📚 Case Study Practice for Year 7 OCR Maths | Year 7 OCR 数学：案例分析实战演练
In Year 7 OCR Mathematics, case study questions bring real-life situations into the exam room. You are asked to read a short scenario, pick out the relevant numbers, and apply your maths skills from all across the course – fractions, percentages, geometry, algebra, statistics, and more. This article walks you through ten carefully designed case studies, showing exactly how to break down each problem, choose the right method, and present a clear solution. Use these examples to build confidence and sharpen your problem-solving toolkit.
在 Year 7 OCR 数学中，案例分析题将真实生活场景带入考场。你需要阅读一段简短的情境描述，提取相关的数字，并运用课程中学到的各种数学技能——分数、百分比、几何、代数、统计等等。本文将通过十个精心设计的案例，详细展示如何分解每一道题目、选择正确的方法并给出清晰的解答。借助这些例子，你可以建立信心，磨练解题本领。
1. Understanding Case Study Questions | 认识案例分析题
OCR case study questions usually contain a short paragraph or two describing a situation – a shopping trip, a school event, a recipe, a journey. Hidden inside the text are the numbers you need. Your first job is to read slowly and underline the key data. Ask yourself: 'What maths topic is this? Am I calculating a percentage, working out a perimeter, or interpreting a chart?' Then plan your steps. Don't rush into calculations; a quick sketch or a note of known facts often saves time later.
OCR 的案例分析题通常包含一段或两段描述某个情境的文字——一次购物、一场学校活动、一份食谱、一段旅程。需要的数据就藏在文字里。你的首要任务是慢慢阅读，划出关键数据。问一问自己：“这属于哪个数学专题？是要计算百分比、求周长，还是解读图表？”然后规划解题步骤。不要急于计算；先快速画个简图或记下已知条件，往往能节省后续的时间。
2. Case Study 1 – Shopping Discounts | 案例一 – 购物折扣
Zara buys a pair of trainers originally priced at £60. The shop offers a 15% discount. She also has a voucher for an extra £5 off the reduced price. How much does she pay? This case test two steps: percentage decrease and simple subtraction. First find 15% of £60: 0.15 × 60 = £9. Subtract to get the sale price: 60 – 9 = £51. Then apply the voucher: 51 – 5 = £46. Zara pays £46.
Zara 买了一双原价 60 英镑的运动鞋。商店提供 15% 的折扣，她还有一张额外减免 5 英镑的优惠券，可在折后价上使用。她实际应付多少钱？这个案例测试两步：百分比减少和简单的减法。首先求 60 英镑的 15%：0.15 × 60 = 9 英镑。减去折扣得售价：60 – 9 = 51 英镑。再用优惠券：51 – 5 = 46 英镑。Zara 需支付 46 英镑。
3. Case Study 2 – Garden Area and Perimeter | 案例二 – 花园面积与周长
. The owner wants to put a fence around it and lay turf over the whole area. Fencing costs £4 per metre and turf costs £ etre. Calculate the total cost. Perimeter = 2 × (8 + 5) = 26 m, so fencing cost = 26 × 4 = £104. Area = 8 × 5 = 40 m², so turf cost = 40 × 6 = £240. Total cost = 104 + 240 = £344. Always check whether the question asks for 'around' (perimeter) or 'cover' (area).
一个长方形草坪长 8 米、宽 5 米。主人想围上篱笆并铺满草皮。篱笆每米 4 英镑，草皮每平方米 6 英镑。计算总费用。周长 = 2 × (8 + 5) = 26 米，篱笆费用 = 26 × 4 = 104 英镑。面积 = 8 × 5 = 40 平方米，草皮费用 = 40 × 6 = 240 英镑。总费用 = 104 + 240 = 344 英镑。务必确认题目是求“围起来”（周长）还是“覆盖”（面积）。
4. Case Study 3 – Travel Time and Averages | 案例三 – 行程时间与平均值
Mia cycles 12 km to a friend' km/h. She stays for 30 minutes and cycles back at 12 km/h. How long is she away from home? ed problems often need converting between hours and minutes; remember 0.1 hour = 6 minutes.
Mia 骑自行车 12 公里去朋友家，平均时速 16 公里。她停留 30 分钟，然后以 12 公里/时的速度骑回家。她离家多长时间？去程时间 = 距离 ÷ 速度 = 12 ÷ 16 = 0.75 小时 = 45 分钟。回程时间 = 12 ÷ 12 = 1 小时 = 60 分钟。总时间 = 45 + 30 + 60 = 135 分钟 = 2 小时 15 分钟。速度问题常常需要在小时和分钟之间转换；记住 0.1 小时 = 6 分钟。
5. Case Study 4 – Recipe Scaling with Fractions | 案例四 – 食谱分量缩放与分数
ctor is 6 ÷ 4 = 1.5. Multiply: flour = 200 × 1.5 = 300 g; eggs = 2 × 1.5 = 3; milk = 300 × 1.5 = 450 ml. Alternatively, you could find the amount per person first. This case reinforces ratio and proportion – a core Year 7 skill.
一份可供 4 人食用的煎饼食谱需要 200 克面粉、2 个鸡蛋和 300 毫升牛奶。Tom 需要做 6 人份。求每种食材的用量。比例因子 = 6 ÷ 4 = 1.5。相乘得：面粉 = 200 × 1.5 = 300 克；鸡蛋 = 2 × 1.5 = 3 个；牛奶 = 300 × 1.5 = 450 毫升。也可以先求出每人用量。这个案例巩固了比和比例——Year 7 的核心技能。
6. Case Study 5 – Pocket Money Savings with Algebra | 案例五 – 零花钱储蓄与代数
Ali saves £5 each week. He already has £28. He wants to buy a game costing £ 7 more weeks. This shows how a simple linear equation models a real saving plan. Always define your variable clearly.
Ali 每周存 5 英镑，他已经有 28 英镑。他想买一款售价 63 英镑的游戏。他列出方程 5w + 28 = 63，其中 w 表示所需的周数。解方程：两边减 28，得 5w = 35；两边除以 5，得 w = 7。Ali 还需要 7 周。这展示了如何用简单的一次方程给实际储蓄计划建模。务必清晰地定义变量。
7. Case Study 6 – School Canteen Data | 案例六 – 学校食堂数据统计
A canteen records meal choices one day: 45 jacket potatoes, 60 pasta, 35 salad, 40 soup. Draw a bar chart and find the mode. The mode is the most frequent: pasta (60). To draw a bar chart, label the horizontal axis with meal types and vertical axis with frequency (scale up to 60). Use equal width bars with gaps. Check that the heights match the frequencies. Charts must always have a title and labelled axes.
食堂记录了一天的午餐选择：45 份烤土豆、60 份意面、35 份沙拉、40 份汤。绘制条形图并找出众数。众数是出现次数最多的类别：意面（60）。绘制条形图时，横轴标上餐食种类，纵轴标上频数（坐标最大到 60）。条形的宽度要一致，条形之间留有空隙。确保高度与频数相符。图表必须有标题和轴标签。
8. Case Study 7 – Temperature Drop and Negative Numbers | 案例七 – 温度下降与负数
ay the temperature is 4°C. By midnight it drops by 11 degrees. What is the new temperature? 4 – 11 = –7°C. This is straightforward subtraction crossing zero. A number line or 'coldness' analogy helps. If the temperature then rises 9°C, the new value is – pear frequently in weather, bank balances, and lifts; always note the direction of change.
中午气温为 4°C，到午夜下降了 11 度。新的温度是多少？4 – 11 = –7°C。这是一道跨越零点的简单减法题。数轴或“寒冷程度”的类比有助于理解。如果随后温度上升 9°C，新温度是 –7 + 9 = 2°C。负数常出现在天气、银行余额和电梯等情境中；务必注意变化的方向。
9. Case Study 8 – Angle Adventures in a Triangle | 案例八 – 三角形中的角度探险
43° and angle B = 67°. Find angle C. Sum of angles in a triangle = 180°, so angle C = 180 – (43 + 67) = 180 – angles carefully. If the triangle is isosceles with base angles 52°, the top angle is 180 – 2×52 = 76°. Remember to use the correct angle fact for the shape.
在三角形 ABC 中，角 A = 43°，角 B = 67°。求角 C。三角形内角和为 180°，所以角 C = 180 – (43 + 67) = 180 – 110 = 70°。有些案例分析题会附有图形；仔细读取已知角度。如果等腰三角形的底角均为 52°，则顶角为 180 – 2×52 = 76°。要为不同的形状选用正确的角度性质。
10. Case Study 9 – Probability Spinner Game | 案例九 – 转盘游戏中的概率
equal sectors: red, blue, blue, yellow, green. What is the probability of landing on blue? mber of favourable outcomes) ÷ (total outcomes) = 2/5. What is the probability of not blue? 1 – l outcomes on a sample space diagram, but in Year 7 most cases involve only single events.
一个均匀转盘等分为 5 个扇区：红、蓝、蓝、黄、绿。转到蓝色的概率是多少？概率 = （有利结果数目）÷（总结果数目）= 2/5。不是蓝色的概率是多少？1 – 2/5 = 3/5。OCR 常要求用最简分数表示概率。如果问到两次旋转，可能需要用样本空间图列出所有结果，不过在 Year 7 通常只涉及单一事件。
11. Case Study 10 – Budgeting for a School Trip | 案例十 – 学校旅行预算
A trip costs £ f the total, and the rest is collected from families. How much does each family contribute? Total cost = 28 × 14 = £392. School contribution = 3/8 × 392 = (3 × 392) ÷ 8 = 1176 ÷ 8 = £147. Families pay £392 – 147 = £245. Divide equally: £245 ÷ 28 = £ action, and division. Always show each step clearly and write the final answer in context.
一次旅行每位学生 14 英镑，班级有 28 人。学校支付总费用的 3/8，余下由家庭分摊。每个家庭需交多少钱？总费用 = 28 × 14 = 392 英镑。学校支付 = 3/8 × 392 = (3 × 392) ÷ 8 = 1176 ÷ 8 = 147 英镑。家庭支付 = 392 – 147 = 245 英镑。平均分摊：245 ÷ 28 = 8.75 英镑/人。这个多步骤问题综合了乘法、求一个数的几分之几、减法和除法。务必清晰地写出每一步，并在情境中写出最终答案。
12. Tips for Mastering Case Studies | 攻克案例分析的技巧
Always read the scenario twice before writing any calculation. Highlight keywords like 'each', 'total', 'discount', 'per', 'share equally'. Convert units when necessary (e.g. minutes to hours, pence to pounds). Show your working – even if your final answer is slightly off, method marks can save you. After solving, check your answer makes sense in the real-world context (could a backpack really cost £400?). Practise with a variety of case studies, and you'll soon spot the maths hiding in everyday stories.
每次解题前，先阅读情境两遍。圈出关键词，如“每个”、“总共”、“折扣”、“每”、“平均分配”。必要时换算单位（如分钟转小时，便士转英镑）。写出解题过程——即使最终答案稍有偏差，过程分也能帮你拿分。解完后，检查答案放在真实情境中是否合理（一个背包真的需要 400 英镑吗？）。通过多种案例进行练习，你很快就能发现藏在日常故事中的数学。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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