📚 Year 10 Eduqas Mathematics: Cross-Curricular Problem Solving | 跨学科综合题型训练
In the Eduqas GCSE Mathematics course, you are regularly asked to apply mathematical skills to contexts borrowed from science, geography, economics, art and technology. These cross-curricular problems help you see how maths works beyond the textbook, while also preparing you for the multi-step reasoning typical of higher-tier questions. This article brings together twelve topic areas, each with a worked example and clear strategy, so you can practise transferring core algebra, ratio, geometry and statistics to real-world scenarios.
在 Eduqas GCSE 数学课程中，你经常会遇到将数学技能应用到科学、地理、经济、艺术和技术等背景中的题目。这些跨学科问题不仅让你看到数学在课本之外的用途，也帮助你为高等级考试中常见的多步骤推理做好准备。本文汇集了十二个专题，每个都配有例题和清晰的解题策略，以便你练习把核心代数、比例、几何和统计知识迁移到真实情境中。
1. Physics: Speed, Distance and Time | 物理：速度、距离与时间
The relationship between speed, distance and time is fundamental in kinematics. The average speed v can be found using v = d ÷ t, where d is distance travelled and t is the time taken. Rearranging gives d = v × t and t = d ÷ v. In cross-curricular problems, make sure all units are consistent and convert where necessary.
速度、距离和时间的关系是运动学的基础。平均速度 v 可用 v = d ÷ t 求得，其中 d 为行驶距离，t 为所用时间。变形后可得 d = v × t 和 t = d ÷ v。在处理跨学科问题时，务必保持单位一致并在必要时进行换算。
Example: A delivery van travels 210 km in 2 hours 20 minutes. Calculate its average speed in km/h.
例题：一辆送货车行驶 210 km 耗时 2 小时 20 分钟。计算其平均速度 (km/h)。
Convert 20 minutes to hours: 20 min = 1/3 h, so total time t = 2⅓ = 7/3 hours. Then v = 210 ÷ (7/3) = 210 × 3/7 = 90 km/h.
将 20 分钟化成小时：20 min = 1/3 h，总时间 t = 2 ⅓ = 7/3 小时。于是 v = 210 ÷ (7/3) = 210 × 3/7 = 90 km/h。
2. Chemistry: Concentrations and Mixtures | 化学：浓度与混合物
Chemists often mix solutions of different strengths to obtain a desired concentration. The amount of solute remains constant, so mixtures can be modelled with simple linear equations. If a volume a of p% solution is combined with volume b of q% solution to make a total volume V of r% solution, we can set up two equations: a + b = V and (p/100)a + (q/100)b = (r/100)V. This style of problem regularly appears on non-calculator papers as a test of proportional reasoning.
化学家常常混合不同浓度的溶液以获得目标浓度。溶质总量保持不变，因此可以用简单的一次方程对混合过程建模。若将体积为 a、浓度为 p% 的溶液与体积为 b、浓度为 q% 的溶液混合，得到总体积为 V、浓度为 r% 的溶液，则可列出方程组：a + b = V 以及 (p/100)a + (q/100)b = (r/100)V。这类题目经常出现在非计算器试卷中，以考查比例推理能力。
Example: A lab technician has a 10% salt solution and a 25% salt solution. How many millilitres of each must be mixed to obtain 400 ml of a 16% salt solution?
例题：一位实验员有 10% 盐溶液和 25% 盐溶液。要配制 400 ml 浓度为 16% 的盐溶液，需各取多少毫升？
Let x be the volume (ml) of 10% solution and y the volume of 25% solution. Then x + y = 400 and (0.10)x + (0.25)y = 0.16 × 400 = 64. Subtracting 0.10×(x + y = 400) from the second gives 0.15y = 24, so y = 160 ml and x = 240 ml.
设 x 为 10% 溶液的体积(ml)，y 为 25% 溶液的体积。则有 x + y = 400，且 0.10x + 0.25y = 0.16×400 = 64。用第二式减去 0.10 倍的第一式得到 0.15y = 24，解得 y = 160 ml，x = 240 ml。
3. Biology: Population Growth and Decay | 生物：种群增长与衰减
In biology, populations can grow by repeated multiplication, which mathematics treats as geometric sequences or compound increases. The general formula for exponential growth is P = P₀ × mⁿ, where P₀ is the initial population, m is the growth factor per time interval, and n is the number of intervals. Year 10 problems typically use whole‑number intervals and simple doubling or halving factors.
在生物学中，种群通过重复乘法实现增长，这在数学中被视为等比数列或复利增长。指数增长的通用公式为 P = P₀ × mⁿ，其中 P₀ 为初始种群数量，m 为每个时间间隔的增长因子，n 为间隔数。Year 10 的题目通常使用整数时间间隔和简单的翻倍或减半因子。
Example: A colony of bacteria doubles every 30 minutes. If the colony starts with 500 cells, how many cells will there be after 3 hours?
例题：某细菌菌落每 30 分钟数量翻倍。若初始有 500 个细胞，3 小时后将有多少个细胞？
Number of doubling intervals n = 3 hours ÷ 0.5 hours = 6. Final population P = 500 × 2⁶. Since 2⁶ = 64, P = 500 × 64 = 32 000 cells.
翻倍间隔数 n = 3 小时 ÷ 0.5 小时 = 6。最终数量 P = 500 × 2⁶。由于 2⁶ = 64，P = 500 × 64 = 32 000 个细胞。
4. Geography: Map Scales and Bearings | 地理：地图比例尺与方位
Map scales are expressed as ratios such as 1:25 000, meaning 1 cm on the map represents 25 000 cm (or 0.25 km) on the ground. Bearings are measured clockwise from north and require trigonometry to resolve into east/west and north/south components. GCSE questions often combine scale conversion with the use of sine and cosine rules in right‑angled triangles.
地图比例尺用诸如 1:25 000 的比值表示，即图上 1 cm 代表实地 25 000 cm（即 0.25 km）。方位角从正北顺时针测量，需要借助三角函数分解为东西分量和南北分量。GCSE 试题常将比例尺换算与直角三角形中的正弦、余弦定理结合考查。
Example: On a map of scale 1:50 000, two villages are 4.8 cm apart. What is the actual distance in kilometres? If a hiker walks from one village to the other on a bearing of 065° for a straight‑line distance of 6 km, find the eastward and northward displacement.
例题：在一幅 1:50 000 的地图上，两个村庄相距 4.8 cm。实地距离为多少公里？若一位徒步者沿 065° 方位直线行走 6 km 从一村到达另一村，求向东和向北的位移分量。
Actual distance = 4.8 cm × 50 000 = 240 000 cm = 2.4 km. For the bearing, 065° is clockwise from north. North component = 6 × cos 65° ≈ 6 × 0.4226 = 2.54 km. East component = 6 × sin 65° ≈ 6 × 0.9063 = 5.44 km.
实地距离 = 4.8 cm × 50 000 = 240 000 cm = 2.4 km。对于方位角 065°，向北分量 = 6 × cos 65° ≈ 6 × 0.4226 = 2.54 km；向东分量 = 6 × sin 65° ≈ 6 × 0.9063 = 5.44 km。
5. Economics: Profit, Loss and Interest | 经济：利润、亏损与利息
Business applications in mathematics revolve around percentages, ratios and iterative growth. Profit percentage can be expressed as a fraction of the cost price, profit% = (profit ÷ cost price) × 100. Compound interest uses the multiplier method: A = P × (1 + r/100)ⁿ, where P is the principal, r the annual interest rate and n the number of years.
数学在经济中的应用围绕着百分数、比率和迭代增长。利润率可表示为成本价的分数：利润% = (利润 ÷ 成本价) × 100。复利采用乘数法：A = P × (1 + r/100)ⁿ，其中 P 为本金，r 为年利率，n 为年数。
Example: A shop buys a jacket for £45 and sells it for £63. Find the profit percentage based on the cost price. Separately, £2000 is invested at 3.5% compound interest per annum. Calculate the amount after 2 years.
例题：一家商店以 £45 购得一件夹克，以 £63 售出。求基于成本价的利润率。另外，将 £2000 以 3.5% 的年复利投资，计算 2 年后的总额。
Profit = 63 − 45 = £18. Profit % = (18 ÷ 45) × 100 = 40%. Compound amount: multiplier = 1 + 3.5/100 = 1.035. After 2 years, A = 2000 × 1.035² = 2000 × 1.071225 = £2142.45.
利润 = 63 − 45 = £18。利润率 = (18 ÷ 45) × 100 = 40%。复利总额：乘数 = 1 + 3.5/100 = 1.035。2 年后，A = 2000 × 1.035² = 2000 × 1.071225 = £2142.45。
6. Art & Design: Symmetry, Patterns and Tessellations | 艺术与设计：对称、图案与镶嵌
Tessellations are patterns of shapes that fit together without gaps. A regular polygon can tessellate on its own only if its interior angle is a factor of 360°. The interior angle of a regular polygon with n sides is interior angle = (n − 2) × 180° ÷ n. This simple geometric property links closely to design, tiling and architecture, making it a favourite cross‑curricular topic.
镶嵌是指形状之间无缝隙地拼接成图案。单独的正多边形能够密铺平面的条件是它的内角是 360° 的因数。正 n 边形的内角为 内角 = (n − 2) × 180° ÷ n。这一简单的几何性质与设计、瓷砖铺贴和建筑密切相关，因而成为受欢迎的跨学科主题。
Example: Show that a regular pentagon cannot tessellate by itself, but a regular hexagon can.
例题：证明正五边形不能单独镶嵌，而正六边形可以。
For a pentagon (n=5): interior angle = (5−2)×180÷5 = 108°. Since 360 ÷ 108 = 3.33, not an integer, pentagons leave gaps. For a hexagon (n=6): interior angle = (6−2)×180÷6 = 120°. And 360 ÷ 120 = 3, so three hexagons meet neatly at a point.
对五边形 (n=5)：内角 = (5−2)×180÷5 = 108°。360 ÷ 108 = 3.33，不是整数，因此会留下空隙。对六边形 (n=6)：内角 = (6−2)×180÷6 = 120°。360 ÷ 120 = 3，三个六边形恰好在一个顶点汇合。
7. Engineering: Area and Volume in Real-Life Contexts | 工程：实际应用中的面积与体积
Engineers routinely calculate volumes of tanks, pipes and containers. The volume of a cylinder is V = π r² h, where r is the radius and h is the height. The challenge often lies in unit conversion, especially between cubic metres and litres (1 m³ = 1000 litres). Working with π can be done with the π button or rounded to 3.14, depending on the paper.
工程师经常需要计算水箱、管道和容器的体积。圆柱体的体积公式为 V = π r² h，其中 r 为半径，h 为高。难点往往在于单位换算，尤其是立方米与升之间的换算（1 m³ = 1000 升）。对 π 的处理可根据试卷要求使用 π 键或四舍五入为 3.14。
Example: A cylindrical water butt has an internal radius of 0.4 m and a height of 1.2 m. Find its capacity in litres, using π ≈ 3.14.
例题：一个圆柱形集水桶的内半径为 0.4 m，高为 1.2 m。求其容积（升），取 π ≈ 3.14。
Volume in m³ = π × (0.4)² × 1.2 = 3.14 × 0.16 × 1.2 = 3.14 × 0.192 = 0.60288 m³. In litres: 0.60288 × 1000 ≈ 603 litres (to 3 s.f.).
体积(m³) = π × (0.4)² × 1.2 = 3.14 × 0.16 × 1.2 = 3.14 × 0.192 = 0.60288 m³。换算为升：0.60288 × 1000 ≈ 603 升（保留3位有效数字）。
8. Food Technology: Ratios and Proportions in Recipes | 食品技术：食谱中的比率与比例
Scaling recipes is a direct application of ratio and proportion. Given a recipe for a certain number of servings, the amount of each ingredient must be multiplied by the scale factor (new servings ÷ original servings). The same method can check whether the ratio of two ingredients remains constant when adjusting a single quantity.
调整食谱的份数是比率和比例的直接应用。当食谱为特定人数设计时，每种配料的用量需乘以缩放因子（新份数 ÷ 原始份数）。同样的方法也可用于在调整某单一用量时，检查两种配料的比例是否保持不变。
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