📚 Case Study Practical Exercises for Year 8 CCEA Additional Mathematics | Year 8 CCEA 进阶数学：案例分析实战演练
In Additional Mathematics, the ability to apply mathematical concepts to real-world situations is essential. This article presents a series of carefully designed case studies that allow Year 8 pupils to explore topics such as algebra, geometry, statistics, and number through practical problems. Each case study includes a clear problem statement, a step-by-step analysis, and a fully worked solution or investigation – exactly the style required by CCEA for developing reasoning and problem-solving skills. By working through these examples, students will sharpen their ability to interpret a scenario, choose the correct mathematics, and communicate findings effectively.
在进阶数学中，将数学概念应用于现实情境的能力至关重要。本文提供了一系列精心设计的案例研究，让八年级学生通过实际问题探索代数、几何、统计和数论等主题。每个案例都包含清晰的问题陈述、分步分析和完整的解答或探究过程——这正是CCEA培养学生推理和问题解决能力所要求的风格。通过练习这些实例，学生将提高解读情景、选择正确数学方法并有效表达结论的能力。
1. Case Study 1: Fencing a Rectangular Garden | 案例一：围建长方形花园
A gardener has 32 metres of fencing and plans to build a rectangular vegetable patch against a long brick wall. Only three sides need fencing: two equal widths (x metres) and one length (L metres) opposite the wall. The gardener wants to maximise the planting area. What width x achieves the greatest area, and what is that maximum area?
一个园丁有32米的围栏，打算靠着一面长砖墙建一个长方形菜地。只需要围三边：两侧等宽（x 米）和与墙相对的一边长度（L 米）。园丁想使种植面积最大化。宽度 x 是多少时面积最大？最大面积是多少？
ression. We can test integer values of x to find the maximum area, or complete the square.
我们用 x 表示长度。因为围栏总长为32米：2x + L = 32 → L = 32 − 2x。面积 A = x × L，所以 A = x(32 − 2x) = 32x − 2x²。这是一个二次表达式。我们可以尝试 x 的整数值来寻找最大面积，或者用配方法。
Using tabular investigation for whole numbers (keeping L positive, so x < 16):
用表格检验整数值（需保持 L 为正，即 x < 16）：
| Width x (m) | Length L = 32 − 2x (m) | Area A = x × L (m²) |
|---|---|---|
| 5 | 22 | 110 |
| 6 | 20 | 120 |
| 7 | 18 | 126 |
| 8 | 16 | 128 |
| 9 | 14 | 126 |
| 10 | 12 | 120 |
rly shows a maximum area of 128 m² when x = 8 m. Algebraically, completing the square: A = –2(x² − 16x) = –2[(x − 8)² − 64] = – l width is 8 metres, giving a length of 16 metres and a maximum area of 128 square metres.
表格清楚显示当 x = 8 米时面积最大，为128平方米。用代数配方法验证：A = –2(x² − 16x) = –2[(x − 8)² − 64] = –2(x − 8)² + 128。由于平方项被减去，面积最大值为128，此时 x = 8。因此最佳宽度是8米，对应长度为16米，最大面积为128平方米。
2. Case Study 2: The Mystery Number | 案例二：神秘的数字
I am thinking of a two-digit number. The sum of its digits is 11. If I reverse the digits, the new number is 27 more than the original. What is the original number?
我在想一个两位数。它的各位数字之和是11。如果将数字颠倒，得到的新数比原数大27。原数是多少？
Let the tens digit be t and the units digit be u. The original number is 10t + u, and the reversed number is 10u + t. We are given:
t + u = 11 (Equation 1)
10u + t = (10t + u) + 27 (Equation 2)
设十位数字为 t，个位数字为 u。原数为 10t + u，颠倒后的数为 10u + t。已知：
t + u = 11 （方程1）
10u + t = (10t + u) + 27 （方程2）
4. The original number is 47. Checking: 4+7=11, 74 − 47 = 27. So the mystery number is 47.
化简方程2：10u + t = 10t + u + 27 → 10u − u + t − 10t = 27 → 9u − 9t = 27 → u − t = 3。现在我们有方程组：u − t = 3 和 t + u = 11。两式相加得 2u = 14 → u = 7，继而 t = 4。原数为47。验证：4+7=11，74 − 47 = 27。因此神秘数字是47。
3. Case Study 3: School Trip Fundraiser | 案例三：学校旅行筹款
Students run a cake stall to raise money for a school trip. Each cake costs £2 to make and is sold for £5. The stall hire fee is £10. The class wants to make a profit of at least £50. How many cakes must they sell?
学生们摆蛋糕摊为学校旅行筹款。每个蛋糕的制作成本为2英镑，售价为5英镑。摊位租金为10英镑。班级希望获得至少50英镑的利润。他们需要卖出多少个蛋糕？
60 → n ≥ 20. The students must sell at least 20 cakes. If they sell exactly 20, profit is £50; if they sell 25, profit is 3×25−10 = £65. Representing the solution on a number line shows n is an integer from 20 upwards.
设卖出的蛋糕数为 n。总收入 = 5n，总成本 = 2n + 10。利润 = 5n − (2n + 10) = 3n − 10。要求 3n − 10 ≥ 50。求解：3n ≥ 60 → n ≥ 20。学生们必须至少卖出20个蛋糕。若正好卖出20个，利润为50英镑；若卖出25个，利润为3×25−10 = 65英镑。在数轴上表示解集，n 为20及以上的整数。
4. Case Study 4: Designing an Open Box | 案例四：设计无盖盒子
olume V of the box is given by V = x(30 − 2x)(21 − 2x). We investigate which x gives the largest volume by testing sensible x values (x must be less than 10.5).
一张30厘米×21厘米的纸板被用来制作一个无盖盒子。从每个角剪去边长为x厘米的正方形，然后将四边折起。盒子的容积 V = x(30 − 2x)(21 − 2x)。我们通过尝试合理的x值（x必须小于10.5）来探究哪个x能使容积最大。
We calculate V for x = 1, 2, 3, 4, 5, 6. Using a table:
我们计算 x = 1, 2, 3, 4, 5, 6 时的容积。用表格：
| x (cm) | 30−2x (cm) | 21−2x (cm) | Volume V (cm³) |
|---|---|---|---|
| 1 | 28 | 19 | 532 |
| 2 | 26 | 17 | 884 |
| 3 | 24 | 15 | 1080 |
| 4 | 22 | 13 | 1144 |
| 5 | 20 | 11 | 1100 |
| 6 | 18 | 9 | 972 |
.2) suggests the maximum occurs close to 4 cm. This case study shows how a simple table can guide optimisation before calculus is introduced.
从测试值看，当 x = 4 cm 时体积最大，为1144 cm³。在 x = 4 附近更精细的测试（如 3.8、4.2）表明最大值出现在4 cm附近。此案例展示了在引入微积分之前，如何用简单的表格指导优化问题。
5. Case Study 5: Temperature Patterns | 案例五：气温模式
During one week in May, the highest daily temperatures (°C) recorded at noon were: Monday 15, Tuesday 17, Wednesday 19, Thursday 21, Friday 23. The sequence appears to follow a linear pattern. Predict the temperature for the following Sunday, and find a formula for the nth day.
五月里一周的中午最高气温记录（°C）为：星期一15，星期二17，星期三19，星期四21，星期五23。这个数列似乎呈线性规律。预测接下来星期天的气温，并找出第n天的公式。
Observing the sequence: 15, 17, 19, 21, 23 – +13=23, correct. Saturday would be day 6: T = 2×6+13 = 25 °C. Sunday (day 7): T = 2×7+13 = 27 °C. The arithmetic sequence formula allows predictions beyond the data. This is a simple example of linear modelling.
观察数列：15, 17, 19, 21, 23——每天的公差为+2。若设n=1代表星期一，则第n天的气温T为：T = 15 + 2(n − 1) = 2n + 13。检查：n=5得T=2×5+13=23，正确。星期六为第6天：T=2×6+13=25°C。星期天（第7天）：T=2×7+13=27°C。等差数列公式可用于预测已知范围外的值。这是线性建模的简单实例。
6. Case Study 6: Fair Games with Two Dice | 案例六：两颗骰子的公平游戏
Two fair six-sided dice are rolled. A game has the rule: if the sum is exactly 7, the player wins £1; otherwise the player loses £1. Is this a fair game? If not, how could the winning condition be changed to make the game fair?
同时掷两颗公平的六面骰子。一个游戏的规则是：如果点数和正好为7，玩家赢1英镑；否则输1英镑。这个游戏公平吗？如果不公平，如何改变获胜条件使之公平？
likely sums. The favourable pairs for sum 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) – ose) = 30/36 = 5/6. Expected gain per game = (1)×(1/6) + (−1)×(5/6) = 1/6 − 5/6 = −4/6 ≈ −£0.67. The player loses on average, so the game is not fair; it favours the organiser.
首先列出所有可能结果：共有6×6=36种等可能的总和点。点数和为7的组合有：(1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1)——共6种。因此P(赢)=6/36=1/6，P(输)=30/36=5/6。每局预期收益 = (1)×(1/6) + (−1)×(5/6) = 1/6 − 5/6 = −4/6 ≈ −0.67英镑。玩家平均会输钱，所以游戏不公平；它对庄家有利。
To make it fair, we need expected gain = 0. If the win prize is £W and loss remains £1, then W × (1/6) − 1 × (5/6) = 0 ⇒ W/6 = 5/6 ⇒ W = 5. So paying £5 for a sum of 7 would make the game fair (expected return zero). Alternatively, keep £ , 4, 10, 11, 12, balancing probabilities. This investigation reinforces the concept of expectation.
为使游戏公平，需要预期收益为零。若赢时奖金为£W，输时仍失1英镑，则W × (1/6) − 1 × (5/6) = 0 ⇒ W/6 = 5/6 ⇒ W = 5。因此和为7时支付5英镑就能让游戏公平（预期回报为零）。或者保持1英镑奖金但改变获胜的点和。例如，若赢的点数为6、7、8，概率为16/36，并非恰好使预期为零。一个公平游戏可以设计为：若和为2、3、4、10、11、12时赢，以此平衡概率。本探究强化了期望值概念。
7. Case Study 7: Scaling a Recipe | 案例七：食谱的比例缩放
A cookie recipe requires 200 g plain flour, 100 g sugar, 50 g butter and 1 egg to make 12 cookies. You want to bake 30 cookies for a party. However, you only have 450 g flour at home. Do you have enough flour, and if not, what is the maximum number of cookies you can make while keeping the ingredients in the same ratio?
一份饼干食谱需要200克中筋面粉、100克糖、50克黄油和1个鸡蛋，可做12块饼干。你想为聚会烤30块饼干。但家里只有450克面粉。面粉够吗？如果不够，在保持原料比例相同的情况下，最多可以做多少块饼干？
= 500 g. Since you have only 450 g, it is not enough. The limiting factor is flour. Set up the ratio: 200 g flour ↔ 12 cookies. With 450 g flour, number of cookies = (450 / 200) × 12 = 2.25 × 12 = 27. So maximum cookies is 27. Check sugar: 100 g → (27/12)×100 = 225 g needed; butter: 50 g → 112.5 g; egg: 1 egg → 2.25 eggs, so you would need 3 eggs practically. The recipe scale must respect all ingredients, but flour dictates the limit here.
首先计算30块饼干需要的面粉量。每块饼干需面粉 = 200 g ÷ 12 ≈ 16.67 g。30块需面粉 = 30 × (200/12) = 30 × 50/3 = 500克。而家中只有450克，因此不够。限制因素是面粉。建立比例：200克面粉 ↔ 12块饼干。用450克面粉可得饼干数 = (450 / 200) × 12 = 2.25 × 12 = 27块。所以最多可做27块。检查糖：100克 → (27/12)×100 = 225克需要；黄油：50克 → 112.5克；鸡蛋：1个 → 2.25个，所以实际需要3个鸡蛋。食谱缩放必须考虑所有原料，但此处面粉决定了上限。
8. Case Study 8: Saving for a Games Console | 案例八：为游戏机存钱
Emma has £30 in savings. She receives £25 pocket money each month and spends £15 on snacks and magazines. She wants to buy a console costing £230. How many full months must she save before she can afford it?
艾玛有30英镑存款。她每月得到25英镑零花钱，并花去15英镑买零食和杂志。她想买一台230英镑的游戏机。她需要存几个完整月份才能买得起？
Monthly savings = income − spending = 25 − 15 = £ ≥ 20. So Emma must save for 20 whole months. Checking: after 20 months, total = 30 + 200 = £ waiting one extra month for a small buffer. The inequality models the financial planning neatly.
每月储蓄 = 收入 − 支出 = 25 − 15 = 10英镑。经过m个月后，总储蓄 = 30 + 10m。需要 30 + 10m ≥ 230。求解：10m ≥ 200 → m ≥ 20。因此艾玛必须存20个完整月份。验证：20个月后，总金额 = 30 + 200 = 230英镑，刚好够。如果第20个月她仍需花掉零花钱，则月底正好凑齐。现实情况下她可能会考虑多存一个月留出一点余量。该不等式清晰地为理财计划建立了模型。
Published by TutorHao | Additional Mathematics Revision Series | aleveler.com
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