📚 Year 9 Cambridge Mathematics: Cross-Curricular Problem-Solving Training | 剑桥Year 9数学：跨学科综合题型训练
In Year 9 Cambridge Mathematics, the ability to apply mathematical skills across different subjects is highly valued. This integrated problem-solving training will help you connect algebra, geometry, statistics and ratio with real-world contexts from physics, chemistry, biology, geography, economics and beyond. Each section presents a typical interdisciplinary problem, paired with a worked solution, so you can see how to transfer mathematical reasoning flexibly.
在剑桥Year 9数学中，能够将数学技能应用到不同学科中是非常重要的。本次跨学科综合题型训练将帮助你把代数、几何、统计和比例等知识与物理、化学、生物、地理、经济学等真实场景联系起来。每个小节展示一个典型的跨学科问题并给出解析，让你学会灵活运用数学推理。
1. Physics: Speed, Distance and Time | 物理：速度、距离与时间
Speed is the distance travelled per unit of time. The formula v = d / t is fundamental, and unit conversions such as km/h to m/s are frequently tested.
速度是单位时间内行进的距离。基本公式 v = d / t，并且经常考察单位换算，如从千米/小时转换为米/秒。
Example: A train covers 210 km in 2 hours 30 minutes. Calculate its average speed in km/h and m/s.
例题：一列火车在2小时30分钟内行驶210千米。计算平均速度，分别以千米/小时和米/秒为单位。
Solution: Time = 2.5 h. v = 210 km / 2.5 h = 84 km/h. To convert: 84 km/h = 84 × (1000 m / 3600 s) = 84 × 5/18 m/s = 23.33 m/s (to 2 d.p.).
解答：时间 = 2.5 小时。v = 210 km ÷ 2.5 h = 84 km/h。换算：84 km/h = 84 × (1000 m / 3600 s) = 84 × 5/18 m/s ≈ 23.33 m/s（保留两位小数）。
2. Chemistry: Mixtures and Ratio | 化学：混合物与比例
In chemistry, mixing solutions of different concentrations often leads to ratio problems. Setting up an equation using the amount of solute is a reliable algebraic method.
在化学中，混合不同浓度的溶液常引发出比例问题。利用溶质质量建立方程是一种可靠的代数方法。
Example: A 15% salt solution is mixed with a 40% salt solution to make 5 litres of a 25% salt solution. How many litres of each are needed?
例题：用15%的盐水与40%的盐水混合，配制5升25%的盐水。问每种需要多少升？
Solution: Let x litres be the 15% solution, then (5 – x) litres is the 40% solution. Salt mass: 0.15x + 0.40(5 – x) = 0.25 × 5. So 0.15x + 2 – 0.40x = 1.25, -0.25x = -0.75, x = 3. Thus 3 L of 15% and 2 L of 40% solution.
解答：设15%的溶液为 x 升，则40%的溶液是 (5 – x) 升。溶质质量：0.15x + 0.40(5 – x) = 0.25 × 5。计算得 0.15x + 2 – 0.40x = 1.25，-0.25x = -0.75，x = 3。所以需要3升15%的溶液和2升40%的溶液。
3. Biology: Population Growth and Percentage Change | 生物：种群增长与百分比变化
Population models often use repeated percentage increases. Understanding multipliers like 1 + r% helps calculate growth over several years without extensive tables.
种群模型常使用反复的百分比增长。理解乘数因子（如1 + r%）可以不用列表就能计算多年后的数量。
Example: A colony of 500 bacteria increases by 8% each hour. Estimate the population after 3 hours, giving your answer to the nearest whole number.
例题：一个500个细菌的群落每小时增长8%。估算3小时后的数量，答案保留整数。
Solution: Multiplier per hour = 1.08. After 3 hours: population = 500 × (1.08)³ = 500 × 1.259712 = 629.856 ≈ 630 bacteria.
解答：每小时增长乘数 = 1.08。3小时后：数量 = 500 × (1.08)³ = 500 × 1.259712 = 629.856 ≈ 630 个细菌。
4. Geography: Map Scales and Areas | 地理：地图比例尺与面积
Map scales express the ratio of a distance on a map to the actual distance on the ground. Calculating real area requires careful conversion of both length and width, and then units such as km².
地图比例尺表示图上距离与实际地面距离之比。计算真实面积需要仔细换算长和宽，并处理好单位（例如平方千米）。
Example: A map has a scale of 1 : 25 000. A rectangular park is shown as 6 cm by 4 cm. Find its actual area in square kilometres.
例题：一幅地图的比例尺为 1 : 25 000。一个长方形公园在图上为 6 厘米 × 4 厘米。求其实际面积，以平方千米为单位。
Solution: 1 cm on map = 25 000 cm in reality = 250 m = 0.25 km. Length = 6 × 0.25 km = 1.5 km, width = 4 × 0.25 km = 1.0 km. Area = 1.5 km × 1.0 km = 1.5 km².
解答：图上1厘米 = 实际25 000厘米 = 250米 = 0.25千米。实际长 = 6 × 0.25 km = 1.5 km，实际宽 = 4 × 0.25 km = 1.0 km。面积 = 1.5 km × 1.0 km = 1.5 km²。
5. Economics: Profit, Loss and Discount | 经济学：利润、亏损与折扣
Business applications involve calculating profit percentage, discount and successive price changes. Students need to work confidently with decimals and percentage multipliers.
商业应用涉及计算利润百分比、折扣和连续价格变动。学生需要熟练运用小数和百分比乘数。
Example: A shop buys a jacket for $80 and wants a 25% profit on cost. At a sale, the jacket is sold with a 15% discount on the marked price. Find the final selling price and the actual profit percentage.
例题：商店以80美元购入一件夹克，希望获得成本25%的利润。在促销中，夹克按标价打八五折出售。求最终售价和实际利润率。
Solution: Marked price = $80 × 1.25 = $100. Discount = 15%, so selling price = $100 × 0.85 = $85. Profit = $85 – $80 = $5. Profit % on cost = (5 / 80) × 100 = 6.25%.
解答：标价 = 80 × 1.25 = 100美元。折扣15%，售价 = 100 × 0.85 = 85美元。利润 = 85 – 80 = 5美元。成本利润率 = (5 / 80) × 100 = 6.25%。
6. Physics: Density, Mass and Volume | 物理：密度、质量与体积
Density links mass and volume through the formula ρ = m / V. Interdisciplinary problems may require calculating one quantity given the other two, or converting between units such as g/cm³ and kg/m³.
密度通过公式 ρ = m / V 将质量和体积联系起来。跨学科问题可能要求根据两个量求第三个，或者在单位间换算，例如 g/cm³ 和 kg/m³。
Example: A metal block has mass 540 g and volume 200 cm³. Find its density in g/cm³ and kg/m³.
例题：一个金属块质量为540克，体积为200立方厘米。求其密度，分别以 g/cm³ 和 kg/m³ 表示。
Solution: ρ = m / V = 540 g / 200 cm³ = 2.7 g/cm³. To convert to kg/m³: 1 g/cm³ = 1000 kg/m³, so 2.7 g/cm³ = 2700 kg/m³.
解答：ρ = 540 g / 200 cm³ = 2.7 g/cm³。换算为 kg/m³：1 g/cm³ = 1000 kg/m³，因此 2.7 g/cm³ = 2700 kg/m³。
7. Environmental Science: Data Analysis from Tables | 环境科学：表格数据分析
Environmental studies often provide raw data in tables. Year 9 students should be able to calculate mean, median, mode and interpret trends while linking to real-world issues like energy consumption.
环境研究常提供表格形式的原始数据。Year 9 学生应能计算平均数、中位数、众数，并结合能源消耗等现实问题分析趋势。
Example: The table below shows electricity used (kWh) by a household over 5 days: 18, 22, 19, 25, 21. Find the mean, median and suggest one reason for the variation.
例题：下表显示某家庭5天的用电量（千瓦时）：18, 22, 19, 25, 21。求平均值、中位数，并说明一个可能造成差异的原因。
Solution: Mean = (18+22+19+25+21)/5 = 105/5 = 21 kWh. Ordered: 18,19,21,22,25; median = 21 kWh. Variation could be due to different appliance usage or weather.
解答：平均数 = (18+22+19+25+21) ÷ 5 = 21 kWh。排序后：18,19,21,22,25；中位数 = 21 kWh。差异可能因电器使用不同或天气变化引起。
8. Engineering: Pythagoras’ Theorem in Construction | 工程：建筑中的勾股定理
Pythagoras’ theorem is used in construction to ensure right angles and to calculate lengths in structures. It is one of the most common real-world geometry applications.
勾股定理在建筑中用于确保直角并计算结构中的长度，是最常见的现实世界几何应用之一。
Example: A ladder is 5.0 m long. It leans against a vertical wall with its foot 1.4 m away from the wall. How high up the wall does the ladder reach? Give your answer to 2 decimal places.
例题：一架梯子长5.0米，斜靠在竖直墙上，梯脚离墙1.4米。梯子顶端能达到墙多高？结果保留两位小数。
Solution: Let height be h. Then h² + 1.4² = 5.0². h² = 25 – 1.96 = 23.04. h = √23.04 = 4.80 m.
解答：设高度为 h。则 h² + 1.4² = 5.0²。h² = 25 – 1.96 = 23.04。h = √23.04 = 4.80 米。
9. Art & Design: Symmetry and Tessellation | 艺术与设计：对称与镶嵌
Geometry in art involves symmetry, angles and shapes that tile a plane. Recognising properties of regular polygons helps in designing tessellating patterns.
艺术中的几何涉及对称、角度和能够密铺平面的形状。认识正多边形的性质有助于设计镶嵌图案。
Example: A regular hexagon is used to create a floor pattern. How many lines of symmetry does a regular hexagon have? What is the interior angle? Explain why hexagons tessellate.
例题：用正六边形设计地板图案。正六边形有几条对称轴？内角是多少？解释为什么六边形可以密铺。
Solution: A regular hexagon has 6 lines of symmetry. Interior angle = (6 – 2) × 180° / 6 = 120°. At a point where three hexagons meet, the sum of angles = 3 × 120° = 360°, so they fit without gaps.
解答：正六边形有6条对称轴。内角 = (6 – 2) × 180° ÷ 6 = 120°。在三个六边形相遇的顶点处，角度之和 = 3 × 120° = 360°，因此可以无缝隙地密铺。
10. Music: Fractions and Rhythm Patterns | 音乐：分数与节奏模式
Musical time signatures are fractions in action. A 4/4 bar fills with notes that total 4 quarter-note beats. Understanding addition and equivalence of fractions is essential for reading rhythms.
音乐中的拍号就是分数在起作用。一小节4/4拍需要用总时值为4个四分音符的音符填满。理解分数的加法与等值是阅读节奏的关键。
Example: In a 4/4 bar, a composition uses one half note, one quarter note, and two eighth notes. Has the bar been filled completely? What fraction of the bar remains, if any?
例题：在一个4/4拍的小节中，一首曲子使用了一个二分音符、一个四分音符和两个八分音符。这小节被填满了吗？如果没有，还剩下多少拍（用分数表示）？
Solution: Half note = 2 beats, quarter note = 1 beat, each eighth note = 1/2 beat. Total = 2 + 1 + 1/2 + 1/2 = 4 beats. The bar is exactly full; 0 beats remain.
解答：二分音符 = 2拍，四分音符 = 1拍，每个八分音符 = 1/2 拍。总计 = 2 + 1 + ½ + ½ = 4拍。小节刚好填满，剩余0拍。
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