📚 Year 8 CCEA Further Mathematics: Case Study Practical Exercises | Year 8 CCEA 进阶数学：案例分析实战演练
In Year 8 CCEA Further Mathematics, applying mathematical concepts to real‑world scenarios is a key skill. Case studies help you move beyond textbook calculations and see how area, percentages, proportions, statistics, probability and algebra are used every day. This article presents a series of practical exercises designed to strengthen your problem‑solving abilities and build confidence for assessments. Each case study is broken down into clear steps, with explanations provided in both English and Chinese to support bilingual learners.
在 CCEA Year 8 进阶数学中，将数学概念应用到实际情境中是一项关键技能。案例分析能帮助你超越课本计算，看到面积、百分数、比例、统计、概率和代数如何应用在日常生活里。本文通过一系列实战演练来加强你的解题能力，为评估建立信心。每个案例都分解成清晰的步骤，并提供中英双语解说，帮助双语学习者。
1. Redesigning a Bedroom – Area and Perimeter | 卧室重新设计——面积与周长
A family wants to replace the carpet and install a new skirting board in a rectangular bedroom measuring 4.2 m by 3.8 m. The carpet costs £12.50 per square metre, and the skirting board costs £4.80 per metre. We need to calculate the total cost, including a 10% extra allowance for carpet wastage.
一个家庭想为长 4.2 米、宽 3.8 米的长方形卧室更换地毯并安装新的踢脚板。地毯每平方米售价 £12.50，踢脚板每米 £4.80。我们需要计算总费用，包括地毯额外 10% 的损耗余量。
First, work out the area of the bedroom floor using the formula Area = length × width.
首先，使用公式 面积 = 长 × 宽 计算卧室地板的面积。
Area = 4.2 m × 3.8 m = 15.96 m²
With 10% ded becomes 15.96 × 1.10 = 17.556 m². Rounding up to 17.56 m² is practical for purchasing.
加上 10% 的裁剪和安装余量，所需地毯总面积变为 15.96 × 1.10 = 17.556 m²。在实际购买中四舍五入到 17.56 m²。
The carpet cost is 17.56 m² × £12.50/m².
地毯费用为 17.56 m² × £12.50/m²。
17.56 × 12.50 = £219.50
Next, calculate the perimeter for the skirting board: P = 2 × (length + width).
接着，计算踢脚板所需的周长：P = 2 × (长 + 宽)。
P = 2 × (4.2 + 3.8) = 2 × 8.0 = 16.0 m
The skirting cost is 16.0 m × £4.80/m = £76.80. The total budget will be £219.50 + £76.80 = £296.30.
踢脚板费用为 16.0 m × £4.80/m = £76.80。总预算为 £219.50 + £76.80 = £296.30。
Always remember to add a wastage factor for materials that are cut, and use the correct units throughout.
一定要记住，对需要裁剪的材料添加损耗系数，并始终使用正确的单位。
2. Budgeting a Class Party – Percentages and Decimals | 班级聚会预算——百分数与小数
A class of 28 students plans an end‑of‑term party. The total cost is £210, but the school contributes 35% of the amount. The remaining cost is shared equally among the students. How much does each student pay?
一个 28 名学生的班级计划举办期末聚会。总费用为 £210，但学校承担其中的 35%。剩余费用由学生平均分担。每位学生需支付多少？
First, convert the percentage to a decimal to find the school’s contribution: 35% = 0.35.
首先，将百分数转换为小数以求出学校承担的金额：35% = 0.35。
School contribution = 0.35 × £210 = £73.50
Subtract this from the total to find the amount left for the students.
从总额中减去这个数，得出学生需要支付的金额。
£210 – £73.50 = £136.50
Now divide the remaining amount equally among 28 students.
现在把剩余金额平均分给 28 名学生。
£136.50 ÷ 28 = £4.875
Rounded to the nearest penny, each student pays £4.88. A quick check: 28 × £4.88 = £136.64, which is slightly above £136.50, so collecting £4.87 might leave a small deficit. Organisers would probably ask for £4.88.
四舍五入到最接近的便士，每位学生支付 £4.88。快速检验：28 × £4.88 = £136.64，比 £136.50 略高，因此收取 £4.87 会导致小额亏空。组织者很可能会要求 £4.88。
Understanding percentage calculations is essential for fair cost sharing and simple financial decisions.
理解百分数计算对公平分摊费用和简单的财务决策至关重要。
3. The Cookie Recipe – Ratio and Proportion | 饼干食谱——比率与比例
A basic cookie recipe for 12 cookies uses 200 g of flour, 120 g of butter and 80 g of sugar. You need to bake 30 cookies for a school fair. How much of each ingredient is required?
一份制作 12 块饼干的基础食谱需要 200 克面粉、120 克黄油和 80 克糖。你需要为学校义卖烘焙 30 块饼干。需要每种配料多少？
he scaling factor from 12 to 30 cookies is 30 ÷ 12 = 2.5. Multiply each ingredient by 2.5.
从 12 块饼干扩大到 30 块的放大倍数为 30 ÷ 12 = 2.5。将每种配料乘以 2.5。
Flour: 200 g × 2.5 = 500 g
面粉：200 克 × 2.5 = 500 克
Butter: 120 g × 2.5 = 300 g
黄油：120 克 × 2.5 = 300 克
Sugar: 80 g × 2.5 = 200 g
糖：80 克 × 2.5 = 200 克
Ratios remain the same. The original ratio of flour to butter to sugar is 200:120:80, which simplifies to 5:3:2. Check: 500:300:200 also simplifies to 5:3:2.
比率保持不变。原配方中面粉、黄油、糖的比例为 200:120:80，简化为 5:3:2。检验：500:300:200 同样简化为 5:3:2。
seful for reducing recipes. If you only have 150 g of flour, how many cookies can you make? r 12 cookies, so the unit ratio is 200 ÷ 12 ≈ 16.67 g per cookie. 150 ÷ 16.67 ≈ 9 cookies.
比例对缩小食谱同样有用。如果你只有 150 克面粉，你能做多少块饼干？食谱中用 200 克做 12 块，因此单位比率为 200 ÷ 12 ≈ 16.67 克/块。150 ÷ 16.67 ≈ 9 块饼干。
Mastering ratio and proportion allows flexible scaling in cooking, mixing paints and making models.
掌握比和比例可以在烹饪、调漆和制作模型时灵活缩放。
4. Planning a Cycling Trip – Distance, Speed and Time | 规划骑行之旅——距离、速度和时间
A group of friends cycle from Belfast to Lisburn, a distance of 18 km. They maintain an average speed of 12 km/h, but take a 20‑minute rest halfway. What is the total journey time?
一群朋友从贝尔法斯特骑行到里斯本，全程 18 公里。他们保持 12 公里/小时的平均速度，但在中途休息 20 分钟。整个行程耗时多少？
Use the formula: Time = Distance ÷ Speed.
使用公式：时间 = 距离 ÷ 速度。
Cycling Time = 18 km ÷ 12 km/h = 1.5 hours
Add the 20‑te break. Remember to convert minutes into hours: 20 minutes = 20/60 = 1/3 hour ≈ 0.333 h.
加上 20 分钟休息。记得把分钟转换为小时：20 分钟 = 20/60 = 1/3 小时 ≈ 0.333 小时。
Total Time = 1.5 h + 0.333 h = 1.833 h
That is 1 hour 50 minutes (since 0.833 × 60 = 50 minutes).
即 1 小时 50 分钟（因为 0.833 × 60 = 50 分钟）。
If they need to arrive by 11:00 am, they must set off by 11:00 – 1 h 50 min = 9:10 am.
如果他们需要在上午 11:00 到达，就必须在 11:00 – 1 小时 50 分 = 上午 9:10 出发。
What if they increase their speed to 15 km/h? ), plus 20 min rest, total 1 h 32 min. Small increases in speed can save significant time.
如果他们提速到 15 公里/小时呢？骑行时间变为 18 ÷ 15 = 1.2 小时（1 小时 12 分），加上 20 分钟休息，总计 1 小时 32 分钟。速度小幅提升能节省不少时间。
Distance‑speed‑time relationships are vital for journey planning and interpreting timetables.
距离–速度–时间关系对行程规划和解读时刻表至关重要。
5. Analysing Survey Data – Mean, Median, Mode and Range | 分析调查数据——平均数、中位数、众数和极差
A class surveyed the number of books read by 15 students in one month: 3, 5, 4, 6, 5, 2, 5, 7, 4, 5, 8, 3, 6, 5, 4. Analyse this data set using measures of central tendency and spread.
一个班级调查了 15 名学生一个月内阅读书籍的数量：3, 5, 4, 6, 5, 2, 5, 7, 4, 5, 8, 3, 6, 5, 4。使用集中趋势和离散程度的度量来分析这组数据。
First, arrange the numbers in ascending order: 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 7, 8. The mode is the most frequent value.
首先，把数字按升序排列：2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 7, 8。众数是出现最频繁的值。
Mode = 5 (appears 5 times)
The median is the middle value. With 15 data points, the 8th value is the median.
中位数是中间的数值。数据点有 15 个，第 8 个数是中位数。
Median = 5
To find the mean, sum all values and divide by 15.
要求平均数，先把所有数值相加再除以 15。
Sum = 2+3+3+4+4+4+5+5+5+5+5+6+6+7+8 = 72
Mean = 72 ÷ 15 = 4.8
The range (max – min) = 8 – 2 = 6, showing a spread of 6 books.
极差（最大值 – 最小值）= 8 – 2 = 6，表明数据跨度为 6 本书。
Using all three averages gives a fuller picture. Here, the mean is slightly below the median and mode because of the lower values at 2 and 3.
同时使用三种平均数可以呈现更全面的情况。在这里，平均数略低于中位数和众数，是因为 2 和 3 这些低值的影响。
Presenting data in a frequency table can also help spot patterns quickly.
用频数表呈现数据也有助于快速发现模式。
| Books | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 1 | 2 | 3 | 5 | 2 | 1 | 1 |
6. Probability in a Board Game – Experimental and Theoretical Probability | 棋盘游戏中的概率——实验概率与理论概率
In a board game, a player rolls two fair dice and needs a sum of 7 to win a bonus. The theoretical probability of getting a sum of 7 is 6 out of 36, or 1/6. But is that what actually happens in a short game?
在一个棋盘游戏中，玩家掷两个公平的骰子，需要掷出和为 7 才能赢得奖励。掷出和为 7 的理论概率是 36 分之 6，即 1/6。但在短游戏中实际情况是这样吗？
List all combinations that give 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That’s 6 possibilities out of 36 total outcomes.
列出所有和为 7 的组合：(1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1)。共 6 种可能，总共 36 种结果。
P(sum of 7) = 6/36 = 1/6 ≈ 0.167
probability is 8/30 ≈ 0.267. This is higher than the theoretical probability due to short‑run variation.
假设一名学生掷骰子 30 次并记录结果。出现和为 7 的次数是 8 次。实验概率是 8/30 ≈ 0.267。由于短期波动，这高于理论概率。
xperiment were repeated 300 or 3000 times, the relative frequency would get closer to 4/24? Wait, 1/6 ≈ 0.167. So larger trials tend to match the theoretical value.
如果实验重复 300 次或 3000 次，相对频数会逐渐趋近于 1/6 ≈ 0.167。因此试验次数越多，越接近理论值。
Understanding the difference between experimental and theoretical probability helps in evaluating real‑life risks and in designing fair games.
理解实验概率与理论概率之间的区别，有助于评估现实生活中的风险以及设计公平的游戏。
When using probability to predict, always consider sample size. A single game may deviate, but over many games the odds even out.
用概率进行预测时，要始终考虑样本大小。单次游戏可能会出现偏差，但在多次游戏中，胜率会趋于平均。
7. Filling a Fish Tank – Volume and Capacity | 填满鱼缸——体积与容积
, 35 cm in width and 40 cm in height. How many litres of water are needed to fill it to 90% of its capacity? (1 litre = 1000 cm³)
一个长方形鱼缸长 80 厘米、宽 35 厘米、高 40 厘米。把它装到 90% 的容量需要多少升水？（1 升 = 1000 立方厘米）
First, find the volume of the tank in cubic centimetres.
首先，以立方厘米为单位求出鱼缸的体积。
Volume = length × width × height = 80 × 35 × 40
80 × 35 = 2800; 2800 × 40 = 112 000 cm³
Convert to litres: 112 000 ÷ 1000 = 112 litres. This is the full capacity.
换算为升：112 000 ÷ 1000 = 112 升。这是满容量。
90% of capacity = 0.90 × 112 L = 100.8 L. So the tank needs 100.8 litres of water.
容量的 90% = 0.90 × 112 升 = 100.8 升。因此鱼缸需要 100.8 升水。
f gravel and decorations displace 8 litres of water, the actual water needed to fill to 90% would be 100.8 – 8 = 92.8 L. Always account for displacement.
如果底砂和装饰品排开 8 升水，那么实际需要的水量为 100.8 – 8 = 92.8 升。永远要考虑排量。
Volume calculations also apply to packing boxes, swimming pools, and storage containers.
体积计算同样适用于打包箱、游泳池和储物容器。
8. Scale Drawing of a Playground – Scale Factors and Similarity | 操场比例图——比例因子与相似性
A rectangular playground measures 48 m by 30 m. It is drawn to a scale of 1 : 200 on a plan. What are the dimensions on the drawing, and what is the area of the playground on the plan in cm²?
一个长方形操场长 48 米、宽 30 米。在一张平面图上用 1 : 200 的比例绘制。图纸上的尺寸是多少？平面图上操场的面积是多少平方厘米？
ntimetres first, since the drawing will be measured in cm. 48 m = 4800 cm, 30 m = 3000 cm.
先将实际尺寸转换为厘米，因为图纸通常以厘米计量。48 m = 4800 cm，30 m = 3000 cm。
Divide by the scale factor 200 to obtain drawing dimensions.
除以比例因子 200，得到图纸尺寸。
Length on plan = 4800 cm ÷ 200 = 24 cm
Width on plan = 3000 cm ÷ 200 = 15 cm
tor is 200² = 40 000. 1440 m² = 14 400 000 cm², and 14 400 000 ÷ 40 000 = 360 cm², confirming the result.
平面图上操场的面积为 24 cm × 15 cm = 360 cm²。注意，实际面积（48 × 30 = 1440 m²）并不是简单地除以 200；面积的比例因子是 200² = 40 000。1440 m² = 14 400 000 cm²，14 400 000 ÷ 40 000 = 360 cm²，印证了结果。
Scale drawings are used in maps, architectural plans and model making. Always check whether the scale refers to linear dimensions or area.
比例图用于地图、建筑平面图和模型制作。务必检查比例指的是线性尺寸还是面积。
9. Investigating Number Patterns – Algebraic Expressions | 探究数字模式——代数表达式
Matchsticks are used to form a sequence of squares. One square uses 4 matchsticks, two adjacent squares use 7, three squares use 10, and so on. Find an algebraic rule for the number of matchsticks needed for n squares, and use it to predict the 50th term.
用火柴棍摆出一排正方形。1 个正方形用 4 根火柴，2 个相邻正方形用 7 根，3 个用 10 根，依此类推。找出 n 个正方形所需火柴棍数量的代数规则，并用其预测第 50 项。
Observe the pattern: 4, 7, 10, … The difference between consecutive terms is 3, so the rule is linear. Start with the term‑to‑ ionship: first term = 4 = 3×1 + 1, second = 7 = 3×2 + 1. In general, matchsticks = 3n + 1.
观察模式：4, 7, 10, … 相邻项的差是 3，因此规则是线性的。通过项对项关系：第 1 项 = 4 = 3×1 + 1，第 2 项 = 7 = 3×2 + 1。因此，通项公式为 火柴棍数 = 3n + 1。
Term n: T(n) = 3n + 1
For n = 50, T(50) = 3×50 + 1 = 151 matchsticks.
当 n = 50，T(50) = 3×50 + 1 = 151 根火柴。
We can also represent the sequence in a table to reinforce the concept of the coefficient and constant.
我们也可以用表格表示序列，以强化系数与常数的概念。
| n | 1 | 2 | 3 | … | n |
| Matchsticks | 4 | 7 | 10 | … | 3n+1 |
Writing rules algebraically allows you to make predictions without drawing the whole pattern.
用代数形式写出规则，可以让你不必画出整个模式就能做出预测。
10. Constructing a Triangular Frame – Pythagoras’ Theorem | 搭建三角形框架——勾股定理
A gardener wants to construct a triangular support for climbing plants. The two shorter sides are 2.5 m and 6 m. What length must the sloping beam be to fit exactly, assuming a right‑angled triangle?
一位园丁想为爬藤植物搭建一个三角形支架。两个较短的边分别是 2.5 米和 6 米。如果这是直角三角形，斜梁需要多长才能恰好吻合？
Use Pythagoras’ theorem: in a right‑angled triangle, a² + b² = c², where c is the hypotenuse.
使用勾股定理：在直角三角形中，a² + b² = c²，其中 c 是斜边。
c² = (2.5)² + 6² = 6.25 + 36 = 42.25
Take the square root of both sides.
两边取平方根。
c = √42.25 = 6.5 m
So the sloping beam must be 6.5 m long. Always check your working: (6.5)² = 42.25 matches.
因此斜梁的长度必须为 6.5 米。一定验算：(6.5)² = 42.25 吻合。
Pythagoras’ theorem is extremely useful in construction, navigation, and any situation requiring right
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