📚 Case Study Practice for Year 9 Edexcel Further Maths | Year 9 Edexcel 进阶数学：案例分析实战演练
Welcome to a collection of real-world case studies designed to strengthen your problem-solving skills in Year 9 Edexcel Further Mathematics. Each scenario invites you to model a situation mathematically, apply algebraic, geometric, statistical or probabilistic techniques, and interpret your findings in context. Working through these examples will help you see how advanced mathematical concepts come alive beyond the textbook.
欢迎来到一系列真实案例研究，旨在提升你在 Year 9 Edexcel 进阶数学中的问题解决能力。每个情景都要求你将实际情况进行数学建模，运用代数、几何、统计或概率技巧，并在具体情境中解读结果。逐一攻克这些案例，会让你体会到高阶数学概念如何在课本之外鲜活起来。
1. Optimising a Rectangular Enclosure | 优化矩形围场
A farmer has 60 metres of fencing and wishes to create a rectangular feeding area against a long straight barn wall. The wall acts as one side, so only three sides require fencing. The goal is to maximise the enclosed area. Let the side parallel to the wall be x metres and each perpendicular side be y he total fencing gives the constraint: x + 2y = 60. The area to be maximised is A = x × y.
一位农民拥有 60 米围栏，想靠着一条笔直的谷仓墙壁圈出一块矩形喂食区。墙壁充当其中一边，因此只需围三面。目标是让围起来的面积最大化。设平行于墙壁的边长为 x 米，每一条垂直边长为 y 米。围栏总长给出约束条件：x + 2y = 60。待最大化的面积为 A = x × y。
gative coefficient of x², so its graph is a downward-opening parabola. The maximum area occurs at the vertex. Using the axis of symmetry x = -b/(2a) with a = -½ and b = 30, we find x = 30.
将 y 用 x 表示：y = (60 − x)/2。代入面积表达式得到 A(x) = x·(60 − x)/2 = 30x − ½x²。这是一个二次函数，由于 x² 系数为负，其图像是一条开口向下的抛物线。最大面积出现在顶点处。利用对称轴公式 x = −b/(2a)，其中 a = −½、b = 30，求得 x = 30。
he wall and width 15 m. Completing the square also confirms the result: A = -½(x - 30)² + 450.
当 x = 30 米时，y = (60 − 30)/2 = 15 米。最大面积为 A = 30 × 15 = 450 平方米。农民应当建造一个与墙壁平行的边长为 30 米、宽度为 15 米的喂食区。通过配方法同样可验证此结果：A = −½(x − 30)² + 450。
2. Analysing Social Media Usage Data | 分析社交媒体使用数据
A school conducted a survey among 50 Year 9 students, asking how many hours they spend on social media per day. The raw data are summarised in the frequency table below.
某校对 50 名 Year 9 学生进行了一项调查，询问他们每天在社交媒体上花费的小时数。原始数据汇总于下面的频率表中。
| Hours (h) | Frequency (f) |
|---|---|
| 0 ≤ h < 1 | 6 |
| 1 ≤ h < 2 | 14 |
| 2 ≤ h < 3 | 18 |
| 3 ≤ h < 4 | 8 |
| 4 ≤ h < 5 | 4 |
+ 14×1.5 + 18×2.5 + 8×3.5 + 4×4.5 = 3 + 21 + 45 + 28 + 18 = 115. The estimated mean is 115 ÷ 50 = 2.3 hours per day.
我们可以通过组中值来估算平均使用时长。设 xₘ 代表每个区间的中点：0.5、1.5、2.5、3.5、4.5。计算 Σ(f × xₘ) = 6×0.5 + 14×1.5 + 18×2.5 + 8×3.5 + 4×4.5 = 3 + 21 + 45 + 28 + 18 = 115。估算的平均值为 115 ÷ 50 = 2.3 小时每天。
25th value) is 2 ≤ h < 3. The lower quartile (12.5th) lies in 1 ≤ h < 2; the upper quartile (37.5th) lies in 2 ≤ h < 3. Using linear interpolation, Q₁ ≈ 1.46 h, Q₃ ≈ 2.81 h, so IQR ≈ 1.35 h. A box plot can be drawn to visualise the distribution.
为描述离散程度，我们计算四分位距。数据已经分组；累计频数依次为：6、20、38、46、50。中位数组（第 25 个值）落在 2 ≤ h < 3 区间。下四分位数（第 12.5 个）落在 1 ≤ h < 2；上四分位数（第 37.5 个）落在 2 ≤ h < 3。利用线性插值，Q₁ ≈ 1.46 小时，Q₃ ≈ 2.81 小时，因此四分位距 IQR ≈ 1.35 小时。可以绘制箱形图以直观展示分布情况。
3. Saving Plan with Compound Interest | 复利储蓄计划
Mia deposits £2,000 into a savings account that pays 3% nce formula A = 2000 × (1.03)ⁿ. She wants to know how long it will take for her money to double.
米娅将 2000 英镑存入一个年利率为 3% 的复利储蓄账户，利息在每年年末计入。她不再追加存款。第 n 年末的金额由等比数列公式给出：A = 2000 × (1.03)ⁿ。她想知道自己的钱何时可以翻倍。
t or logarithms. Trying n = 23: (1.03)²³ ≈ 1.9736; n = 24: (1.03)²⁴ ≈ 2.0328. Therefore, after 24 years the amount exceeds £4,000 for the first time.
建立不等式 2000 × (1.03)ⁿ ≥ 4000，化简得 (1.03)ⁿ ≥ 2。由于这是一个指数不等式，我们可以使用试错法或对数求解。尝试 n = 23：(1.03)²³ ≈ 1.9736；n = 24：(1.03)²⁴ ≈ 2.0328。因此，24 年后金额将首次超过 4000 英镑。
lly, we round up to 24 years. This case study highlights the long-term effect of compound growth, even at modest rates.
为进行更精确的建模，可以取自然对数：n ln(1.03) ≥ ln 2，于是 n ≥ ln 2 / ln(1.03) ≈ 0.6931 / 0.02956 ≈ 23.45。由于利息按年支付，我们向上取整为 24 年。这个案例凸显了复利增长的长期效应，即便利率不高也是如此。
4. Measuring Height Using Similar Triangles | 利用相似三角形测量高度
On a sunny day, a group of students wants to determine the height of a tall tree without climbing it. They place a 1.5 m vertical stick in the ground and measure its shadow length as 2.4 m. At the same time, the tree’s shadow measures 18 m. Assuming the sun’s rays are parallel, the stick and its shadow form a triangle similar to that formed by the tree and its shadow.
在一个晴朗的日子里，一群学生想在不攀爬的情况下测定一棵大树的高度。他们把一根 1.5 米长的垂直杆子插在地上，测得杆子的影子长度为 2.4 米。在同一时刻，大树的影子长度为 18 米。假设太阳光线是平行的，那么杆子和其影子构成的三角形与大树和其影子构成的三角形相似。
s-multiplication: h = 18 × (1.5 / 2.4) = 18 × 0.625 = 11.25. So the tree is approximately 11.25 metres tall.
设树的高度为 h 米。由于三角形相似，对应边的比例相等：h / 18 = 1.5 / 2.4。通过交叉相乘求解 h：h = 18 × (1.5 / 2.4) = 18 × 0.625 = 11.25。因此，这棵树大约有 11.25 米高。
The students verify the method by measuring a second object of known height and checking the shadow ratio. This practical application of similar triangles is used in surveying and architecture. It reinforces the proportional reasoning required for solving geometric problems in Further Maths.
学生们通过测量另一个已知高度的物体并验证影子比例来确认这一方法的可靠性。这种相似三角形的实际应用被广泛用于测量学和建筑学中。它巩固了在进阶数学中解决几何问题所需的比例推理能力。
5. Designing a Fair Carnival Game | 设计公平的嘉年华游戏
A charity stall wants to create a simple game: pay £1 to draw one card from a standard 52-card deck. If the card is a heart, the player wins £2; if it is a diamond, they win £1; otherwise, they win nothing. Organisers need to check whether the game is fair, meaning the expected value for the player is zero (after paying the entry fee).
一个慈善摊位想设计一个简单游戏：支付 1 英镑从一副标准的 52 张扑克牌中抽一张。如果抽到红桃，玩家赢得 2 英镑；如果是方块，赢得 1 英镑；否则没有奖金。组织者需要检验游戏是否公平，也就是剔除入场费后玩家的期望值为零。
f heart = 13/52 = 1/4; diamond = 13/52 = 1/4; others = 26/52 = 1/2. Expected gross win = (£2 × 1/4) + (£1 × 1/4) + (£0 × 1/2) = £0.50 + £0.25 = £0.75. After paying £1 to play, the player’s net expectation is £0.75 − £1 = −£0.25. The game favours the stall.
首先计算抽牌的期望奖金。红桃概率 = 13/52 = 1/4；方块 = 13/52 = 1/4；其他 = 26/52 = 1/2。期望毛收益 = (2 英镑 × 1/4) + (1 英镑 × 1/4) + (0 英镑 × 1/2) = 0.50 英镑 + 0.25 英镑 = 0.75 英镑。支付 1 英镑后，玩家的净期望值为 0.75 英镑 − 1 英镑 = −0.25 英镑。游戏对摊位有利。
adjust the payouts or the entry fee. For instance, if the prize for heart is increased to £3, expected gross = 0.75 + 0.25 = £ entry fee to 75p. This case encourages the use of probability to evaluate real-life risks.
为使游戏公平，组织者可以调整奖金或入场费。例如，若将红桃奖金提高到 3 英镑，期望毛收益 = 0.75 + 0.25 = 1.00 英镑，从而实现公平游戏。另一种选择是将入场费降至 75 便士。这一案例鼓励运用概率来评估现实生活中的风险。
6. Consecutive Integer Puzzle | 连续整数谜题
A classic number puzzle states: 'The difference between the squares of two consecutive odd numbers is 48. Find the numbers.' t consecutive odd is n + 2. The condition translates into the equation (n + 2)² − n² = 48.
一个经典的数字谜题是：“两个连续奇数的平方差为 48。求这两个数。”设较小的奇数为 n（整数），则下一个连续的奇数为 n + 2。条件转化为方程：(n + 2)² − n² = 48。
3. Check: 13² − 11² = 169 − 121 = 48. The solution is unique within the positive integers.
展开等式左边：(n² + 4n + 4) − n² = 4n + 4。于是 4n + 4 = 48。解得 4n = 44，因此 n = 11。这两个数为 11 和 13。检验：13² − 11² = 169 − 121 = 48。在正整数范围内，解是唯一的。
What if the difference were, say, 72? 7, giving 17 and 19. This algebraic approach generalises the puzzle and shows that for any multiple of 4, you can find a pair of consecutive odds. It reinforces expanding brackets and solving linear equations.
假如平方差是 72 呢？方程变为 4n + 4 = 72，因此 n = 17，得到 17 和 19 这一对。这种代数方法推广了这个谜题，并表明对于任何 4 的倍数，都可以找到一对连续奇数。这加强了展开括号和求解线性方程的能力。
7. Interpreting a Distance-Time Graph | 解读距离时间图
A cyclist sets off from home, rides for 2 hours covering 30 km, rests for 30 minutes, then rides back home at a constant speed, arriving 1.5 hours after the rest ends. The journey is represented on a distance-time graph. The key task is to sketch the graph and calculate the average speed for the whole trip.
一名自行车手从家出发，骑行 2 小时，行驶 30 公里，然后休息 30 分钟，接着以恒定速度骑行回家，在休息结束后 1.5 小时到家。整个行程用距离—时间图表示。核心任务是绘制图形并计算全程平均速度。
zontal axis, distance from home (km) on the vertical axis. Segment 1: from (0,0) to (2,30) – straight line with gradient 15 km/h. Segment 2: horizontal line from t=2 to t=2.5 at 30 km – rest. Segment 3: from (2.5, 30) to (4, 0) – descent of 30 km in 1.5 h, giving speed 30/1.5 = 20 km/h. Gradient is negative.
在图中，横轴为时间（小时），纵轴为离家距离（公里）。第一段：从 (0,0) 到 (2,30)——斜率为 15 公里/小时的直线。第二段：从 t=2 到 t=2.5 在 30 公里处的水平线——休息。第三段：从 (2.5, 30) 到 (4, 0)——在 1.5 小时内下降 30 公里，得出速度为 30/1.5 = 20 公里/小时。斜率为负。
otal time = 60 / 4 = 15 km/h. Notice this is slower than the outward speed because of the rest period. Graph interpretation is vital for understanding motion.
总路程 = 去程 30 公里 + 回程 30 公里 = 60 公里。全程总时间 = 4 小时。因此，全程平均速度 = 总路程 / 总时间 = 60 / 4 = 15 公里/小时。请注意，由于休息时间，平均速度低于去程速度。图形解读对于理解运动至关重要。
8. Choosing the Best Mobile Phone Plan | 选择最佳手机套餐
Rohan is comparing two monthly mobile plans. Plan A has a fixed charge of £12 plus £0.03 per minute of calls. Plan B has no fixed charge but charges £0.08 per minute. He wants to find the number of minutes for which the two plans cost the same, and recommend the cheaper plan under different usage levels.
罗翰正在比较两个手机月租套餐。套餐 A 有 12 英镑的固定费用，外加每分钟 0.03 英镑的通话费。套餐 B 没有固定费用，但每分钟收费 0.08 英镑。他想找出使两个套餐花费相同的通话分钟数，并针对不同使用水平推荐更便宜的套餐。
represent the number of minutes used. Cost for Plan A: Cₐ = 12 + 0.03t. Cost for Plan B: CB = 0.05t, hence t = 12 / 0.05 = 240 minutes. The break-even point is 240 minutes per month.
设 t 表示使用的分钟数。套餐 A 的费用为：Cₐ = 12 + 0.03t。套餐 B 的费用为：CB = 0.08t。令两者相等：12 + 0.03t = 0.08t。两边同时减去 0.03t：12 = 0.05t，因此 t = 12 / 0.05 = 240 分钟。盈亏平衡点为每月 240 分钟。
ion can also be visualised by plotting both linear functions: the intersection point corresponds to t = 240. This is a clear application of simultaneous equations.
如果罗翰通话少于 240 分钟，套餐 B 更便宜，因为 0.08t < 12 + 0.03t。超过 240 分钟时，套餐 A 更经济。这一解法也可以通过绘制两个线性函数来直观展示：交点对应 t = 240。这是联立方程的一个清晰应用。
9. Fibonacci in Nature: A Modelling Task | 自然界中的斐波那契数列：建模任务
The Fibonacci sequence appears in the arrangement of leaves around a stem, known as phyllotaxis. The sequence begins 1, 1, 2, 3, 5, 8, 13, 21, … where each term is the sum of the two preceding terms. A student collection shows that the number of spirals on a pinecone often matches consecutive Fibonacci numbers. Your task is to explore the ratio of successive terms.
斐波那契数列出现在叶子绕茎的排列中，被称为叶序。该数列以 1, 1, 2, 3, 5, 8, 13, 21, … 开始，每一项都是前两项之和。一项学生采集活动显示，松果上的螺旋数目常常与连续的斐波那契数吻合。你的任务是探索连续项之间的比值。
Published by TutorHao | Year 9 进阶数学 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图