📚 Year 8 SQA Advanced Maths: Case Study in Action | Year 8 SQA 进阶数学：案例分析实战演练
In this article, we will work through a real-world case study that brings together many of the skills you have learnt in Year 8 Advanced Mathematics. You will be planning a class celebration party, using area and perimeter calculations, budgeting, ratios, percentages, statistical surveys and even basic probability. This kind of integrated problem-solving is exactly what SQA assessments value — showing that you can apply mathematical techniques in practical contexts.
在本文中，我们将通过一个真实世界的案例研究，把你 Year 8 进阶数学中学到的许多技能综合运用起来。你将策划一场班级庆祝派对，用到面积与周长计算、预算规划、比例、百分数、统计调查，甚至基础概率。这种综合性的问题解决能力正是 SQA 评估所看重的——证明你能在实际情境中应用数学方法。
1. The Case Study Scenario | 案例背景
A Year 8 class of 30 students has been given permission to organise an end-of-term party. The event will be held in the school's multipurpose hall, a rectangular space measuring 12 metres in length and 8 metres in width. Each student has agreed to contribute £5.00 towards the party fund, creating a total budget to cover decorations, food, drinks and simple entertainment. Your task is to use mathematics to plan the layout, manage costs, analyse food preferences and predict financial outcomes. The decisions you make must be based on accurate calculations, just as you would in a real project.
一个 30 名学生的 Year 8 班级获准组织一场期末派对。活动将在学校多功能厅举行，这是一个长 12 米、宽 8 米的长方形空间。每位学生同意捐款 5.00 英镑作为派对基金，形成一笔总预算，用于装饰、食物、饮料和简单的娱乐活动。你的任务是运用数学来规划场地布局、管理成本、分析食物偏好并预测财务结果。你所做的每个决定都必须基于精确的计算，就像在真实项目中一样。
2. Finding the Floor Area | 计算场地面积
Before arranging tables and seating, we need to know the total available area. The hall is rectangular, so the area is found by multiplying length by width. Using the measurements given:
Area = length × width = 12 m × 8 m = 96 m2
This 96 square metres is the gross space, but we will later need to subtract space for a buffet table, a small stage and walkways to ensure safe movement.
在布置桌椅之前，我们需要知道可用的总面积。大厅是长方形，所以面积等于长乘以宽。利用给出的测量值：
面积 = 长 × 宽 = 12 m × 8 m = 96 m2
这 96 平方米是总面积，但我们后续需要减去自助餐桌、小型舞台和通道的空间，以确保安全通行。
3. Table and Chair Arrangement | 桌椅排列与空间规划
ameter of 1.5 metres. The area a single table covers on the floor is approximately its own ‘ eed 5 tables. The total floor area required for seating is therefore 5 × (2 m × 2 m) = 20 m2. If we also need to fit a buffet table (2 m × 1 m) and leave at least 15 m2 aining area for decorations and any additional activity (e.g., a small games corner) is 96 – 20 – 2 – 15 = 59 m2, which is plenty.
我们计划让学生每 6 人围坐一张圆桌。每张桌子直径为 1.5 米。一张桌子在地面上占据的面积大约是它自身的‘占有区’加上必要的椅子和走动空间。为了规划方便，我们给每套桌椅留出 2 m × 2 m 的方形区域。30 名学生需要 5 张桌子。因此座位所需的总地面面积为 5 × (2 m × 2 m) = 20 m2。如果我们还需要放置一张自助餐桌(2 m × 1 m)并留出至少 15 m2的走动空间，便可以检验座位是否舒适地放下。用于装饰和其他活动(如小型游戏角)的剩余面积为 96 – 20 – 2 – 15 = 59 m2，非常宽裕。
4. Budgeting and Fundraising | 预算与筹款
The class has a simple income model: 30 students × £5.00 = £ drinks and prizes. A quick survey of party shops shows that a basic decorations pack costs £18.50. We decide to set aside £20.00 for decorations to be safe. The remaining £ d, drinks and any game prizes. To track spending carefully, we will maintain a cost table.
这个班级有一个简单的收入模型：30 名学生 × 5.00 英镑 = 150.00 英镑。这笔总额必须分配到装饰、食物、饮料和奖品上。快速调查派对用品商店后得知，一套基础装饰包售价 18.50 英镑。为安全起见，我们决定预留 20.00 英镑用于装饰。剩余的 130.00 英镑可用于食物、饮料和游戏奖品。为谨慎追踪支出，我们将维护一张成本表。
| Budget Item | Amount (£) |
|---|---|
| Total Income (30 × £5) | 150.00 |
| Decorations | 20.00 |
| Remaining for food, drinks, prizes | 130.00 |
This structured budget keeps us from overspending and makes it easy to adjust if we find cheaper suppliers.
这种结构化的预算能防止超支，而且如果有更便宜的供应商，也容易进行调整。
5. Food Cost Calculation | 食物采购成本计算
Pizzas are chosen as the main food item. A large pizza costs £ izzas required is 60 ÷ 8 = 7.5, so we must order 8 pizzas to have enough. The total pizza cost is 8 × £8.00 = £64.00. We will also buy a few sharing bags of crisps, estimated at £6.00. So the total food outlay is £70.00.
比萨被选为主食。一个大号比萨售价 8.00 英镑，可切成 8 等份。根据班级投票，每位学生平均想要 2 片。这意味着我们需要 30 × 2 = 60 片。所需比萨数量为 60 ÷ 8 = 7.5，因此我们必须订购 8 个比萨才够。比萨总成本为 8 × £8.00 = £64.00。我们还会购买几袋分享装薯片，估计花费 6.00 英镑。因此食物总支出为 70.00 英镑。
| Food Item | Unit Cost | Quantity | Total |
|---|---|---|---|
| Large Pizza | £8.00 | 8 | £64.00 |
| Crisps (multipack) | £6.00 | 1 | £6.00 |
| Total Food | £70.00 |
We now have £130.00 – £70.00 = £60.00 left for drinks and entertainment, which gives us flexibility.
现在我们还有 130.00 – 70.00 = 60.00 英镑用于饮料和娱乐，这给了我们灵活的空间。
6. Drink Mixing Ratios | 饮料比例混合
itres in total. Using the ratio, the amount of orange juice needed is (2/3) × 12 L = 8 L, and apple juice is (1/3) × 12 L = 4 L. Orange juice costs £1.20 per litre, and apple juice costs £1.50 per litre. So the drink costs are:
Orange juice: 8 × £1.20 = £9.60
Apple juice: 4 × £1.50 = £6.00
Total drink cost = £15.60
sing variables: let x be the total volume; then orange juice = ²⁄₃ x, apple juice = ¹⁄₃ x.
学生们想要提供一种用橙汁和苹果汁按 2:1 比例混合的水果宾治。我们估计 30 名学生大约共饮用 12 升。根据比例，所需橙汁量为 (2/3) × 12 L = 8 L，苹果汁为 (1/3) × 12 L = 4 L。橙汁每升 1.20 英镑，苹果汁每升 1.50 英镑。因此饮料成本为：
橙汁: 8 × £1.20 = £9.60
苹果汁: 4 × £1.50 = £6.00
饮料总成本 = £15.60
这远在剩余预算之内。我们也可以用变量写出混合规则：设 x 为总体积，则橙汁 = ²⁄₃ x，苹果汁 = ¹⁄₃ x。
7. Survey on Favourite Snacks | 最受欢迎零食调查
To decide on extra snacks, we conduct a survey of the 30 students. Each student picks one favourite from: crisps, chocolate, sweets and biscuits. The results are as follows:
- Crisps: 10 students
- Chocolate: 8 students
- Sweets: 7 students
- Biscuits: 5 students
These can be expressed as fractions out of 30 (10/30, 8/30, etc.) and simplified. We will use them to calculate percentages and later to draw a pie chart.
为了决定额外的零食，我们对 30 名学生进行了一项调查。每人从薯片、巧克力、糖果和饼干中选出自己最喜欢的一样。结果如下：
- 薯片: 10 名学生
- 巧克力: 8 名学生
- 糖果: 7 名学生
- 饼干: 5 名学生
这些数据可以表示为以 30 为分母的分数 (10/30, 8/30 等) 并化简。我们将用它们来计算百分数，并随后绘制饼图。
8. Data Visualisation: Pie and Bar Charts | 数据可视化：饼图与柱状图
nd a bar chart. For the pie chart, the angle for each category is (frequency / 30) × 360°:
- Crisps: (10/30)×360° = 120°
- Chocolate: (8/30)×360° = 96°
- Sweets: (7/30)×360° = 84°
- Biscuits: (5/30)×360° = 60°
These angles sum to 360°. A bar chart can also be drawn with frequency on the vertical axis and snack type on the horizontal axis. Representing data in multiple forms is an important skill, as it allows different audiences to quickly see the patterns. For instance, the pie chart immediately shows that crisps take up ¹⁄₃ of the whole, which matches the simplified fraction 1/3.
调查结果最好用饼图和柱状图展示。对于饼图，每个类别的角度为 (频数 / 30) × 360°：
- 薯片: (10/30)×360° = 120°
- 巧克力: (8/30)×360° = 96°
- 糖果: (7/30)×360° = 84°
- 饼干: (5/30)×360° = 60°
这些角度总和为 360°。柱状图也可以绘制出来，纵轴为频数，横轴为零食种类。以多种形式呈现数据是一项重要技能，因为它能帮助不同受众迅速看出规律。例如，饼图立刻显示出薯片占据了整体的 ¹⁄₃，这与化简后的分数 1/3 吻合。
9. Profit and Loss Calculation | 盈亏计算
Let’s finalise the party accounts. So far we have:
| Category | Spent (£) |
|---|---|
| Decorations | 20.00 |
| Food (pizza + crisps) | 70.00 |
| Drinks (juices) | 15.60 |
| Total Costs | 105.60 |
With initial income of £150.00, the surplus is £150.00 – £105.60 = £ aise additional money for a class charity. Each raffle strip of 5 tickets will be sold for £1.00, and we aim to sell 40 strips. This brings in an extra £40.00. The raffle prizes will cost £10.00. The final profit will be the original surplus plus raffle net income: £44.40 + (£40.00 – £10.00) = £74.40. All calculations must be clearly set out to ensure openness.
让我们最后结算派对账目。目前我们已有：
| 类别 | 支出 (£) |
|---|---|
| 装饰 | 20.00 |
| 食物(比萨+薯片) | 70.00 |
| 饮料(果汁) | 15.60 |
| 总成本 | 105.60 |
初始收入为 150.00 英镑，盈余为 150.00 – 105.60 = 44.40 英镑。然而，我们计划在派对上推出小型抽奖活动，为班级慈善基金筹款。每条 5 张的抽奖券条售价 1.00 英镑，我们目标售出 40 条。这会额外带来 40.00 英镑。抽奖奖品成本为 10.00 英镑。最终利润将是原始盈余加上抽奖净收入：44.40 + (40.00 – 10.00) = 74.40 英镑。所有计算必须清晰列出，以保证透明度。
10. A Probability Game: Lucky Spinner | 概率游戏：幸运转盘
g the party. The spinner is divided into 8 equal sectors: 1 sector is gold (grand prize, a £2.00 voucher), 2 sectors are silver (small prize, a £ e (no prize). The probability that a single spin wins the grand prize is 1/8 = 0.125 (12.5%). The probability of a small prize is 2/8 = 1/4 = 0.25 (25%). If we charge 50p per spin, the expected gain per spin from entry fees is £0.50, while the expected payout per spin is (1/8)×£2.00 + (2/8)×£0.50 = £0.25 + £0.125 = £0.375. The expected profit per spin is £0.50 – £0.375 = £0.125. If 80 spins are taken across the party, we can expect an additional profit of 80 × £0.125 = £10.00. This aligns with our fundraising goal.
作为娱乐，我们在派对的一个摊位上设计了一个转盘游戏。转盘被分成 8 个相等的扇形：1 个金色区（大奖，一张 2.00 英镑的代金券），2 个银色区（小奖，一枚 0.50 英镑的徽章），5 个白色区（无奖）。一次旋转赢得大奖的概率为 1/8 = 0.125（12.5%）。获得小奖的概率为 2/8 = 1/4 = 0.25（25%）。若每次旋转收费 50 便士，每次旋转的门票收入期望为 0.50 英镑，而每次旋转的期望奖金支出为 (1/8)×2.00 + (2/8)×0.50 = 0.25 + 0.125 = 0.375 英镑。每次旋转的期望利润为 0.50 – 0.375 = 0.125 英镑。如果整个派对期间有 80 次旋转，我们可以期望获得额外利润 80 × 0.125 = 10.00 英镑。这与我们的筹款目标一致。
11. Reflection and Application | 反思与应用
This case study demonstrates how seemingly separate topics — area, ratio, budgeting, statistics and probability — work together to solve a single, real-life project. You have seen that a clear plan, supported by careful arithmetic, leads to a successful event while avoiding costly mistakes. In SQA assessments, you will often encounter multi-step problems exactly like this, where you must select the appropriate mathematical tool, perform calculations accurately and interpret your results. Practice similar case studies, such as planning a school trip or designing a garden, to strengthen your problem-solving skills.
这个案例研究展示了看似分离的主题——面积、比例、预算、统计和概率——如何协同工作以解决一个真实的项目。你已经看到，一个清晰的计划加上仔细的算术运算，会带来成功的活动，同时避免代价高昂的错误。在 SQA 评估中，你经常会遇到与此完全类似的多步骤问题，你必须选择合适的数学工具，准确地进行计算并解释结果。通过练习类似的案例研究，比如规划一次学校旅行或设计一个花园，来强化你的问题解决能力。
Published by TutorHao | Advanced Maths Revision Series | aleveler.com
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