📚 Year 9 AQA Maths: Case Study Practice | AQA 九年级数学：案例分析实战演练
In this article, we work through a detailed real‑world case study built around a school carnival. Each step shows how the topics you learn in Year 9 AQA Mathematics – area, scale, percentages, algebra, probability, statistics, volume, best buys and currency conversion – are the exact tools needed to plan a successful event. Follow along with the calculations and deepen your problem‑solving skills.
本文通过一个围绕学校嘉年华的详细案例研究，展示九年级 AQA 数学课程中的各个知识点——面积、比例尺、百分数、代数、概率、统计、体积、最优购买和货币换算——如何成为策划一场成功活动所必需的工具。请跟随计算过程，提高你的问题解决能力。
1. The Carnival Case Study | 嘉年华案例简介
The Year 9 mathematics class at Greenwood Academy has been asked to lead the planning of the annual school carnival. Their task is to use the maths they have studied all year to design the layout, manage the money, order supplies, run games and analyse data from the day. The whole project is a genuine example of ‘mathematics in action’.
格里伍德学院九年级数学班受邀牵头策划一年一度的学校嘉年华。他们的任务是把全年所学的数学知识用于场地设计、资金管理、物资采购、游戏运营和当日数据分析。整个项目就是“数学在行动”的真实案例。
Working through this case study will help you see how topics like percentages and probability appear outside the classroom, and how they connect to each other when tackling a larger challenge.
通过本案例研究，你将看到百分数和概率等知识如何出现在课堂之外，以及它们是如何在应对更大挑战时相互关联的。
2. Planning the Layout: Area Calculations | 场地规划：面积计算
The main activity field is a rectangle measuring 45 m by 30 m. A square stage with side length 8 m will be placed in one corner. All the remaining space is available for game stalls, food tents and seating. The first job is to work out exactly how much area the stall organisers can use.
主活动场地是一个长 45 米、宽 30 米的长方形。一个边长为 8 米的正方形舞台将被放置在场地的一角。剩余空间全部用于游戏摊位、食品帐篷和休息区。第一项任务就是准确计算摊位组织者可以使用的面积。
Step 1: Area of the whole field = length × width = 45 × 30 = 1350 m².
步骤 1：整个场地的面积 = 长 × 宽 = 45 × 30 = 1350 平方米。
Step 2: Area taken by the stage = 8 × 8 = 64 m².
步骤 2：舞台占用的面积 = 8 × 8 = 64 平方米。
Step 3: Remaining area for stalls = 1350 − 64 = 1286 m².
步骤 3：摊位可用面积 = 1350 − 64 = 1286 平方米。
This value will be used every time a new stall is added, making sure the space is never overbooked.
每次新增摊位时都会用到这个数值，以确保空间不会被超量预订。
3. Scaling the Map | 比例绘图
The team needs a clear site map so that each stall holder knows exactly where to set up. They decide to use a scale of 1 cm to represent 5 m on the ground. This is a proportional reasoning skill straight from the ratio and proportion topic.
策划团队需要一张清晰的场地地图，以便每位摊主准确知道自己的位置。他们决定用 1 厘米代表实际 5 米的比例尺。这直接运用了比与比例章节中的比例推理技能。
Width on the map = 30 ÷ 5 = 6 cm. So the field will be drawn as a 9 cm by 6 cm rectangle.
地图上的长度 = 实际长度 ÷ 5 = 45 ÷ 5 = 9 厘米。地图上的宽度 = 30 ÷ 5 = 6 厘米。因此场地将被画成一个 9 厘米 × 6 厘米的长方形。
awing. Every detail on the map must follow this scale, which lets the team check clearances and paths without visiting the field.
舞台边长 8 米，在图中表现为边长 8 ÷ 5 = 1.6 厘米的正方形。地图上的每个细节都必须遵循此比例，这使团队无需亲临现场就能核查间距和通道。
4. Food Stall Finances: Percentages | 食品摊位财务：百分数
One of the food stalls plans to sell homemade cakes. The ingredients cost £30 in total. The team wants to apply a 40% profit on the cost price to cover other expenses and raise money for the school. What total revenue should the cake stall aim for?
其中一个食品摊位计划售卖自制蛋糕。原料成本合计 30 英镑。团队希望在成本价的基础上获取 40% 的利润，以支付其他开支并为学校筹款。蛋糕摊位应以多少总收入为目标？
Profit = 40% of £30 = 0.40 × 30 = £12. Total selling price = cost + profit = 30 + 12 = £42. If they cut the cakes into 60 equal slices, the price per slice must be £42 ÷ 60 = £0.70.
利润 = 30 英镑的 40% = 0.40 × 30 = 12 英镑。总销售额 = 成本 + 利润 = 30 + 12 = 42 英镑。如果把蛋糕切成 60 等份，每份售价为 42 ÷ 60 = 0.70 英镑。
The same stall also sells drinks. A can of lemonade usually costs £1.20, but on carnival day they offer a 15% discount. The discounted price is 85% of the original: 0.85 × 1.20 = £1.02. Working both with percentage increases and decreases is a core skill in any budget.
该摊位还售卖饮料。一罐柠檬水通常售价 1.20 英镑，但在嘉年华当天打八五折。折后价为原价的 85%：0.85 × 1.20 = 1.02 英镑。无论是百分数增加还是减少，都是预算中的核心技能。
5. Ticket Sales and Price Strategy | 门票销售与定价策略
The entrance ticket model is a classic setting for forming and solving equations. Adult tickets are priced at £5 and child tickets at £3. The school expects around 300 visitors and wants to raise exactly £1200 from ticket sales. How many adults and children should attend if equal numbers of each type are sold?
门票模型是建立和求解方程的经典场景。成人票定价 5 英镑，儿童票 3 英镑。学校预计约有 300 名访客，并希望门票收入恰好为 1200 英镑。如果成人票和儿童票售出数量相同，各应售出多少张？
8x. Set 8x = 1200, giving x = 150. Therefore the school needs 150 adults and 150 children.
设成人票售出 x 张，儿童票也为 x 张。总收入 = 5x + 3x = 8x。令 8x = 1200，解得 x = 150。因此学校需要 150 名成人和 150 名儿童参加。
.1, which needs rounding in practice. This highlights how real constraints shape the maths.
若预期儿童人数是成人的两倍，方程变为：设成人 = a，儿童 = 2a。则 5a + 3(2a) = 1200 → 11a = 1200 → a ≈ 109.1，实际中需要取整。这突显了实际约束如何影响数学处理。
6. Game Probability: Spinner Fun | 游戏概率：旋转轮盘
The carnival includes a prize spinner. The wheel is divided into 8 equal sectors coloured: 2 red, 3 blue, 2 yellow and 1 gold. A player wins a small prize if the spinner lands on red, and a large prize if it lands on gold. The probability of each outcome is a simple fraction of the whole.
嘉年华设有一个抽奖转盘。转盘被均匀分成 8 个扇形：2 个红色、3 个蓝色、2 个黄色和 1 个金色。指针停在红色区得小奖，停在金色区得大奖。每种结果的概率都是一个简单的分数。
1/4. P(gold) = 1/8. The probability of winning any prize (red or gold) is 1/4 + 1/8 = 3/8.
P(红色) = 2/8 = 1/4；P(蓝色) = 3/8；P(黄色) = 2/8 = 1/4；P(金色) = 1/8。赢得任意奖品（红色或金色）的概率为 1/4 + 1/8 = 3/8。
le, if 400 people are expected to spin, about 400 × 3/8 = 150 are likely to win something.
理解这些分数有助于团队决定购买多少奖品。例如，如果预计有 400 人参与转盘游戏，大约 400 × 3/8 = 150 人可能获奖。
7. Surveying Visitor Preferences | 调查游客偏好
A week before the event, the team surveys 50 students to find out which carnival activity they most look forward to: music, games or food. The results are recorded in the table below.
活动前一周，团队调查了 50 名学生，了解他们最期待的嘉年华活动：音乐、游戏还是美食。结果记录在下表中。
| Activity | Frequency | Sector angle for pie chart |
| Music | 20 | (20/50)×360 = 144° |
| Games | 18 | (18/50)×360 = 129.6° |
| Food | 12 | (12/50)×360 = 86.4° |
draw a pie chart to present the data visually. They also calculate percentages: Music = 40%, Games = 36%, Food = 24%. This information helps them allocate more budget to the most popular area.
团队将绘制饼图来直观展示数据。他们还计算了百分比：音乐 40%，游戏 36%，美食 24%。这些信息有助于他们将更多预算分配给最受欢迎的区域。
8. Building a Display: 3D Shapes | 搭建展示：立体图形
For a prize display, students construct a closed box in the shape of a rectangular prism (cuboid) with internal dimensions 50 cm long, 30 cm wide and 25 cm high. They need to know the volume to check it can hold enough items, and the surface area to buy wrapping paper.
作为奖品陈列，学生们制作了一个长方体形状的封闭盒子，内部尺寸为长 50 厘米、宽 30 厘米、高 25 厘米。他们需要知道体积以确认能容纳足够的物品，还需要表面积来购买包装纸。
= 2(lw + lh + wh) = 2(50×30 + 50×25 + 30×25) = 2(1500 + 1250 + 750) = 2 × 3500 = 7000 cm².
体积 = 长 × 宽 × 高 = 50 × 30 × 25 = 37,500 立方厘米。长方体表面积 = 2(lw + lh + wh) = 2(50×30 + 50×25 + 30×25) = 2(1500 + 1250 + 750) = 2 × 3500 = 7000 平方厘米。
If one roll of wrapping paper covers 2000 cm², they will need at least 4 rolls to cover the box completely. The volume check confirms that the box can store around 30 small prize boxes of 1200 cm³ each.
如果一卷包装纸能覆盖 2000 平方厘米，他们至少需要 4 卷才能完全包好盒子。体积核算确认该盒子可储存约 30 个每个 1200 立方厘米的小奖品盒。
9. Managing the Budget: Discounts and Best Buys | 预算管理：折扣与最优购买
The decorations team must buy 240 party balloons. Two local shops offer different deals. Shop A sells packs of 30 balloons for £4.50; Shop B sells packs of 50 balloons for £7.00. Which shop gives the better value, and how much can they save?
装饰小组需要购买 240 个派对气球。两家本地商店提供不同的套餐。A 店每包 30 个气球售价 4.50 英镑；B 店每包 50 个气球售价 7.00 英镑。哪家店性价比更高，能节省多少钱？
Price per balloon at Shop A = £4.50 ÷ 30 = £0.15. At Shop B = £7.00 ÷ 50 = £ balloons at Shop B they would need 240 ÷ 50 = 4.8 packs, so they must buy 5 packs, costing £35.00. At Shop A they would need exactly 8 packs, costing £36.00. The saving is £1.00.
A 店每个气球单价 = 4.50 ÷ 30 = 0.15 英镑。B 店单价 = 7.00 ÷ 50 = 0.14 英镑。因此 B 店更便宜。在 B 店购买 240 个气球需要 240 ÷ 50 = 4.8 包，必须买 5 包，花费 35.00 英镑。在 A 店则正好需要 8 包，花费 36.00 英镑。可节省 1.00 英镑。
Although the unit cost points to B, the need to buy whole packs means the ‘best buy’ also depends on the total quantity required. This is a realistic layer to value‑for‑money problems.
尽管单价指向 B 店，但需要整包购买的事实意味着“最优购买”也取决于所需的总数量。这为性价比较问题增添了现实层面的思考。
10. Exchange Rates for International Snacks | 国际零食汇率换算
ng back some American sweets for the carnival. The total cost quoted by the US supplier is $24. If the exchange rate is £1 = $1.20, how much must the school pay in pounds?
一位家长从美国旅行归来，提议为嘉年华带回一些美国糖果。美国供应商的报价总额为 24 美元。如果汇率为 £1 = $1.20，学校需要支付多少英镑？
Amount in pounds = dollar amount ÷ exchange rate = 24 ÷ 1.20 = £20. The school pays £20. If the payment needed to be in euros instead, and £1 = €1.15, the cost would be 20 × 1.15 = €23.
英镑金额 = 美元金额 ÷ 汇率 = 24 ÷ 1.20 = 20 英镑。学校支付 20 英镑。如果换作欧元支付，且 £1 = €1.15，则费用为 20 × 1.15 = 23 欧元。
Exchange rate work reinforces the idea that multiplying and dividing by the conversion factor moves you correctly between currencies, depending on which direction you are converting.
汇率换算强化了一个概念：根据换算方向，乘以或除以转换因子即可在不同货币之间正确换算。
11. Reviewing the Event with Averages | 用平均数回顾活动
After the carnival, the team looks at the number of visitors recorded each hour: 55, 68, 72, 61, 49, 55. They want to summarise these figures with measures of central tendency to report back to the headteacher.
嘉年华结束后，团队查看了每小时记录的访客人数：55, 68, 72, 61, 49, 55。他们希望用集中趋势量数来总结这些数据，以便向校长汇报。
Mean: (55+68+72+61+49+55) ÷ 6 = 360 ÷ 6 = 60 visitors per hour. Median: order the data — lies between the 3rd and 4th: (55+61) ÷ 2 = 58. Mode: 55 appears twice, so the mode is 55.
平均数：(55+68+72+61+49+55) ÷ 6 = 360 ÷ 6 = 60 人/小时。中位数：将数据排序——49, 55, 55, 61, 68, 72。中间值位于第 3 和第 4 个之间：(55+61) ÷ 2 = 58。众数：55 出现两次，因此众数为 55。
The mean gives an overall average, while the median is not affected by the low 49, and the mode shows the most common hourly count. Using all three together paints a fuller picture.
平均数反映整体平均水平，中位数不受最低值 49 的影响，众数则显示最常见的每小时人数。三者结合使用可呈现更全面的情况。
12. Reflecting on the Maths Used | 反思所用数学
Across the carnival project, the Year 9 class applied: area and perimeter for the site layout, scale drawing for the map, percentages for profit and discounts, algebraic equations for ticket targets, probability for game design, statistical graphs and averages for surveys and reviews, volume and surface area for 3D models, value‑for‑money calculations for best buys, and exchange rates for international purchases. Every strand of their AQA syllabus was touched in one coherent task.
在整个嘉年华项目中，九年级学生运用了：场地布置的面积与周长，地图的比例绘图，利润与折扣的百分数，门票目标的代数方程，游戏设计的概率，调查与回顾的统计图表和平均数，三维模型的体积与表面积，最优购买的性价比较计算，以及国际采购的汇率换算。AQA 教学大纲的每个分支都在一个连贯的任务中得到了实践。
Case studies like this show that mathematics is not an isolated set of techniques but a toolkit for solving real problems. Practise constructing your own scenarios – perhaps a sports day or a charity fundraiser – to strengthen your reasoning and revision.
像这样的案例研究表明，数学并非一组孤立的技巧，而是解决现实问题的工具箱。试着构建你自己的场景——比如运动会或慈善筹款——来强化你的推理能力和复习效果。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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