📚 Year 8 CIE Further Maths: Case Study Practical Workout | 八年级CIE进阶数学：案例分析实战演练
Advanced mathematics thrives on real‑world problem‑solving. This article presents a series of case studies designed to sharpen analytical thinking, algebraic manipulation, geometric reasoning, and data interpretation exactly as expected in a Year 8 CIE Further Maths classroom. Each case is built around a practical scenario, followed by guided questions that mirror exam‑style challenges. Working through these examples will help you connect abstract concepts to tangible situations.
进阶数学的精髓在于解决真实世界的问题。本文通过一系列案例分析，锻炼你的分析思维、代数运算、几何推理和数据解读能力，完全贴合八年级CIE进阶数学课堂的要求。每个案例都围绕一个实际场景展开，并配有模拟考试风格的引导性问题。通过这些练习，你将学会把抽象概念与具体情境联系起来。
1. Case 1: Budgeting a School Trip | 案例一：学校旅行预算
A school plans a trip to a science museum. The entry fee is £6 per student for the first 40 students, and £4.50 per student for any additional student. The coach costs £180 regardless of the number of passengers, and each teacher supervising is free but there must be at least one teacher for every 15 students. The school aims to keep the total cost per student below £10. Let n be the number of students going on the trip.
一所学校计划组织学生参观科学博物馆。前40名学生的门票为每人6英镑，超出40人的部分每人4.50英镑。巴士租赁费固定为180英镑，不论乘客人数多少；带队老师免费，但必须保证每15名学生至少配一位老师。学校希望将每名学生的总费用控制在10英镑以下。设参加旅行的学生人数为n。
- r the total cost of entry tickets in terms of n, distinguishing between n ≤ 40 and n > 40.
- 用n表示门票总费用的表达式，需区分n ≤ 40 和 n > 40 两种情况。
- Form an inequality that represents the total cost per student being less than £10, including coach hire and tickets.
- 列出不等式，表示每名学生总费用（包含车费和门票）低于10英镑。
- Solve the inequality for n > 40 and determine the minimum number of students required to satisfy the budget condition.
- 在n > 40的情况下求解不等式，并确定满足预算条件所需的最小学生人数。
ich contradicts n ≤ 40. Thus no solution for n ≤ 40. For n > 40, ticket cost = 6×40 + 4.5(n–40) = 240 + 4.5n – 40, so n ≥ 44. Minimum 44 students needed. Also check teacher requirements: for 44 students, teachers needed = ceil(44/15) = 3, they travel free.
解题提示：当n ≤ 40时，总门票费 = 6n，总费用 = 6n + 180，人均费用 (6n+180)/n < 10 化简得6 + 180/n < 10，推出n > 45，与n ≤ 40矛盾，故无解。当n > 40时，门票费 = 6×40 + 4.5(n–40) = 4.5n + 60，总费用 = 4.5n + 240，人均 (4.5n+240)/n < 10 → 4.5 + 240/n < 10 → 240/n < 5.5 → n > 240/5.5 ≈ 43.636，所以n最小取44。同时检查师生比：44名学生需3位教师免费随行。
2. Case 2: Optimising a Rectangular Enclosure | 案例二：优化矩形围场面积
A farmer has 100 metres of fencing and wants to create a rectangular enclosure against an existing long barn wall, so only three sides need fencing. The width perpendicular to the wall is x metres. The length parallel to the wall is y metres.
一位农民有100米长的围栏，想沿着已有的长谷仓墙壁围一个矩形的场地，因此只需围三面。垂直于墙的宽度为x米，平行于墙的长度为y米。
- Express y in terms of x using the fact that the total length of fencing used is 100 m.
- 利用围栏总长为100米这一条件，用x表示y。
- Show that the area A of the enclosure can be expressed as A = 100x – 2x².
- 证明场地的面积A可以表示为 A = 100x – 2x²。
- Find the value of x that gives the maximum area, and calculate that maximum area.
- 求使得面积最大的x值，并计算该最大面积。
- Discuss why the domain of possible x is constrained by practical considerations, and state the realistic domain.
- 讨论实际条件下x的取值范围受限的原因，并给出符合现实的定义域。
Working: With one side (length) against the barn, fencing used = 2x + y = 100 → y = 100 – 2x. Area A = x × y = x(100 – 2x) = 100x – 2x². This is a quadratic with a maximum at x = –b/(2a) for A = –2x² + 100x. Here a = –2, b = 100, so x = –100/(2×(–2)) = 100/4 = 25. Then y = 100 – such that the enclosure is practical; if x is too small, y becomes very long and wasteful.
解题过程：因为一边靠墙，围栏用量为2x + y = 100，所以y = 100 – 2x。面积A = x × y = x(100 – 2x) = 100x – 2x²。这是一个二次函数，最大值出现在对称轴 x = –b/(2a) 处，其中a = –2, b = 100，得x = 25。此时y = 100 – 50 = 50，最大面积 = 1250平方米。宽度x必须大于0，且为保证y > 0，需有x < 50。因此实际定义域为 0 < x < 50。若x过小，长度会过长浪费空间。
3. Case 3: Mobile Phone Tariff Comparison | 案例三：手机套餐对比
Two mobile phone companies offer monthly plans. Company A: £10 fixed charge plus 5p per minute of calls. Company B: £6 fixed charge plus 8p per minute. A student estimates she uses between 200 and 600 minutes per month. Determine which plan is cheaper for different usage levels, and find the exact number of minutes where the costs are equal.
两家移动通信公司推出月度套餐。A公司：固定月费10英镑，通话每分钟0.05英镑。B公司：固定月费6英镑，通话每分钟0.08英镑。一位学生估计每月通话时长在200到600分钟之间。针对不同使用量确定哪种套餐更便宜，并求出花费相等时的精确通话分钟数。
- Write equations for the monthly cost CA and CB as functions of minutes m.
- 写出月费用 CA 和 CB 关于通话分钟数 m 的表达式。
- Plot both cost functions on a graph (use Desmos or graph paper) and visually identify the region where each plan is cheaper.
- 绘制两个费用函数的图像（可使用 Desmos 或坐标纸），并直观判断每种套餐更便宜的区间。
- Solve CA = CB algebraically, and interpret the solution in the context of the student's usage range.
- 通过代数方法求解 CA = CB，并结合学生的使用时长范围解释该解的意义。
Algebra: CA = 10 + 0.05m, CB m = 400/3 ≈ 133.33 minutes. Since the student uses at least 200 minutes, at 200 minutes: CA = 20, CB = 22, so B is more expensive. For m > 133.33, CA < CB because the lower per‑minute rate of A outweighs the higher fixed charge after the break‑even point. Therefore for all usage between 200 and 600 minutes, Company A is cheaper. The break‑even point is outside the relevant range.
代数计算：CA = 10 + 0.05m，CB = 6 + 0.08m。令两者相等：10 + 0.05m = 6 + 0.08m → 4 = 0.03m → m = 400/3 ≈ 133.33 分钟。由于学生至少使用200分钟，当 m=200 时，CA = 20，CB = 22，B更贵。当 m > 133.33 时，CA < CB，因为A的每分钟费率更低，在盈亏平衡点后即使固定月费更高也总体更划算。因此在200至600分钟的范围内，A公司套餐始终更便宜。盈利平衡点不在关注范围内。
4. Case 4: Statistical Analysis of Heights | 案例四：身高数据分析
A Year 8 class measured their heights to the nearest centimetre. The data for 24 students (in cm) is: 148, 155, 162, 149, 157, 163, 150, 158, 165, 152, 159, 166, 153, 160, 167, 154, 161, 168, 155, 162, 170, 156, 164, 172. They want to construct a grouped frequency table, calculate the mean, and draw a histogram with unequal class widths to highlight the distribution.
某八年级班级测量了身高（精确到厘米）。24名学生的数据（单位：cm）如下：148, 155, 162, 149, 157, 163, 150, 158, 165, 152, 159, 166, 153, 160, 167, 154, 161, 168, 155, 162, 170, 156, 164, 172。他们希望制作分组频数表、计算均值，并绘制不等组距的直方图来展示分布。
- 145 ≤ h < 150, 150 ≤ h < 155, 155 ≤ h < 160, 160 ≤ h < 165, 165 ≤ h < 170, 170 ≤ h < 175.
- 将数据分组：145 ≤ h < 150, 150 ≤ h < 155, 155 ≤ h < 160, 160 ≤ h < 165, 165 ≤ h < 170, 170 ≤ h < 175。
- Calculate the frequency for each interval and estimate the mean using midpoints.
- 计算每组的频数，并用组中值估算平均值。
- Construct a frequency density table for a histogram where the first two intervals have width 5, and the remaining have width 5 as well (equal widths here, but discuss how to adjust if widths differ).
- 制作频数密度表，以便绘制直方图（本例组距相同均为5，但需讨论如果组距不同应如何调整）。
- Comment on the spread and identify any possible outliers.
- 分析数据分布情况，并指出可能的异常值。
Solution: Frequencies: 145–: 2 (148,149); 150–: 5 (150,152,153,154,155? careful: 155 belongs to 155–); Actually 155 is in 155≤h<160, so 150–: 150,152,153,154 → 4 values? Re‑ 155,155,156,157,158,159 / 160,161,162,162,163,164 / 165,166,167,168 / 170,172. Counts: 145–:2; 150–:4; 155–:6; 160–:6; 165–:4; 170– (295 + 610 + 945 + 975 + 670 + 345)/24 = 3840/24 = 160 cm. For equal widths, frequency density = frequency (since width=5, FD = freq/5, but can keep as freq for shape). Normal distribution‑like spread. Lowest 148, highest 172, range 24 cm. No extreme outliers.
解答：频数分布：145–: 2；150–: 4；155–: 6；160–: 6；165–: 4；170–: 2。组中值依次为147.5, 152.5, 157.5, 162.5, 167.5, 172.5。估算均值 = 3840/24 = 160 cm。由于所有组距均为5 cm，频数密度可用频率本身。分布近似对称，呈钟形。最低148 cm，最高172 cm，全距24 cm，无显著异常值。
5. Case 5: Exploring Number Patterns and Sequences | 案例五：数列与模式探索
Consider the sequence of triangular numbers: 1, 3, 6, 10, 15, ... which follows the rule Tn he sum of two consecutive triangular numbers and notices it always produces a perfect square. Prove this algebraically, and use the insight to find which two consecutive triangular numbers sum to 225.
考察三角形数序列：1, 3, 6, 10, 15, … 其通项公式为 Tn = n(n+1)/2。一位学生发现两个连续的三角形数之和总是一个完全平方数。请用代数方法证明这一性质，并利用该结论找出哪两个连续三角形数之和等于225。
- Write expressions for Tk and Tk+1.
- 写出 Tk 和 Tk+1 的表达式。
- Add them algebraically and simplify to a perfect square form.
- 将它们相加并化简，表示为完全平方形式。
- Hence solve (k+1)² = 225 to identify the required triangular numbers.
- 由此解方程 (k+1)² = 225 找出所求的两个三角形数。
Proof: Tk + Tk+1 deed, perfect square. If sum = 225, then (k+1)² = 225 → k+1 = 15 → k=14. The numbers are T14 and T15. T14 = 14×15/2 = 105, T15 = 15×16/2 = 120, sum 225.
证明：Tk + Tk+1 = k(k+1)/2 + (k+1)(k+2)/2 = (k+1)(2k+2)/2 = (k+1)²。完全正确。若和为225，则(k+1)² = 225，k+1 = 15（取正），k=14。因此T14 = 105，T15 = 120，和为225。
6. Case 6: Probability in Game Design | 案例六：游戏设计中的概率
A game involves rolling two fair six‑sided dice and adding their scores. Players win a prize if the sum is a prime number. The designer wants to calculate the experimental probability from a simulation and compare it with the theoretical value. She also wants to adjust the winning condition to make the probability approximately 0.5 for a more balanced game.
某游戏需要掷两颗均匀的六面骰子，并求点数之和。如果和为质数，玩家赢得奖品。设计者希望通过模拟计算实验概率，并与理论值进行比较。她还希望调整获胜条件，使概率接近0.5，使游戏更加平衡。
- t are prime (2,3,5,7,11). Verify that theoretical probability P(prime sum) = 15/36 = 5/12.
- 绘制6×6的点数和网格，计数所有质数和（2,3,5,7,11）的结果。验证理论概率 P(质数和) = 15/36 = 5/12。
- Simulate 50 rolls using random numbers or a spreadsheet, and calculate the relative frequency. Explain why the experimental probability might differ.
- 使用随机数或电子表格模拟50次掷骰，计算相对频率，并解释实验概率为何可能与理论值不同。
- Propose a new winning condition (e.g., sum is even, or sum less than 7) that gives a probability close to 1/2, and justify with combinatorial counting.
- 提出一个新的获胜条件（如和为偶数，或和小于7），使其概率接近1/2，并通过组合计数说明理由。
5. Probability 15/36 = 5/12 ≈ 0.4167. For balance, e.g., sum ≥ 7 gives 21/36 = 7/12 ≈ 0.583; sum ≤ 6 gives 15/36 again; sum is even: 18/36 = 0.5 exactly (since half of the sums are even). Even sum condition works: there are 18 outcomes with even sum. So the designer could change winning condition to 'sum is even' for fairness. Experimental probability may vary due to small sample size.
理论质数和统计：2（1种），3（2种），5（4种），7（6种），11（2种），共15种。理论概率=15/36=5/12≈0.4167。为达到平衡，可选择“和为偶数”，其包含18种结果，概率正好0.5。实验概率因样本较小而可能波动。模拟50次后计算频率，可观察到与理论值的偏差。
7. Case 7: 3D Visualisation and Surface Area | 案例七：三维可视化与表面积
A designer creates a gift box in the shape of a right regular hexagonal prism. The side length of the hexagonal base is 5 cm, and the height of the prism is 12 cm. She needs to calculate the amount of wrapping paper required, ignoring overlaps, and also the volume. This requires understanding of the area of a regular hexagon and the lateral surface area.
一位设计师制作一个正六棱柱形状的礼品盒。六边形底面边长为5 cm，棱柱高为12 cm。她需要计算所需包装纸的面积（忽略重叠部分）以及盒子的体积。这需要理解正六边形的面积和侧面积的计算。
- g it into 6 equilateral triangles of side 5 cm. Recall area of an equilateral triangle = (√3/4)×side². Compute exact area in surd form, then approximate to decimal.
- 通过将正六边形分割为6个边长为5 cm的等边三角形，计算一个底面的面积。等边三角形面积公式为 (√3/4)×边长²。先给出根号形式的结果，再计算近似值。
- s lateral rectangle that wraps around). The lateral surface is perimeter of base × height.
- 计算总表面积（两个底面加侧面展开的矩形）。侧面积等于底面周长乘以高。
- Compute the volume = base area × height.
- 计算体积 = 底面积 × 高。
- Discuss how the amount of paper changes if the prism is replaced by a cylinder of equal volume but different surface area, touching on optimisation.
- 讨论若用体积相等但表面积不同的圆柱体替代该棱柱，所需纸张如何变化，并简要涉及优化思想。
Area of one equilateral triangle = (√3/4)×25 = (25√3)/4 cm². Hexagon area = 6 × (25√3)/4 = (150√3)/4 = (75√3)/2 cm² ≈ (75×1.732)/2 ≈ 129.9/2? Wait: 75×1.732 = 129.9, divided by 2 = 64.95 cm². Better: 75√3/2 ≈ 75×1.73205/2 ≈ 64.95 cm². Two bases = 75√3 : perimeter = 6×5 = 30 cm, height 12 cm, so 30×12 = 360 cm². Total surface area = 360 + 75√3 ≈ 360 + 129.9 = 489.9 cm². Volume = base area × height = (75√3/2)×12 = 450√3 ≈ 450×1.732 = 779.4 cm³. A cylinder of same volume: V = πr²h = 450√3. Choose h=12, then πr² = 450√3/12 = 37.5√3, r = √37.5√3/π h. Compare values; prism may have larger surface area due to corners, illustrating shape efficiency.
一个等边三角形面积 = (25√3)/4 cm²，六边形底面积 = 6倍 = (75√3)/2 ≈ 64.95 cm²。两个底面总面积 = 75√3 ≈ 129.9 cm²。侧面积 = 周长30 cm × 高12 cm = 360 cm²。总表面积 = 360 + 75√3 ≈ 489.9 cm²。体积 = 底面积×高 = (75√3/2)×12 = 450√3 ≈ 779.4 cm³。体积相同的圆柱体若高同样为12 cm，底半径需满足 πr² = 37.5√3，经计算其表面积会小于棱柱，说明圆柱在节省材料上更优。
8. Case 8: Rational Numbers and Proportional Reasoning | 案例八：有理数与比例推理
A recipe for 8 scones requires 300 g of self‑raising flour, 75 g of butter, and 2 eggs. A baker wants to make 30 scones for a school fair, but she only has 11 eggs. She must adjust the quantities proportionally and determine whether she has enough eggs. This leads to a multi‑step proportional reasoning task involving fractions and ratios.
一份制作8个司康饼的食谱需要300克自发面粉、75克黄油和2个鸡蛋。一位烘焙师为学校义卖活动准备制作30个司康饼，但她只有11个鸡蛋。她必须按比例调整配料用量，并判断鸡蛋是否够用。这涉及多步骤的比例推理和分数计算。
- Write the standard ratio of flour : butter : eggs for one batch, simplify if possible.
- 写出每批料的面粉、黄油、鸡蛋的标准比例，尽量化简。
- Determine the scaling factor needed to go from 8 scones to 30 scones. Express as a fraction in simplest form.
- 确定从8个扩展到30个所需的比例因子，用最简分数表示。
- Calculate the required amount of each ingredient for 30 scones, and check the egg requirement.
- 计算制作30个司康饼所需的每种配料的用量，并检查鸡蛋需求。
- If the baker only has 11 eggs, what is the maximum number of whole scones she can make, assuming all other ingredients are sufficient?
- 假设面粉和黄油充足，但只有11个鸡蛋，她最多能制作多少个完整的司康饼？
Ratio: flour:butter:eggs = 300:75:2 = dividing by common factor? 300:75 = 4:1, so 4:1:0.0267? butter = 75 × 15/4 = 281.25 g; eggs = 2 × 15/4 = 30/4 = 7.5 eggs. Since eggs must be whole? It' .5 eggs, which is less than 11, so yes, enough. But if we consider whole eggs per batch, maybe recipe uses 2 eggs per 8 scones, so for 30 scones you could use 7 eggs + some other binder? gg → 4 scones, thus 11 eggs → 44 scones max, provided other ingredients scale accordingly.
比例：每份8个司康饼需面粉300 g、黄油75 g、鸡蛋2个。放大因子 = 30/8 = 15/4 = 3.75。面粉要求 = 1125 g，黄油 = 281.25 g，鸡蛋 = 7.5个。11个鸡蛋足够。若仅受鸡蛋限制，11个鸡蛋最多可制作 (11/2)×8 = 44个司康饼。
9. Case 9: Direct and Inverse Proportion in Physics | 案例九：物理中的正比与反比
ank to empty through a tap. The time T (in minutes) is inversely proportional to the cross‑ o find the relationship, predict T for a smaller tap, and also understand how a second tank with a different initial volume would affect the situation.
一位学生研究水箱通过水龙头放空所需的时间。时间 T（分钟）与龙头出口的横截面积 A 成反比。当 A = 2 cm² 时，T = 18 分钟。她希望找出关系式，预测更小口径龙头所需的时间，并理解若初始水量不同的第二个水箱会如何影响结果。
- Write down the equation linking T and A, including the constant of proportionality k.
- 写出 T 与 A 的关系式，并包含比例常数 k。
- Use the given data to find k, and hence express T in terms of A.
- 利用给定数据求出 k，并写出 T 关于 A 的表达式。
- Predict the emptying time for a tap with A = 1.2 cm².
- 预测当 A = 1.2 cm² 时的放空时间。
- If the tank were twice as high (so volume doubles), explain how the relationship would change, considering that time is directly proportional to volume for a given A.
- 如果水箱高度加倍（因此体积翻倍），解释这一关系会如何变化，须知对于给定的 A，时间与体积成正比。
oportional to volume, so for same A, T would be twice as long: the new relationship would be T = (2×36)/A = 72/A. This shows the constant k incorporates volume. Generally T ∝ Volume / A.
反比关系：T = k/A。代入数据得 k = 36，故 T = 36/A。当 A=1.2 cm²，T = 36/1.2 = 30分钟。若体积翻倍，则时间与体积成正比，所以对于相同的 A，新时间将是原来的两倍，关系变为 T = 72/A。这表明比例常数 k 实际上包含了体积因素，普遍公式为 T ∝ 容积 / A。
10. Case 10: Algebraic Identities and Geometric Proof | 案例十：代数恒等式与几何证明
are of side (a+b) into two smaller squares and two rectangles. Your task is to recreate this proof with a concrete numerical example, then generalise to algebraic symbols, and extend to the difference of two squares using a similar geometric argument.
经典的几何证明通过将一个边长为 (a+b) 的大正方形分割为两个小正方形和两个长方形，直观展示了 (a+b)² = a² + 2ab + b²。你的任务是用具体的数值例子重现该证明，然后推广到代数符号，并用类似的几何论证说明平方差公式 a² – b² = (a–b)(a+b)。
- f side 10, subdivide, and calculate areas of the parts to verify the identity numerically.
- 选择 a=7, b=3。画一个边长为10的正方形，进行分割，计算各部分面积，用数值验证恒等式。
- Write the general algebraic derivation and label each term with its geometric meaning.
- 写出一般的代数推导，并为每一项标记其几何含义。
- de a, remove a smaller square of side b from one corner. Cut and rearrange the remaining L‑shaped region into a rectangle of dimensions (a–b) by (a+b), proving a² – b² = (a–b)(a+b).
- 对于平方差：从一个边长为 a 的大正方形中切下一个边长为 b 的小正方形（位于一角）。将剩下的L形区域切割并重新拼成一个 (a–b) × (a+b) 的长方形，从而证明 a² – b² = (a–b)(a+b)。
the larger square, b² the smaller square, ab each rectangle. For difference of squares: a² – b² area L‑ ner to the outer edge, creating two pieces that form a rectangle of length a+b and width a–b, hence formula.
数值验证：10×10正方形划分出7×7=49，3×3=9，两个7×3长方形各21，总和100。代数：四项对应几何区域。平方差几何证明：从 a² 正方形移除角上的 b²，余下L形，切割后可拼成一个长为 a+b、宽为 a–b 的长方形，面积即为 (a–b)(a+b)。
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