📚 Year 7 CCEA Further Maths: Case Study Practice | Year 7 CCEA 进阶数学：案例分析实战演练
Case studies are a powerful way to bring Further Maths to life. They challenge you to apply number, algebra, geometry, data handling and probability in real-world contexts. In this article we work through eight carefully designed scenarios, building the problem‑solving mindset that CCEA expects at this level.
案例学习是将进阶数学学活的有力方法。它要求你在真实情境中运用数、代数、几何、数据处理和概率。本文将演练八个精心设计的场景，帮你培养 CCEA 阶段所期望的问题解决思维。
1. Shopping Discounts and Multi‑buys | 购物折扣与多买优惠
A fashion store advertises 30% off every item. You pick a £45 backpack and an £8 water bottle. The till also applies an extra 10% off the total after the first discount has been taken. Find the final amount you pay.
一家时装店全场七折。你选中了一个45英镑的背包和一个8英镑的水瓶。收银台还会在第一次折扣后对总额再打九折。计算你最终支付的金额。
Step 1 – Work out the sale price of each item individually. Backpack: £45 × (1 − 0.30) = £45 × 0.7 = £31.50. Water bottle: £8 × 0.7 = £5.60.
步骤1——分别算出单件折扣价。背包：£45 × (1 − 0.30) = £45 × 0.7 = £31.50。水瓶：£8 × 0.7 = £5.60。
Step 2 – Add these two amounts to get the sub‑total before the extra discount: £31.50 + £5.60 = £37.10.
步骤2——将两件价格相加得到额外折扣前的合计数：£31.50 + £5.60 = £37.10。
Step 3 – Apply the extra 10% off. Multiplying by 0.90 gives the final payment: £37.10 × 0.90 = £33.39.
步骤3——再打九折。乘以 0.90 得到最终付款额：£37.10 × 0.90 = £33.39。
A common mistake is to add the percentages and think you save 40%. The discounts are applied sequentially, so the overall saving is a bit less. This principle often appears in multi‑buy offers.
常见错误是把百分比直接相加，以为自己节省了40%。因为折扣是依次计算的，总节省会略少。这个原理经常出现在多买优惠中。
2. Planning a School Trip | 规划学校旅行
A year group hires a coach for £120. Each student pays £4 for the trip. The organiser wants the trip to make at least £20 profit so that the surplus can be donated to a charity. How many students need to go?
一个年级组租用大巴花费120英镑，每位学生缴付4英镑车费。组织者希望这次出行至少获得20英镑利润，以便把盈余捐给慈善机构。需要多少名学生参加？
Step 1 – tion for profit. Let n be the number of students. Income = 4n, cost = 120. Profit = Income – Cost = 4n − 120.
步骤1——列出利润的简单方程。设学生人数为 n。收入 = 4n，成本 = 120。利润 = 收入 − 成本 = 4n − 120。
Step 2 – Set up the inequality for the minimum profit: 4n − 120 ≥ 20.
步骤2——根据最低利润要求建立不等式：4n − 120 ≥ 20。
Step 3 – Add 120 to both sides: 4n ≥ 140. Divide by 4: n ≥ 35. So at least 35 students must travel.
步骤3——求解。两边加上120：4n ≥ 140。除以4：n ≥ 35。因此至少需要35名学生出行。
n. This shows how inequalities turn a planning problem into a straightforward algebra task.
如果只有30名学生报名，利润就是 4×30 − 120 = 0，刚好收支平衡。这展示出不等式如何把一个规划问题变成简单的代数任务。
3. Building a Vegetable Patch | 搭建菜园
A rectangular vegetable patch measures 8 m by 5 m. You need to buy fencing that costs £12 per metre, and fertiliser that requires 2 kg per square metre. Fertiliser is sold at £1.50 per kg. Find the total cost.
一块矩形菜园长8米、宽5米。你需要购买围栏，每米12英镑；还需要施肥，每平方米需要2公斤肥料。肥料售价每公斤1.5英镑。计算总费用。
Step 1 – e perimeter for fencing. Perimeter = 2 × (8 + 5) = 2 × 13 = 26 m. Fencing cost = 26 × 12 = £312.
步骤1——计算围栏所需的周长。周长 = 2 × (8 + 5) = 2 × 13 = 26 米。围栏费用 = 26 × 12 = £312。
Step 2 – rea = 8 × 5 = 40 m². Fertiliser mass needed = 40 × 2 = 80 kg. Fertiliser cost = 80 × 1.5 = £120.
步骤2——计算施肥所需的面积。面积 = 8 × 5 = 40 平方米。肥料质量 = 40 × 2 = 80 公斤。肥料费用 = 80 × 1.5 = £120。
Step 3 – Total cost = fencing + fertiliser = £312 + £120 = £432.
步骤3——总费用 = 围栏 + 肥料 = £312 + £120 = £432。
Linking perimeter and area in one task is typical of outdoor design projects. It reinforces the difference between metres for length and square metres for area.
把周长和面积结合在一个任务中是户外设计项目的典型特征。这能加深学生对长度单位米和面积单位平方米区别的理解。
4. Analysing Sports Scores | 分析体育比分
A basketball team's points from six matches are: 72, 68, 85, 79, 92, 81. Calculate the mean, median and range. Then work out what score is needed in the next game to raise the mean to exactly 80 points.
一支篮球队六场比赛的得分分别是：72, 68, 85, 79, 92, 81。计算平均数、中位数和极差。然后计算下一场需要得多少分才能让平均数刚好上升到80分。
Step 1 – Total the points: 72 + 68 + 85 + 79 + 92 + 81 = 477. Mean = 477 ÷ 6 = 79.5 points.
步骤1——总分：72 + 68 + 85 + 79 + 92 + 81 = 477。平均数 = 477 ÷ 6 = 79.5 分。
Step 2 – 81, 85, 92. With six values, median is the average of the 3rd and 4th: (79 + 81) ÷ 2 = 80.
步骤2——求中位数，将得分排序：68, 72, 79, 81, 85, 92。六个数据的中位数是第3和第4个数的平均值：(79 + 81) ÷ 2 = 80。
Step 3 – Range = highest − lowest = 92 − 68 = 24 points.
步骤3——极差 = 最高分 − 最低分 = 92 − 68 = 24 分。
Step 4 – games. For mean = 80: (477 + x)/7 = 80 → 477 + x = 560 → x = 83. The team needs 83 points.
步骤4——设下一场得分为 x。新总分 = 477 + x，场次为7。若平均数 = 80，则 (477 + x)/7 = 80 → 477 + x = 560 → x = 83。球队需要得到83分。
Using a target mean introduces simple algebra into data handling, a skill CCEA students are expected to handle confidently.
在数据处理中引入目标平均数，将简单代数与统计结合起来，这正是 CCEA 要求学生熟练掌握的技能。
5. Designing a Logo with Shapes | 用形状设计标志
otal area of the logo, taking π ≈ 3.14 and giving the answer correct to one decimal place.
一个标志由一个边长为6厘米的等边三角形和一个半圆组成，半圆的直径边恰好与三角形的一条边重合。求该标志的总面积，π 取 3.14，并将答案精确到一位小数。
Step 1 – Height of the equilateral triangle: using Pythagoras, height = √6² − 3² = √27 = 3√3 ≈ 5.196 cm. Area of triangle = ½ × base × height = 0.5 × 6 × 5.196 = 15.588 cm².
步骤1——等边三角形的高：利用勾股定理，高 = √6² − 3² = √27 = 3√3 ≈ 5.196 厘米。三角形面积 = ½ × 底 × 高 = 0.5 × 6 × 5.196 = 15.588 平方厘米。
Step 2 – m. Area of full circle = π × 3² = 9π ≈ 28.26 cm². Semicircle area = ½ × 28.26 = 14.13 cm².
步骤2——半圆的直径为6厘米，因此半径 = 3 厘米。整圆面积 = π × 3² = 9π ≈ 28.26 平方厘米。半圆面积 = ½ × 28.26 = 14.13 平方厘米。
Step 3 – Total area = 15.588 + 14.13 = 29.718 ≈ 29.7 cm² (to 1 decimal place).
步骤3——总面积 = 15.588 + 14.13 = 29.718 ≈ 29.7 平方厘米（精确到一位小数）。
Combining a triangle and a semicircle revises both Pythagoras and circle area. The task also emphasises the importance of accurate rounding.
将三角形与半圆组合，同时复习了勾股定理和圆的面积。这个任务还强调了正确取近似值的重要性。
6. Probability in Card Games | 纸牌游戏中的概率
A standard pack of 52 playing cards is shuffled. One card is drawn at random. Calculate the probability that the card is (a) a Heart, (b) an Ace, (c) the Ace of Hearts, and (d) a Heart or an Ace.
一副标准的52张扑克牌洗匀后，随机抽取一张。计算抽到的牌是以下情况的概率：(a) 红心，(b) A，(c) 红心A，(d) 红心或A。
Probability (Heart) = number of Hearts / total cards = 13/52 = 1/4.
概率（红心）= 红心张数 / 总张数 = 13/52 = 1/4。
Probability (Ace) = 4/52 = 1/13.
概率（A）= 4/52 = 1/13。
Probability (Ace of Hearts) = 1/52, since only one card satisfies both conditions.
概率（红心A）= 1/52，因为只有一张牌同时满足两个条件。
For 'Heart or Ace' Heart or Ace) = P(Heart) + P(Ace) − P(Heart and Ace) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
对于“红心或A”，使用加法法则：P(红心或A) = P(红心) + P(A) − P(红心且A) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13。
The addition rule prevents double‑counting the Ace of Hearts. Understanding this rule is fundamental for tackling combined events in CCEA probability questions.
加法法则避免了红心A被重复计算。理解这一法则是解决 CCEA 概率题中组合事件的基础。
7. Saving for a Bike | 存钱买自行车
A bicycle costs £250. You already have £70 saved. You receive £30 pocket money each month and decide to save exactly 70% of it. How many full months will it take until you can afford the bike?
一辆自行车售价250英镑。你已经存了70英镑。你每月得到30英镑零花钱，并决定把其中的70%存起来。需要存多少个完整的月份才买得起这辆自行车？
Step 1 – Monthly saving amount: 70% of £30 = 0.7 × 30 = £21.
步骤1——每月存款金额：£30 的 70% = 0.7 × 30 = £21。
Step 2 – Set up the inequality: 70 + 21m ≥ 250, where m is the number of months.
步骤2——建立不等式：70 + 21m ≥ 250，其中 m 代表月数。
Step 3 – by 21: m ≥ 180/21 ≈ 8.57. Since m must be a whole number of months, m must be at least 9.
步骤3——减去70：21m ≥ 180。除以21：m ≥ 180/21 ≈ 8.57。由于 m 必须是整月数，m 至少为9。
After 9 months you will have saved 70 + 9×21 = £259, which is more than £ g uses percentages and inequalities and shows how mathematics helps with personal finance.
9个月后你将存到 70 + 9×21 = 259 英镑，超出250英镑。这种预算问题运用了百分比和不等式，展示了数学如何帮助管理个人财务。
8. Interpreting Bus Timetables | 解读公交车时刻表
The table below shows a morning bus schedule from Central Station to North Park. Use it to answer the questions that follow.
下表显示了从中央车站到北公园的早班公交车时刻表。请根据它回答后面的问题。
| Bus | Departure | Arrival at North Park |
|---|---|---|
| A | 06:30 | 06:52 |
| B | 06:45 | 07:07 |
| C | 07:00 | 07:22 |
Question (a): How long does Bus A take for the journey? Answer: From 06:30 to 06:52 is 22 minutes.
问题(a)：A 路公交车全程需要多长时间？答：从 06:30 到 06:52，共22分钟。
stion (b): Bus B is delayed by 10 minutes. What time does it actually arrive at North Park? Answer: Scheduled arrival 07:07 + 10 minutes = 07:17.
问题(b)：B 路公交车晚点10分钟。它实际到达北公园的时间是几点？答：原定到达 07:07 + 10 分钟 = 07:17。
e = 22 minutes = 22/60 hours. Speed = distance/time = 11 ÷ (22/60) = 11 × 60/22 = 30 km/h.
问题(c)：两站之间的距离为11公里。计算 A 路公交车的平均速度（公里/小时）。时间 = 22 分钟 = 22/60 小时。速度 = 距离 / 时间 = 11 ÷ (22/60) = 11 × 60/22 = 30 公里/小时。
Timetable questions blend time arithmetic with the speed‑distance‑time relationship. They also develop skills in converting minutes to hours, a basic ratio skill that underpins many travel problems.
时刻表问题将时间计算与速度-距离-时间关系结合起来。它们还能锻炼将分钟转换为小时的能力，这是许多行程问题中的基本比率技能。
Published by TutorHao | CCEA Year 7 Further Maths Revision Series | aleveler.com
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