📚 Year 9 CCEA Mathematics: Case Study Practical Exercises | Year 9 CCEA 数学：案例分析实战演练
In Year 9 CCEA Mathematics, students are expected to move beyond routine exercises and apply their knowledge to more complex, real-world situations. Case study practical exercises are an excellent way to develop problem-solving, reasoning, and modelling skills. The following examples cover a range of topics from the Key Stage 3 curriculum, including number, algebra, geometry, statistics, and probability, all woven into engaging scenarios.
在 Year 9 CCEA 数学中，学生需要超越常规练习，将所学知识应用于更复杂的现实情境。案例分析实战演练是培养解决问题、推理和建模能力的绝佳方式。以下案例涵盖了关键阶段 3 课程中的一系列主题，包括数字、代数、几何、统计和概率，都被融入引人入胜的情境之中。
1. Planning a School Trip | 规划学校旅行
A school is organising a trip to a museum for 180 students. The coach hire costs £420, and the museum charges an entry fee of £6.80 per student. The school plans to provide a packed lunch, which costs £4.25 per student. Calculate the total cost of the trip and the cost per student.
一所学校正为 180 名学生组织一次博物馆之旅。大巴租赁费用为 420 英镑，博物馆向每名学生收取 6.80 英镑的门票费。学校计划提供盒装午餐，每份成本为 4.25 英镑。计算旅行的总费用和每名学生的平均费用。
We first find the total entrance fees: 180 × 6.80 = £1224. The total lunch cost is 180 × 4.25 = £765. Adding the coach cost: 420 + 1224 + 765 = £2409. The cost per student is 2409 ÷ 180 = £13.38 (to the nearest penny).
首先计算总门票费：180 × 6.80 = 1224 英镑。午餐总费用为 180 × 4.25 = 765 英镑。加上大巴费用：420 + 1224 + 765 = 2409 英镑。每名学生费用为 2409 ÷ 180 ≈ 13.38 英镑（精确到便士）。
If the school decides to subsidise £2.50 per student from its funds, how much would each student actually pay? Subtract 2.50 from 13.38, giving £10.88 per student.
如果学校决定从基金中给每名学生补贴 2.50 英镑，每名学生实际需支付多少？从 13.38 中减去 2.50，得到每名学生 10.88 英镑。
This case study practises multiplication, division, and money calculations, as well as interpreting real-world constraints.
这个案例练习了乘除法、货币计算，以及解读现实约束条件。
2. Redecorating a Bedroom | 重新装修卧室
Jamie wants to paint the four walls of his bedroom. The room is 3.6 m long, 2.8 m wide, and 2.4 m high. There is a door measuring 2.0 m by 0.9 m and a window measuring 1.5 m by 1.2 m. One litre of paint covers 12 m². How many litres of paint should Jamie buy?
杰米想粉刷卧室的四壁。房间长 3.6 米、宽 2.8 米、高 2.4 米。有一扇门，尺寸为 2.0 米 × 0.9 米，和一扇窗户，尺寸为 1.5 米 × 1.2 米。一升油漆可涂刷 12 平方米。杰米需要买多少升油漆？
2.8 m × 2.4 m each: 2 × 2.8 × 2.4 = 13.44 m². Total wall area = 17.28 + 13.44 = 30.72 m².
首先计算四壁的面积。两面墙各为 3.6 米 × 2.4 米：2 × 3.6 × 2.4 = 17.28 平方米。另两面墙各为 2.8 米 × 2.4 米：2 × 2.8 × 2.4 = 13.44 平方米。墙壁总面积 = 17.28 + 13.44 = 30.72 平方米。
otal area not to be painted = 1.8 + 1.8 = 3.6 m². Paintable area = 30.72 − 3.6 = 27.12 m².
门的面积：2.0 × 0.9 = 1.8 平方米。窗的面积：1.5 × 1.2 = 1.8 平方米。不需涂刷的总面积 = 1.8 + 1.8 = 3.6 平方米。需涂刷面积 = 30.72 − 3.6 = 27.12 平方米。
27.12 ÷ 12 = 2.26. Jamie needs to buy 3 litres of paint, as paint is sold in whole litres.
所需升数 = 27.12 ÷ 12 = 2.26。杰米需要买 3 升油漆，因为油漆按整升出售。
This exercise combines perimeter and area, subtraction of areas, and rounding up in a practical context.
这个练习结合了周长和面积、面积相减，以及在实际情境中的向上取整。
3. Analysing Sports Data | 分析体育数据
A basketball team recorded the points scored by a player in 10 matches: 18, 22, 15, 30, 22, 18, 26, 22, 19, 28. Find the mean, median, mode, and range of the scores.
一支篮球队记录了一名球员在 10 场比赛中的得分：18, 22, 15, 30, 22, 18, 26, 22, 19, 28。求这些得分的平均数、中位数、众数和极差。
e of the 5th and 6th values: (22+22)/2 = 22. The mode is 22 (appears three times). Range = 30 − 15 = 15.
首先将得分排序：15, 18, 18, 19, 22, 22, 22, 26, 28, 30。平均数 = (15+18+18+19+22+22+22+26+28+30) ÷ 10 = 220 ÷ 10 = 22。中位数是第 5 和第 6 位的平均值：(22+22)/2 = 22。众数是 22（出现三次）。极差 = 30 − 15 = 15。
How consistent is the player? A lower range indicates more consistency. The range of 15 with a mean of 22 shows some variation, but the mode and median equal the mean, suggesting a fairly typical performance around 22 points.
这名球员的表现有多稳定？较低的极差表明更稳定。极差为 15，平均值为 22，显示了一些波动，但众数和中位数与平均数相等，表明表现通常稳定在 22 分左右。
You could also represent the data in a bar chart or stem-and-leaf diagram to visualise the distribution.
你也可以用条形图或茎叶图来表示这些数据，以直观展现分布情况。
4. Budgeting for a Party | 派对预算
Maria is organising a birthday party with a budget of £150. She wants to buy sandwiches at £3.50 each, drinks at £1.20 per bottle, and party favours costing £2.75 each. She needs one sandwich and one drink per guest, and one party favour per table of 6 guests. If she invites 24 guests, how much will she spend, and does she stay within budget?
玛丽亚正在组织一场生日派对，预算为 150 英镑。她想购买三明治（每个 3.50 英镑）、饮料（每瓶 1.20 英镑）以及派对小礼物（每个 2.75 英镑）。每位客人需要一个三明治和一瓶饮料，每 6 位客人需要一张桌子的派对小礼物。如果她邀请了 24 位客人，她将花费多少？是否在预算之内？
Total sandwiches: 24 × £3.50 = £84. Total drinks: 24 × £1.20 = £28.80. Number of tables: 24 ÷ 6 = 4 tables, so party favours cost 4 × £2.75 = £11. Total cost = 84 + 28.80 + 11 = £123.80.
三明治总费用：24 × 3.50 = 84 英镑。饮料总费用：24 × 1.20 = 28.80 英镑。桌子数量：24 ÷ 6 = 4 张，因此派对小礼物费用为 4 × 2.75 = 11 英镑。总费用 = 84 + 28.80 + 11 = 123.80 英镑。
Maria has £150 − £123.80 = £26.20 left. She is within budget. If she also adds a tip of 10% on the total food and drink cost, what would the new total be? Food and drink cost = 84 + 28.80 = 112.80, 10% tip = £11.28, new total = 123.80 + 11.28 = £135.08, still within budget.
玛丽亚剩余 150 − 123.80 = 26.20 英镑。她在预算之内。如果她额外加上餐饮总费用的 10% 作为小费，新的总费用是多少？餐饮成本 = 84 + 28.80 = 112.80，10% 小费 = 11.28 英镑，新总费用 = 123.80 + 11.28 = 135.08 英镑，仍在预算之内。
This case study reinforces percentages, multiplication, and budgeting.
这个案例巩固了百分数、乘法和预算编制。
5. Designing a Garden | 设计花园
A rectangular garden measures 8.5 m by 6.2 m. The owner wants to cover three-quarters of the area with lawn and the rest with a patio. Lawn turf costs £5.80 per square metre, and patio slabs cost £12 per square metre. Estimate the total cost to the nearest £10.
一个矩形花园的尺寸为 8.5 米 × 6.2 米。主人想把四分之三的面积铺上草坪，其余部分铺上露台。草坪草皮每平方米 5.80 英镑，露台铺板每平方米 12 英镑。估算总费用，精确到 10 英镑。
× 52.7 = 39.525 m². Patio area = 52.7 − 39.525 = 13.175 m². Cost of lawn = 39.525 × 5.80 = £229.245. Cost of patio = 13.175 × 12 = £158.10. Total cost = 229.245 + 158.10 = £387.345, which rounds to £390 to the nearest £10.
总面积 = 8.5 × 6.2 = 52.7 平方米。草坪面积 = 52.7 的 ¾ = 0.75 × 52.7 = 39.525 平方米。露台面积 = 52.7 − 39.525 = 13.175 平方米。草坪费用 = 39.525 × 5.80 = 229.245 英镑。露台费用 = 13.175 × 12 = 158.10 英镑。总费用 = 229.245 + 158.10 = 387.345 英镑，四舍五入到最接近的 10 英镑为 390 英镑。
This problem uses fraction-to-decimal conversion, area calculation, and rounding in a practical landscaping scenario.
这个问题在实用的庭院美化场景中使用了分数转小数、面积计算和四舍五入。
6. Interpreting Weather Data | 解读天气数据
The table below shows the minimum and maximum temperatures (in °C) in Belfast for one week in January. Calculate the daily temperature range and the mean minimum temperature.
下表显示了贝尔法斯特一月份一周的最低和最高气温（单位：°C）。计算每日温度极差以及平均最低气温。
| Day | Min (°C) | Max (°C) |
|---|---|---|
| Mon | −2 | 5 |
| Tue | −4 | 3 |
| Wed | 0 | 7 |
| Thu | 1 | 8 |
| Fri | −1 | 6 |
| Sat | −3 | 4 |
| Sun | −5 | 2 |
°C. Saturday: 4 − (−3) = 7°C. Sunday: 2 − (−5) = 7°C. The daily range is consistently 7°C.
周一极差：5 − (−2) = 7°C。周二：3 − (−4) = 7°C。周三：7 − 0 = 7°C。周四：8 − 1 = 7°C。周五：6 − (−1) = 7°C。周六：4 − (−3) = 7°C。周日：2 − (−5) = 7°C。每日极差一致为 7°C。
7 = (−14) ÷ 7 = −2°C. Working with negative numbers is essential here, a key Year 9 skill.
平均最低气温 = (−2 + (−4) + 0 + 1 + (−1) + (−3) + (−5)) ÷ 7 = (−14) ÷ 7 = −2°C。这里必须处理负数，这是一项 Year 9 的关键技能。
7. Comparing Mobile Phone Plans | 比较手机套餐
Two mobile networks offer the following deals. Network A: £10 monthly fee plus 5p per minute of calls. Network B: £15 monthly fee plus 3p per minute. Write an algebraic expression for the monthly cost for each network if you talk for m minutes. For what number of minutes do both networks cost the same?
两家移动网络提供商推出以下套餐。网络 A：月费 10 英镑，外加每分钟通话费 5 便士。网络 B：月费 15 英镑，外加每分钟 3 便士。如果通话时长为 m 分钟，写出每家网络月费用的代数表达式。通话多少分钟时，两家网络的费用相同？
+ 0.03m. To find when they are equal, set the expressions equal: 10 + 0.05m = 15 + 0.03m.
网络 A 的费用（英镑） = 10 + 0.05m。网络 B 的费用 = 15 + 0.03m。求相等时的 m，令两个表达式相等：10 + 0.05m = 15 + 0.03m。
0.02m = 5. Divide by 0.02: m = 5 ÷ 0.02 = 250 minutes. So at 250 minutes, both plans cost £22.50.
两边减去 0.03m：10 + 0.02m = 15。再减去 10：0.02m = 5。除以 0.02：m = 5 ÷ 0.02 = 250 分钟。因此，250 分钟时，两个套餐费用均为 22.50 英镑。
This case study uses forming and solving linear equations, bridging to real financial decisions.
这个案例使用建立和求解线性方程，连接到真实的财务决策。
8. Probability in Games | 游戏中的概率
A fair six-sided die is rolled 60 times. The results are recorded: 1 appeared 8 times, 2 appeared 12 times, 3 appeared 9 times, 4 appeared 11 times, 5 appeared 10 times, and 6 appeared 10 times. What is the experimental probability of rolling a 2? How does it compare to the theoretical probability?
一个公平的六面骰子投掷了 60 次。结果记录如下：1 出现 8 次，2 出现 12 次，3 出现 9 次，4 出现 11 次，5 出现 10 次，6 出现 10 次。掷出 2 的实验概率是多少？它与理论概率相比如何？
xperimental result is slightly higher but is still a reasonable outcome given the sample size.
掷出 2 的实验概率 = 2 出现的次数 / 总投掷次数 = 12/60 = 1/5 = 0.2。理论概率为 1/6 ≈ 0.167。实验结果略高，但考虑到样本量，仍是一个合理的结果。
If you roll the die 600 times, you would expect the experimental probability to be closer to 1/6. This illustrates the law of large numbers.
如果你投掷骰子 600 次，你会预期实验概率更接近 1/6。这体现了大数定律。
These exercises build a strong foundation for handling probability in practical situations and understanding the difference between experimental and theoretical probability.
这些练习为处理实际情境中的概率以及理解实验概率与理论概率之间的差异奠定了坚实基础。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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