📚 Case Study Practical Exercises for Year 7 OCR Advanced Mathematics | 七年级OCR进阶数学案例分析实战演练
In Year 7 OCR Advanced Mathematics, case study practical exercises are designed to help students apply mathematical concepts to real-world scenarios. These exercises encourage logical reasoning, problem-solving, and the ability to interpret data. By working through case studies, learners can see how topics such as algebra, geometry, statistics, and probability are used in everyday situations, reinforcing their understanding and preparing them for more advanced study.
在七年级OCR进阶数学中，案例分析实战演练旨在帮助学生将数学概念应用于现实情境。这些练习鼓励逻辑推理、问题解决以及解读数据的能力。通过案例学习，学生可以看到代数、几何、统计和概率等主题如何在日常生活中运用，从而巩固理解，为更深入的学习做好准备。
1. Understanding Case Study Approaches | 理解案例分析方法
A case study in mathematics is not just a simple word problem; it is a multi-step scenario that requires you to identify the relevant information, choose the correct mathematical techniques, and interpret your results. The key is to read the case carefully, underline key data, and note what the question is asking you to find. Often, you will need to combine skills from different areas of mathematics.
数学中的案例研究不仅仅是一个简单的文字题；它是一个多步骤的情境，要求你找出相关信息，选择正确的数学技巧，并解释你的结果。关键是仔细阅读案例，划出关键数据，并注意题目要求你求解的内容。通常，你需要综合运用不同数学领域的知识。
2. Case 1: Solving Linear Equations in Context | 案例1：情境中的线性方程求解
Imagine a scenario: A school is planning a trip. The cost of hiring a coach is a fixed fee of £120, plus £8 per student. If the total budget for the trip is £400, how many students can attend? We can model this as the equation 120 + 8x = 400, where x is the number of students. To solve, subtract 120 from both sides to get 8x = 280, then divide both sides by 8, giving x = 35.
想象一个场景：学校正在计划一次旅行。租用大巴的固定费用为120英镑，再加上每位学生8英镑。如果旅行总预算为400英镑，最多可以有多少名学生参加？我们可以将其建模为方程 120 + 8x = 400，其中x代表学生人数。求解时，两边减去120得到 8x = 280，然后两边除以8，得到 x = 35。
This type of case study shows how algebra helps in budget planning. You must interpret the solution: 35 students can attend. Always check your answer: 120 + 8 × 35 = 120 + 280 = 400, which matches the budget. This reinforces the idea that equations model real-world constraints.
这类案例展示了代数如何帮助预算规划。你必须解释解：35名学生可以参加。务必检验答案：120 + 8 × 35 = 120 + 280 = 400，与预算相符。这强化了方程模拟现实约束条件的理念。
3. Case 2: Area and Perimeter in Design | 案例2：设计中的面积与周长
A gardener wants to create a rectangular flower bed with a length that is twice its width. If she has 36 metres of fencing to enclose the bed, what dimensions should the bed have? Let the width be w metres; then length = 2w. The perimeter is 2(length + width) = 2(2w + w) = 6w. Set this equal to 36: 6w = 36, so w = 6 m. The length is 12 m.
一位园丁想建一个长方形的花坛，其长度是宽度的两倍。如果她有36米的围栏来围住花坛，花坛的尺寸应该是多少？设宽度为w米，则长度=2w。周长是 2(长+宽) = 2(2w + w) = 6w。令其等于36：6w = 36，所以 w = 6 米。长度为12米。
Now consider the area of this flower bed: Area = length × width = 12 × 6 = 72 m². The gardener might also need to buy topsoil; this case connects perimeter (fencing) with area (soil). In Year 7, you learn to apply both formulas and see how changing dimensions affects these measurements.
现在考虑花坛的面积：面积 = 长×宽 = 12×6 = 72 平方米。园丁可能还需要购买表土；这个案例将周长（围栏）与面积（土壤）联系起来。在七年级，你将学习如何应用这两个公式，并看到尺寸变化如何影响这些度量。
4. Case 3: Using Averages to Compare Data | 案例3：使用平均数比较数据
Two basketball players recorded their points scored in five matches. Player A scored: 12, 15, 10, 20, 13. Player B scored: 14, 14, 14, 14, 14. Which player is more consistent, and who scores more on average? Calculate the mean (average) for each. For Player A: (12+15+10+20+13) ÷ 5 = 70 ÷ 5 = 14. For Player B: 70 ÷ 5 = 14. Both have the same mean.
两名篮球运动员记录了他们在五场比赛中的得分。球员A得分：12, 15, 10, 20, 13。球员B得分：14, 14, 14, 14, 14。哪位球员更稳定，谁的平均得分更高？计算每人的平均数（均值）。球员A：(12+15+10+20+13) ÷ 5 = 70 ÷ 5 = 14。球员B：70 ÷ 5 = 14。两者均值相同。
To measure consistency, we can look at the range: A’s range is 20 – 10 = 10; B’s range is 0. This shows Player B is perfectly consistent, while A has more variation. In an advanced Year 7 case, you might also calculate the median. For A, ordered: 10, 12, 13, 15, 20 → median = 13. For B, median = 14. This deepens data interpretation skills.
为了衡量稳定性，我们可以看极差：A的极差是20-10=10；B的极差是0。这表明球员B非常稳定，而A波动较大。在七年级进阶案例中，你可能还需要计算中位数。对于A，排序后：10, 12, 13, 15, 20 → 中位数=13。对于B，中位数=14。这深化了数据解读能力。
5. Case 4: Ratio and Proportion in Recipes | 案例4：食谱中的比例与比率
A recipe for 8 pancakes requires 200 g of flour, 2 eggs, and 300 ml of milk. How much of each ingredient is needed for 20 pancakes? First find the amount for 1 pancake: flour per pancake = 200 ÷ 8 = 25 g; eggs per pancake = 2 ÷ 8 = 0.25; milk per pancake = 300 ÷ 8 = 37.5 ml. Then multiply by 20: flour = 25 × 20 = 500 g; eggs = 0.25 × 20 = 5 eggs; milk = 37.5 × 20 = 750 ml.
一份制作8个煎饼的食谱需要200克面粉、2个鸡蛋和300毫升牛奶。制作20个煎饼需要多少每种原料？首先求出每个煎饼的用量：每个煎饼面粉量 = 200 ÷ 8 = 25克；每个煎饼鸡蛋量 = 2 ÷ 8 = 0.25；每个煎饼牛奶量 = 300 ÷ 8 = 37.5毫升。然后乘以20：面粉 = 25 × 20 = 500克；鸡蛋 = 0.25 × 20 = 5个；牛奶 = 37.5 × 20 = 750毫升。
You could also use a scaling factor: 20 ÷ 8 = 2.5. Multiply all original quantities by 2.5: flour 200 × 2.5 = 500 g; eggs 2 × 2.5 = 5; milk 300 × 2.5 = 750 ml. Ratio problems like this are common in catering and help you practice proportional reasoning. In Year 7, you also explore simplifying ratios and dividing amounts in a given ratio.
你也可以使用比例因子：20 ÷ 8 = 2.5。将所有原始数量乘以2.5：面粉 200 × 2.5 = 500克；鸡蛋 2 × 2.5 = 5个；牛奶 300 × 2.5 = 750毫升。这类比率问题在餐饮业中很常见，有助于练习比例推理。在七年级，你还将学习简化比以及按给定比率分配数量。
6. Case 5: Number Sequences and Patterns | 案例5：数列与模式
A staircase is built using square blocks. The first step uses 1 block, the second step uses 3 blocks, the third uses 5 blocks, forming a pattern. How many blocks are needed for the 10th step? The sequence of block counts is 1, 3, 5, 7, … which is odd numbers. The nth term is given by 2n – 1. For the 10th step, substitute n = 10: 2(10) – 1 = 19 blocks.
一个楼梯用正方形积木搭建。第一级用1块积木，第二级用3块，第三级用5块，形成一个模式。第10级台阶需要多少块积木？积木数量的序列是1, 3, 5, 7, … 即奇数。第n项由 2n – 1 给出。对于第10级，代入 n = 10：2(10) – 1 = 19 块积木。
If the pattern were cumulative (total blocks for a staircase of 10 steps), we would sum the sequence. The sum of the first n odd numbers is n². So total blocks for 10 steps = 10² = 100. This case links term-to-term rules with position-to-term rules, and introduces quadratic relationships in a visual way suitable for Year 7 advanced learners.
如果模式是累积的（10级台阶的楼梯所需总积木数），我们需要求序列的和。前n个奇数的和为 n²。所以10级台阶的总积木数 = 10² = 100。这个案例将项到项的规则与位置到项的规则联系起来，并以视觉化的方式引入二次关系，适合七年级进阶学生。
7. Case 6: Coordinates and Straight-Line Graphs | 案例6：坐标与直线图
A mobile phone plan charges a monthly fee of £10 plus £0.05 per text message. Plot the relationship between the number of texts (t) and the total monthly cost (C). We can write the equation: C = 10 + 0.05t. Choose values for t: 0, 100, 200. When t = 0, C = £10; t = 100, C = 10 + 5 = £15; t = 200, C = £20. Plot points (0,10), (100,15), (200,20) and draw a straight line.
一个手机套餐每月固定收费10英镑，外加每条短信0.05英镑。绘制短信数量(t)与月总费用(C)的关系图。我们可以写出方程：C = 10 + 0.05t。选取t值：0, 100, 200。当 t = 0 时，C = 10英镑；t = 100，C = 10 + 5 = 15英镑；t = 200，C = 20英镑。绘制点(0,10)、(100,15)、(200,20)并画出直线。
This is a real-life linear function. From the graph, you can estimate the cost for any number of texts, or find how many texts you can send for a given budget. For example, if you have £25, read across from C = 25 to the line, then down to t ≈ 300 texts. This case study consolidates work on substituting values, plotting coordinates, and interpreting gradients and intercepts.
这是一个现实生活中的线性函数。从图中，你可以估算任意短信数量的费用，或者对于给定的预算，找出可以发送的短信条数。例如，如果你有25英镑，从C=25水平移动到直线上，再向下找到t≈300条短信。这个案例巩固了代入数值、绘制坐标以及解释斜率和截距的知识。
8. Case 7: Introduction to Probability with Dice | 案例7：骰子概率初步
You roll two fair six-sided dice and add the scores. What is the probability of getting a total of 7? List all possible outcomes (36 equally likely combinations). The pairs that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six favourable outcomes. So P(sum = 7) = 6/36 = 1/6.
你同时投掷两个公平的六面骰子，将点数相加。得到总和为7的概率是多少？列出所有可能的结果（36种等可能组合）。总和为7的数对是：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) —— 六种有利结果。因此 P(和为7) = 6/36 = 1/6。
Now consider a game: you win if the sum is 7 or 11. Sum 11 pairs: (5,6), (6,5) — two outcomes. Total favourable = 6 + 2 = 8. Probability of winning = 8/36 = 2/9. This case encourages systematic listing and introduces the addition rule for mutually exclusive events. You can also explore experimental probability by conducting trials.
现在考虑一个游戏：如果和为7或11则获胜。和为11的组合：(5,6), (6,5) ——两种结果。总有利结果 = 6+2=8。获胜概率 = 8/36 = 2/9。这个案例鼓励系统化列举，并引入互斥事件的加法规则。你还可以通过进行试验来探索实验概率。
9. Case 8: Currency Conversion and Percentages | 案例8：货币兑换与百分比
You are travelling from the UK to Europe. The exchange rate is £1 = €1.15. A souvenir costs €46. How much is that in pounds? Divide by the exchange rate: 46 ÷ 1.15. This can be tricky; notice that 1.15 = 23/20, so divide by 1.15 is equivalent to multiplying by 20/23: 46 × 20/23 = (46/23) × 20 = 2 × 20 = £40. So the souvenir costs £40.
你正从英国前往欧洲旅行。汇率为 1英镑 = 1.15欧元。一件纪念品售价46欧元，相当于多少英镑？除以汇率：46 ÷ 1.15。这可能有点棘手；注意到 1.15 = 23/20，因此除以1.15等价于乘以20/23：46 × 20/23 = (46/23) × 20 = 2 × 20 = 40英镑。所以纪念品价格为40英镑。
If the shop offers a 15% discount on the €46 price for paying in cash, the new price is 85% of €46: 0.85 × 46 = €39.10. Then convert: 39.10 ÷ 1.15 ≈ £34. This multi-step case combines percentages and exchange rates, requiring careful order of operations and understanding of multiplier methods. Year 7 advanced students should practice applying percentage increase/decrease in financial contexts.
如果商店对现金支付提供15%折扣，新价格为46欧元的85%：0.85 × 46 = 39.10欧元。然后兑换：39.10 ÷ 1.15 ≈ 34英镑。这个多步骤案例结合了百分比和汇率，需要仔细安排运算顺序，并理解乘数方法。七年级进阶学生应练习在金融情境中应用百分比增减。
10. Review and Practical Exercise Tips | 复习与实战技巧
When tackling case study exercises, always follow these steps: first, identify what you know and what you need to find. Second, decide which mathematical tool (equation, formula, graph, ratio, probability) fits the situation. Third, carry out calculations step by step, showing your working clearly. Fourth, interpret your result in the context of the case—does it make sense?
处理案例分析练习时，始终遵循以下步骤：第一步，明确已知信息和求解目标。第二步，决定哪种数学工具（方程、公式、图表、比率、概率）适合这种情况。第三步，逐步执行计算，清晰地展示计算过程。第四步，在案例背景下解释你的结果 —— 它合理吗？
Practice by creating your own mini case studies from daily life: splitting a restaurant bill, calculating travel time, or mixing paint colours. The more you connect mathematics to real situations, the more confident you will become. Remember that the OCR advanced path encourages exploring multiple representations—words, tables, graphs, and symbols—to deepen understanding.
通过从日常生活中创建自己的迷你案例来练习：分摊餐厅账单、计算旅行时间或混合颜料。你将数学与现实情境联系得越多，就越自信。请记住，OCR进阶路线鼓励探索多种表示方法 —— 文字、表格、图形和符号 —— 以加深理解。
Published by TutorHao | 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导