📚 Year 8 Edexcel Further Mathematics: Case Study Practical Drills | 八年级爱德思进阶数学：案例分析实战演练
This article presents a series of targeted case studies designed for Year 8 students following the Edexcel Further Mathematics curriculum. Each case applies key concepts – from linear equations and ratios to geometry, probability and logic – in real-life contexts. By working through these practical drills, you will strengthen your problem-solving skills and build deeper understanding of advanced topics.
本文为学习爱德思进阶数学课程的八年级学生提供一系列针对性强的案例分析。每个案例将线性方程、比例、几何、概率和逻辑等核心概念应用到实际场景中。通过这些实战演练，你将提升解题能力，加深对高阶主题的理解。
1. Linear Equations and Discounts | 线性方程与折扣
A jacket originally costs £60. A store offers a 20% discount, and Laura applies a £5 voucher on top of the reduced price. Let the final price be p pounds. Write a linear equation for p and solve it.
一件夹克原价60英镑。商店提供20%折扣，Laura再使用一张5英镑的优惠券。设最终价格为p英镑。为p写出线性方程并求解。
The discount reduces the original price to 80% of its value. The voucher subtracts £5 from that amount. The equation is: p = 60 × 0.8 − 5.
折扣后价格为原价的80%，优惠券再减去5英镑。方程为：p = 60 × 0.8 − 5。
Now evaluate: p = 48 − 5 = 43. Therefore, Laura pays £43.
计算可得：p = 48 − 5 = 43。因此，Laura需支付43英镑。
To check, you could work backwards: adding the £5 voucher gives £48, which is indeed 80% of £60. This method of linking real-world discounts to one-step equations is key in Further Mathematics.
我们可以反向检验：加上5英镑得48英镑，恰好是60英镑的80%。这种将现实折扣与一步方程联系的方法在进阶数学中至关重要。
2. Ratio and Mixed Drinks | 比例与混合饮品
A fruit punch is made by mixing orange juice and lemonade in the ratio 2 : 3. If 600 ml of lemonade is used, how much orange juice is needed to keep the same strength?
一种水果宾治由橙汁和柠檬水按2:3的比例混合而成。如果使用了600毫升柠檬水，需要多少毫升橙汁才能保持相同的浓度？
orange juice) / (lemonade) = 2/3. Let x be the volume of orange juice. Then x / 600 = 2/3.
列出比例式：(橙汁量)/(柠檬水量) = 2/3。设橙汁体积为x，则x/600 = 2/3。
Multiply both sides by 600: x = (2/3) × 600 = 400. So, 400 ml of orange juice is required.
两边同乘600：x = (2/3) × 600 = 400。因此，需要400毫升橙汁。
ts 3 parts → 600 ml, so 1 part = 200 ml. Orange juice takes 2 parts → 2 × 200 ml = 400 ml.
你也可以从总份数考虑：混合物共有2+3=5份。柠檬水占3份→600毫升，因此1份=200毫升。橙汁占2份→2×200毫升=400毫升。
3. Geometry: Triangle Angle Chase | 几何：三角形角度求解
+ 20)° and angle C = (4x − 30)°. Find the value of x and determine the size of each angle.
在三角形ABC中，角A = (2x + 10)°，角B = (3x + 20)°，角C = (4x − 30)°。求x的值，并计算每个角的大小。
nterior angles in any triangle is 180°, we write: (2x + 10) + (3x + 20) + (4x − 30) = 180.
由于任意三角形内角和为180°，列出方程：(2x + 10) + (3x + 20) + (4x − 30) = 180。
Simplify: 9x + 0 = 180 (since 10 + 20 − 30 = 0). Thus 9x = 180, giving x = 20.
化简得：9x + 0 = 180（因为10+20−30=0）。所以9x = 180，解得x = 20。
back into each expression: A = 2×20 + 10 = 50°, B = 3×20 + 20 = 80°, C = 4×20 − 30 = 50°.
将x = 20代入各表达式：A = 2×20 + 10 = 50°，B = 3×20 + 20 = 80°，C = 4×20 − 30 = 50°。
Notice that the triangle is isosceles because two angles are equal. Identifying such properties helps when solving geometry problems in Further Mathematics.
注意到该三角形是等腰三角形，因为两个角相等。识别此类性质有助于解决进阶数学中的几何问题。
4. Statistics: Mean, Mode, Median and Range | 统计：平均数、众数、中位数及极差
Five students achieved the following scores in a test: 78, 85, 92, 78, 88. Calculate the mean, mode, median and range of this data set.
五名学生在一次测试中取得以下分数：78, 85, 92, 78, 88。计算这组数据的平均数、众数、中位数和极差。
First, arrange the scores in ascending order: 78, 78, 85, 88, 92.
首先将分数按升序排列：78, 78, 85, 88, 92。
The mean = (78 + 85 + 92 + 78 + 88) ÷ 5 = 421 ÷ 5 = 84.2.
平均数 = (78 + 85 + 92 + 78 + 88) ÷ 5 = 421 ÷ 5 = 84.2。
The mode is the value that appears most frequently: 78 occurs twice, so the mode is 78.
众数是出现频率最高的数值：78出现两次，因此众数为78。
The median is the middle value in the ordered list. With five scores, the third value is the median: 85.
中位数是排序后位于中间位置的数值。五个分数中，第三个值是85。
The range is the difference between the highest and lowest scores: 92 − 78 = 14.
极差是最大值与最小值之差：92 − 78 = 14。
Summarising data with these measures gives a clear picture of its distribution, a fundamental skill in statistical analysis.
用这些指标概括数据可以清晰描绘其分布情况，这是统计分析中的一项基本技能。
5. Probability: Coin Tosses | 概率：抛硬币
A fair coin is tossed twice. List all possible outcomes and find the probability of getting at least one head.
投掷一枚均匀硬币两次。列出所有可能的结果，并求至少出现一次正面的概率。
The sample space is: HH, HT, TH, TT. Each outcome is equally likely when the coin is fair.
样本空间为：HH, HT, TH, TT。硬币均匀时，每种结果发生的可能性相同。
Outcomes with at least one head are HH, HT and TH – three favourable outcomes out of four.
至少包含一次正面的结果是HH、HT和TH——四个结果中有三个有利结果。
So, P(at least one head) = 3/4.
因此，P(至少一次正面) = 3/4。
ovide a quicker route: P(at least one head) = 1 − P(no heads) = 1 − P(TT) = 1 − 1/4 = 3/4.
互补事件提供了一条更快的途径：P(至少一次正面) = 1 − P(没有正面) = 1 − P(TT) = 1 − 1/4 = 3/4。
Recognising when to use complements can simplify probability calculations, a valuable skill in advanced problem sets.
判断何时使用互补事件可简化概率计算，这是解决高阶习题时的一项宝贵技能。
6. Sequences and Patterns | 数列与规律
The nth term of a sequence is given by the formula 3n² − 2. Write down the first three terms and describe the pattern.
一个数列的第n项由公式3n² − 2给出。写出前三项并描述其规律。
Substitute n = 1: 3 × (1)² − 2 = 3 − 2 = 1.
代入n = 1：3 × (1)² − 2 = 3 − 2 = 1。
For n = 2: 3 × (2)² − 2 = 3 × 4 − 2 = 12 − 2 = 10.
当n = 2时：3 × (2)² − 2 = 3 × 4 − 2 = 12 − 2 = 10。
For n = 3: 3 × (3)² − 2 = 3 × 9 − 2 = 27 − 2 = 25.
当n = 3时：3 × (3)² − 2 = 3 × 9 − 2 = 27 − 2 = 25。
The first three terms are 1, 10, 25. The differences between consecutive terms are 9 and 15, and the second differences are constant (6), confirming a quadratic sequence.
前三项为1, 10, 25。相邻项之差为9和15，而二阶差恒为6，确认这是一个二次序列。
Finding nth term rules from patterns is a key part of algebraic reasoning. Here the 3n² term dominates growth, while the constant −2 shifts the whole sequence down.
从模式中找出第n项通式是代数推理的关键部分。此处3n²项主导增长，而常数−2将整个数列向下平移。
7. Inequalities and Budget Planning | 不等式与预算规划
A cinema charges £8 for an adult ticket and £5 for a student ticket. A group of friends has a budget of at most £50 for tickets. If they buy 3 student tickets, how many adult tickets can they afford?
一家电影院成人票价为8英镑，学生票价为5英镑。一群朋友购买电影票的预算不超过50英镑。如果他们购买3张学生票，最多能买多少张成人票？
ets. The total cost is 3 × 5 + 8a = 15 + 8a pounds. This must be less than or equal to 50.
设a为成人票张数。总花费为3 × 5 + 8a = 15 + 8a英镑。该金额必须小于或等于50。
Write the inequality: 15 + 8a ≤ 50.
列出不等式：15 + 8a ≤ 50。
Subtract 15 from both sides: 8a ≤ 35.
两边减去15：8a ≤ 35。
a ≤ 4.375. Since we can only buy whole tickets, the maximum number of adult tickets is 4.
两边除以8：a ≤ 4.375。因为只能购买整数张票，成人票的最大数量为4张。
a = 5 gives 15 + 40 = 55 > 50. Always check the boundary condition in inequality problems.
代入a = 4验证总花费为15 + 32 = 47 ≤ 50，而a = 5时15 + 40 = 55 > 50。在不等式问题中，务必检验边界情况。
8. Functions and Mappings | 函数与映射
Given the functions f(x) = 2x² + 1 and g(x) = x − 3, find f(g(4)) and g(f(1)).
给定函数f(x) = 2x² + 1和g(x) = x − 3，求f(g(4))和g(f(1))。
First, find g(4): g(4) = 4 − 3 = 1.
先求g(4)：g(4) = 4 − 3 = 1。
Then apply f to this result: f(g(4)) = f(1) = 2 × 1² + 1 = 2 × 1 + 1 = 3.
再将结果代入f：f(g(4)) = f(1) = 2 × 1² + 1 = 2 × 1 + 1 = 3。
d part, calculate f(1): f(1) = 2 × 1² + 1 = 3 (as above). Then g(f(1)) = g(3) = 3 − 3 = 0.
对于第二部分，计算f(1)：f(1) = 2 × 1² + 1 = 3（同上）。然后g(f(1)) = g(3) = 3 − 3 = 0。
Notice that f(g(4)) and g(f(1)) gave different results. Composition of functions is not commutative – order matters. This concept is essential when dealing with inverse functions later on.
注意f(g(4))与g(f(1))得到了不同的结果。复合函数不满足交换律——顺序至关重要。这一概念在日后学习反函数时不可或缺。
9. Basics of Permutations | 排列基础
From a team of 6 players, a captain and a vice-captain need to be chosen. How many different ways can these two roles be assigned?
从一支6名队员的队伍中，需要选出一名队长和一名副队长。有多少种不同的分配这两个职务的方式？
Because the roles are distinct, the order of selection matters. We use a permutation: first choose the captain (6 options), then the vice-captain from the remaining 5 players (5 options).
因为职务不同，选择顺序事关重要。我们使用排列：先选择队长（6种选择），然后从剩余5人中选副队长（5种选择）。
By the multiplication principle, total ways = 6 × 5 = 30.
根据乘法原理，总方式数 = 6 × 5 = 30。
We can also write this as ⁶P₂ = 6! / (6−2)! = (6×5×4!)/4! = 30.
我们也可以写作⁶P₂ = 6! / (6−2)! = (6×5×4!)/4! = 30。
se 2 = 15). Distinguishing permutations from combinations is crucial in counting problems.
如果只是无差别地选出两名代表，顺序无关紧要，则应当使用组合（6选2 = 15）。在计数问题中，区分排列与组合至关重要。
10. Logical Reasoning Puzzle | 逻辑推理谜题
Three friends – Alice, Bob and Charlie – each have a different favourite colour: red, blue or green. Alice does not like red. Bob's favourite colour is blue. Who likes green?
三位朋友——Alice、Bob和Charlie——各自最喜欢一种不同的颜色：红色、蓝色或绿色。Alice不喜欢红色。Bob最喜欢的颜色是蓝色。谁最喜欢绿色？
Start by assigning the known facts: Bob takes blue. That leaves red and green for Alice and Charlie.
从已知事实入手：Bob对应蓝色。剩下红色和绿色分配给Alice和Charlie。
Alice does not like red, so she must have green. Therefore, Charlie gets the remaining colour, red.
Alice不喜欢红色，因此她必须拥有绿色。于是，Charlie得到剩下的红色。
So, Alice likes green. This simple deduction uses the idea of elimination, a logical tool that appears frequently in mathematics, from set theory to proof writing.
因此，Alice最喜欢绿色。这一简单推理运用了排除法的思想，这是一种在数学中——从集合论到证明书写——频繁出现的逻辑工具。
Puzzles like this train your ability to systematically handle constraints, an invaluable skill in Further Mathematics problem solving.
此类谜题训练你有条理地处理约束条件的能力，这在进阶数学问题解决中是一项无价的技能。
Published by TutorHao | Further Mathematics 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