📚 KS3 CCEA Further Mathematics: Case Study Practical Exercises | KS3 CCEA 进阶数学：案例分析实战演练
Further Mathematics at KS3 level extends your problem-solving skills by applying familiar concepts to unfamiliar situations. This article presents a series of real-world case studies designed to sharpen your analytical thinking, logical reasoning, and mathematical fluency. Each case study encourages you to break down a complex scenario, identify the underlying mathematics, and communicate your solution clearly—exactly the skills assessed in CCEA examinations.
KS3 阶段的进阶数学通过将熟悉的概念应用于陌生情境来提升解决问题的能力。本文提供一系列真实世界的案例研究，旨在锻炼你的分析思维、逻辑推理和数学流畅度。每个案例都鼓励你拆解复杂场景，识别背后的数学原理，并清晰地展示解答过程——这些正是 CCEA 考试所考查的核心技能。
1. Solving Symbol Puzzles with Algebraic Expressions | 用代数表达式破解符号谜题
Case: A secret code uses shapes to represent numbers. Three triangles and two circles total 29, while one triangle and four circles total 23. Determine the value of each shape. This classic puzzle tests your ability to translate word conditions into algebraic equations, forming a system of two unknowns.
案例：一套密码用形状表示数字。三个三角形加两个圆的和是 29，一个三角形加四个圆的和是 23。判定每个形状所代表的数值。这个经典谜题考查你将文字条件转化为代数方程、建立二元方程组的能力。
Let the value of a triangle be t and a circle be c. We write the equations:
设三角形数值为 t，圆为 c。我们写出方程：
3t + 2c = 29
1t + 4c = 23
3t + 12c = 69. Subtract the first equation from this new equation to eliminate t: (3t+12c) – (3t+2c) = 69 – + 4×4 = 23 → t + 16 = 23 → t = 7. The triangle represents 7, and the circle represents 4.
将第二个方程乘以 3 使 t 的系数对齐：3t + 12c = 69。用该新方程减去第一个方程以消去 t：(3t+12c) – (3t+2c) = 69 – 29 → 10c = 40，得 c = 4。将 c = 4 代入 t + 4×4 = 23 → t + 16 = 23 → t = 7。三角形代表 7，圆代表 4。
2. Comparing Mobile Phone Plans with Linear Equations | 用线性方程比较手机话费套餐
Case: Two network providers offer monthly contracts. Plan A charges a £10 fixed fee plus 5p per minute of calls. Plan B has no fixed fee but charges 15p per minute. Determine the number of call minutes for which the two plans cost the same, and advise which plan is cheaper for a user who talks 200 minutes per month. This scenario models real-life cost analysis using linear functions.
案例：两家运营商提供月套餐。套餐 A 收 10 英镑固定费加每分钟 5 便士通话费。套餐 B 无固定费但每分钟收 15 便士。求通话多少分钟时两套餐费用相同，并为每月通话 200 分钟的用户推荐更便宜的方案。该场景用一次函数模拟现实费用分析。
Let the number of minutes be m. Cost for Plan A: 1000 + 5m pence (since £ al: 1000 + 5m = 15m. Solve: 1000 = 10m → m = 100 minutes. At 100 minutes both cost 1500p = £15. For m = 200, Plan A costs 1000 + 5×200 = 2000p = £20, Plan B costs 15×200 = 3000p = £30. Thus, Plan A is cheaper for 200 minutes.
设通话时长为 m 分钟。套餐 A 费用：1000 + 5m 便士（£10 = 1000p）。套餐 B 费用：15m 便士。令两者相等：1000 + 5m = 15m。求解：1000 = 10m → m = 100 分钟。在 100 分钟时两者均为 1500p = £15。当 m = 200，A 花费 1000 + 5×200 = 2000p = £20，B 花费 15×200 = 3000p = £30。因此，通话 200 分钟时 A 套餐更便宜。
Representing both cost lines on a graph would show Plan A starting higher but with a shallower gradient. The intersection point (100, 1500) is the break-even point. For any usage beyond 100 minutes, Plan A remains the better value.
在坐标图中绘制两条费用直线，会看到 A 的起点更高但斜率更小。交点 (100, 1500) 就是盈亏平衡点。任何超过 100 分钟的通话量，A 方案都是更划算的选择。
3. Ratio and Proportion in a Fruit Punch Recipe | 果汁配方中的比与比例
Case: A fruit punch recipe mixes orange juice, pineapple juice, and lemonade in the ratio 3 : 2 : 5 by volume. You need to prepare 4.5 litres of punch for a party. Calculate the required amount of each ingredient. Additionally, you only have 600 ml of pineapple juice; what is the maximum punch you can make, keeping the ratio? This reinforces the concept of unitary method and scaling ratios.
案例：一款果汁配方按体积将橙汁、菠萝汁和柠檬水以 3 : 2 : 5 的比例混合。你需要为派对准备 4.5 升果汁。计算每种配料的需求量。此外，你只有 600 毫升菠萝汁；在保持比例的前提下，最多能制作多少果汁？这强化了单位法和比例缩放的概念。
2 × 0.45 = 0.90 L; Lemonade: 5 × 0.45 = 2.25 L. The check sum: 1.35 + 0.90 + 2.25 = 4.5 L.
总份数 = 3 + 2 + 5 = 10。对于 4.5 升，每份为 4.5 ÷ 10 = 0.45 升。橙汁：3 × 0.45 = 1.35 升；菠萝汁：2 × 0.45 = 0.90 升；柠檬水：5 × 0.45 = 2.25 升。验算：1.35 + 0.90 + 2.25 = 4.5 升。
eded: 3 × 300 = 900 ml; lemonade: 5 × 300 = 1500 ml. The limiting ingredient restricts the total volume.
菠萝汁仅有 600 毫升，代表 2 份，因此每份 = 600 ml ÷ 2 = 300 ml。最大果汁总量 = 10 份 × 300 ml = 3000 ml = 3 升。所需橙汁：3 × 300 = 900 ml；柠檬水：5 × 300 = 1500 ml。限制性配料决定了最终的总量。
4. Geometry: Optimising Pen Perimeter with Fixed Area | 几何：固定面积下优化围栏周长
Case: A farmer wants to build a rectangular pen against an existing barn wall, so only three sides need fencing. The pen must have an area of 50 m². Explore the dimensions that minimise the total length of fencing required, testing from width 4 m to 10 m. This case integrates area, perimeter expressions, and simple optimisation without calculus.
案例：一位农场主想借助已有的谷仓墙修建一个矩形围栏，因此只需围三边。围栏面积必须为 50 m²。探究在宽度从 4 米到 10 米的范围内，使得所需围栏总长最小的尺寸。这个案例结合了面积、周长表达式和无须微积分的简单优化。
Let the width perpendicular to the barn be w metres, and the length parallel to the barn be l rea: w × l = 50, so l = 50 ÷ w. Fencing needed for three sides: F = w + w + l = 2w + 50/w.
设垂直于谷仓的宽度为 w 米，平行于谷仓的长度为 l 米。面积：w × l = 50，得 l = 50 ÷ w。三边所需围栏：F = w + w + l = 2w + 50/w。
Now evaluate F for integer widths from 4 to 10: - w = 4 m: l = 12.5 m, F = 2×4 + 12.5 = 20.5 m - w = 5 m: l = 10 m, F = 2×5 + 10 = 20 m - w = 6 m: l ≈ 8.33 m, F = 2×6 + 8.33 = 20.33 m - w = 7 m: l ≈ 7.14 m, F = 2×7 + 7.14 = 21.14 m - w = 8 m: l = 6.25 m, F = 2×8 + 6.25 = 22.25 m - w = 10 m: l = 5 m, F = 2×10 + 5 = 25 m. s when the overall shape is as close to a square as possible considering the missing side.
现在计算 w 从 4 到 10（整数）时的 F 值： - w = 4 m: l = 12.5 m, F = 2×4 + 12.5 = 20.5 m - w = 5 m: l = 10 m, F = 2×5 + 10 = 20 m - w = 6 m: l ≈ 8.33 m, F = 2×6 + 8.33 = 20.33 m - w = 7 m: l ≈ 7.14 m, F = 2×7 + 7.14 = 21.14 m - w = 8 m: l = 6.25 m, F = 2×8 + 6.25 = 22.25 m - w = 10 m: l = 5 m, F = 2×10 + 5 = 25 m。 最小围栏长度出现在 w = 5 m 时，为 20 m。当形状考虑缺失边时尽可能接近正方形时达到最优。
5. Data Handling: Analysing Two Bookstore Sales | 数据处理：分析两家书店的销售数据
Case: Two bookstores record their daily sales of a popular novel over 10 days. Store X: 8, 9, 7, 10, 12, 6, 9, 11, 8, 10. Store Y: 5, 15, 7, 8, 14, 6, 9, 10, 7, 13. Compare their performance using mean, median, mode, and range. Which store would you recommend as more consistent? This case applies standard statistical measures to inform business decisions.
案例：两家书店记录了一本畅销小说连续 10 天的日销量。书店 X：8, 9, 7, 10, 12, 6, 9, 11, 8, 10。书店 Y：5, 15, 7, 8, 14, 6, 9, 10, 7, 13。用平均数、中位数、众数和极差比较它们的表现。你会推荐哪家书店更稳定？这个案例用标准统计量帮助商业决策。
0+10+11+12)/10 = 90/10 = 9. Median = (9+9)/2 = 9. Mode = 8, 9, 10 (multimodal). Range = 12 – 6 = 6.
书店 X：排序数据：6,7,8,8,9,9,10,10,11,12。平均数 = (6+7+8+8+9+9+10+10+11+12)/10 = 90/10 = 9。中位数 = (9+9)/2 = 9。众数 = 8, 9, 10（多众数）。极差 = 12 – 6 = 6。
= (5+6+7+7+8+9+10+13+14+15)/10 = 94/10 = 9.4. Median = (8+9)/2 = 8.5. Mode = 7. Range = 15 – e X shows a tighter cluster around the median and a smaller range, so it is more reliable.
书店 Y：排序数据：5,6,7,7,8,9,10,13,14,15。平均数 = (5+6+7+7+8+9+10+13+14+15)/10 = 94/10 = 9.4。中位数 = (8+9)/2 = 8.5。众数 = 7。极差 = 15 – 5 = 10。尽管书店 Y 的平均数略高，但它的极差大得多，显示日销量不稳定。书店 X 的数据更紧密地聚集在中位数附近且极差较小，因此更可靠。
6. Probability: Fairness of a Spinner Game | 概率：转盘游戏的公平性
Case: A game uses a spinner divided into five equal sectors: two red, one blue, one green, one yellow. Players pay £1 to spin. If the spinner lands on red, they win £0.50; on blue, win £1.50; on green, win £2; on yellow, win £1. Is the game profitable for the organiser over many plays? Determine the expected win per game and assess fairness.
案例：一个游戏使用分为五个相等扇区的转盘：两红、一蓝、一绿、一黄。玩家支付 £1 旋转一次。若转到红色，赢得 £0.50；蓝色 £1.50；绿色 £2；黄色 £1。从长远来看，该游戏对组织者是否有利可图？计算每局期望赢利并评估公平性。
(1/5)×2 + (1/5)×1 = (1/5)(2×0.50 + 1.50 + 2 + 1) = (1/5)(1 + 1.50 + 2 + 1) = (1/5)×5.50 = £1.10. Since the player pays £1, the expected profit for the organiser = £1 – £1.10 = –£ make the game fair or profitable, the prize values or the entry fee would need adjustment.
概率：P(红) = 2/5, P(蓝) = 1/5, P(绿) = 1/5, P(黄) = 1/5。玩家的期望支出 (1/5)×2 + (1/5)×1 = (1/5)(2×0.50 + 1.50 + 2 + 1) = (1/5)(1 + 1.50 + 2 + 1) = (1/5)×5.50 = £1.10。由于玩家支付 £1，组织者的期望利润 = £1 – £1.10 = –£0.10。游戏对玩家有利；组织者平均每局亏损 10 便士。要使游戏公平或盈利，需调整奖金金额或入场费。
Alternatively, setting the entry fee to £1.10 would create a fair game. This case shows how probability and expectation directly influence decision-making in games of chance.
或者，将入场费设为 £1.10 可使游戏公平。这个案例展示了概率与期望值如何直接影响随机游戏中的决策。
7. Sequences and Patterns: Triangular Numbers in Tournament Scheduling | 数列与模式：锦标赛赛程中的三角形数
Case: In a round-robin tournament where each team plays every other team exactly once, find the total number of matches for n teams. Use a pattern for small numbers of teams to deduce a formula, then calculate the matches for 12 teams. This introduces triangular numbers and combinatorial reasoning.
案例：在一场单循环锦标赛中，每支队伍与其他各队恰好比赛一次。求 n 支队伍的总比赛场数。用小规模队伍的模式推导公式，然后计算 12 支队伍的比赛数。这引入了三角形数和组合推理。
ch. 3 teams: 3 matches. 4 teams: 6 matches. 5 teams: 10 matches. The sequence 1, 3, 6, 10, … are triangular numbers: Tₙ₋₁ = (n–1)n/2. Reasoning: each of the n teams plays (n–1) games, but that counts each match twice, so total = n(n–1)/2.
2 支队伍：1 场。3 支队伍：3 场。4 支队伍：6 场。5 支队伍：10 场。数列 1, 3, 6, 10, … 是三角形数：Tₙ₋₁ = (n–1)n/2。推理：每支队伍赛 (n–1) 场，但每场比赛被计两次，因此总数 = n(n–1)/2。
. This formula can also be understood as choosing 2 teams from n: C(n,2). The triangular numbers appear naturally in many contexts, from handshakes at a party to the arrangement of bowling pins.
当 n = 12，总场数 = 12 × 11 ÷ 2 = 66。该公式也可理解为从 n 中选 2 的组合数 C(n,2)。三角形数在许多情境中自然出现，从派对上的握手次数到保龄球瓶的排列。
To check the pattern: adding teams one by one, the number of new matches increases by 1 each time (team 3 adds 2 matches, team 4 adds 3, etc.). This recursive pattern reinforces the triangular growth.
验证规律：逐一增加队伍，新增比赛场数每次增加 1（第 3 队增加 2 场，第 4 队增加 3 场，以此类推）。这种递推规律强化了三角形增长。
8. Co-ordinate Geometry: Distance and Midpoint in Map Reading | 坐标几何：地图阅读中的距离与中点
Case: On a grid map, a treasure is located at point A(2, 3) and a clue at B(8, 11). Find the straight-line distance between them, the midpoint for burying a second clue, and the equation of the line joining A and B. A unit represents 1 km. This exercises Pythagoras, midpoint, and linear equation skills.
案例：在网格地图上，宝藏位于点 A(2, 3)，线索在 B(8, 11)。求两者间的直线距离、埋设第二条线索的中点坐标以及连接 A 和 B 的直线方程。一个单位代表 1 公里。这训练了毕达哥拉斯定理、中点和线性方程技能。
Distance AB: difference in x = 8 – 2 = 6, difference in y = 11 – 3 = 8. Using Pythagoras: distance = √6² + 8² = √36+64 = √100 = 10 km.
距离 AB：x 坐标差 = 8 – 2 = 6，y 坐标差 = 11 – 3 = 8。用毕达哥拉斯定理：距离 = √6² + 8² = √36+64 = √100 = 10 km。
Midpoint M: x-coordinate = (2+8)/2 = 5, y-coordinate = (3+11)/2 = 7. So M(5, 7).
中点 M：x 坐标 = (2+8)/2 = 5，y 坐标 = (3+11)/2 = 7。即 M(5, 7)。
Gradient m = (11 – 3)/(8 – 2) = 8/6 = 4/3. Using point A(2,3), equation: y – 3 = (4/3)(x – 2). Multiply out: y = (4/3)x – 8/3 + 3 = (4/3)x + 1/3. In standard integer form: 3y = 4x + 1 or 4x – 3y + 1 = 0.
斜率 m = (11 – 3)/(8 – 2) = 8/6 = 4/3。利用点 A(2,3)，方程：y – 3 = (4/3)(x – 2)。展开得 y = (4/3)x – 8/3 + 3 = (4/3)x + 1/3。整数标准形式：3y = 4x + 1 或 4x – 3y + 1 = 0。
These calculations are essential for scaling distances on maps, finding meeting points, or setting up coordinate-based puzzles.
这些计算对于在地图上缩放距离、寻找汇合点或设置基于坐标的谜题至关重要。
9. Logical Reasoning: The Island of Knights and Knaves | 逻辑推理：骑士与无赖之岛
Case: On a fictional island, inhabitants are either knights (always tell the truth) or knaves (always lie). You meet two residents, Alex and Blake. Alex says, 'At least one of us is a knave.' Blake says, 'Alex is a knight.' Determine who is what. This puzzle cultivates careful logical deduction and the use of truth tables.
案例：在一个虚构的岛上，居民要么是骑士（总说真话），要么是无赖（总说谎）。你遇到两个居民 Alex 和 Blake。Alex 说：“我们中至少有一个是无赖。” Blake 说：“Alex 是骑士。” 判断各自的身份。这个谜题培养严谨的逻辑推理和真值表的使用。
Assume Alex is a knight. Then his statement must be true: at least one is a knave. Since Alex is a knight, Blake must be a knave. Blake, as a knave, would lie, so his statement 'Alex is a knight' must be false—but if Alex is a knight, that statement is actually true, which contradicts Blake being a knave. So the assumption that Alex is a knight leads to contradiction. Therefore Alex must be a knave.
假设 Alex 是骑士。那么他的话必为真：至少有一个无赖。由于 Alex 是骑士，Blake 必须是无赖。Blake 作为无赖会说谎，因此他的话“Alex 是骑士”为假——但若 Alex 是骑士，那句话实际上是真，这与 Blake 是无赖矛盾。因此 Alex 是骑士的假设导致矛盾。所以 Alex 必定是无赖。
If Alex is a knave, his statement 'at least one is a knave' is false. But if it's false, then it means neither is a knave, i.e., both are knights. This creates another contradiction because Alex is a knave. So we need to carefully re-evaluate: A knave always lies. Therefore, the statement Alex makes, 'at least one of us is a knave,' must be false. The negation of 'at least one is a knave' is 'neither is a knave', meaning both are knights. But Alex himself is a knave in this scenario, so it's impossible for both to be knights. Thus, the only consistent solution from our earlier contradiction is that our initial assumption in the first case was wrong? Wait, we must systematically check.
若 Alex 是无赖，他的话“至少有一个无赖”为假。如果它为假，则意味着“没有无赖”，即两人都是骑士。这又产生矛盾，因为 Alex 正是无赖。因此我们需要仔细重新评估：无赖总是说谎，因此 Alex 所说“至少有一个无赖”必为假。该陈述的否定是“两个都不是无赖”，即两人都是骑士。但在该情境下 Alex 本人就是无赖，所以不可能两人都是骑士。那么，唯一的解是我们在第一种情况下推导出的？其实我们应先做真值分析：设 A 为 Alex 的身份（Knight 为真，Knave 为假），B 为 Blake。Alex 的话逻辑等价于 (¬A ∨ ¬B)。若 A 为真，则 (¬A ∨ ¬B) 为真，即 (false ∨ ¬B) => ¬B 为真，所以 B 为假。Blake 说 A 为真，若 B 为假，他该说谎，所以“A 为真”必须为假，但 A 为真，矛盾。若 A 为假，他的话 (¬A ∨ ¬B) 必须为假，即 true ∨ ¬B = true，不可能为假，矛盾？不对，若 A 假，Alex 是无赖，他说谎，所以其陈述真实值应为假。但其陈述是“至少一个无赖”，即 ¬A ∨ ¬B。A 假 → ¬A 真，所以 ¬A ∨ ¬B 必然真，无论如何不能为假。因此 Alex 不可能是无赖，因为那样他的陈述会因 ¬A 真而必然真，但无赖不能说真话。此矛盾说明该情境无解？实际上常见版本是 Alex 说“我们俩都是无赖”或类似。但原题可能有误，让我们调整：Alex 说“我们俩中恰好有一个是骑士”，那样更经典。为了文章合理，我应使用经典逻辑谜题。修改案例：Alex 说：“我们俩是不同类型的。” Blake 说：“Alex 是骑士。”这样可以得出 Alex 是骑士，Blake 是无赖。我重新构造。
修正案例：Alex says, 'We are of different types.' Blake says, 'Alex is a knight.' 推断。这样：若 Alex 是骑士，他的话真，两人不同类型，Blake 是无赖。Blake 的话“Alex 是骑士”为真，但无赖不能说真话，矛盾。若 Alex 是无赖，他的话假，意味着两人相同类型，因此 Blake 也是无赖。Blake 是无赖，他的话“Alex 是骑士”为假，与 Alex 是无赖相符，无矛盾。由此 Alex 和 Blake 都是无赖。这样可以。但原句“至少一个是无赖”导致矛盾，还是修改为经典版本。我将在文章中直接使用经典谜题，调整表述。
因此，调整案例为：Alex 说：“我们类型不同。” Blake 说：“Alex 是骑士。” 这样推理：若 Alex 为骑士（真），则两人不同，Blake 为无赖。Blake 说的“Alex 是骑士”为真，但无赖不能说真话，矛盾。因此 Alex 必为无赖。若 Alex 为无赖（假），则他的话假，即两人类型相同，所以 Blake 也是无赖。Blake 为无赖，他的话“Alex 是骑士”为假，符合 Alex 是无赖。因此两人均为无赖。这个逻辑成立。我将用这个版本重写英文和中文字段。确保文章流畅。我在前面已经写了两个居民的部分，但现在我会融合进去，将最初不一致的描述改为一致。由于是长文，为保持准确，我将修改第9节内容。现在继续。
最终第9节：
9. Logical Reasoning: Truth-Tellers and Liars Puzzle | 逻辑推理：说真话与说谎者谜题
Case: On an island, each person is either a knight (always tells the truth) or a knave (always lies). Alex says, 'We are of different types.' Blake says, 'Alex is a knight.' Can you determine who is what? This classic puzzle builds skills in propositional logic and contradiction.
案例：在一个岛上，每个人要么是骑士（永远说真话），要么是无赖（永远说谎）。Alex
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