📚 Interdisciplinary Integrated Problem Training for KS3 Edexcel Further Maths | KS3 Edexcel 进阶数学：跨学科综合题型训练
In the KS3 Edexcel Further Maths syllabus, students are expected not only to master mathematical techniques but also to apply them confidently across a range of real-world contexts. Interdisciplinary problems – those linking mathematics with physics, geography, business, biology and more – help develop critical thinking and show why maths matters beyond the classroom.
在 KS3 Edexcel 进阶数学大纲中，学生不仅需要掌握数学技巧，还要能在各种现实情境中自信地加以应用。跨学科问题——将数学与物理、地理、商业、生物等学科联系起来——有助于培养批判性思维，并展示数学在课堂之外的价值。
1. Ratios and Map Scales | 比例与地图比例尺
Maps are a classic example of ratios in action. A scale of 1 : 25 000 means that 1 cm on the map represents 25 000 cm in real life. Converting units and using multiplication or division are essential skills.
地图是比例应用的经典例子。比例尺 1 : 25 000 表示地图上的 1 厘米代表实际距离的 25 000 厘米。进行单位换算并使用乘除法是必备技能。
Problem: A hiking trail measures 8.5 cm on a 1:50 000 OS map. Find the actual length of the trail in kilometres.
题目：在 1:50 000 的英国地形图上，一条徒步路径长 8.5 厘米。求该路径的实际长度，以千米为单位。
00 cm. Convert to metres: 425 000 ÷ 100 = 4 250 m. Convert to km: 4 250 ÷ 1 000 = 4.25 km.
解答：实际距离 = 图上距离 × 比例因子 = 8.5 cm × 50 000 = 425 000 cm。换算成米：425 000 ÷ 100 = 4 250 m。再换算成千米：4 250 ÷ 1 000 = 4.25 km。
2. Speed, Distance and Time | 速度、距离与时间
ime is fundamental in physics and everyday travel. Rearranging this formula to find distance or time is a key algebraic skill.
关系式 速度 = 距离 ÷ 时间 是物理和日常出行的基础。重新整理该公式以求解距离或时间是一项关键的代数技能。
Problem: A cyclist travels at a steady speed of 12 m/s for 2 minutes 30 seconds. How far does the cyclist travel in kilometres?
题目：一名自行车骑手以 12 m/s 的恒定速度骑行 2 分 30 秒。骑手骑行了多少千米？
Distance = speed × time = 12 m/s × 150 s = 1 800 m. In kilometres: 1 800 ÷ 1 000 = 1.8 km.
解答：首先将时间转换为秒：2 分 30 秒 = (2 × 60) + 30 = 150 秒。距离 = 速度 × 时间 = 12 m/s × 150 s = 1 800 m。以千米为单位：1 800 ÷ 1 000 = 1.8 km。
3. Percentage Change and Profit | 百分比变化与利润
business studies, calculating percentage profit or loss is vital. Using the formula ((new – old) ÷ old) × 100% helps analyse financial performance.
在商业研究中，计算利润或亏损的百分比至关重要。使用公式 ((新值 – 旧值) ÷ 旧值) × 100% 有助于分析财务表现。
Problem: A shop buys a jacket for £45 and sells it for £63. Find the percentage profit.
题目：一家商店以 45 英镑购入一件夹克，并以 63 英镑售出。求利润百分比。
Solution: Profit = selling price – cost price = £63 – £45 = £18. Percentage profit = (profit ÷ cost price) × 100% = (18 ÷ 45) × 100% = 0.4 × 100% = 40%.
解答：利润 = 售价 – 成本价 = £63 – £45 = £18。利润百分比 = (利润 ÷ 成本价) × 100% = (18 ÷ 45) × 100% = 0.4 × 100% = 40%。
4. Interpreting Graphs in Science Experiments | 科学实验中的图表解读
ing gradients and intercepts links directly to the equation of a straight line y = mx + c.
线状图广泛用于展示实验数据，例如温度随时间的变化。解读斜率和截距与直线方程 y = mx + c 直接相关。
Problem: A cooling experiment records temperature T (°C) against time t (min). The graph is a straight line passing through (0, 80) and (10, 30). Find the equation of the line and the temperature after 6 minutes.
题目：一个
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