KS3 WJEC Further Maths: Case Study Practical Exercises | 案例分析实战演练

📚 KS3 WJEC Further Maths: Case Study Practical Exercises | 案例分析实战演练

Welcome to this practical revision guide, where we tackle real-world problems using the key skills from the KS3 WJEC Further Maths curriculum. Case studies help you see how algebra, geometry, statistics, and ratio appear in everyday situations, building your confidence for assessment.

欢迎进入这份实战复习指南，我们将运用 KS3 WJEC 进阶数学课程中的关键技能，来解决真实世界的问题。案例分析能帮你看清代数、几何、统计和比例是如何出现在日常情境中的，从而为测评建立信心。

---

1. Introduction: What is a Case Study in Further Maths? | 引言：进阶数学中的案例分析是什么？

A case study is a multi-step problem set in a realistic context. It requires you to identify the mathematics involved, choose the right techniques, and communicate your reasoning clearly. Typical topics include linear equations, area and perimeter, data analysis, ratio, and probability.

案例分析是一个设定在真实情境中的多步骤问题。它要求你识别其中涉及的数学、选择合适的技巧，并清晰地表达你的推理过程。常见的主题包括线性方程、面积与周长、数据分析、比例和概率等。

In WJEC assessments, marks are awarded not only for the final answer but also for showing your working step by step. This guide will walk you through several examples, demonstrating how to break down a case study and apply your further maths knowledge successfully.

在 WJEC 的测评中，得分点不仅在于最终答案，还在于你逐步展示的计算过程。本指南将带你梳理多个示例，演示如何拆解案例分析并成功应用你的进阶数学知识。

---

2. Case 1: Designing a Rectangular Garden | 案例一：设计矩形花园

Emily wants to enclose a rectangular garden with fencing. The length is 3 metres longer than the width, and the perimeter is 26 metres. She will then cover the garden with turf costing £5 per square metre.

艾米丽想用篱笆围出一个矩形花园。长度比宽度多 3 米，周长为 26 米。之后她打算在花园里铺设草皮，每平方米成本为 5 英镑。

First, form an equation for the perimeter. Let the width be w metres. Then length = w + 3. Perimeter = 2(w + w + 3) = 26 → 4w + 6 = 26 → 4w = 20 → w = 5 m. Length = 8 m.

首先，建立周长的方程。设宽度为 w 米，则长度 = w + 3。周长 = 2(w + w + 3) = 26 → 4w + 6 = 26 → 4w = 20 → w = 5 米。长度 = 8 米。

Area = length × width = 8 × 5 = 40 m². Cost = 40 × £5 = £200. Always write units in your final statement.

面积 = 长 × 宽 = 8 × 5 = 40 平方米。成本 = 40 × 5 英镑 = 200 英镑。务必在最终陈述中写出单位。

---

3. Case 2: Travel Costs and Budgeting | 案例二：旅行成本与预算

A family of two adults and three children plans a day trip. An adult train ticket costs £15, and a child ticket costs £10. The family has a total budget of £100 for transport and souvenirs.

一个由两名成人和三名儿童组成的家庭计划一次一日游。成人火车票每张 15 英镑，儿童票每张 10 英镑。这个家庭的交通和纪念品总预算为 100 英镑。

Total ticket cost = 2 × 15 + 3 × 10 = 30 + 30 = £60. Money left for souvenirs = £100 − £60 = £40. If each souvenir costs £4, they can buy 40 ÷ 4 = 10 souvenirs.

总车票费用 = 2 × 15 + 3 × 10 = 30 + 30 = 60 英镑。剩余购买纪念品的钱 = 100 − 60 = 40 英镑。如果每件纪念品 4 英镑，他们可以购买 40 ÷ 4 = 10 件纪念品。

The mathematical operations here involve multiplication, addition, subtraction, and division. Presenting the steps in a clear table can help organise your working.

这里的数学运算涉及乘法、加法、减法和除法。用清晰的表格来展示步骤有助于理顺你的计算过程。

Category	Calculation	Amount (£)
Adult tickets	2 × 15	30
Child tickets	3 × 10	30
Remaining budget	100 − 60	40

---

4. Case 3: Analysing Sports Data | 案例三：运动数据分析

A basketball team’s points scored over six matches were: 12, 15, 8, 20, 10, 14. Calculate the mean, median, mode, and range to summarise their performance.

一支篮球队在六场比赛中的得分分别为：12, 15, 8, 20, 10, 14。请计算平均数、中位数、众数和极差，以总结他们的表现。

Mean = (12 + 15 + 8 + 20 + 10 + 14) ÷ 6 = 79 ÷ 6 ≈ 13.2 (to one decimal place). To find the median, order the values: 8, 10, 12, 14, 15, 20. The median is the middle of an even set: (12 + 14) ÷ 2 = 13.

平均数 = (12 + 15 + 8 + 20 + 10 + 14) ÷ 6 = 79 ÷ 6 ≈ 13.2（保留一位小数）。要找到中位数，先将数值排序：8, 10, 12, 14, 15, 20。偶数个数据的中位数是中间两个数的平均值：(12 + 14) ÷ 2 = 13。

There is no mode because all values appear once. Range = highest − lowest = 20 − 8 = 12. These statistics help coaches quickly understand consistency and typical scores.

由于所有数值都只出现一次，因此没有众数。极差 = 最大值 − 最小值 = 20 − 8 = 12。这些统计量可以帮助教练快速了解球队表现的稳定性和典型得分。

---

5. Case 4: Ratios in Cooking | 案例四：烹饪中的比例

A recipe for pancakes uses flour and sugar in the ratio 3 : 1. Sam needs 480 g of the flour-sugar mixture. How much of each ingredient is required?

一份煎饼配方中面粉和糖的比例为 3 : 1。山姆需要 480 克面粉与糖的混合物。每种原料各需要多少？

The ratio 3:1 means there are 4 parts in total. One part = 480 ÷ 4 = 120 g. Flour = 3 parts = 3 × 120 = 360 g. Sugar = 1 part = 120 g.

比例 3:1 意味着总共有 4 份。一份 = 480 ÷ 4 = 120 克。面粉 = 3 份 = 3 × 120 = 360 克。糖 = 1 份 = 120 克。

You can check your answer: 360 + 120 = 480 g, and 360:120 simplifies to 3:1. Ratio problems often appear in scaling up or down recipes in catering.

你可以检验答案：360 + 120 = 480 克，且 360:120 化简后为 3:1。比例问题经常出现在餐饮业中食谱的放大或缩小中。

---

6. Case 5: Surface Area and Volume of a Storage Box | 案例五：储物盒的表面积与体积

A rectangular storage box has length 30 cm, width 20 cm, and height 15 cm. Calculate the volume in cm³ and the total surface area in cm². Express the volume in litres.

一个长方体储物盒的长为 30 厘米、宽为 20 厘米、高为 15 厘米。请计算体积（cm³）和总表面积（cm²），并将体积以升为单位表示。

Volume V = l × w × h = 30 × 20 × 15 = 9000 cm³

Since 1 litre = 1000 cm³, the volume is 9000 ÷ 1000 = 9 litres.

因为 1 升 = 1000 立方厘米，所以体积为 9000 ÷ 1000 = 9 升。

Surface area A = 2(lw + lh + wh) = 2(30×20 + 30×15 + 20×15) = 2(600 + 450 + 300) = 2 × 1350 = 2700 cm²

This is useful when determining how much material is needed to make the box or how much paint is required to cover it.

这在确定制作盒子所需的材料量或涂刷所需的油漆量时非常有用。

---

7. Case 6: Probability with Dice Games | 案例六：骰子游戏中的概率

Two fair six-sided dice are rolled. What is the probability that the sum of the numbers shown is exactly 7?

投掷两个均匀的六面骰子。出现点数之和恰好为 7 的概率是多少？

There are 6 × 6 = 36 equally likely outcomes. The pairs that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — a total of 6 outcomes.

共有 6 × 6 = 36 种等可能的结果。点数之和为 7 的组合有 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) —— 共 6 种结果。

P(sum = 7) = 6/36 = 1/6

Understanding sample spaces and writing out outcomes systematically is a core skill in probability case studies. Always check that your fraction is in its simplest form.

理解样本空间并系统地列出所有结果是概率案例分析中的核心技能。务必检查你的分数是否为最简形式。

---

8. Case 7: Linear Equations and Mobile Phone Plans | 案例七：线性方程与手机套餐

Jake compares two mobile phone plans. Plan A costs £10 per month and gives 100 free minutes; extra minutes cost £0.10 each. Plan B costs £15 per month with 200 free minutes; extra minutes cost £0.05 each. If Jake uses 150 minutes in a month, which plan is cheaper?

杰克正在比较两个手机套餐。套餐 A 每月 10 英镑，包含 100 分钟免费通话；超出部分每分钟 0.10 英镑。套餐 B 每月 15 英镑，包含 200 分钟免费通话；超出部分每分钟 0.05 英镑。如果杰克某个月使用了 150 分钟，哪个套餐更便宜？

For Plan A: included minutes = 100, so extra minutes = 50. Cost = 10 + 50 × 0.10 = 10 + 5 = £15. For Plan B: 150 minutes are entirely within the 200 minutes, so cost = £15. Both plans cost £15 this month.

对于套餐 A：包含分钟数 = 100，因此超出分钟数 = 50。费用 = 10 + 50 × 0.10 = 10 + 5 = 15 英镑。对于套餐 B：150 分钟完全在 200 分钟之内，因此费用 = 15 英镑。本月两个套餐费用相同。

If usage changes, you can model costs with linear expressions: Cost_A = 10 + 0.10(x − 100) for x > 100; Cost_B = 15 + 0.05(x − 200) for x > 200. Solving the equation 10 + 0.10(x − 100) = 15 helps find the break-even point.

如果使用量发生变化，你可以用线性表达式来建立费用模型：当 x > 100 时，Cost_A = 10 + 0.10(x − 100)；当 x > 200 时，Cost_B = 15 + 0.05(x − 200)。解方程 10 + 0.10(x − 100) = 15 有助于找到盈亏平衡点。

---

9. Case 8: Interpreting Graphs and Charts | 案例八：解读图表

A line graph shows the temperature in a greenhouse from 8 am to 4 pm. At 8 am it was 14 °C, rising steadily to 26 °C at 12 noon, then falling to 20 °C by 4 pm. Calculate the average rate of temperature increase between 8 am and 12 noon in °C per hour.

一张折线图展示了一个温室从早上 8 点到下午 4 点的温度变化。早上 8 点为 14 °C，之后稳步上升，到中午 12 点达到 26 °C，然后到下午 4 点下降至 20 °C。请计算从早上 8 点到中午 12 点温度的平均升高速率（°C/小时）。

Temperature rise = 26 − 14 = 12 °C. Time interval = 4 hours. Rate of increase = 12 ÷ 4 = 3 °C per hour. For the afternoon, rate of decrease = (26 − 20) ÷ 4 = 6 ÷ 4 = 1.5 °C per hour.

温度上升幅度 = 26 − 14 = 12 °C，时间间隔 = 4 小时。升高速率 = 12 ÷ 4 = 3 °C/小时。下午的下降速率 = (26 − 20) ÷ 4 = 6 ÷ 4 = 1.5 °C/小时。

Being able to extract information from line graphs, bar charts, and pie charts is essential. Always label your rates with units, and round sensibly when reading values between grid lines.

能够从折线图、条形图和饼图中提取信息是至关重要的。务必为速率标注单位，当读取网格线之间的数值时要合理舍入。

---

10. Case 9: Compound Measures – Speed and Density | 案例九：复合量度 – 速度与密度

A car travels 150 miles in 2.5 hours. Calculate its average speed. A metal block has a mass of 500 g and a volume of 200 cm³. Find its density. Will it float in water? (Density of water = 1 g/cm³)

一辆汽车在 2.5 小时内行驶了 150 英里。计算其平均速度。一块金属的质量为 500 克，体积为 200 立方厘米。求其密度。这块金属在水中会浮起来吗？（水的密度 = 1 g/cm³）

Speed = distance ÷ time = 150 ÷ 2.5 = 60 miles per hour (mph)

Density = mass ÷ volume = 500 ÷ 200 = 2.5 g/cm³

Since 2.5 > 1, the metal is denser than water, so it will sink. Always use the formula triangle (D = M/V) to check your rearrangement.

由于 2.5 > 1，该金属的密度大于水，因此它会沉没。可借助公式三角形（D = M/V）来验证你的变形是否正确。

---

11. Case 10: Mixed Practice Problems | 案例十：综合练习

Now try a blended case study: A cylindrical candle has a diameter of 8 cm and a height of 12 cm. It burns at a rate of 15 cm³ of wax per hour. (i) Calculate the volume of the candle. (ii) How long will the candle last? (iii) The candle’s label wraps the curved surface exactly once; find the area of the label.

现在来尝试一个混合案例分析：一支圆柱形蜡烛的直径为 8 厘米，高为 12 厘米。它以每小时 15 立方厘米的速度燃烧。（i）计算蜡烛的体积。（ii）蜡烛能燃烧多长时间？（iii）蜡烛上的标签恰好围绕侧面一周；求标签的面积。

Use π ≈ 3.14. Radius r = diameter / 2 = 4 cm.

π 取近似值 3.14。半径 r = 直径 / 2 = 4 厘米。

Volume V = πr²h = 3.14 × 4² × 12 = 3.14 × 16 × 12 = 602.88 cm³

Burn time = total volume ÷ burn rate = 602.88 ÷ 15 ≈ 40.19 hours, or about 40 hours and 11 minutes.

燃烧时间 = 总体积 ÷ 燃烧速率 = 602.88 ÷ 15 ≈ 40.19 小时，约合 40 小时 11 分钟。

Curved surface area = circumference × height = 2πr × h = 2 × 3.14 × 4 × 12 = 301.44 cm²

These multi-step problems combine geometry, unit conversions, and arithmetic, just like a real WJEC case study question.

这些多步骤问题融合了几何、单位换算和算术，就像一道真正的 WJEC 案例分析题一样。

---

12. Conclusion: Tips for Tackling Case Studies | 结论：攻克案例分析的技巧

Successful case study answers share common features: read the whole scenario before calculating, identify the maths topics involved, show all working step by step, and always state the answer in context with correct units. Underline key numbers and check your final answer makes sense in the real-world situation.

成功的案例分析答案具有共同的特点：先通读整个情境再开始计算，识别所涉及的数学主题，逐步展示所有的计算过程，并始终在上下文中用正确的单位陈述答案。划出关键数字，并检查最终答案在真实情境中是否合理。

Practice by writing your own short case studies based on everyday situations – shopping discounts, sports scores, or DIY projects. The more you connect maths to real life, the easier these problem-solving tasks become.

你可以通过编写基于日常情境——如购物折扣、运动成绩或手工项目——的简短案例来进行练习。你将数学与现实生活联系得越多，这类问题解决任务就会变得越容易。

Published by TutorHao | Further Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）