KS3 AQA Further Mathematics: Case Study Practice | KS3 AQA 进阶数学：案例分析实战演练

📚 KS3 AQA Further Mathematics: Case Study Practice | KS3 AQA 进阶数学：案例分析实战演练

Case studies are the bridge between abstract mathematical techniques and the real world. In this article, we tackle ten carefully selected scenarios that demand precise reasoning, proportional thinking, algebraic modelling and graphical interpretation — all core skills in the KS3 AQA Further Mathematics specification. Each case walks you through a genuine problem, exposes the underlying structure and shows how to apply advanced topics systematically.

案例分析是抽象数学技巧与真实世界之间的桥梁。本文将针对十个精选情境，要求进行精确推理、比例思维、代数建模和图像解读——这些都是 KS3 AQA 进阶数学考试大纲的核心技能。每个案例都会带你走过一个真实问题，揭示隐含的结构，并展示如何系统地应用进阶数学课题。

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1. Case Study 1: Party Planning | 案例一：派对策划

Ahmed is organising an end-of-year party for 32 classmates. He has a budget of £120. Pizzas cost £8.50 each and serve 4 people; 2-litre drink bottles cost £1.80 each and serve 6 people. Decorations come in a pack costing £22.50. He also considers adding a photo booth that rents for £30. What is the minimum order of pizzas and drinks? Can he afford the booth?

Ahmed 正在为 32 位同学组织年终派对，预算为 120 英镑。每个披萨售价 8.50 英镑，可供 4 人食用；2 升装饮料每瓶 1.80 英镑，可供 6 人饮用。装饰品一包 22.50 英镑。他还考虑增加一个租用费用为 30 英镑的照相亭。最少需要订购多少披萨和饮料？他能负担照相亭吗？

Step 1 – Calculate the number of pizzas needed. 32 ÷ 4 = 8 pizzas. Since pizzas are sold whole, 8 must be ordered.

步骤一 – 计算所需披萨数量。32 ÷ 4 = 8 个披萨。因为披萨是整只出售，必须订购 8 个。

Step 2 – Calculate the number of drink bottles. 32 ÷ 6 ≈ 5.33, so 6 bottles must be purchased to ensure everyone is served.

步骤二 – 计算所需饮料瓶数。32 ÷ 6 ≈ 5.33，因此必须购买 6 瓶以确保每人有份。

Step 3 – Work out total cost without the booth. Pizzas: 8 × £8.50 = £68.00. Drinks: 6 × £1.80 = £10.80. Decorations: £22.50. Subtotal = £68.00 + £10.80 + £22.50 = £101.30.

步骤三 – 计算不含照相亭的总费用。披萨：8 × £8.50 = £68.00。饮料：6 × £1.80 = £10.80。装饰：£22.50。小计 = £68.00 + £10.80 + £22.50 = £101.30。

Step 4 – Check if the booth fits the budget. Remaining money = £120 − £101.30 = £18.70. The booth costs £30, so Ahmed cannot afford it unless he reduces other items or increases the budget.

步骤四 – 检查照相亭是否在预算内。剩余资金 = £120 − £101.30 = £18.70。照相亭费用为 £30，因此除非减少其他支出或增加预算，否则 Ahmed 无法负担。

This case reinforces proportion, rounding up for practical constraints and budget balancing – a powerful mix of number and reasoning skills.

此案例强化了比例、为了满足实际约束而向上取整以及预算平衡——融合了数感与推理技能。

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2. Case Study 2: Designing a Vegetable Patch | 案例二：菜园设计

A gardener has 20 m of fencing to create a rectangular vegetable patch. She wants the area to be at least 24 m². Find all possible integer widths of the rectangle that satisfy both the perimeter constraint and the area requirement.

一位园丁有 20 米长的围栏，用来围一个长方形菜园。她希望面积至少为 24 平方米。试找出满足周长限制和面积要求的所有可能的整数宽度。

From the perimeter, 2(l + w) = 20 ⇒ l + w = 10, so length l = 10 − w. The area is A = l × w = w(10 − w).

由周长可得 2(l + w) = 20 ⇒ l + w = 10，因此长 l = 10 − w。面积 A = l × w = w(10 − w)。

The condition A ≥ 24 gives the inequality w(10 − w) ≥ 24. Expanding: 10w − w² ≥ 24 ⇒ −w² + 10w − 24 ≥ 0. Multiply by −1 and reverse the sign: w² − 10w + 24 ≤ 0.

条件 A ≥ 24 给出不等式 w(10 − w) ≥ 24。展开得：10w − w² ≥ 24 ⇒ −w² + 10w − 24 ≥ 0。两边乘以 −1 并翻转不等号：w² − 10w + 24 ≤ 0。

Factorising: (w − 4)(w − 6) ≤ 0. The quadratic is negative or zero between the roots, so 4 ≤ w ≤ 6.

分解因式：(w − 4)(w − 6) ≤ 0。二次式在两根之间为负或零，因此 4 ≤ w ≤ 6。

Since l = 10 − w must also be positive and longer or equal to w for convention, the possible integer widths are 4 m, 5 m and 6 m, giving lengths 6 m, 5 m and 4 m respectively. All satisfy the area requirement.

由于 l = 10 − w 必须为正，且通常长不小于宽，可行的整数宽度为 4 m、5 m 和 6 m，对应长度分别为 6 m、5 m 和 4 m。三者均满足面积要求。

Here, algebraic manipulation of inequalities and factorisation are key Further Mathematics techniques.

在此，不等式的代数变形与因式分解是关键进阶数学技巧。

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3. Case Study 3: Cracking Codes with Modular Arithmetic | 案例三：用模运算破解密码

Alice wants to send the word ‘CASE’ to Bob using a Caesar cipher with a shift of +3. Each letter is converted to a number (A=0, B=1, …, Z=25), shifted by 3, and then reduced modulo 26 to ensure it wraps around. Decode the received word ‘FDVH’ to confirm the method.

Alice 想用移位量为 +3 的凯撒密码将单词 ‘CASE’ 发送给 Bob。先将每个字母转换为数字（A=0, B=1, …, Z=25），加 3，然后模 26 以确保回绕。对收到的 ‘FDVH’ 进行解码来验证该方法。

Encoding ‘CASE’: C → 2, A → 0, S → 18, E → 4. Apply shift: (2+3) mod 26 = 5 → F, (0+3)=3 → D, (18+3)=21 → V, (4+3)=7 → H. So ciphertext is ‘FDVH’.

对 ‘CASE’ 编码：C → 2, A → 0, S → 18, E → 4。应用移位：(2+3) mod 26 = 5 → F, (0+3)=3 → D, (18+3)=21 → V, (4+3)=7 → H。因此密文为 ‘FDVH’。

Decoding: reverse by shifting −3. F → 5, 5−3=2 → C; D → 3, 3−3=0 → A; V → 21, 21−3=18 → S; H → 7, 7−3=4 → E. The original word is recovered.

解码：反向移动 −3。F → 5, 5−3=2 → C; D → 3, 3−3=0 → A; V → 21, 21−3=18 → S; H → 7, 7−3=4 → E。还原出原词。

This shows how modular arithmetic ensures the alphabet cycles smoothly. The general encryption function is E(x) = (x + k) mod 26, where k is the shift key.

这展示了模运算如何确保字母表平滑循环。一般加密函数为 E(x) = (x + k) mod 26，其中 k 为移位密钥。

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4. Case Study 4: The Bouncing Ball Sequence | 案例四：弹跳球序列

A ball is dropped from a height of 2 metres. After each bounce it rises to ¾ of its previous height. Calculate the height reached after the third bounce and the total vertical distance travelled by the ball up to the moment it hits the ground for the fourth time.

一个球从 2 米高度落下。每次弹跳后它会反弹到前一次高度的 ¾。计算第三次弹跳后所达到的高度，以及到球第四次撞击地面前所经过的总垂直距离。

Heights after each bounce: h₁ = 2 × ¾ = 1.5 m; h₂ = 1.5 × ¾ = 1.125 m; h₃ = 1.125 × ¾ = 0.84375 m. So after the third bounce it rises to 0.84375 m (approx 0.844 m).

每次弹跳后的高度：h₁ = 2 × ¾ = 1.5 m；h₂ = 1.5 × ¾ = 1.125 m；h₃ = 1.125 × ¾ = 0.84375 m。因此第三次弹跳后上升至约 0.844 m。

Total distance: initial drop = 2 m. Between first and fourth ground impacts, the ball travels up and down for the first three bounces: 2 × (1.5 + 1.125 + 0.84375) = 2 × 3.46875 = 6.9375 m. Add the initial drop: total = 2 + 6.9375 = 8.9375 m.

总距离：初始下落 = 2 m。从第一次到第四次撞击地面，球经历了前三次弹跳的上下运动：2 × (1.5 + 1.125 + 0.84375) = 2 × 3.46875 = 6.9375 m。加上初始下落：总计 = 2 + 6.9375 = 8.9375 m。

General term-to-term rule: hₙ = (¾)ⁿ × 2, forming a geometric sequence. This bridges multiplicative reasoning and series summation – excellent practice for algebraic thinking.

通用递推规则：hₙ = (¾)ⁿ × 2，构成一个等比数列。这连接了乘法推理与数列求和——是代数思维的绝佳训练。

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5. Case Study 5: Sports Day Timing | 案例五：运动会时间安排

Runner A completes 100 m in 12.5 s. Runner B runs 150 m in 19.0 s. Who has the higher average speed? If Runner A is given a 5-metre head start in a 100-metre race, who would cross the finish line first?

运动员 A 用 12.5 秒跑完 100 米。运动员 B 用 19.0 秒跑完 150 米。谁的平均速度更高？如果在 100 米比赛中让 A 提前 5 米起跑，谁会先到终点？

Speed A = 100 ÷ 12.5 = 8.0 m/s. Speed B = 150 ÷ 19.0 ≈ 7.8947 m/s. So A is marginally faster.

A 的速度 = 100 ÷ 12.5 = 8.0 m/s。B 的速度 = 150 ÷ 19.0 ≈ 7.8947 m/s。因此 A 稍快一些。

With a 5 m head start, A must run only 95 m. Time for A = 95 ÷ 8.0 = 11.875 s. Time for B over full 100 m = 100 ÷ 7.8947 ≈ 12.666 s. Even with the advantage, A already finishes earlier; the head start makes the gap even larger.

在 5 米领先优势下，A 只需跑 95 米。所需时间 = 95 ÷ 8.0 = 11.875 s。B 跑完整 100 米的时间 = 100 ÷ 7.8947 ≈ 12.666 s。即使不考虑起跑优势，A 也能更快完成；领先起跑更拉大了差距。

Using speed = distance ÷ time and rearranging for time hones proportional reasoning and the concept of inverse relationships.

使用速度 = 距离 ÷ 时间并变形求时间，有助于锻炼比例推理和反比关系的概念。

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6. Case Study 6: Scaling Up a Recipe | 案例六：食谱缩放

A cupcake recipe for 6 servings requires 120 g flour, 2 eggs, 80 g sugar and 60 g butter. You need to bake enough for 15 people. Determine the required quantities using the unitary method.

一份供 6 人享用的纸杯蛋糕食谱需要 120 克面粉、2 个鸡蛋、80 克糖和 60 克黄油。你需要烤制足够 15 人吃的分量。请用归一法确定各原料的用量。

The scaling factor is 15 ÷ 6 = 2.5. Multiply each ingredient by 2.5:

缩放因子为 15 ÷ 6 = 2.5。每项原料乘以 2.5：

• Flour: 120 × 2.5 = 300 g
• Eggs: 2 × 2.5 = 5 eggs
• Sugar: 80 × 2.5 = 200 g
• Butter: 60 × 2.5 = 150 g

Alternatively, using the unitary method: for 1 person, flour = 120 ÷ 6 = 20 g; then for 15 people, 20 × 15 = 300 g, etc. Both approaches reinforce the concept of direct proportion.

也可以用归一法：每人面粉 = 120 ÷ 6 = 20 克；15 人就需要 20 × 15 = 300 克，其他同理。两种方法均强化了正比例概念。

Understanding multiplicative scaling without changing the ratios is fundamental in both pure and applied Further Mathematics.

在不改变比例的前提下进行乘法缩放，是纯数学和应用进阶数学的基础。

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7. Case Study 7: Building a Safe Ramp | 案例七：修建安全斜坡

A wheelchair ramp has a sloping length of 5.0 m and must achieve a vertical rise of 1.5 m. Regulations require the horizontal run to be at least 4.5 m and the gradient (rise ÷ run) not to exceed 0.35. Does this ramp meet the safety standards?

一条轮椅坡道的斜面长度为 5.0 米，必须实现 1.5 米的垂直升高。规范要求水平跨度至少为 4.5 米，且坡度（升高÷水平跨度）不得超过 0.35。该坡道符合安全标准吗？

Use Pythagoras’ theorem: horizontal run = √(5.0² − 1.5²) = √(25 − 2.25) = √22.75 ≈ 4.77 m. This is greater than the minimum 4.5 m, so the length condition is satisfied.

运用勾股定理：水平跨度 = √(5.0² − 1.5²) = √(25 − 2.25) = √22.75 ≈ 4.77 m。这大于最低要求的 4.5 米，因此长度条件满足。

Gradient = rise ÷ run = 1.5 ÷ 4.77 ≈ 0.314. This is less than 0.35, so the steepness condition is also met. The ramp passes both checks.

坡度 = 升高 ÷ 水平跨度 = 1.5 ÷ 4.77 ≈ 0.314。这小于 0.35，因此陡度条件也满足。该坡道两项检查均通过。

This problem blends geometry, surd manipulation and ratio interpretation — typical of the multi-step reasoning expected in Further Mathematics.

此题融合了几何、根式处理与比率解读——正是进阶数学所要求的典型多步推理。

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8. Case Study 8: Population Growth Model | 案例八：人口增长模型

A colony of bacteria doubles every hour. The initial count at 9:00 AM is 50. Find the population at 2:00 PM. Later, a second colony triples every two hours; after 6 hours its population is 4050. What was its starting count?

一个细菌菌落每小时翻倍。上午 9:00 的初始数量为 50。求下午 2:00 时的数量。之后，另一个菌落每两小时增至三倍；6 小时后其数量为 4050。它的起始数量是多少？

From 9:00 AM to 2:00 PM is 5 hours. Population = 50 × 2⁵ = 50 × 32 = 1600.

从上午 9:00 到下午 2:00 共 5 小时。数量 = 50 × 2⁵ = 50 × 32 = 1600。

For the second colony: tripling every 2 hours means the growth factor per 2-hour period is 3. In 6 hours there are 6 ÷ 2 = 3 growth periods. Let initial be P. Then P × 3³ = 4050 ⇒ P × 27 = 4050 ⇒ P = 150.

对于第二个菌落：每 2 小时增至三倍意味着每 2 小时的生长因子为 3。6 小时内包含 6 ÷ 2 = 3 个生长周期。设起始数为 P，则 P × 3³ = 4050 ⇒ P × 27 = 4050 ⇒ P = 150。

Exponential models like these appear in sequences and introduce the power of repeated multiplication, laying foundations for geometric progressions.

像这样的指数模型出现在数列中，并引入了重复乘法的力量，为等比数列打下基础。

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9. Case Study 9: Is the Game Fair? | 案例九：游戏公平吗？

In a coin-toss game, you flip two fair coins. If both show heads, you win £2; otherwise you lose £1. Calculate the expected gain per game. Determine if the game is fair and suggest a change to make the expected value zero.

在一个抛硬币游戏中，抛两枚公平硬币。如果均为正面，你赢 2 英镑；否则输 1 英镑。计算每次游戏的期望收益。判断游戏是否公平，并建议一个使期望值为零的修改方案。

Sample space: HH, HT, TH, TT (each ¼). Win £2 on HH only. Lose £1 on the other three outcomes.

样本空间：HH、HT、TH、TT（各概率 ¼）。只在 HH 上赢 2 英镑，其他三种结果输 1 英镑。

Expected value = (¼ × 2) + (¾ × (−1)) = 0.50 − 0.75 = −0.25. The negative expected value means the player loses 25p per game on average, so the game favours the organiser.

期望值 = (¼ × 2) + (¾ × (−1)) = 0.50 − 0.75 = −0.25。负的期望值意味着玩家平均每次游戏亏损 25 便士，因此游戏对组织者有利。

To make it fair, set win amount W such that ¼ × W − ¾ × 1 = 0 ⇒ ¼W = ¾ ⇒ W = 3. Paying £3 for two heads would yield expected value zero.

为使游戏公平，设赢取金额为 W，使得 ¼ × W − ¾ × 1 = 0 ⇒ ¼W = ¾ ⇒ W = 3。为两个正面支付 3 英镑可使期望值为零。

This case sharpens probability reasoning and introduces financial expectation – a practical extension of combined events.

此案例强化了概率推理，并引入了金融期望——组合事件的实用延伸。

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10. Case Study 10: Interpreting a Distance–Time Graph | 案例十：解读距离―时间图

The distance–time graph of a short journey shows: from 0 to 10 minutes, a straight line from 0 km to 2 km; from 10 to 20 minutes, a horizontal segment at 2 km; from 20 to 30 minutes, a straight line back to 0 km. Describe the motion and calculate the speeds for each segment.

一段短途旅程的距离―时间图显示：0 到 10 分钟，从 0 公里到 2 公里的直线段；10 到 20 分钟，位于 2 公里处的水平线段；20 到 30 分钟，直线回到 0 公里。请描述运动情况并计算各段的速度。

First segment: constant positive gradient → moving away from start. Speed = (2 − 0) km ÷ (10/60) h = 2 ÷ (1/6) = 12 km/h.

第一段：恒定正斜率 → 远离出发点的运动。速度 = (2 − 0) 公里 ÷ (10/60) 小时 = 2 ÷ (1/6) = 12 公里/小时。

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