh + 2πr². So lateral area = 2 × 3.14 × 1.2 × 3.5 = 2 × 3.14 × 4.2 = 26.376 m². Ends: 2 × 3.14 × 1.44 = 9.0432 m². Total = 35.4192 m². This task applies circle geometry and unit conversions.
圆柱体积 = πr²h = 3.14 × (1.2)² × 3.5 = 3.14 × 1.44 × 3.5。逐步计算：3.14 × 1.44 = 4.5216，然后 4.5216 × 3.5 = 15.8256 m³。换算为升：15.8256 × 1000 = 15,825.6 L。封闭圆柱表面积 = 2πrh + 2πr²。侧面积 = 2 × 3.14 × 1.2 × 3.5 = 2 × 3.14 × 4.2 = 26.376 m²。两个底面积 = 2 × 3.14 × 1.44 = 9.0432 m²。总面积 = 35.4192 m²。该任务应用了圆几何与单位换算。
6. Music – Frequency Ratios and Scales | 音乐——频率比率与音阶
In a just intonation scale, a perfect fifth has a frequency ratio of 3:2. If the note A4 is tuned to 440 Hz, find the frequency of E5, which is a perfect fifth above A4. Then find the frequency of B4, which is a perfect fourth above E4 (ratio 4:3 relative to E4). Use multiplicative reasoning with ratios.
在纯律音阶中，纯五度的频率比为3:2。若音符A4调至440 Hz，求高一个纯五度的E5的频率。再求B4的频率，它是E4上方纯四度（相对于E4的频率比为4:3）。使用比率乘法进行推理。
For a perfect fifth up, multiply by 3/2: E5 = 440 × (3/2) = 440 × 1.5 = 660 Hz. Now, to find B4, we need E4. E4 is an octave below E5? Actually, E4 is one octave lower than E5, so E4 = 660 ÷ 2 = 330 Hz (since octave ratio is 2:1). Then B4, a perfect fourth above E4, multiply by 4/3: B4 = 330 × (4/3) = 330 × 1.333… = 440 Hz. Interestingly, B4 equals A4 in this just intonation chain, demonstrating the circle of fifths concept. This interweaves fractions, multiplication, and the physics of sound.
升纯五度，乘以3/2：E5 = 440 × (3/2) = 440 × 1.5 = 660 Hz。现在求B4，需要E4。E4比E5低一个八度，所以E4 = 660 ÷ 2 = 330 Hz（因为八度比为2:1）。然后B4，即E4上方纯四度，乘以4/3：B4 = 330 × (4/3) = 330 × 1.333… = 440 Hz。有趣的是，在这条纯律链中B4等于A4，展示了五度圈的概念。这交织了分数、乘法与声音物理。
Frequency ratio relationship: f₂ = f₁ × (interval ratio)
频率比关系：f₂ = f₁ ×（音程比）
7. Art – Golden Ratio and Proportions | 艺术——黄金比例
The golden ratio, often denoted by the Greek letter φ (phi), appears in art, architecture, and nature. It is defined as a ratio such that (a + b) / a = a / b = φ. Algebraically, φ satisfies φ² = φ + 1. Find the positive solution of this quadratic equation and use it to check the proportions of a rectangle with sides 34 cm and 21 cm. Are they approximately in the golden ratio?
黄金比例，常用希腊字母φ（phi）表示，出现在艺术、建筑与自然界中。其定义为 (a + b) / a = a / b = φ。从代数看，φ满足 φ² = φ + 1。求该二次方程的正数解，并用它验证边长34 cm和21 cm的矩形比例是否接近黄金比。
Solve φ² – φ – 1 = 0 using the quadratic formula: φ = [1 ± √(1 + 4)] / 2 = (1 ± √5) / 2. The positive solution is (1 + √5) / 2 ≈ (1 + 2.236) / 2 = 3.236 / 2 = 1.618. For the rectangle, ratio = 34 ÷ 21 ≈ 1.6190, which is extremely close to φ. Therefore, it possesses golden proportions. This problem integrates quadratics, surds, and aesthetic design.
解 φ² – φ – 1 = 0 用二次公式：φ = [1 ± √(1 + 4)] / 2 = (1 ± √5) / 2。正数解为 (1 + √5) / 2 ≈ (1 + 2.236) / 2 = 3.236 / 2 = 1.618。对于矩形，比例 = 34 ÷ 21 ≈ 1.6190，与φ极其接近，因此其具备黄金比例。此题整合了二次方程、根式与美学设计。
φ = (1 + √5) ÷ 2 ≈ 1.618
黄金比例 φ = (1 + √5) ÷ 2 ≈ 1.618
8. Environmental Science – Carbon Emission Data | 环境科学——碳排放数据分析
A school conducts a carbon footprint audit. Monthly CO₂ emissions from electricity and heating over a year, in tonnes, are: 2.4, 2.1, 1.9, 1.8, 1.6, 1.5, 1.7, 1.9, 2.2, 2.5, 2.7, 2.8. Draw a line graph and calculate the mean, median, and range. Then determine the percentage reduction from the highest month to the lowest. Interpret the trend and suggest measures.
某学校开展碳足迹审计。一年中电力和供暖的月度CO₂排放量（吨）为：2.4, 2.1, 1.9, 1.8, 1.6, 1.5, 1.7, 1.9, 2.2, 2.5, 2.7, 2.8。绘制折线图并计算平均数、中位数和极差。再计算从最高月份到最低月份的下降百分比。解释趋势并提出措施。
Arrange data in order: 1.5, 1.6, 1.7, 1.8, 1.9, 1.9, 2.1, 2.2, 2.4, 2.5, 2.7, 2.8. Mean = sum ÷ 12. Sum = 1.5+1.6+1.7+1.8+1.9+1.9+2.1+2.2+2.4+2.5+2.7+2.8 = 25.1. Mean ≈ 2.09 tonnes. Median = (1.9 + 2.1) ÷ 2 = 2.0 tonnes. Range = 2.8 – 1.5 = 1.3 tonnes. Highest month (December) = 2.8 t, lowest (June) = 1.5 t. Reduction percentage = (1.3 ÷ 2.8) × 100 ≈ 46.4%. The line graph shows higher emissions in winter months, suggesting heating demand. The analysis uses statistics, percentages, and graph interpretation.
将数据排序：1.5, 1.6, 1.7, 1.8, 1.9, 1.9, 2.1, 2.2, 2.4, 2.5, 2.7, 2.8。平均数 = 总和 ÷ 12。总和 = 25.1，平均数 ≈ 2.09吨。中位数 = (1.9 + 2.1) ÷ 2 = 2.0吨。极差 = 2.8 – 1.5 = 1.3吨。最高月（12月）= 2.8 t，最低月（6月）= 1.5 t。下降百分比 = (1.3 ÷ 2.8) × 100 ≈ 46.4%。折线图显示冬季排放较高，反映取暖需求。该分析运用了统计量、百分比与图表解读。
9. Health – Calories and Exercise | 健康——卡路里与运动
A person consumes a snack of 350 kilocalories. To burn this energy, they can choose between cycling, which burns 8 kcal per minute, or running, which burns 12 kcal per minute. Write linear equations for remaining calories after t minutes for each activity. Find the time needed to burn off the entire snack with each exercise. Also, compare the gradients.
某人摄入了一份350千卡的小吃。为消耗这些能量，他可以选择骑车（每分钟消耗8千卡）或跑步（每分钟消耗12千卡）。为每种活动建立 t 分钟后剩余卡路里的线性方程。求用每种运动消耗全部小吃所需的时间。并比较斜率。
Cycling: remaining calories R = 350 – 8t. Running: R = 350 – 12t. To find when R = 0, set 350 – 8t = 0 → t = 350 ÷ 8 = 43.75 minutes for cycling. For running, t = 350 ÷ 12 ≈ 29.17 minutes. The gradient for cycling is –8, and for running is –12. The steeper gradient (more negative) indicates faster energy expenditure. Plotted on the same axes, the running line reaches zero more quickly. This connects linear functions, slope interpretation, and health science.
骑车：剩余卡路里 R = 350 – 8t。跑步：R = 350 – 12t。求 R = 0 时的 t：令 350 – 8t = 0 → t = 350 ÷ 8 = 43.75 分钟（骑车）。跑步，t = 350 ÷ 12 ≈ 29.17 分钟。骑车方程的斜率为 –8，跑步为 –12。斜率更陡（更负）表示能量消耗更快。绘制在同一坐标系上，跑步的直线更快到达零点。这联系了线性函数、斜率解读与健康科学。
R = R₀ – k t, where k = burn rate per minute
R = R₀ – k t，其中 k 为每分钟消耗率
10. Computer Science – Binary and Data Size | 计算机科学——二进制与数据容量
In computer systems, memory is often expressed in powers of 2. A standard 1 KB (kilobyte) = 2¹⁰ bytes = 1024 bytes, 1 MB = 2²⁰ bytes, 1 GB = 2³⁰ bytes. A digital photo is 3.5 MB in size. Calculate its size in bytes and express it in scientific notation. Also, determine how many such photos can fit on an 8 GB memory card, assuming no other data.
在计算机系统中，内存容量常以2的幂表示。标准 1 KB（千字节）= 2¹⁰ 字节 = 1024 字节，1 MB = 2²⁰ 字节，1 GB = 2³⁰ 字节。一张数码照片大小为3.5 MB。计算其以字节为单位的大小，并用科学记数法表示。同时，若无其他数据，一张8 GB存储卡能存放多少张这样的照片？
1 MB = 2²⁰ = 1,048,576 bytes. Photo size = 3.5 × 1,048,576 = 3,670,016 bytes. In scientific notation: 3.67 × 10⁶ bytes (approx). Card capacity = 8 GB = 8 × 2³⁰ bytes = 8 × 1,073,741,824 = 8,589,934,592 bytes. Number of photos = total bytes ÷ bytes per photo = 8,589,934,592 ÷ 3,670,016 ≈ 2340.5. Therefore, about 2340 photos can be stored. This problem uses indices, multiplication, and unit conversions within the binary system.
1 MB = 2²⁰ = 1,048,576 字节。照片大小 = 3.5 × 1,048,576 = 3,670,016 字节。科学记数法：约 3.67 × 10⁶ 字节。卡容量 = 8 GB = 8 × 2³⁰ 字节 = 8 × 1,073,741,824 = 8,589,934,592 字节。照片数量 = 总字节数 ÷ 每张照片字节数 = 8,589,934,592 ÷ 3,670,016 ≈ 2340.5。因此，大约可存储2340张照片。此题使用了指数、乘法以及二进制中的单位换算。
Data size in bytes = value × 2^(10n), where n for KB=1, MB=2, GB=3
字节大小 = 数值 × 2^(10n)，其中 KB 对应
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