📚 KS3 OCR Further Mathematics: Case Study Practical Exercises | KS3 OCR 进阶数学：案例分析实战演练
Case studies in OCR KS3 Further Mathematics bridge the gap between textbook exercises and real-world problem-solving. By working through carefully designed scenarios, you develop the ability to select and combine mathematical tools — a skill that distinguishes strong mathematicians. This article presents three detailed case studies covering geometry, algebra and number theory, each with step-by-step reasoning and bilingual commentary.
OCR KS3 进阶数学中的案例研究，在课本练习与现实问题解决之间架起了一座桥梁。通过精心设计的场景，你能培养选择并组合数学工具的能力，这是优秀数学家的标志。本文提供三个涵盖几何、代数与数论的详细案例，每一个都配有逐步推理与双语解说。
1. Overview of the Case Study Approach | 案例学习方法概述
In OCR Further Mathematics at Key Stage 3, case studies ask you to interpret a situation, identify relevant mathematics, and justify your reasoning. Unlike standard questions, there is often more than one valid pathway to a solution.
在关键阶段三的 OCR 进阶数学中，案例研究要求你解读情境、识别相关数学内容并论证推理。与标准题目不同，通常不止一条正确的解题路径。
You are expected to communicate your thinking clearly. This includes defining variables, writing equations or inequalities, showing substitutions, and interpreting results in context. The emphasis is on process as much as on final answers.
你需要清晰地表达思考过程，包括定义变量、写出方程或不等式、展示代入过程，以及在情境中解读结果。重点既在最终答案，也在解题过程。
Each case study in this article follows a pattern: a scenario is introduced, a problem is posed, mathematics is applied, and the outcome is evaluated. Work through them actively — pause to attempt each step before reading the solution.
本文的每个案例都遵循一种模式：引入场景、提出问题、应用数学、评估结果。请主动练习——在读解答之前先尝试每一步。
2. Case Study 1: Designing a School Garden | 案例一：设计学校花园
A school has a rectangular plot of land with a perimeter of 36 metres. The gardening club wants the length to be twice the width. They need to know the exact dimensions to order fencing and plan planting beds. This is a typical geometric reasoning task that blends measurement with algebra.
学校有一块周长为 36 米的矩形土地。园艺俱乐部希望长度是宽度的两倍。他们需要知道精确尺寸以订购围栏并规划种植床。这是一道融合测量与代数的典型几何推理题。
Let the width be w metres. Then the length is 2w metres. The perimeter formula for a rectangle is P = 2(length + width). Substituting the known perimeter gives 2(2w + w) = 36.
设宽为 w 米，则长为 2w 米。矩形周长公式为 P = 2 × (长 + 宽)。代入已知周长得到 2(2w + w) = 36。
Simplify the bracket: 2(3w) = 36 → 6w = 36. Dividing both sides by 6 gives w = 6. Therefore width = 6 m, length = 12 m.
化简括号：2(3w) = 36 → 6w = 36。两边除以 6 得 w = 6。所以宽为 6 米，长为 12 米。
The club can now calculate the area for planting: 12 × 6 = 72 m². They also verify the perimeter: 2(12+6)=36 m, which matches the requirement. This double-checking step is essential in case studies.
俱乐部现在可以计算种植面积：12 × 6 = 72 平方米。他们也验证了周长：2(12+6)=36 米，符合要求。这种双重检验步骤在案例研究中至关重要。
3. Extending the Garden Problem: Cost Analysis | 案例延伸：成本分析
The school decides to build a path of uniform width around the garden inside the fence. The total area of the path must not exceed 40 m². If the path width is x metres, write an expression for the area of the path and find the maximum possible width to one decimal place.
学校决定在围栏内部围绕花园修建一条宽度均匀的小路。小路的总面积不得超过 40 平方米。设小路宽度为 x 米，写出小路面积的表达式，并求出准确到一位小数的最大可能宽度。
The outer rectangle formed by the fence has dimensions 12 m by 6 m. The inner rectangle (garden bed) has lengths (12 – 2x) and (6 – 2x). Path area = outer area – inner area = 72 – (12-2x)(6-2x).
由围栏形成的外部矩形尺寸为 12 米 × 6 米。内部矩形（花床）的边长为 (12 – 2x) 和 (6 – 2x)。小路面积 = 外部面积 – 内部面积 = 72 – (12-2x)(6-2x)。
Expand the product: (12-2x)(6-2x) = 72 – 24x – 12x + 4x² = 72 – 36x + 4x². So path area = 72 – (72 – 36x + 4x²) = 36x – 4x².
展开乘积：(12-2x)(6-2x) = 72 – 24x – 12x + 4x² = 72 – 36x + 4x²。因此小路面积 = 72 – (72 – 36x + 4x²) = 36x – 4x²。
We require 36x – 4x² ≤ 40. Rearranging: -4x² + 36x – 40 ≤ 0 → divide by -4 (reverse inequality): x² – 9x + 10 ≥ 0.
要求 36x – 4x² ≤ 40。整理：-4x² + 36x – 40 ≤ 0 → 除以 -4（不等号变向）：x² – 9x + 10 ≥ 0。
Solve the quadratic equation x² – 9x + 10 = 0 using the quadratic formula: x = [9 ± √(81 – 40)] / 2 = [9 ± √41] / 2. √41 ≈ 6.403, so roots are (9+6.403)/2 ≈ 7.70 and (9-6.403)/2 ≈ 1.30.
用二次公式解方程 x² – 9x + 10 = 0：x = [9 ± √(81 – 40)] / 2 = [9 ± √41] / 2。√41 ≈ 6.403，因此根为 (9+6.403)/2 ≈ 7.70 和 (9-6.403)/2 ≈ 1.30。
The inequality x² – 9x + 10 ≥ 0 holds for x ≤ 1.30 or x ≥ 7.70. But x must be less than half the smaller dimension (6/2 = 3), so x ≤ 3. Thus the feasible interval is 0 < x ≤ 1.3. Maximum path width ≈ 1.3 m.
不等式 x² – 9x + 10 ≥ 0 在 x ≤ 1.30 或 x ≥ 7.70 时成立。但 x 必须小于较小边长的一半 (6/2=3)，所以 x ≤ 3。因此可行区间为 0 < x ≤ 1.3。最大路径宽度约为 1.3 米。
4. Case Study 2: Fundraising Event Planning | 案例二：筹款活动策划
The student council organises a charity concert. The venue costs £200 to hire, and each ticket includes a £1.50 refreshment charge. Tickets are sold at £8 each. The council wants the profit to be at least £500. Set up and solve an inequality to find the minimum number of tickets to sell.
学生会组织一场慈善音乐会。场地租金为 200 英镑，每张票包含 1.50 英镑的茶点费。门票售价为 8 英镑。学生会希望利润至少为 500 英镑。建立并求解不等式，找出最低售票数量。
Let n be the number of tickets sold. Revenue = 8n. Total costs = 200 + 1.5n. Profit = Revenue – Costs = 8n – (200 + 1.5n) = 6.5n – 200.
设 n 为售出门票数量。收入 = 8n。总成本 = 200 + 1.5n。利润 = 收入 – 成本 = 8n – (200 + 1.5n) = 6.5n – 200。
Require profit ≥ 500: 6.5n – 200 ≥ 500 → 6.5n ≥ 700 → n ≥ 700 / 6.5. Calculate 700 ÷ 6.5 = 107.692… Since you cannot sell a fraction of a ticket, n ≥ 108.
要求利润 ≥ 500：6.5n – 200 ≥ 500 → 6.5n ≥ 700 → n ≥ 700 / 6.5。计算 700 ÷ 6.5 = 107.692… 因为不能卖零散票，所以 n ≥ 108。
The council must sell at least 108 tickets. Check: 108 × 6.5 – 200 = 702 – 200 = 502, which is just over £500. If they sell 107, profit = 695.5 – 200 = 495.5, which is below the target.
学生会必须至少卖出 108 张票。验证：108 × 6.5 – 200 = 702 – 200 = 502，刚刚超过 500 英镑。如果卖出 107 张，利润 = 695.5 – 200 = 495.5，低于目标。
5. Adding a Second Revenue Stream | 增加第二个收入来源
To boost funds, the council plans to sell programmes at £2 each, costing £0.80 to print. They expect one programme per three attendees. Modify the inequality to find the new minimum ticket sales if the profit target remains £500.
为了增加资金，学生会计划以 2 英镑一本的价格出售节目单，印刷成本为 0.80 英镑。他们预计每三名观众购买一本。修改不等式，求出利润目标仍为 500 英镑时的最低售票数量。
Number of programmes sold ≈ n/3. Revenue from programmes = 2 × (n/3) = 2n/3. Programme costs = 0.80 × (n/3) = 0.8n/3. Total profit now includes ticket and programme profit: (8n – 200 – 1.5n) + (2n/3 – 0.8n/3) = 6.5n – 200 + (1.2n/3) = 6.5n – 200 + 0.4n = 6.9n – 200.
节目单销售量约等于 n/3。节目单收入 = 2 × (n/3) = 2n/3。节目单成本 = 0.80 × (n/3) = 0.8n/3。总利润现在包括门票和节目单利润：(8n – 200 – 1.5n) + (2n/3 – 0.8n/3) = 6.5n – 200 + (1.2n/3) = 6.5n – 200 + 0.4n = 6.9n – 200。
Set 6.9n – 200 ≥ 500 → 6.9n ≥ 700 → n ≥ 700 / 6.9 ≈ 101.45. So at least 102 tickets are needed. Notice how extra revenue reduces the required ticket sales.
令 6.9n – 200 ≥ 500 → 6.9n ≥ 700 → n ≥ 700 / 6.9 ≈ 101.45。所以至少需要 102 张票。注意额外收入是如何减少所需售票数量的。
6. Using Graphs to Visualise Constraints | 用图像直观呈现约束条件
In OCR Further Maths, you might be asked to represent inequalities on a graph. For the concert, we can plot two lines: profit from tickets only (y = 6.5n – 200) and profit with programmes (y = 6.9n – 200). The horizontal line y = 500 shows the target.
在 OCR 进阶数学中，可能会要求你在图像上表示不等式。对于音乐会，我们可以画出两条直线：仅门票利润 (y = 6.5n – 200) 和含节目单利润 (y = 6.9n – 200)。水平线 y = 500 表示目标。
The intersection of y=6.5n-200 with y=500 gives n ≈ 107.7, and for y=6.9n-200 it gives n ≈ 101.4. The steeper line (higher profit per ticket) reaches the target sooner. This is an example of using coordinate geometry to interpret financial information.
y=6.5n-200 与 y=500 的交点给出 n ≈ 107.7，而 y=6.9n-200 与 y=500 的交点给出 n ≈ 101.4。更陡的直线（每张票利润更高）能更快达到目标。这是运用坐标几何来解读财务信息的一个例子。
Drawing these graphs by hand reinforces the connection between algebraic and graphical representations. Choose a suitable scale; n on the horizontal axis from 0 to 120, profit on the vertical axis from -200 to 600.
亲手绘制这些图像能加强代数表达与图像表达之间的联系。选择合适的尺度：横轴 n 从 0 到 120，纵轴利润从 -200 到 600。
7. Case Study 3: Prime Number Cryptography | 案例三：质数密码学
Cryptography provides an exciting context for prime numbers. The RSA method, which secures online transactions, relies on the difficulty of factorising large products of two primes. We explore a simplified version suitable for KS3 Further Maths.
密码学为质数提供了一个激动人心的应用场景。保障在线交易安全的 RSA 方法依赖于分解两个质数大乘积的困难性。我们来探究一个适合 KS3 进阶数学的简化版本。
Suppose Alice wants to send a secret number to Bob. She picks two prime numbers, p = 11 and q = 13. Compute n = p × q = 143. This n is part of the public key.
假设爱丽丝想向鲍勃发送一个秘密数字。她选择两个质数 p = 11 和 q = 13。计算 n = p × q = 143。这个 n 是公钥的一部分。
She then computes φ(n) = (p-1)(q-1) = 10 × 12 = 120. Choose a public exponent e that is coprime with 120, such as e = 7. The public key is (143, 7).
然后她计算 φ(n) = (p-1)(q-1) = 10 × 12 = 120。选择一个与 120 互质的公开指数 e，例如 e = 7。公钥就是 (143, 7)。
Bob takes his plaintext message M = 5. He encrypts it using C = Mᵉ mod n = 5⁷ mod 143. Compute 5⁷ = 78,125. Divide by 143: 143 × 546 = 78,078, remainder 47. So ciphertext C = 47.
鲍勃取他的明文消息 M = 5。他用 C = Mᵉ mod n 进行加密，即 5⁷ mod 143。计算 5⁷ = 78,125。除以 143：143 × 546 = 78,078，余数为 47。所以密文 C = 47。
To decrypt, Alice needs her private key d, where e × d ≡ 1 mod φ(n). That is, 7d ≡ 1 mod 120. Find d such that 7d = 120k + 1. Trying k=2 gives 241, not divisible by 7. k=3 gives 361, no. k=4 gives 481, no. k=5 gives 601, 601/7 = 85.857. k=6 gives 721, 721/7 = 103 exactly. So d = 103.
为了解密，爱丽丝需要她的私钥 d，满足 e × d ≡ 1 mod φ(n)，即 7d ≡ 1 mod 120。寻找 d 使得 7d = 120k + 1。尝试 k=2 得 241，不能被 7 整除；k=3 得 361，不能；k=4 得 481，不能；k=5 得 601，601/7=85.857；k=6 得 721，721/7=103 恰好整除。所以 d = 103。
Alice decrypts: M = Cᵈ mod n = 47¹⁰³ mod 143. Computing this directly is huge, but we can use the fact that 47¹⁰³ ≡ 5 mod 143 by the mathematics behind RSA. A simplified check: 47¹⁰³ mod 143 indeed returns 5, showing the message is recovered.
爱丽丝解密：M = Cᵈ mod n = 47¹⁰³ mod 143。直接计算这个量极其巨大，但我们可以依据 RSA 背后的数学原理知道 47¹⁰³ ≡ 5 mod 143。简化验证：47¹⁰³ mod 143 确实返回 5，表明消息被恢复。
8. Factorisation Challenge and Security | 分解挑战与安全性
The security of this system relies on keeping p and q secret. Anyone can see n = 143; if they factorise it into 11 × 13, they can compute φ(n) and find the private key d. In practice, primes are hundreds of digits long.
该系统的安全性依赖于对 p 和 q 的保密。任何人都能看到 n = 143；如果他们将其分解为 11 × 13，就能计算出 φ(n) 并找到私钥 d。在实践中，质数通常有数百位长。
Try a challenge: n = 323 is the product of two primes. Find them. (Hint: both primes are between 10 and 25.) By trial, 323 ÷ 17 = 19 exactly. So p=17, q=19. Then φ(n)=16×18=288.
尝试一个挑战：n = 323 是两个质数的乘积。找出它们。（提示：两个质数都在 10 到 25 之间。）通过试除，323 ÷ 17 = 19 恰好整除。所以 p=17, q=19。然后 φ(n)=16×18=288。
You are now ready to create your own simplified RSA keys. Choose two primes below 30, compute n, choose e coprime to (p-1)(q-1), and encipher a simple message. Exchange with a partner to decrypt.
现在你可以创建自己的简化 RSA 密钥了。选择两个 30 以下的质数，计算 n，选择一个与 (p-1)(q-1) 互质的 e，然后加密一个简单的消息。与同伴交换并解密。
9. Linking Case Studies to Core KS3 Topics | 将案例研究链接到 KS3 核心主题
These case studies are not isolated puzzles; they draw on number (factors, primes, fractions), algebra (linear equations, quadratics, inequalities), geometry (perimeter, area) and working with graphs. OCR Further Mathematics expects you to recognise these connections.
这些案例研究并非孤立的谜题；它们运用了数（因数、质数、分数）、代数（线性方程、二次方程、不等式）、几何（周长、面积）以及图像处理。OCR 进阶数学期望你能识别这些联系。
For instance, the garden design used perimeter and area formulas, linear equations, and quadratic inequalities all in one context. The fundraising case required forming and solving linear inequalities, then interpreting rational results in a discrete context.
例如，花园设计在同一个情境中运用了周长与面积公式、线性方程以及二次不等式。筹款案例需要建立并求解线性不等式，然后在离散情境中解读分数结果。
The cryptography case strengthened fluency with powers, modulo arithmetic, factorisation, and systematic problem-solving. These are all essential skills for GCSE and beyond.
密码学案例增强了对乘方、模运算、因数分解以及系统化问题解决的熟练度。这些都是 GCSE 及更高阶段的核心技能。
10. Tackling Unfamiliar Problems: A Strategy | 应对陌生问题的策略
When presented with a new case study, follow a structured approach: 1) Read the scenario carefully, identifying what is given and what is unknown. 2) Define variables for quantities that change. 3) Translate words into mathematical sentences (equations/inequalities).
当遇到新的案例研究时，遵循结构化的方法：1) 仔细阅读场景，确定已知量和未知量。2) 为变化的量定义变量。3) 将文字表述转化为数学语句（方程/不等式）。
4) Solve the mathematics using appropriate techniques. 5) Interpret the solution in the context of the problem — does the answer make sense? 6) Check your working by substituting back or using an alternative method. 7) Communicate your reasoning in clear steps.
4) 使用适当的方法求解数学问题。5) 在问题情境中解读答案——答案是否合理？6) 通过代回检验或使用其他方法来检查你的运算。7) 用清晰的步骤沟通你的推理过程。
This seven-step framework works for any applied problem. Practice it with the case studies above, then invent your own scenarios: plan a trip, design a box with maximum volume, or code a substitution cipher.
这个七步框架适用于任何应用问题。用上述案例研究进行练习，然后自己创造场景：规划一次旅行、设计一个体积最大的盒子，或编写一套替换密码。
11. Common Pitfalls and How to Avoid Them | 常见陷阱与如何避免
Students often move too quickly to calculation without defining variables. Always write down what each letter represents. For example, ‘Let x be the width of the path in metres’ clarifies the problem for yourself and the examiner.
学生经常过快进入计算却没有定义变量。始终写下每个字母所代表的含义。例如，“设 x 为小路宽度（单位：米）”能让你自己和考官都更清楚问题。
Another mistake is forgetting to check the feasibility of solutions. In the garden case, the quadratic gave x ≈ 7.7, but x could not exceed 3. Always test solutions against real-world constraints.
另一个错误是忘记检验解的可行性。在花园案例中，二次方程给出了 x ≈ 7.7，但 x 不能超过 3。始终将解与现实约束条件进行对照检验。
When working with inequalities, remember to reverse the sign when multiplying or dividing by a negative number. Many marks are lost through sign errors.
在处理不等式时，记住当乘以或除以一个负数时要改变不等号方向。很多分数因符号错误而丢失。
Finally, present your answer in a sentence that refers back to the question. ‘The minimum number of tickets needed is 108’ is better than just ‘n = 108’.
最后，用一句回扣问题的完整句子给出答案。“所需的最低售票数量为 108 张”比仅仅写“n = 108”要好。
12. Practice Case Study: The Cinema Trip | 练习案例：电影院旅行
Try this case study independently. A group of friends wants to go to the cinema. The ticket price is £7.50 per person, and a group discount of 20% applies if there are at least 6 people. The total budget for the group is £60. How many friends can go? Set up an inequality for the number of people, n, and find the maximum n.
独立尝试这个案例研究。一群朋友想去看电影。票价每人 7.50 英镑，若至少 6 人可享团体折扣 20%。团体总预算为 60 英镑。最多能去多少人？为人数 n 建立不等式，并求出最大 n。
For n < 6, cost = 7.5n. For n ≥ 6, cost = 7.5n × 0.80 = 6n. Since budget is £60, when n ≥ 6 we have 6n ≤ 60 → n ≤ 10. When n < 6, 7.5n ≤ 60 → n ≤ 8. But n must be integer. The group could bring 10 friends using the discount, which is within budget (6×10=60). Without discount, max would be 8 (7.5×8=60). Therefore the discount allows 2 extra people. The optimal n is 10.
当 n < 6 时，费用 = 7.5n。当 n ≥ 6 时，费用 = 7.5n × 0.80 = 6n。因为预算为 60 英镑，当 n ≥ 6 时有 6n ≤ 60 → n ≤ 10。当 n < 6 时，7.5n ≤ 60 → n ≤ 8。但 n 必须是整数。利用折扣该团体可以带 10 个人，刚好在预算内 (6×10=60)。没有折扣时最大为 8 人 (7.5×8=60)。因此折扣多容纳了 2 人。最优 n 为 10。
Work through similar problems to build confidence. Combine geometry with cost, or probability with profit. The more you practise, the more natural case study analysis becomes.
通过类似的题目来建立信心。将几何与成本相结合，或将概率与利润相结合。你练习得越多，案例分析就会变得越自然。
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