📚 Year 8 CCEA Maths: Case Study Practice | Year 8 CCEA 数学：案例分析实战演练
Case studies are a vital part of the Year 8 CCEA Maths curriculum because they teach you to apply your mathematical knowledge to real-life situations. Instead of just solving isolated sums, you will read about a scenario, extract the important numbers and relationships, and then decide which mathematical skills you need. This article takes you through a comprehensive school fair case study, showing step by step how to tackle questions involving percentages, ratio, geometry, statistics, probability and algebra. By working through this practice, you will build confidence and sharpen your problem-solving skills for the CCEA end-of-year assessment.
案例分析是 Year 8 CCEA 数学课程的重要组成部分，因为它教会你将数学知识应用到真实生活情境中。你不再是孤立地做计算题，而是阅读一个场景，提取重要的数字和关系，然后决定需要使用哪些数学技能。本文将通过一个全面的学校嘉年华案例，逐步展示如何处理涉及百分数、比和比例、几何、统计、概率和代数的问题。通过完成这一实战演练，你将建立信心，并为 CCEA 年终评估磨炼解决问题的能力。
1. Understanding the Scenario and Extracting Information | 理解情境与提取信息
The first step in any case study is to read the description carefully. Our school fair case study includes details about ticket sales, food stalls, game booths, and prize draws. For example: "Lee Primary School is organising a summer fair. The entrance fee is £3 per person, but a 20% discount is offered if you buy a family ticket for 4 people. The food stall sells pizza slices and juice. Each pizza is cut into 8 equal slices. A recipe for 12 cupcakes requires 300 g of flour and 200 g of sugar. The main game booth has a prize wheel with 6 equal sectors coloured red, blue, green, yellow, purple and orange. The school aims to raise at least £500 after deducting £120 total costs." Highlight all the numbers and units as you read.
任何案例分析的第一步都是仔细阅读情境描述。我们的学校嘉年华案例包含关于门票销售、食品摊位、游戏摊位和抽奖活动的详细信息。例如："Lee 小学正在组织一场夏季嘉年华。入场费为每人 £3，但如果你购买 4 人家庭票，可享受 20% 的折扣。食品摊位出售比萨饼片和果汁。每个比萨饼被切成 8 个相等的扇片。一份制作 12 个纸杯蛋糕的食谱需要 300 克面粉和 200 克糖。主要的游戏摊位有一个奖品转盘，转盘被平分为 6 个扇区，颜色分别为红色、蓝色、绿色、黄色、紫色和橙色。学校的目标是在扣除总额为 £120 的成本后，至少筹得 £500。" 阅读时，将所有数字和单位标出来。
2. Working with Percentages: Ticket Discounts | 百分数运用：门票折扣
One of the first calculations asks: "If a family of 4 buys a family ticket, how much do they pay?" The normal price for 4 people would be 4 × £3 = £12. The discount is 20%, so the reduction is 20% of £12. To find 10% of £12, divide by 10: £1.20. Then 20% is double that: £1.20 × 2 = £2.40. The discounted price is £12 − £2.40 = £9.60. Alternatively, find 80% of £12 directly, because 100% − 20% = 80%. 80% as a decimal is 0.8, so £12 × 0.8 = £9.60. You should be comfortable using both methods.
第一类计算问题是："如果一个 4 口之家购买一张家庭票，他们需要支付多少钱？" 4 个人正常价格为 4 × £3 = £12。折扣是 20%，所以减少的金额是 £12 的 20%。求 £12 的 10%，就除以 10：£1.20。那么 20% 是其两倍：£1.20 × 2 = £2.40。折后价为 £12 − £2.40 = £9.60。另一种方法是直接求 £12 的 80%，因为 100% − 20% = 80%。80% 写成小数是 0.8，因此 £12 × 0.8 = £9.60。这两种方法你都应该熟练掌握。
3. Ratio and Proportion: Adjusting a Recipe | 比和比例：调整食谱
of sugar. The stall manager wants to make 30 cupcakes. How much flour and sugar are needed? . Alternatively, notice 30 is 2.5 times 12, because 30 ÷ 12 = 2.5. So multiply the original amounts by 2.5: flour = 300 g × 2.5 = 750 g, sugar = 200 g × 2.5 = 500 g. Understanding proportional reasoning saves time and is a key skill for CCEA.
纸杯蛋糕食谱要求：12 个蛋糕需要 300 克面粉和 200 克糖。摊位经理想制作 30 个蛋糕，需要多少面粉和糖？首先求 1 个蛋糕所需量：面粉 = 300 g ÷ 12 = 25 g，糖 = 200 g ÷ 12 ≈ 16.67 g。然后乘以 30：面粉 = 25 g × 30 = 750 g，糖 = 16.67 g × 30 ≈ 500 g。另一种方法是注意到 30 是 12 的 2.5 倍，因为 30 ÷ 12 = 2.5。因此将原配料的量分别乘以 2.5：面粉 = 300 g × 2.5 = 750 g，糖 = 200 g × 2.5 = 500 g。理解比例推理可以节省时间，这是 CCEA 的一项关键技能。
4. Geometry: Area and Perimeter of Stalls | 几何：摊位的面积和周长
d the perimeter: P = 2 × (length + width) = 2 × (6 + 4) = 2 × 10 = 20 m. If the tape costs £1.50 per metre, the total cost of the tape is 20 × £1.50 = £ ieces needed is 24 ÷ 2 = 12 pieces. Being able to switch between perimeter and area formulas is essential.
食品摊位设在一块 6 米长、4 米宽的矩形草地上。要用装饰胶带将摊位围起来，你需要计算周长：P = 2 × (长 + 宽) = 2 × (6 + 4) = 2 × 10 = 20 m。如果胶带每米 £1.50，胶带的总成本就是 20 × £1.50 = £30。至于面积，学校想用塑料垫覆盖地面。面积 A = 长 × 宽 = 6 × 4 = 24 m²。如果每块垫子覆盖 2 m²，所需垫子数量为 24 ÷ 2 = 12 块。能够灵活切换周长和面积公式至关重要。
5. Volume and Capacity: Juice Containers | 体积和容积：果汁容器
re = 1000 cm³, each carton holds 480 ÷ 1000 = 0.48 litres. The stall expects to sell 80 cartons. Total juice needed = 80 × 0.48 = 38.4 litres. The juice comes in 2-litre bottles, so the number of bottles required is 38.4 ÷ 2 = 19.2, meaning they must buy 20 bottles to have enough. Here, you also practise rounding up in context.
这个摊位售卖果汁，果汁包装为长方体纸盒，底面尺寸为 8 厘米乘 5 厘米，高 12 厘米。体积 = 长 × 宽 × 高 = 8 × 5 × 12 = 40 × 12 = 480 cm³。如果 1 升 = 1000 cm³，每个纸盒的容量为 480 ÷ 1000 = 0.48 升。摊位预计售出 80 盒，所需果汁总量 = 80 × 0.48 = 38.4 升。果汁以 2 升瓶装供应，因此需要的瓶数为 38.4 ÷ 2 = 19.2，这意味着他们必须购买 20 瓶才够。在此过程中，你还练习了根据情境进行向上取整。
6. Statistics: Collecting and Displaying Data | 统计：数据收集与展示
, 15 chose treasure hunt, and 5 chose face painting. You are asked to draw a bar chart. Remember: label both axes, give the chart a title, and use equal-width bars with gaps between them. The vertical axis could show frequency from 0 to 40. For a pie chart, calculate each angle: hoop toss = 35/100 × 360° = 126°, coconut shy = 25/100 × 360° = 90°, etc. Checking that the total angles sum to 360° is a good habit.
学校调查了 100 名小学生，以确定嘉年华中最受欢迎的游戏。结果为：35 人选了套圈，25 人选了掷椰子，20 人选了钓鸭子，15 人选了寻宝，5 人选了面部彩绘。要求你绘制一幅条形图。记住：要为两轴添上标签，给图表加上标题，使用等宽的条形并且条形之间留有空隙。纵轴可以显示从 0 到 40 的频数。如果要绘制饼图，请计算每个扇区的角度：套圈 = 35/100 × 360° = 126°，掷椰子 = 25/100 × 360° = 90°，依此类推。检查总角度之和是否为 360° 是个好习惯。
7. Probability: The Prize Wheel | 概率：奖品转盘
f landing on red is 1 out of 6, written as 1/6. The probability of landing on a primary colour (red, blue, yellow) is 3/6 = 1/2. To win a large prize, the wheel must land on purple twice in a row. The probability is 1/6 × 1/6 = 1/36. This is a good moment to review that the sum of all probabilities is 1, and that for independent events you multiply probabilities.
游戏摊位有一个奖品转盘，平分为 6 个扇区：红色、蓝色、绿色、黄色、紫色和橙色。如果一名玩家转动转盘一次，落在红色扇区的概率是 6 分之 1，写作 1/6。落在一个主色（红、蓝、黄）上的概率是 3/6 = 1/2。要赢得一份大奖，转盘必须连续两次停在紫色扇区，其概率为 1/6 × 1/6 = 1/36。这是复习所有概率之和为 1，以及对于独立事件应将概率相乘的好时机。
8. Algebra: Solving Simple Equations for Profit | 代数：解简单方程求盈利
rom tickets and stall sales is represented by the expression 3n + 200, and total costs are £ 3n = 420. Divide both sides by 3: n = 140. So the school needs at least 141 visitors (since P ≥ 500 when n > 140, and n must be a whole number). This demonstrates how algebra helps in planning.
学校想检查这次嘉年华是否能达到其利润目标。设 n 为访客人数。如果门票和摊位销售的总收入用表达式 3n + 200 表示，总成本为 £120，那么利润 P = 3n + 200 − 120 = 3n + 80。目标是 P ≥ 500。列出方程 3n + 80 = 500 来求刚好达到目标的访客数。两边同时减去 80：3n = 420。两边同时除以 3：n = 140。因此，学校至少需要 141 名访客（因为当 n > 140 时 P ≥ 500，并且 n 必须是整数）。这展示了代数如何帮助进行规划。
9. Measurement: Unit Conversions in the Fair | 测量：嘉年华中的单位换算
me above was in cm³ but total juice was needed in litres. Another example: the prize wheel' . If the wheel spins 10 times during a game, the total distance covered is 2.64 × 10 = 26.4 m. You may also convert grams to kilograms: 750 g of flour is 0.75 kg. Practising unit changes ensures accurate answers.
许多嘉年华活动都涉及单位换算。例如，上文中果汁纸盒的体积是用 cm³ 表示的，但所需的果汁总量是用升来表示的。再比如，奖品转盘的直径为 84 cm。周长 C = π × d。为简便起见，令 π ≈ 22/7，则 C = 22/7 × 84 = 22 × 12 = 264 cm。若要以米为单位表示，则除以 100：2.64 m。如果在一次游戏中转盘转了 10 圈，所覆盖的总距离为 2.64 × 10 = 26.4 m。你还可能需要将克换算成千克：750 g 面粉即 0.75 kg。练习单位换算可以确保答案准确。
10. Averages and Range: Analysing Game Scores | 平均数和极差：分析游戏得分
ian is the middle number when ordered: 3,4,5,6,7,7,8,9. With 8 numbers, median = (6+7) ÷ 2 = 6.5. The mode is the most frequent score, which is 7. The range = maximum − minimum = 9 − 3 = 6. Comparing these averages helps the school decide if the game is too easy or too hard.
在套圈游戏中，8 名玩家的得分为：3, 7, 5, 9, 7, 4, 8, 6。平均数等于总和除以 8：总和 = 3+7+5+9+7+4+8+6 = 49，因此平均数 = 49 ÷ 8 = 6.125。中位数是排序后位于中间的数：3,4,5,6,7,7,8,9。因为有 8 个数，中位数 = (6+7) ÷ 2 = 6.5。众数是出现频率最高的分数，即 7。极差 = 最大值 − 最小值 = 9 − 3 = 6。比较这些平均数可以帮助学校判断这个游戏是否太容易或太难。
11. Bringing It All Together: Checking and Evaluating | 综合应用：检查与评估
e CCEA exam, you may need to combine several skills in one extended question. For example: "Using the family ticket discount and the target profit of £500, how many family tickets must be sold if only family tickets are bought and the £200 stall income remains unchanged?" lies are needed. Always check your solution makes sense in the context. Does rounding match the situation? Are units consistent? Did you answer the exact question asked? These habits secure top marks.
在 CCEA 考试中，你可能需要在一道扩展题中综合运用多种技能。比如："利用家庭票折扣和 £500 的目标利润，如果所有访客都购买家庭票，且 £200 的摊位收入保持不变，那么必须售出多少张家庭票？" 设家庭数为 f。收入 = 9.6f + 200。利润 = 9.6f + 80。列不等式 9.6f + 80 ≥ 500 → 9.6f ≥ 420 → f ≥ 43.75。因此至少需要 44 个家庭购买家庭票。务必检查你的解在情境中是否合理。取整是否符合实际情况？单位是否一致？你是否回答了题目所问的确切问题？这些习惯能帮你稳拿高分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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