📚 Year 7 AQA Mathematics: Cross-curricular Problem-Solving Training | Year 7 AQA 数学：跨学科综合题型训练
In Year 7, mathematics is not just about numbers in isolation – it connects deeply with science, geography, economics, art, sport and design. This cross-curricular training article helps you see how maths skills are used across different subjects, with real‑world problems and step‑by‑step solutions. Practising these problems will strengthen both your mathematical reasoning and your ability to apply maths in unfamiliar contexts, exactly as the AQA curriculum expects.
在 Year 7，数学不仅仅是孤立的数字运算——它与科学、地理、经济、艺术、体育和设计紧密相连。这篇跨学科综合题型训练文章帮助你了解数学技能如何在不同学科中应用，并提供真实世界的问题和分步解答。练习这些问题将增强你的数学推理能力以及在不熟悉的情境中应用数学的能力，这正是 AQA 课程所要求的。
1. Science & Maths: Measurement and Unit Conversion | 科学与数学：测量与单位转换
0.35 litres of water, but the measuring cylinder shows millilitres. Since 1 litre = 1000 ml, you multiply by 1000: 0.35 × 1000 = 350 ml. Similarly, converting mass from grams to kilograms involves dividing by 1000. Being confident with multiplying and dividing by powers of 10 makes these conversions quick and accurate.
在科学实验中，你经常需要在长度、质量和容量单位之间进行转换。例如，一个化学探究可能需要 0.35 升水，但量筒显示的是毫升。因为 1 升 = 1000 毫升，你需要乘以 1000：0.35 × 1000 = 350 毫升。同样，将克转换为千克需要除以 1000。熟练掌握乘以和除以 10 的幂能使这些转换快速而准确。
Problem: A biology student measures three plant stems: 12.4 cm, 98 mm and 0.05 m. Find the total length in millimetres.
问题：一个生物专业的学生测量了三根植物茎：12.4 厘米、98 毫米和 0.05 米。求总长度，单位为毫米。
Solution: = 124 mm; 98 mm stays 98 mm; 0.05 m = 0.05 × 1000 = 50 mm. Total = 124 + 98 + 50 = 272 mm.
解答：将所有单位转换为毫米：12.4 厘米 = 12.4 × 10 = 124 毫米；98 毫米保持不变；0.05 米 = 0.05 × 1000 = 50 毫米。总长度 = 124 + 98 + 50 = 272 毫米。
2. Geography & Maths: Map Scales and Distance | 地理与数学：地图比例尺和距离
Map scales allow you to calculate real‑life distances from a diagram. A scale such as 1 : 50 000 means that 1 cm on the map represents 50 000 cm (or 0.5 km) on the ground. To find the actual straight‑line distance between two points, measure the map distance in centimetres and multiply by the scale factor. Unit conversion is then needed to express the answer in kilometres or metres.
地图比例尺可以让你从图上计算现实中的距离。像 1 : 50 000 这样的比例尺表示地图上的 1 厘米代表地面上的 50 000 厘米（或 0.5 公里）。要找到两点之间的实际直线距离，先以厘米为单位测量地图距离，然后乘以比例尺系数。然后需要进行单位转换，用公里或米表示答案。
Problem: On a 1 : 25 000 map, two villages are 6.4 cm apart. What is the actual distance in kilometres?
问题：在一张 1 : 25 000 的地图上，两个村庄相距 6.4 厘米。实际距离是多少公里？
Solution: tance: 6.4 × 25 000 = 160 000 cm. Convert cm to km (÷100 000): 160 000 ÷ 100 000 = 1.6 km.
解答：地图距离 → 实际距离：6.4 × 25 000 = 160 000 厘米。将厘米转换为公里（÷100 000）：160 000 ÷ 100 000 = 1.6 公里。
3. Geography & Maths: Population Statistics and Graphs | 地理与数学：人口统计与图表
Geographers collect population data and present it using bar charts, line graphs and pie charts. You might be asked to read values from a graph, compare categories or calculate the mean. For instance, a bar chart showing the populations of five cities requires careful reading of the scale and accurate addition before finding the average.
地理学家收集人口数据，并用条形图、折线图和饼图展示。你可能会被要求从图表中读取数值、比较类别或计算平均数。例如，显示五个城市人口的条形图需要仔细读取刻度，并进行准确的加法运算才能求出平均值。
Problem: million, C = 1.5 million, D = 0.9 million, E = 1.1 million. Calculate the mean population.
问题：条形图显示了城市人口：A = 120 万，B = 80 万，C = 150 万，D = 90 万，E = 110 万。计算平均人口。
Solution: Sum = 1.2 + 0.8 + 1.5 + 0.9 + 1.1 = 5.5 million. Mean = 5.5 ÷ 5 = 1.1 million.
解答：总和 = 1.2 + 0.8 + 1.5 + 0.9 + 1.1 = 550 万。平均 = 5.5 ÷ 5 = 110 万。
4. Science: Speed, Distance and Time | 科学：速度、距离和时间
g one quantity when the other two are known, using consistent units (e.g. metres per second, kilometres per hour). Rearranging the formula helps you solve for distance (distance = speed × time) or time (time = distance ÷ speed).
速度、距离和时间的关系是物理学中的一个关键公式：速度 = 距离 ÷ 时间。在 Year 7，问题通常涉及在已知两个量时计算第三个量，并使用一致的单位（例如米每秒、公里每小时）。重新整理公式可以帮助你求解距离（距离 = 速度 × 时间）或时间（时间 = 距离 ÷ 速度）。
Problem: A cyclist travels at a constant speed of 6 m/s for 2 minutes. How far does the cyclist go in metres?
问题：一名自行车手以 6 米/秒的恒定速度骑行 2 分钟。她骑了多远（以米为单位）？
Solution: Convert time to seconds: 2 min = 120 s. Distance = speed × time = 6 × 120 = 720 m.
解答：将时间转换为秒：2 分钟 = 120 秒。距离 = 速度 × 时间 = 6 × 120 = 720 米。
5. Science: Density Calculations | 科学：密度计算
much mass is packed into a given volume. The formula is density = mass ÷ volume. In cross‑ o helps in explaining why objects float or sink in water (water has a density of 1 g/cm³).
密度衡量一定体积内包含的质量大小。公式为：密度 = 质量 ÷ 体积。在跨学科问题中，你可能需要进行质量（克到千克）或体积（立方厘米到立方米）的单位转换。理解密度还有助于解释为什么物体会在水中浮沉（水的密度为 1 克/立方厘米）。
Problem: A metal block has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³. Will it float or sink in water?
问题：一个金属块的质量为 540 克，体积为 200 立方厘米。计算其密度，单位为克/立方厘米。它在水中是浮还是沉？
Solution: Density = 540 ÷ 200 = 2.7 g/cm³. Since 2.7 > 1, the metal will sink.
解答：密度 = 540 ÷ 200 = 2.7 克/立方厘米。由于 2.7 > 1，金属块会下沉。
6. Economics & Maths: Simple Interest | 经济与数学：单利
rest, P is principal, r is the annual interest rate (as a decimal), and t is time in years. Being able to convert percentages to decimals and perform multi‑step calculations is essential.
单利是原始金额（本金）在一定时间内赚取或支付的一定百分比的利息。公式为 I = P × r × t，其中 I 是利息，P 是本金，r 是年利率（以小数表示），t 是时间（年）。能够将百分比转换为小数并进行多步计算至关重要。
Problem: Sam invests £400 in a savings account that pays 3% simple interest per year. How much interest does Sam earn in 5 years? What is the total amount in the account after 5 years?
问题：山姆将 400 英镑存入一个年利率为 3% 的单利储蓄账户。5 年后山姆能获得多少利息？5 年后账户里的总金额是多少？
Solution: r = 3% = 0.03. I = 400 × 0.03 × 5 = £60. Total amount = £400 + £60 = £460.
解答：r = 3% = 0.03。I = 400 × 0.03 × 5 = 60 英镑。总金额 = 400 + 60 = 460 英镑。
7. Economics & Maths: Budgeting and Costs | 经济与数学：预算与成本
Managing money involves adding expenses, calculating change and comparing prices. Budgeting tasks often use tables showing income and outgoings. You may need to find totals, work out how much is left after spending, or use multiplication to find the cost of multiple items. These problems reinforce addition, subtraction and multiplication of decimals.
理财涉及将开支相加、计算找零和比较价格。预算任务通常使用显示收入和支出的表格。你可能需要计算总额、算出支出后剩余多少，或使用乘法计算多件商品的费用。这些问题强化了小数的加、减和乘法运算。
Problem: A family's monthly income is £2400. Their expenses are: rent £850, food £380, bills £240, transport £165, and entertainment £130. How much money is left after these expenses?
问题：一个家庭的月收入为 2400 英镑。他们的支出为：房租 850 英镑，食品 380 英镑，账单 240 英镑，交通 165 英镑，娱乐 130 英镑。支出后还剩多少钱？
Solution: Total expenses = 850 + 380 + 240 + 165 + 130 = £1765. Money left = £2400 - £1765 = £635.
解答：总支出 = 850 + 380 + 240 + 165 + 130 = 1765 英镑。剩余 = 2400 - 1765 = 635 英镑。
8. PE & Maths: Statistics and Averages | 体育与数学：统计与平均数
In physical education, performance data is recorded and analysed. You might time a 100 m sprint, count goals scored per match or measure heart rates. Calculating the mean, median, mode and range helps identify typical performance and consistency. Reading data from tables and using the correct formula for the mean (sum of values ÷ number of values) are core skills.
在体育课上，成绩数据会被记录和分析。你可能会计时 100 米短跑、计算每场比赛的进球数或测量心率。计算平均值、中位数、众数和极差有助于找出典型表现和一致性。从表格中读取数据并使用正确的平均数公式（数值之和 ÷ 数值个数）是核心技能。
Problem: A basketball player scores the following points over 5 matches: 12, 18, 15, 10, 20. Find the mean and the range.
问题：一名篮球运动员在 5 场比赛中的得分为：12、18、15、10、20。求平均值和极差。
Solution: Mean = (12+18+15+10+20) ÷ 5 = 75 ÷ 5 = 15 points. Range = highest – lowest = 20 – 10 = 10 points.
解答：平均值 = (12+18+15+10+20) ÷ 5 = 75 ÷ 5 = 15 分。极差 = 最高分 – 最低分 = 20 – 10 = 10 分。
9. Art & Maths: Symmetry and Patterns | 艺术与数学：对称与图案
Artists and designers use line symmetry and rotational symmetry to create balanced, visually appealing work. A shape has line symmetry if one half is a mirror image of the other. Rotational symmetry is the number of times a shape looks the same during a full 360° turn. Identifying and drawing lines of symmetry, or determining the order of rotational symmetry, links geometry to creative design.
艺术家和设计师利用线对称和旋转对称来创作平衡、具有视觉吸引力的作品。如果一个形状的一半是另一半的镜像，它就有线对称。旋转对称是指一个形状在旋转一整圈 360° 的过程中看起来与原来重合的次数。识别并画出对称线，或确定旋转对称的阶数，将几何与创意设计联系起来。
Problem: An equilateral triangle has side length 6 cm. How many lines of symmetry does it have? What is its order of rotational symmetry?
问题：一个等边三角形的边长为 6 厘米。它有多少条对称线？它的旋转对称阶数是多少？
Solution: int of the opposite side). Rotational symmetry order = 3 (matches at 120°, 240° and 360°).
解答：等边三角形有 3 条对称线（各从一个顶点到对边中点）。旋转对称阶数 = 3（在 120°、240° 和 360° 时重合）。
10. Design & Technology: Area and Perimeter | 设计与技术：面积与周长
ectangular shapes, area = length × width and perimeter = 2 × (length + width). These calculations are used to determine the amount of material needed, such as paint for a wall or wood for a frame.
在设计产品或规划空间时，你经常要计算面积（表面的量）和周长（围绕边缘的距离）。对于矩形，面积 = 长 × 宽，周长 = 2 × (长 + 宽)。这些计算用于确定所需材料的数量，例如粉刷一面墙所需的油漆或制作一个框架所需的木材。
Problem: A rectangular table top measures 1.2 m by 80 cm. Find the area in square metres and the perimeter in metres.
问题：一个矩形桌面的尺寸为 1.2 米 × 80 厘米。求面积（平方米）和周长（米）。
Solution: ert 80 cm to 0.8 m. Area = 1.2 × 0.8 = 0.96 m². Perimeter = 2 × (1.2 + 0.8) = 2 × 2 = 4 m.
解答：将 80 厘米转换为 0.8 米。面积 = 1.2 × 0.8 = 0.96 平方米。周长 = 2 × (1.2 + 0.8) = 2 × 2 = 4 米。
11. Multi‑step Cross‑curricular Challenge | 多步骤跨学科挑战
Some problems blend several ideas from different subjects. For example, planning a school garden might involve measuring the area (DT), calculating the amount of soil needed (science volume), working out costs from a budget (economics) and recording plant growth over time (science and statistics). Tackling multi‑step problems helps you see the bigger picture of how maths underpins real‑life decisions.
有些问题融合了来自不同学科的多个概念。例如，规划一个学校花园可能涉及测量面积（设计与技术）、计算所需的土壤量（科学体积）、根据预算计算成本（经济）以及记录植物随时间生长的情况（科学和统计）。解决多步骤问题有助于你更全面地了解数学如何支撑现实生活中的决策。
Problem: hool garden is 5 m long and 3 m wide. You need to buy soil to a depth of 30 cm. Soil costs £4.50 per 100 litres. 1 m³ = 1000 litres. Your budget is £ late the volume of soil needed in litres, the total cost and whether the budget is enough.
问题：一个长方形学校花园长 5 米，宽 3 米。你需要铺设 30 厘米厚的土壤。土壤的价格为每 100 升 4.50 英镑。1 立方米 = 1000 升。预算为 250 英镑。计算所需土壤的体积（升）、总成本，并判断预算是否足够。
Solution: e = 5 × 3 × 0.3 = 4.5 m³. Convert to litres: 4.5 × 1000 = 4500 litres. Cost: number of 100‑litre units = 4500 ÷ 100 = 45. Total cost = 45 × £4.50 = £202.50. Budget of £250 is enough, with £47.50 left.
解答：体积 = 长 × 宽 × 深度。深度 = 30 厘米 = 0.3 米。体积 = 5 × 3 × 0.3 = 4.5 立方米。转换为升：4.5 × 1000 = 4500 升。成本：100 升为单位数 = 4500 ÷ 100 = 45。总成本 = 45 × 4.50 英镑 = 202.50 英镑。250 英镑的预算足够，剩余 47.50 英镑。
12. Top Tips for Cross‑curricular Problem Solving | 跨学科问题解决的重要提示
Always read the question carefully to identify which subject context is being used – the maths may be straightforward, but understanding the scenario is key. Convert all units to a consistent system before calculating. Underline the quantities given and what you need to find. Show all working clearly, as method marks are often awarded even if the final answer is slightly off. Finally, check your answer makes sense in the real‑world context (e.g., you cannot have a negative distance or a cost higher than the budget).
一定要仔细阅读问题，明确所使用的学科情境——数学本身可能很简单，但理解场景是关键。在计算之前将所有单位转换为统一的系统。在给定的数量和需要求解的量下面划线。清晰地展示所有步骤，因为即使最终答案稍有偏差，过程分也经常会被给予。最后，检查你的答案在现实情境中是否有意义（例如，你不能有负的距离或超过预算的成本）。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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