📚 Year 9 WJEC Further Mathematics: Case Study Drills and Worked Examples | Year 9 WJEC 进阶数学：案例分析实战演练
Welcome to our collection of real-world case studies designed for Year 9 WJEC Further Mathematics. Each case breaks down a problem into clear steps, blending core skills such as percentages, equations, geometry, Pythagoras, linear functions, probability, sequences and statistics. Use these worked examples to sharpen your reasoning and build exam confidence.
欢迎来到Year 9 WJEC进阶数学的真实案例分析集。每个案例将问题拆解为清晰的步骤，融合百分比、方程、几何、勾股定理、线性函数、概率、数列与统计等核心技能。通过这些详解实例，你可以提升推理能力，建立应考信心。
1. Percentage Discounts and Profit | 案例分析1：百分比折扣与利润
A jacket originally costs £80. The shop offers a 15% discount, then adds 20% VAT on the discounted price. If the shop paid £50 for the jacket, what is the final selling price and the profit percentage?
一件夹克原价80英镑。商店提供15%的折扣，然后在折扣价基础上加收20%的增值税。如果商店进货成本为50英镑，最终售价和利润率是多少？
Step 1: Find the discount amount. Discount = 15% of £80 = 0.15 × 80 = £12. So the discounted price before VAT is £80 − £12 = £68.
步骤1：计算折扣金额。折扣 = 80的15% = 0.15 × 80 = 12英镑。因此税前折扣价为80 − 12 = 68英镑。
discounted price = £80 − (15% × 80) = £68
Step 2: Add 20% VAT. VAT amount = 20% of £68 = 0.20 × 68 = £13.60. Final price = £68 + £13.60 = £81.60.
步骤2：加收20%增值税。增值税额 = 68的20% = 0.20 × 68 = 13.60英镑。最终售价 = 68 + 13.60 = 81.60英镑。
final price = £68 + (20% × 68) = £81.60
Step 3: Calculate profit. Profit = final price − cost = £81.60 − £50 = £31.60. Profit percentage = (31.60 / 50) × 100% = 63.2%.
步骤3：计算利润。利润 = 最终售价 − 成本 = 81.60 − 50 = 31.60英镑。利润率 = (31.60 / 50) × 100% = 63.2%。
Key insight: Always apply percentage changes in the correct order — discount first, then tax. The profit percentage is based on the original cost.
关键提示：务必按正确顺序进行百分比运算——先折扣，后税费。利润率是基于原始成本计算的。
2. Solving Linear Equations in Geometry | 案例分析2：几何中的线性方程求解
The length of a rectangle is 5 cm longer than its width. The perimeter is 38 cm. Set up and solve an equation to find the dimensions, then calculate the area.
一个矩形的长比宽多5厘米，周长为38厘米。建立并求解方程，求出长和宽，然后计算面积。
hen length = w + 5 cm. Perimeter = 2(length + width) = 2(w + 5 + w) = 2(2w + 5) = 4w + 10.
设宽为w cm，则长为w + 5 cm。周长 = 2(长 + 宽) = 2(w + 5 + w) = 2(2w + 5) = 4w + 10。
4w + 10 = 38
Solve: 4w = 28, so w = 7 cm. Length = 7 + 5 = 12 cm.
解方程：4w = 28，因此w = 7 cm。长 = 7 + 5 = 12 cm。
Area = length × width = 12 × 7 = 84 cm².
面积 = 长 × 宽 = 12 × 7 = 84 cm²。
This case shows how to translate a word problem into an algebraic equation and then use the solution to find other properties like area.
本案例展示了如何将文字问题转化为代数方程，并利用解出的值求其他属性如面积。
3. Angle Relationships in Polygons | 案例分析3：多边形中的角度关系
A pentagon has four known interior angles: 112°, 95°, 130° and 108°. Find the missing angle and determine whether the pentagon is concave or convex.
一个五边形有四个已知内角：112°、95°、130°和108°。求缺失的角度，并判断该五边形是凹的还是凸的。
Sum of interior angles of a pentagon = (5 − 2) × 180° = 540°.
五边形内角和 = (5 − 2) × 180° = 540°。
sum = (n − 2) × 180° = 3 × 180° = 540°
Sum of given angles = 112 + 95 + 130 + 108 = 445°. Missing angle = 540° − 445° = 95°.
已知角之和 = 112 + 95 + 130 + 108 = 445°。缺失角 = 540° − 445° = 95°。
All interior angles are less than 180°, so the pentagon is convex. If one angle were greater than 180°, it would be concave.
所有内角均小于180°，因此该五边形是凸的。若有一个角大于180°，则会是凹的。
Always check polygon angle sums with (n−2)×180° and relate to shape classification.
始终用(n−2)×180°检查多边形内角和，并与图形分类关联。
4. Pythagoras' Theorem in Real Life | 案例分析4：勾股定理在现实生活中的应用
A ladder of length 5 metres leans against a vertical wall. The foot of the ladder is 1.5 metres from the base of the wall. How high up the wall does the ladder reach? Give your answer to two decimal places.
一架5米长的梯子斜靠在一面竖直的墙上。梯脚距墙底1.5米。梯子顶端能达到多高？答案保留两位小数。
By Pythagoras' theorem: ladder² = height² + base². Let height = h m. Then 5² = h² + 1.5².
根据勾股定理：梯子长² = 高度² + 底距²。设高度为h m。则5² = h² + 1.5²。
25 = h² + 2.25 → h² = 22.75
h = √22.75 ≈ 4.77 m (to 2 d.p.).
h = √22.75 ≈ 4.77 m（保留两位小数）。
Check: 4.77² + 1.5² ≈ 22.75 + 2.25 = 25.00, confirming accuracy.
验证：4.77² + 1.5² ≈ 22.75 + 2.25 = 25.00，确认无误。
Pythagoras' theorem is widely used to find distances in right‑angled triangles, especially in construction and navigation problems.
勾股定理广泛用于求直角三角形中的距离，尤其在建筑和导航问题中。
5. Linear Functions and Graph Interpretation | 案例分析5：线性函数与图像解读
The cost £C of hiring a bike for t hours is given by C = 4t + 3. Interpret the gradient and y‑intercept, and find the cost for 5 hours.
租赁自行车t小时的费用£C由C = 4t + 3给出。解释斜率和y截距，并求出5小时的费用。
The equation is in the form y = mx + c. Here, gradient m = 4 gives the hourly rate (£4 per hour). The y‑intercept c = 3 is the fixed booking fee (even for 0 hours).
方程形式为y = mx + c。此处斜率m = 4表示每小时费率（每小时4英镑）。y截距c = 3是固定预约费（即使租0小时也需支付）。
For t = 5: C = 4(5) + 3 = 20 + 3 = 23. So the total cost is £23.
当t = 5时：C = 4(5) + 3 = 20 + 3 = 23。总费用为23英镑。
C = 4 × 5 + 3 = £23
Gradient tells the rate of increase; intercept shows the starting value. Always label axes when sketching the graph: time (t) on the horizontal, cost (C) on the vertical.
斜率表示增长率；截距表示初始值。绘制图像时应标注坐标轴：横轴为时间(t)，纵轴为费用(C)。
6. Probability and Tree Diagrams | 案例分析6：概率与树形图
A bag contains 3 red and 2 blue counters. Two counters are drawn at random without replacement. Find the probability that both counters are red.
一个袋子里有3个红色和2个蓝色筹码。不放回地随机抽取两个。求两个都是红色的概率。
Draw a tree diagram. First draw: P(red) = 3/5, P(blue) = 2/5.
绘制树形图。第一次抽取：P(红) = 3/5, P(蓝) = 2/5。
If first is red, remaining counters: 2 red, 2 blue. P(second red | first red) = 2/4 = 1/2.
若第一次为红，剩余筹码：2红，2蓝。P(第二次为红 | 第一次红) = 2/4 = 1/2。
P(both red) = (3/5) × (1/2) = 3/10.
P(两个红) = (3/5) × (1/2) = 3/10。
P(R,R) = 3/5 × 2/4 = 6/20 = 3/10
Without replacement, the probabilities change after each draw. Tree diagrams help visualise all outcomes systematically.
在不放回的情况下，每次抽取后概率会改变。树形图有助于系统地可视化所有可能结果。
7. Sequences and the nth Term | 案例分析7：数列与第n项
Consider the arithmetic sequence: 3, 7, 11, 15, ... Find the nth term formula and use it to determine the 20th term.
考虑等差数列：3, 7, 11, 15, ... 求通项公式，并求出第20项。
term a = 3, common difference d = 4. For an arithmetic sequence, nth term = a + (n − 1)d.
首项a = 3，公差d = 4。等差数列的通项公式为：a + (n − 1)d。
Substitute values: nth term = 3 + (n − 1)×4 = 3 + 4n − 4 = 4n − 1.
代入数值：通项 = 3 + (n − 1)×4 = 3 + 4n − 4 = 4n − 1。
nth term = 4n − 1
To find the 20th term: n = 20 gives 4(20) − 1 = 80 − 1 = 79. So the 20th term is 79.
求第20项：n = 20，得4(20) − 1 = 80 − 1 = 79。故第20项为79。
This method works for any linear (arithmetic) sequence. For WJEC Further Mathematics, you may also explore quadratic sequences.
此方法适用于任何线性（等差）数列。WJEC进阶数学中，你还可能探索二次数列。
8. Comparing Data Sets with Statistics | 案例分析8：统计比较数据集
Class A scores on a test: 12, 15, 11, 14, 16, 13, 15. Class B scores: 10, 18, 12, 11, 14, 16. Calculate the mean, median and range for each class, and decide which class performed better overall.
A班测试成绩：12, 15, 11, 14, 16, 13, 15。B班成绩：10, 18, 12, 11, 14, 16。计算每个班的平均数、中位数和极差，并判断哪个班整体表现更好。
(11+12+13+14+15+15+16) ÷ 7 = 96 ÷ 7 ≈ 13.71. Median = 14 (4th value). Range = 16 − 11 = 5.
A班：排序后：11, 12, 13, 14, 15, 15, 16。平均数 = (11+12+13+14+15+15+16) ÷ 7 = 96 ÷ 7 ≈ 13.71。中位数 = 14（第4个值）。极差 = 16 − 11 = 5。
= (10+11+12+14+16+18) ÷ 6 = 81 ÷ 6 = 13.5. Median = (12+14) ÷ 2 = 13. Range = 18 − 10 = 8.
B班：排序后：10, 11, 12, 14, 16, 18。平均数 = (10+11+12+14+16+18) ÷ 6 = 81 ÷ 6 = 13.5。中位数 = (12+14) ÷ 2 = 13。极差 = 18 − 10 = 8。
Comparison: Class A has a higher mean (13.71 > 13.5) and a higher median (14 > 13). It also has a smaller range (5 vs 8), indicating more consistent scores. Therefore, Class A performed better overall.
比较：A班平均数更高（13.71 > 13.5）且中位数更高（14 > 13）。极差更小（5对8），表明成绩更稳定。因此，A班整体表现更好。
When comparing data sets, always examine both central tendency (mean, median) and spread (range). Consistency matters as much as averages in Further Mathematics.
比较数据集时，要同时考察集中趋势（平均数、中位数）和离散程度（极差）。进阶数学中，稳定性与平均值同等重要。
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