📚 Year 7 OCR Maths: Interdisciplinary Problem-Solving Training | Year 7 OCR 数学：跨学科综合题型训练
Many Year 7 students wonder when they will ever use the maths they learn in class. The answer is everywhere: every time a scientist measures a liquid, an athlete tracks their speed, or a musician counts a beat, maths is at work. In OCR mathematics, interdisciplinary problem-solving questions help you connect number, algebra, geometry and data to real-life contexts across science, geography, history, art and more. This article provides a focused training session to tackle these mixed-style questions with confidence.
许多七年级学生都会问，课堂上学到的数学到底什么时候会用到。答案是：无处不在——每当科学家测量液体，运动员追踪速度，音乐家数着节拍，数学就在起作用。在 OCR 数学中，跨学科综合题型让你将数、代数、几何和数据与现实情境联系起来，涵盖科学、地理、历史、艺术等多个学科。本文提供一次集中的训练，帮助你自信地应对这类混合题型。
1. Science – Measuring and Converting Units | 科学 – 测量与单位换算
centimetres cubed (cm³) and litres (L), then apply formulas like mass = density × volume.
在科学实验中，精确测量至关重要。你需要换算毫升（ml）、立方厘米（cm³）和升（L）等单位，再运用质量 = 密度 × 体积等公式。
Example: A scientist measures a liquid volume of 1250 cm³. Express this in litres. If the density of the liquid is 0.8 g/cm³, find the mass in grams.
例题：科学家测得某液体体积为 1250 cm³。用升表示。若该液体密度为 0.8 g/cm³，求质量为多少克。
Step 1: 1 litre = 1000 cm³, so 1250 cm³ ÷ 1000 = 1.25 L.
步骤1：1 升 = 1000 cm³，因此 1250 cm³ ÷ 1000 = 1.25 升。
Step 2: mass = density × volume = 0.8 × 1250 = 1000 g.
步骤2：质量 = 密度 × 体积 = 0.8 × 1250 = 1000 克。
This type of calculation appears frequently in physics and chemistry, requiring careful unit tracking.
这类计算经常出现在物理和化学中，需要仔细跟踪单位的变化。
2. Geography – Map Scales and Ratios | 地理 – 地图比例尺与比
Geographers use map scales to translate a distance on paper into the real ground distance. A scale such as 1 : 50 000 means that 1 cm on the map represents 50 000 cm in reality. You apply ratio and multiplication skills learned in Year 7.
地理学家利用地图比例尺把图上距离转换成实地距离。像 1 : 50 000 的比例尺表示图上 1 cm 代表实际 50 000 cm。这需要运用七年级所学到的比和乘法技能。
Example: On a 1 : 50 000 map, two villages are 8.4 cm apart. What is the actual distance in kilometres?
例题：在一幅比例尺为 1 : 50 000 的地图上，两村相距 8.4 cm。实际距离是多少千米？
Actual distance = 8.4 × 50 000 = 420 000 cm. Convert to km: 420 000 ÷ 100 000 = 4.2 km.
实际距离 = 8.4 × 50 000 = 420 000 cm。换算为千米：420 000 ÷ 100 000 = 4.2 km。
The key is to multiply by the scale factor and then change centimetres to kilometres by dividing by 100 000.
关键点是先乘以比例因子，再通过除以 100 000 将厘米转换为千米。
3. History – Chronology and Negative Numbers | 历史 – 年代学与负数
Historians use timelines spanning BCE and CE periods. This requires subtraction with negative numbers, an algebraic skill often tested in OCR. Year 7 students learn to order and calculate with negative integers.
历史学家使用跨越公元前和公元后的时间线。这需要负数的减法运算，也是 OCR 常考的代数技能。七年级学生已学习负整数的排序和运算。
Example: The Roman Republic was founded in 509 BCE. Julius Caesar was assassinated in 44 BCE. How many years later did the assassination occur?
例题：罗马共和国建立于公元前 509 年，凯撒遇刺于公元前 44 年。遇刺发生在建国之后多少年？
BCE years as negative: -509 and -44. Difference = (-44) - (-509) = -44 + 509 = 465 years.
把公元前年份看作负数：-509 和 -44。时间差 = (-44) - (-509) = -44 + 509 = 465 年。
Understanding negative number jumps also helps with temperature changes in science or elevation below sea level in geography.
理解负数的跳跃也有助于科学中的温度变化或地理中的海平面以下海拔计算。
4. Design & Technology – Area and Perimeter of Composite Shapes | 设计与技术 – 复合图形的面积与周长
In product design, materials are cut from sheets, so you must calculate the area and perimeter of T‑shapes, L‑shapes and combinations of rectangles. This consolidates geometry skills from Year 7.
在产品设计中，材料从板材上切割下来，你需要计算 T 形、L 形以及矩形组合的面积和周长。这巩固了七年级的几何技能。
Example: An L‑ 0 cm × 60 cm joined to form an L. Find the total area and the perimeter of the outer edge.
例题：一个 L 形桌面由两块矩形组成：120 cm × 60 cm 和 80 cm × 60 cm，拼接成 L 形。求总面积以及外边缘的周长。
ht segments for perimeter: outer edges sum to 120 + 60 + 80 + 60 + (120-80) + 60 = 420 cm.
面积 = (120 × 60) + (80 × 60) = 7200 + 4800 = 12 000 cm²。周长计算需分段：外缘边长为 120 + 60 + 80 + 60 + (120-80) + 60 = 420 cm。
Breaking a composite shape into simpler parts is a valuable strategy in both maths and design.
把复合图形分解为简单部分是数学和设计中都有用的策略。
5. Economics – Percentage Discounts and Profit | 经济 – 百分比折扣与利润
Shopping and business decisions involve percentages: discounts, profit margins, and VAT. Year 7 maths covers finding a percentage of an amount and percentage increase or decrease.
购物和商业决策涉及百分比：折扣、利润率、增值税等。七年级数学要求计算一个数的百分之几以及百分比增减。
Example: A shop buys a jacket for £32 and wants a 40% profit on the cost price. During a sale, the jacket is reduced by 15% of its selling price. Find the sale price.
例题：商店以 32 英镑购入一件夹克，打算加价 40% 作为售价。促销时再打八五折（降价 15%）。求最终售价。
Selling price = 32 + 40% of 32 = 32 + 12.80 = £44.80. Sale reduction = 15% of 44.80 = 6.72, so sale price = 44.80 - 6.72 = £38.08.
售价 = 32 + 32 的 40% = 32 + 12.80 = 44.80 英镑。折扣额 = 44.80 的 15% = 6.72，最终售价 = 44.80 - 6.72 = 38.08 英镑。
This will help you manage pocket money and future financial literacy.
这将帮助你管理零花钱并培养未来的财务素养。
6. Art & Design – Symmetry, Rotation and Tessellation | 艺术与设计 – 对称、旋转与镶嵌
Artists and designers frequently use line symmetry, rotational symmetry and tessellations. Year 7 geometry includes identifying order of rotational symmetry and creating patterns that tile a plane without gaps.
艺术家和设计师经常运用线对称、旋转对称和镶嵌图案。七年级几何要求识别旋转对称的阶数，并创作无缝平铺的图案。
Example: A regular hexagon has rotational symmetry of order 6. What is the angle of each rotation? How many hexagons need to meet at a point to tessellate?
例题：正六边形具有 6 阶旋转对称。每次旋转的角度是多少？要实现镶嵌，几个正六边形会交于一点？
60°. For tessellation, interior angle = 120°, so 360° ÷ 120° = 3 hexagons meet at a point.
旋转角度 = 360° ÷ 6 = 60°。内角 = 120°，所以交于一点的六边形数量 = 360° ÷ 120° = 3 个。
This shows how pure geometry informs the patterns on wallpapers, fabrics and Islamic art.
这展示了纯几何如何影响壁纸、织物和伊斯兰艺术中的图案设计。
7. Music – Fractions and Beats per Minute | 音乐 – 分数与每分钟节拍
Music notation relies on fractions: a crotchet is 1/4 of a semibreve, a quaver is 1/8. Tempo is measured in beats per minute (BPM). Combining fractions and time is a direct application of Year 7 number skills.
音乐记谱依赖分数：四分音符是全音符的 1/4，八分音符是 1/8。速度用每分钟节拍数（BPM）衡量。结合分数与时间直接运用七年级的数的技能。
Example: A piece is in 4/4 time at 120 BPM. How many seconds does one crotchet last? If a bar contains 4 crotchets, how long is one bar?
例题：一首乐曲是 4/4 拍，速度为 120 BPM。每个四分音符持续多少秒？若每小节包含 4 个四分音符，一小节多长？
so 1 beat = 60 ÷ 120 = 0.5 seconds. One crotchet = 0.5 s. One bar = 4 × 0.5 = 2.0 seconds.
120 BPM = 每分钟 120 拍，所以 1 拍 = 60 ÷ 120 = 0.5 秒。一个四分音符 = 0.5 秒。一小节 = 4 × 0.5 = 2.0 秒。
Dotted notes, triplets and rests extend this fractional understanding further.
附点音符、三连音和休止符进一步深化了这种分数理解。
8. Physical Education – Distance‑Time Graphs and Speed | 体育 – 距离–时间图与速度
In PE, coaches analyse athletes’ performance using graphs. A distance‑time graph shows movement; its gradient gives speed. Year 7 covers reading and plotting coordinates and understanding simple line graphs.
在体育中，教练用图表分析运动员表现。距离–时间图展示运动情况，其斜率表示速度。七年级要求读取和绘制坐标，并理解简单的线图。
Example: An athlete runs 400 m in 50 seconds, rests for 20 seconds, then runs 200 m in 30 seconds. Plot the graph and find the average speed for the whole session.
例题：一名运动员用 50 秒跑 400 米，休息 20 秒，再用 30 秒跑 200 米。画出距离–时间图，并计算全程平均速度。
al distance = 600 m, total time = 50 + 20 + 30 = 100 s. Average speed = 600 ÷ 100 = 6 m/s.
总距离 = 600 米，总时间 = 50 + 20 + 30 = 100 秒。平均速度 = 600 ÷ 100 = 6 米/秒。
Using the formula speed = distance ÷ time helps you interpret split times in athletics.
使用 速度 = 距离 ÷ 时间 的公式，能帮助你解读田径中的分段计时。
9. Computing – Binary Numbers and Data Storage | 计算机 – 二进制数与数据存储
Computers represent data with binary digits (bits), where each place value is a power of 2. Converting between binary and decimal reinforces place value and multiplication by powers of 2, a key Year 7 concept.
计算机用二进制数字（比特）表示数据，每一位都是 2 的幂。二进制与十进制之间的转换巩固了位值概念和 2 的幂乘法，这是七年级的重要概念。
Example: it to a decimal number. If 1 byte = 8 bits, how many different values can 1 byte represent?
例题：二进制数 1101 被存储。将其转换为十进制数。若 1 字节 = 8 比特，1 字节能表示多少个不同数值？
²) + (0×2¹) + (1×2⁰) = 8 + 4 + 0 + 1 = 13. With 8 bits, number of combinations = 2⁸ = 256.
1101₂ = (1×2³) + (1×2²) + (0×2¹) + (1×2⁰) = 8 + 4 + 0 + 1 = 13。8 比特的组合数 = 2⁸ = 256。
Understanding binary is the basis for coding and digital technology, where maths meets logic.
理解二进制是编程和数字技术的基础，数学在此与逻辑相遇。
10. Environmental Science – Averages and Data Interpretation | 环境科学 – 平均数与数据解读
Environmental scientists collect data on rainfall, temperature and pollution. They calculate mean, median and mode to summarise findings. Year 7 statistics covers these measures of central tendency and simple bar charts.
环境科学家收集降雨量、温度和污染数据，通过计算平均数、中位数、众数来概括研究结果。七年级统计部分涵盖这些集中趋势度量和简单的条形图。
Example: Weekly rainfall (mm): 12, 8, 9, 14, 12, 10, 9. Find the mean, median and mode.
例题：一周降雨量（毫米）：12, 8, 9, 14, 12, 10, 9。求平均数、中位数和众数。
7 ≈ 10.6 mm. Ordered: 8, 9, 9, 10, 12, 12, 14. Median = 10 mm. Mode = 9 and 12 (bimodal).
总和 = 74，7 个数据，平均数 = 74 ÷ 7 ≈ 10.6 mm。排序：8, 9, 9, 10, 12, 12, 14，中位数 = 10 mm，众数为 9 和 12（双众数）。
These statistics aid in predicting water usage and understanding climate patterns.
这些统计量有助于预测用水量并理解气候模式。
11. Food Technology – Ratio and Proportion in Recipes | 食品技术 – 食谱中的比与比例
Scaling recipes up or down uses direct proportion. If a recipe for 4 people requires 250 g of flour, you can find the amount for 10 people by setting up a ratio. Year 7 covers solving proportion problems with the unitary method or multipliers.
按人数调整食谱用量是正比例的应用。若 4 人份需要 250 克面粉，可通过比来求出 10 人份的用量。七年级涉及用单一单位法或乘数解决比例问题。
Example: A cake recipe for 6 people uses 3 eggs and 150 g sugar. Adapt the recipe for 15 people. What is the weight of sugar per person in the original recipe?
例题：一份 6 人份的蛋糕食谱需要 3 个鸡蛋和 150 克糖。改为 15 人份。原食谱中人均糖量是多少？
× 2.5 = 7.5 (use 7 or 8); sugar = 150 × 2.5 = 375 g. Original per person: 150 ÷ 6 = 25 g.
倍数 = 15 ÷ 6 = 2.5。15 人份：鸡蛋 = 3 × 2.5 = 7.5（用 7 或 8 个）；糖 = 150 × 2.5 = 375 克。原人均糖量：150 ÷ 6 = 25 克。
Proportional reasoning is used daily in cooking and catering industries.
比例推理广泛应用于日常烹饪和餐饮业。
12. Biology – Estimating Populations with Sampling | 生物 – 用抽样估算种群数量
Biologists cannot count every organism, so they use the capture‑recapture method, which involves ratios and equations. This practical use of algebra connects to solving simple equations in Year 7.
生物学家无法计数每一个生物，因此采用标记重捕法，其中涉及比与方程。这种对代数的实际应用与七年级解简单方程相联系。
Example: In a pond, 20 fish are caught, tagged and released. Later, 30 fish are caught and 5 are tagged. Estimate the total fish population.
例题：在池塘中捕获了 20 条鱼，标记后放回。随后又捕获 30 条鱼，其中 5 条有标记。估算鱼的总数。
total tagged / total population. So 5/30 = 20/P, giving 5P = 20×30, P = 600÷5 = 120 fish.
假设比例：样本中标记数 / 样本总数 = 总标记数 / 总数量。所以 5/30 = 20/P，即 5P = 20×30，P = 600÷5 = 120 条。
This equation-based estimation shows how algebra models real ecosystems.
这种基于方程的估计展示了代数如何模拟真实生态系统。
Published by TutorHao | 数学 Revision Series | aleveler.com
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