📚 School Funfair Project: A Case Study in Year 7 Advanced Mathematics | 学校游园会项目：七年级进阶数学案例研究
In this case study, you will step into the shoes of a Year 7 student planning a school funfair to raise money for charity. You will apply your skills in geometry, arithmetic, algebra, probability, and data handling to solve real-life problems. Every decision you make relies on mathematical reasoning, and this project will show you how to combine different topics into one complete plan.
在本案例研究中，你将扮演一名七年级学生，策划一场为慈善筹款的学校游园会。你需要运用几何、算术、代数、概率和数据处理技能来解决实际问题。每一步决策都依赖于数学推理，这个项目将向你展示如何将不同主题融合成一个完整方案。
1. Getting Started: Understanding the Brief | 起步：理解任务要求
Your team has been given a budget of £200 to organize a funfair on the school field. You must design the layout, buy supplies, set ticket prices, run games and food stalls, and end the day with a clear profit. The headteacher expects a final report showing all calculations and a reflection on what you would improve.
你的团队获得了200英镑预算，在学校操场上组织游园会。你必须设计场地布局、采购物资、设定门票价格、运营游戏和餐饮摊位，并在活动结束时获得明确盈利。校长要求提交一份最终报告，展示所有计算过程，并反思可以改进之处。
2. Designing the Ground Layout: Area and Perimeter | 场地布局设计：面积与周长
The available field is a rectangle measuring 20 metres in length and 15 metres in width. Before placing any stalls, you need to know the total area and how much fencing is required to mark the boundary.
可用场地是一个长20米、宽15米的矩形。在布置任何摊位之前，你需要知道总面积以及围栏需要多长。
To find the area, multiply length by width. To find the perimeter, add the length and width, then multiply by two. Both measurements are essential: area tells you how many stalls can fit, and perimeter tells you the length of rope or barrier needed.
计算面积时，用长乘以宽。计算周长时，将长和宽相加，再乘以2。这两个测量值都很重要：面积告诉你能够容纳多少摊位，周长则告诉你需要多长的绳子或围栏。
Area = 20 × 15 = 300 m²
Perimeter = 2 × (20 + 15) = 2 × 35 = 70 m
3. Setting Up Stalls: Multiplication and Division | 布置摊位：乘除运算
Each stall requires a rectangular space of 2 m by 1.5 m, and you must leave a 1-metre-wide path along one side for queues. The field is 20 m long and 15 m wide. If stalls are arranged in rows with the longer side parallel to the field length, how many whole stalls can you fit?
每个摊位需要一块2米×1.5米的矩形区域，并且你必须在一侧留下1米宽的通道供排队使用。场地长20米、宽15米。如果摊位按排布置，长边与场地长边平行，你最多能放置多少个完整摊位？
. Along the 15 m width, each stall uses 1.5 m, allowing ten rows exactly. Multiply the rows and columns: 6 × 10 = 60 stalls.
沿20米长度方向，每个摊位占用2米，再加1米通道，每个摊位占地3米。20除以3得6个完整摊位，余数很小。沿15米宽度方向，每个摊位占用1.5米，正好能放10排。排数和列数相乘：6×10=60个摊位。
Number of stalls = 20 ÷ 3 ≈ 6 (whole stalls) → 6 × (15 ÷ 1.5) = 6 × 10 = 60
4. Budgeting Basics: Adding and Subtracting Decimals | 预算基础：小数加减
You start with £200. The main costs are stall decorations (£47.65), craft materials (£28.90), prizes (£34.50), and snack ingredients (£53.75). You also need to hire a popcorn machine for £15. Calculate the total spending and the money left before any ticket sales.
你的初始资金为200英镑。主要支出包括：摊位装饰47.65英镑、手工材料28.90英镑、奖品34.50英镑，以及点心原料53.75英镑。你还需要以15英镑租用一台爆米花机。计算总支出，以及门票销售前剩余的资金。
Add all costs carefully, aligning the decimal points. Then subtract the total cost from the budget. This shows whether you can afford everything or need to adjust your plan.
仔细将所有成本相加，小数点对齐。然后从预算中减去总成本。这显示你是否能负担所有开支，或者需要调整计划。
| Decorations | £47.65 |
| Craft materials | £28.90 |
| Prizes | £34.50 |
| Snack ingredients | £53.75 |
| Popcorn machine hire | £15.00 |
| Total cost | £179.80 |
Remaining money = £200.00 − £179.80 = £20.20
5. Ticket Pricing and Revenue: Introducing Algebra | 门票定价与收入：代数入门
You decide to charge £3 per adult and £2 per child. You want to predict the total revenue if exactly x adults and y children attend. Write an expression for the total ticket money and evaluate it when 45 adults and 110 children come.
你决定成人票价为3英镑，儿童票价为2英镑。你希望预测如果有x名成人和y名儿童到场时的总收入。请你写出总票款表达式，并计算当有45名成人和110名儿童时的数值。
l reusable for any crowd size. When x = 45 and y = 110, substitute the values and compute.
总收入等于成人数的三倍加上儿童数的两倍。使用字母可以使模型适用于任何规模的人群。当x=45且y=110时，代入数值并计算。
Revenue (£) = 3x + 2y
For x=45, y=110: 3×45 + 2×110 = 135 + 220 = £355
6. Food and Drink Sales: Ratios and Proportions | 餐饮销售：比与比例
You sell hot dogs and cups of juice. A recipe shows that to make 15 hot dogs you need 1.5 kg of sausages and 750 ml of ketchup. If you expect to sell 60 hot dogs, use ratios to find the required amounts. Juice is mixed in the ratio 1 part concentrate to 4 parts water. How much water is needed to make 12.5 litres of drink?
你出售热狗和果汁。一份食谱显示，制作15个热狗需要1.5公斤香肠和750毫升番茄酱。如果你预计卖出60个热狗，请用比例计算出所需分量。果汁按1份浓缩液比4份水的比例混合调制。制作12.5升饮料需要多少水？
Scaling up: 60 is four times 15, so multiply each ingredient by four. For the juice, the total parts are 5, with water making up 4 out of 5 parts. So water needed is four-fifths of the total volume.
比例放大：60是15的四倍，所以每种成分乘以4。对于果汁，总份数为5，水占5份中的4份。因此所需水量为总体积的4/5。
Sausages: 1.5 kg × 4 = 6 kg; Ketchup: 750 ml × 4 = 3000 ml
Water: 4/5 × 12.5 L = 10 L
7. Game Stall Probability: Simple Fractions and Percentages | 游戏摊位概率：简单分数与百分比
Your spin-the-wheel game has 8 equal sectors: 2 yellow (win a large prize), 2 blue (win a small prize), and 4 white (no prize). A player spins once. Find the probability of winning any prize, and express it as a fraction, decimal, and percentage.
你的转盘游戏设有8个等分扇区：2个黄色（赢大奖）、2个蓝色（赢小奖）、4个白色（无奖）。玩家旋转一次。求出赢得任意奖品的概率，并将其表示为分数、小数和百分比。
Simplify the fraction, then convert to a decimal by dividing numerator by denominator. Multiply by 100 for the percentage.
中奖扇区总数为2+2=4，总扇区为8。化简分数，然后用分子除以分母转化为小数，再乘以100得到百分比。
Probability = 4/8 = ½
As decimal: 0.5; as percentage: 50%
To predict how many winners you might get out of 240 plays, multiply the probability by the number of plays. This helps you order enough prizes.
预测240次游戏中可能有多少中奖者，用概率乘以游戏次数。这有助于你预订足够的奖品。
Expected winners = ½ × 240 = 120 winners
8. Creating a Timetable: Working with Time | 制定时间表：时间计算
The funfair runs from 10:00 to 16:00. You need three 20-minute team breaks for the volunteers, plus 45 minutes for set-up before opening and 30 minutes for clearing afterwards. Calculate the total non-selling time and the actual hours available for selling.
游园会从上午10:00进行到下午16:00。志愿者需要三次20分钟的休息时间，外加开业前45分钟布置和结束后30分钟清理。计算非销售总时间以及实际可用于销售的小时数。
utes. Subtract that from the 6-hour selling window. Convert everything to minutes for accuracy.
整个活动窗口为6小时。布置和清理合计75分钟，在窗口之外。当天时间拆分：总休息时间为3×20=60分钟。从6小时销售窗口中减去这个时间。为求精确，将所有时间转换为分钟。
Total open time = 6 h = 360 min; Break time = 60 min
Selling time = 360 − 60 = 300 min = 5 hours
9. Recording Data: Tally Charts and Bar Graphs | 数据记录：频数表和条形图
16:00 → 28. Organise the data into a frequency table and describe what a bar chart would show. Which hour was busiest and why might attendance drop later?
活动当天，你每小时统计了游客数量：10:00-11:00→52人；11:00-12:00→78人；12:00-13:00→64人；13:00-14:00→41人；14:00-15:00→33人；15:00-16:00→28人。将数据整理成频数表，并描述条形图会展示什么。哪一小时最繁忙？为什么晚些时候人数可能下降？
| Time slot | Frequency |
| 10:00–11:00 | 52 |
| 11:00–12:00 | 78 |
| 12:00–13:00 | 64 |
| 13:00–14:00 | 41 |
| 14:00–15:00 | 33 |
| 15:00–16:00 | 28 |
A bar chart would have time slots on the horizontal axis and visitor frequency on the vertical axis, with bars touching to show continuous time intervals. The tallest bar is for 11:00–12:00, making it the busiest hour, likely because families arrived after morning chores. Numbers fall after lunch as people leave for other activities.
条形图以时间段为横轴，游客频数为纵轴，条形之间紧挨着以表示连续的时间区间。最高的条形是11:00-12:00，这是最繁忙的时段，可能是因为家庭在完成早间事务后抵达。午饭后人数减少，因为人们离开去参加其他活动。
10. Profit and Reflection: Making Decisions | 利润与反思：做出决策
Total ticket revenue was £355 from 45 adults and 110 children. Food sales brought in £142.50, and games generated £87.30. Your total costs were £179.80 plus an extra £25 spent on extra prizes on the day. Calculate the overall profit and reflect: would you change ticket prices, run different games, or alter the timetable next time?
门票总收入为355英镑，来自45名成人和110名儿童。餐饮销售带来142.50英镑，游戏收入为87.30英镑。你的总成本为179.80英镑，外加当天额外花费25英镑添购奖品。计算总体利润，并反思：下次你会改变票价、更换游戏还是调整时间表？
Add all income sources, then add all costs. Subtract total costs from total income to find profit. A positive profit means the funfair was a success, but analysing which part earned the most helps you plan future events more efficiently.
将所有收入来源相加，再将所有成本相加。从总收入中减去总成本以得出利润。正利润意味着游园会成功了，但分析哪部分赚得最多有助于你更有效地计划未来活动。
Total income = £355 + £142.50 + £87.30 = £584.80
Total costs = £179.80 + £25 = £204.80
Profit = £584.80 − £204.80 = £380.00
Reflection: The game stall had a winning probability of 50%, which might have been too generous, reducing potential profit. Next time, you could adjust the wheel to have fewer winning sectors or raise the price per spin. The timetable could start later to capture the lunchtime crowd.
反思：游戏摊位的中奖概率为50%，可能过于慷慨，降低了潜在利润。下次你可以调整转盘，减少中奖扇区或提高每次旋转的价格。时间表可以稍晚开始，以抓住午餐时段的人流。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图