Year 7 Cambridge Advanced Mathematics: Interdisciplinary Problem-Solving Training | Year 7 Cambridge 进阶数学：跨学科综合题型训练

📚 Year 7 Cambridge Advanced Mathematics: Interdisciplinary Problem-Solving Training | Year 7 Cambridge 进阶数学：跨学科综合题型训练

In the Cambridge Lower Secondary programme, Advanced Mathematics extends beyond routine calculations. Students are challenged to interpret real-life scenarios, extract mathematical relationships, and solve problems that span multiple subjects. This type of interdisciplinary training not only deepens understanding but also prepares learners for higher-level thinking.

在 Cambridge 初中课程中，进阶数学突破常规计算。学生需要解读现实情境、提取数学关系，并解决跨越多个学科的问题。这种跨学科训练不仅能加深理解，还能为高级思维奠定基础。

1. What Are Interdisciplinary Problems? | 什么是跨学科问题？

Interdisciplinary problems combine mathematical skills with knowledge from science, geography, technology, or finance. For example, calculating the density of a substance uses ratio and measurement; reading a map involves scale and proportion; analysing climate data requires graph interpretation and statistics.

跨学科问题将数学技能与科学、地理、科技或金融知识相结合。例如，计算物质密度需要用到比和测量；阅读地图涉及比例尺和比例；分析气候数据则需要图表解读和统计。

In Year 7 Advanced Mathematics, you will frequently see questions that start with a real-world context. The key is to identify the mathematical structure hidden in the words – that is your first step towards a solution.

在 Year 7 进阶数学中，你会经常看到以真实世界为背景的题目。关键是识别隐藏在文字中的数学结构——这是你走向解答的第一步。

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2. Ratio and Proportion in Science | 科学中的比与比例

Many scientific formulas involve ratios. A classic example is density: Density = Mass ÷ Volume. If a metal block has a mass of 400 g and a volume of 50 cm³, its density is 8 g/cm³. You must be comfortable rearranging the formula to find missing values.

许多科学公式都包含比值。一个经典例子是密度：密度 = 质量 ÷ 体积。如果一块金属的质量为 400 克，体积为 50 立方厘米，它的密度就是 8 克/立方厘米。你必须能熟练地变形公式，求出缺失的数值。

Similarly, speed (physics), concentrations (chemistry), and population density (geography) all use proportional reasoning. Practice writing ratios in simplest form and solving proportion equations such as 3:7 = x:21.

类似地，速度（物理）、浓度（化学）和人口密度（地理）都运用比例推理。练习将比化为最简形式，并求解如 3:7 = x:21 的比例方程。

Example: A solution contains 15 g of salt dissolved in 200 cm³ of water. Find the concentration in g per litre.

例题： 某溶液在 200 立方厘米水中溶解了 15 克盐。求以克/升为单位的浓度。

Solution: 1 litre = 1000 cm³, so mass per litre = (15 ÷ 200) × 1000 = 75 g/L. This uses unitary method and conversion.

解答：1 升 = 1000 立方厘米，因此每升质量 = (15 ÷ 200) × 1000 = 75 克/升。这里运用了归一法和单位换算。

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3. Speed, Distance and Time in Physics | 物理中的速度、距离与时间

The relationship between speed, distance and time is fundamental in kinematics. The formula is often written as:

速度、距离与时间的关系是运动学的基础。公式通常写作：

Speed = Distance ÷ Time

速度 = 距离 ÷ 时间

When solving interleaved problems, units must be consistent. If distance is in km and time in hours, speed is in km/h. You may need to convert minutes to hours by dividing by 60.

解决交织问题时，单位必须一致。如果距离以千米计、时间以小时计，速度单位就是千米/时。你可能需要将分钟转换为小时，即除以 60。

Example: A cyclist travels 45 km in 2 hours and 30 minutes. Calculate the average speed.

例题： 一名骑行者 2 小时 30 分钟骑行 45 千米。求平均速度。

Convert 30 minutes to 0.5 hours, total time = 2.5 h. Speed = 45 ÷ 2.5 = 18 km/h. Always present the answer with correct units.

将 30 分钟转换为 0.5 小时，总时间 = 2.5 小时。速度 = 45 ÷ 2.5 = 18 千米/时。回答时务必附上正确的单位。

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4. Scale Drawings and Maps in Geography | 地理中的比例尺与地图

Maps use scale to represent real distances. A scale of 1:50,000 means that 1 cm on the map represents 50,000 cm (or 0.5 km) in reality. Measuring lengths on a map and multiplying by the scale factor is a direct application of ratio.

地图利用比例尺表示实际距离。1:50,000 的比例尺意味着地图上 1 厘米代表实际 50,000 厘米（或 0.5 千米）。在地图上量取长度再乘以比例因子，是比例的直接应用。

Area scale factor requires more care: if the linear scale is 1:50,000, the area scale factor is 1²:50,000² = 1:2,500,000,000. For Year 7, focus on linear distances and simple area conversions.

面积比例因子需多加留意：若线性比例为 1:50,000，则面积比例因子为 1²:50,000² = 1:2,500,000,000。对于 Year 7，重点是线性距离和简单的面积换算。

Example: On a 1:25,000 map, two villages are 8.4 cm apart. Find the actual distance in km.

例题： 在一幅 1:25,000 的地图上，两村庄相距 8.4 厘米。求实际距离（千米）。

Actual distance = 8.4 × 25,000 cm = 210,000 cm. Convert to km: 210,000 ÷ 100,000 = 2.1 km.

实际距离 = 8.4 × 25,000 厘米 = 210,000 厘米。转化为千米：210,000 ÷ 100,000 = 2.1 千米。

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5. Mixtures and Concentrations in Chemistry | 化学中的混合物与浓度

Chemists often express concentration as a percentage by mass or volume. For instance, a 5% saline solution means 5 g of salt in 100 g of solution. Understanding mass and percentage calculations is essential.

化学家常用质量分数或体积分数表示浓度。例如，5% 的盐溶液表示 100 克溶液中含有 5 克盐。掌握质量与百分数计算至关重要。

Problems may ask: ‘How much solute is needed to make 250 g of a 20% sugar solution?’ This involves finding 20% of 250 g = 50 g. Reverse problems require solving simple equations.

问题可能要求：“配制 250 克 20% 的糖溶液需要多少溶质？” 这需要计算 250 克的 20%，即

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