📚 Year 7 CAIE Advanced Mathematics: Interdisciplinary Integrated Question Training | Year 7 CAIE 进阶数学：跨学科综合题型训练
This article explores how Year 7 advanced mathematics concepts are applied in real-world contexts across science, geography, economics, and technology. By practising integrated problems, you will strengthen your ability to transfer mathematical skills to unfamiliar situations – exactly the kind of thinking CAIE assessments encourage. Each section below pairs an interdisciplinary scenario with step-by-step reasoning, using only the symbolic language you already know: fractions, ratios, simple equations, graphs, and statistics.
本文探讨 Year 7 进阶数学概念如何应用于科学、地理、经济和科技等真实情境。通过练习跨学科综合题型，你将增强把数学技能迁移到陌生场景的能力——这正是 CAIE 评估所鼓励的思维方式。以下每个小节都将一个跨学科情境与逐步推理配对，仅使用你已掌握的符号语言：分数、比、简单方程、图表和统计。
1. Measurement and Unit Conversion in Science | 科学中的测量与单位换算
ns under a microscope. If a cell has a diameter of 0.045 mm, express this in micrometres (µm) given that 1 mm = 1000 µm.
在生物学中，用显微镜观察标本时，常常需要在米、厘米和毫米之间进行换算。如果一个细胞的直径是 0.045 mm，请将其表示为微米（µm），已知 1 mm = 1000 µm。
Multiply the millimetre value by 1000: 0.045 × 1000 = 45 µ lies on understanding place value and multiplication by powers of 10, a core Year 7 skill.
将毫米值乘以 1000：0.045 × 1000 = 45 µm。这种换算依赖于对位值和乘以 10 的幂的理解，这是 Year 7 的核心技能。
n. Remember: 1 kg = 1000 g, so dividing by 1000 moves the decimal point three places left.
类似地，在研究生态系统时，你可能以克为单位记录一片叶子的质量，并需要将其换算为千克以进行生物量计算。记住：1 kg = 1000 g，因此除以 1000 会将小数点向左移动三位。
2. Algebra in Economics: Cost and Revenue | 经济学中的代数：成本与收入
A school tuck shop sells one type of healthy snack bar. The cost to buy each bar is £0.40, and the shop sells it for £0.70. Write an expression for the profit made when n bars are sold, and calculate the profit for 150 bars.
学校小卖部售卖一种健康零食棒。每条成本为 0.40 英镑，售价为 0.70 英镑。请写出售出 n 条时的利润表达式，并计算售出 150 条时的利润。
Profit per bar = selling price − cost price = £0.70 − £0.40 = £0.30. So total profit = 0.30n. For n = 150, profit = 0.30 × 150 = £ sing a letter to represent a variable is an early algebraic model of a business situation.
每条利润 = 售价 − 成本价 = 0.70 − 0.40 = 0.30 英镑。所以总利润 = 0.30n。当 n = 150 时，利润 = 0.30 × 150 = 45 英镑。使用字母表示变量是商业情境的早期代数模型。
Now consider a fixed cost: the shop pays a £5 weekly licence fee. Write a new expression for total profit after selling n bars, and find how many bars must be sold to just cover all costs (break-even).
现在考虑固定成本：小卖部每周支付 5 英镑的许可费。写出售出 n 条后的总利润新表达式，并计算需要售出多少条才能刚好覆盖所有成本（盈亏平衡）。
666..., so 17 bars must be sold (rounding up because you cannot sell a fraction of a bar).
新利润 = 0.30n − 5。盈亏平衡意味着利润为零，因此 0.30n − 5 = 0 → 0.30n = 5 → n = 5 ÷ 0.30 = 16.666...，所以必须售出 17 条（向上取整，因为不能卖出部分零食棒）。
3. Geometry in Geography: Map Scales and Area | 地理学中的几何：地图比例尺与面积
25 000. Two villages are 8 cm apart on the map. What is the actual distance in kilometres? Use the fact that 1 km = 100 000 cm.
一幅地图的比例尺为 1 : 25 000。两个村庄在地图上相距 8 cm。实际距离是多少千米？请利用 1 km = 100 000 cm。
÷ 100 000 = 2 km. This shows how ratio and proportion link map reading to real distances.
实际距离 = 8 × 25 000 cm = 200 000 cm。换算为 km：200 000 ÷ 100 000 = 2 km。这表明比和比例如何将地图阅读与实际距离联系起来。
A nature reserve is shown as a rectangle on the map measuring 4 cm by 3 cm. Calculate its actual area in square kilometres.
一个自然保护区在地图上显示为一个 4 cm × 3 cm 的矩形。计算其实际面积（以 km² 为单位）。
lternatively, convert sides first: 4 cm → 1 km, 3 cm → 0.75 km, area = 1 × 0.75 = 0.75 km².
地图面积 = 4 × 3 = 12 cm²。地图上每 1 cm 代表 25 000 cm = 0.25 km。因此 1 cm² 代表 (0.25 km)² = 0.0625 km²。实际面积 = 12 × 0.0625 = 0.75 km²。或者，先转换边长：4 cm → 1 km，3 cm → 0.75 km，面积 = 1 × 0.75 = 0.75 km²。
4. Statistics in Environmental Science: Data Interpretation | 环境科学中的统计：数据解读
The table shows the average monthly rainfall (mm) in a rainforest region for the first six months of the year.
| Month | Jan | Feb | Mar | Apr | May | Jun |
|---|---|---|---|---|---|---|
| Rainfall (mm) | 260 | 240 | 310 | 280 | 220 | 190 |
Calculate the mean monthly rainfall for this six-month period and identify the month with the greatest deviation from the mean.
上表显示了某雨林地区前六个月的平均月降雨量（mm）。计算这六个月的平均月降雨量，并找出与平均值偏差最大的月份。
mm) is 60 mm above the mean, and June (190 mm) is 60 mm below the mean. Both have the same absolute deviation of 60 mm. Interpreting such data helps ecologists understand seasonal patterns.
总和 = 260 + 240 + 310 + 280 + 220 + 190 = 1500 mm。平均值 = 1500 ÷ 6 = 250 mm。三月（310 mm）比平均值高 60 mm，六月（190 mm）比平均值低 60 mm。两者的绝对偏差均为 60 mm。解读此类数据有助于生态学家了解季节模式。
Now draw a bar chart to represent the rainfall data. Which type of chart would best show the trend over months? A line graph is most suitable because it displays changes over time clearly.
现在绘制一个条形图来表示降雨量数据。哪种图表最适合显示各月的变化趋势？折线图最合适，因为它能清晰地展示随时间的变化。
5. Number Theory in Computer Science: Binary Basics | 计算机科学中的数论：二进制基础
Computers use binary (base-2) numbers, composed only of bits 0 and 1. Convert the decimal number 29 into an 8-bit binary representation.
计算机使用二进制（base-2），仅由位 0 和 1 组成。将十进制数 29 转换为 8 位二进制表示。
itions for 2⁴ (16), 2³ (8), 2² (4), 2⁰ (1), and 0 elsewhere. From 2⁷ down to 2⁰: 0 0 0 1 1 1 0 1. So 29 in 8-bit binary is 00011101. This is a Year 7 introduction to number systems.
找出加起来等于 29 的最大 2 的幂：16 + 8 + 4 + 1 = 29。在 2⁴ (16)、2³ (8)、2² (4)、2⁰ (1) 的位置放 1，其他位置放 0。从 2⁷ 到 2⁰：0 0 0 1 1 1 0 1。因此 29 的 8 位二进制为 00011101。这是 Year 7 对数制的入门。
Now add the binary numbers 00101100 (44) and 00010101 (21) using column addition. Verify your answer by converting back to decimal.
现在用列加法计算二进制数 00101100 (44) 和 00010101 (21) 的和。通过转换回十进制验证答案。
in decimal (44 + 21 = 65). Learning binary strengthens your understanding of place value.
加法：00101100 + 00010101 = 01000001，对应的十进制为 64 + 1 = 65（44 + 21 = 65）。学习二进制可以加深对位值的理解。
6. Ratio and Proportion in Physics: Speed, Time, and Distance | 物理中的比与比例：速度、时间与距离
The relationship between speed, distance, and time is given by the formula speed = distance ÷ time. If a cyclist travels 36 km at a constant speed of 18 km/h, how long does the journey take? Express your answer in hours and minutes.
速度、距离和时间之间的关系由公式 速度 = 距离 ÷ 时间 给出。如果一名自行车运动员以 18 km/h 的恒定速度骑行 36 km，这趟旅程需要多长时间？答案用小时和分钟表示。
s. This is a straightforward application of a physics formula that requires rearranging a simple equation.
时间 = 距离 ÷ 速度 = 36 ÷ 18 = 2 小时。这是物理公式的直接应用，需要简单方程变形。
ours. Speed = 150 ÷ 2.5 = 60 km/h. The ability to manipulate time units is essential here.
现在，一辆汽车在 2 小时 30 分钟内行驶了 150 km。计算其平均速度（km/h）。首先，将 2 小时 30 分转换为 2.5 小时。速度 = 150 ÷ 2.5 = 60 km/h。这里关键在于处理时间单位的换算。
7. Probability in Sports: Expected Outcomes | 体育中的概率：预期结果
A basketball player has a free-throw success rate of 70%. In a match, she attempts 20 free throws. How many successful throws can she expect? Express this as an expected value calculation.
一名篮球运动员的罚球命中率为 70%。在一场比赛中，她尝试 20 次罚球。她可以期望命中多少次？请表示为期望值计算。
= 14. This uses the concept of expectation from probability, applied to a sports context.
期望次数 = 概率 × 试验次数 = 0.7 × 20 = 14。这运用了概率中的期望概念，应用于体育场景。
Now consider a football team that wins 45% of its matches, draws 30% 0-match season, how many points can they expect if a win gives 3 points and a draw 1 point? 40 matches: 1.65 × 40 = 66 points. This mixes probability with real-world scoring systems.
现在考虑一支足球队，45% 的比赛获胜，30% 打平，其余输球。在一个 40 场比赛的赛季中，如果赢一场得 3 分，平一场得 1 分，他们可以期望获得多少分？首先计算每场比赛的期望得分：(0.45 × 3) + (0.30 × 1) + (0.25 × 0) = 1.35 + 0.30 = 1.65。总的期望分数：1.65 × 40 = 66 分。这混合了概率与现实世界的计分系统。
8. Functions in Engineering: Simple Formulas | 工程学中的函数：简单公式
An engineer uses the formula F = ma (Newton's second law) to calculate force (F, in newtons) from mass (m, in kg) and acceleration (a, in m/s²). If a 12 kg object accelerates at 5 m/s², find the force.
一位工程师使用公式 F = ma（牛顿第二定律）根据质量（m，kg）和加速度（a，m/s²）计算力（F，牛顿）。如果一个 12 kg 的物体以 5 m/s² 加速，求力。
= 60 N. This is direct substitution into a functional relationship, a key algebraic skill.
F = 12 × 5 = 60 N。这是直接代入函数关系式，是一项关键的代数技能。
m/s². Rearranging simple formulas prepares you for later work with more complex equations.
重新排列公式，求当 200 N 的力作用在 40 kg 的质量上时的加速度。a = F ÷ m = 200 ÷ 40 = 5 m/s²。公式变形为你日后处理更复杂的方程做好准备。
9. Sequences in Biology: Population Growth | 生物学中的数列：种群增长
A population of bacteria doubles every hour. Starting with 50 bacteria, write the first four terms of the sequence representing the population at the end of each hour. Identify whether it is arithmetic or geometric.
一群细菌每小时数量翻倍。从 50 个细菌开始，写出表示每小时结束时种群数量的数列的前四项。判断它是等差数列还是等比数列。
Terms: 50, 100, 200, 400. It is a geometric sequence with a common ratio of 2. This models exponential growth, frequently observed in microbiology.
项：50, 100, 200, 400。这是一个公比为 2 的等比数列。它模拟了微生物学中常见的指数增长。
uence: term n = a × rⁿ⁻¹, where a = 50, r = 2. Term 6 = 50 × 2⁵ = 50 × 32 = 1600 bacteria.
使用等比数列第 n 项公式求 6 小时后的细菌数量：第 n 项 = a × rⁿ⁻¹，其中 a = 50，r = 2。第 6 项 = 50 × 2⁵ = 50 × 32 = 1600 个细菌。
10. Graphs in Business: Break-even Analysis | 商业中的图表：盈亏平衡分析
l business makes handmade candles. The cost to produce x candles is given by C = 2x + 30 (£), and the revenue from selling x candles is R = 5x (£ lot both linear functions on the same axes, and use the graph to find the break-even point – the number of candles that must be sold so that cost equals revenue.
一家小企业制作手工蜡烛。生产 x 根蜡烛的成本为 C = 2x + 30（£），销售 x 根蜡烛的收入为 R = 5x（£）。在同一坐标系中绘制这两个线性函数的图像，并利用图像找到盈亏平衡点——即必须售出的蜡烛数量，使得成本等于收入。
10,50), (20,70), (30,90). For R: (0,0), (10,50), (20,100), (30,150). The lines intersect at (10,50), so 10 candles must be sold to break even. This visual approach links algebra to business decision-making.
绘制时，为 x = 0, 10, 20, 30 创建数值表。对于 C：点 (0,30), (10,50), (20,70), (30,90)。对于 R：(0,0), (10,50), (20,100), (30,150)。两直线相交于 (10,50)，因此必须售出 10 根蜡烛才能达到盈亏平衡。这种可视化方法将代数与商业决策联系起来。
rms the graphical solution and demonstrates how graphs and equations reinforce each other.
用代数检验：令 5x = 2x + 30 → 3x = 30 → x = 10。这证实了图像解，并展示了图像与方程如何相互印证。
11. Percentages in Geography: Population Change | 地理中的百分数：人口变化
The population of a town was 20 000 in 2020. It increased by 12% by 2025. Calculate the new population. Then, if it decreases by 8% from 2025 to 2030, what is the final population?
某镇 2020 年人口为 20 000。到 2025 年增加了 12%。计算新人口。然后，如果从 2025 年到 2030 年人口减少 8%，最终人口是多少？
Increase: 12% of 20 000 = 0.12 × 20 000 = 2400. New population = 20 000 + 2400 = 22 400. Decrease: 8% 22 400 = 1792. Final = 22 400 − 1792 = 20 608. Percentage changes are not symmetric: a 12% rise followed by an 8% fall does not return to the original number.
增加：20 000 的 12% = 0.12 × 20 000 = 2400。新人口 = 20 000 + 2400 = 22 400。减少：22 400 的 8% = 0.08 × 22 400 = 1792。最终 = 22 400 − 1792 = 20 608。百分数变化并非对称：先升 12% 再降 8% 并不会回到原数。
This is crucial in demography and economics, teaching you to apply multipliers: a 12% increase multiplies by 1.12, an 8% rease by 0.92. Overall multiplier = 1.12 × 0.92 = 1.0304, giving 20 000 × 1.0304 = 20 608.
这在人口学和经济学中至关重要，教会你使用乘数：增加 12% 即乘以 1.12，减少 8% 即乘以 0.92。总的乘数 = 1.12 × 0.92 = 1.0304，得出 20 000 × 1.0304 = 20 608。
12. Interpreting Data in Health Science | 健康科学中的数据解读
A survey records the number of hours 30 Year 7 students sleep per night. The results are summarised in a frequency table. Estimate the mean sleep duration and discuss whether the data suggests most students meet the recommended 9 hours.
一项调查记录了 30 名 Year 7 学生每晚的睡眠小时数。结果汇总在频数表中。估计平均睡眠时长，并讨论数据是否表明大多数学生达到了推荐的 9 小时。
| Hours | 7- | 8- | 9- | 10-11 |
|---|---|---|---|---|
| Frequency | 4 | 8 | 12 | 6 |
, but the mode is the 9- group (12 students), and the median lies in the 9- group as well. While the average is healthy, 4 + 8 = 12 students sleep less than 9 hours, so a significant minority are below the recommendation.
使用组中值：7.5, 8.5, 9.5, 10.5。总和 = (7.5×4) + (8.5×8) + (9.5×12) + (10.5×6) = 30 + 68 + 114 + 63 = 275。平均 = 275 ÷ 30 ≈ 9.17 小时。平均值高于 9，但众数是 9- 组（12 名学生），中位数也落在 9- 组。虽然平均数是健康的，但有 4 + 8 = 12 名学生睡眠少于 9 小时，因此有相当一部分人未达到推荐值。
This exercise shows how statistics from health science require careful interpretation: a single average can hide the distribution of values.
这个练习表明，来自健康科学的统计数据需要仔细解读：单一的平均值可能掩盖数值的分布情况。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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