📚 Case Study Analysis: Practical Exercises for Year 9 SQA Maths | Year 9 SQA 数学：案例分析实战演练
Case study analysis is an essential skill in SQA Year 9 Mathematics. It requires you to apply mathematical concepts and techniques to real-world situations, breaking down problems into manageable steps. This article presents a series of practical case studies covering numbers, algebra, geometry, statistics, probability, and more. Each case study will help you develop problem-solving strategies and deepen your understanding of how mathematics is used in everyday life.
案例分析是 SQA 九年级数学中的一项重要技能。它要求你将数学概念和技巧应用到现实情境中，把问题分解成易于处理的步骤。本文呈现了一系列实战案例分析，涵盖算术、代数、几何、统计、概率等多个领域。每一个案例都将帮助你培养解决问题的策略，加深你对数学如何在日常生活中运用的理解。
1. Case Study 1: Planning a Garden – Perimeter, Area and Pythagoras | 案例一：规划花园——周长、面积与勾股定理
You are designing a rectangular garden with a length of 8 m and a width of 5 m. The entire garden needs to be fenced, and turf must be laid over the whole area. In one corner, you plan to create a right-angled triangular flower bed.
你需要设计一个长 8 米、宽 5 米的矩形花园。整个花园需要围上篱笆，并且全面铺设草皮。你计划在其中一个角落建造一个直角三角形的花坛。
Problem: Calculate the perimeter of the garden needed for fencing, the area for turf, and show how you can verify that the corner is a true right angle using Pythagoras' theorem.
问题：计算围篱笆所需的周长、铺设草皮所需的面积，并说明如何用勾股定理验证那个角落确实是直角。
= 2 × 13 = 26 m. The area of the rectangular garden is found by multiplying length by width: A = length × width = 8 × 5 = 40 m². To check that a corner is a right angle, measure 3 m along one side and 4 m along the adjacent side. If the diagonal between those two points measures exactly 5 m, then by Pythagoras' theorem (3² + 4² = 5²) the angle is 90°.
周长是花园四周边界的总长度。对于矩形，我们用公式 P = 2 × (长 + 宽)。代入已知数值，得到 P = 2 × (8 + 5) = 2 × 13 = 26 米。矩形花园的面积则由长乘以宽得出：A = 长 × 宽 = 8 × 5 = 40 平方米。要检查一个角落是否为直角，可沿一条边量出 3 米，沿邻边量出 4 米。如果这两点之间的对角线恰好是 5 米，那么根据勾股定理 (3² + 4² = 5²)，该角就是 90°。
Key mathematics used: perimeter of a rectangle, area of a rectangle, and the converse of Pythagoras' theorem with a 3-4-5 triangle.
所用到的关键数学知识：矩形的周长、矩形的面积，以及用 3-4-5 三角形验证的勾股定理的逆定理。
2. Case Study 2: Currency Conversion and Holiday Budgeting | 案例二：货币兑换与度假预算
You are going on a school trip to France. The exchange rate is £1 = €1.14. The flight costs £180, and the hotel charges €85 per night for 5 nights. You also want to budget €50 per day for food and activities for 5 days.
你将参加一次去法国的学校旅行。当前汇率为 £1 = €1.14。机票价格是 180 英镑，酒店每晚收取 85 欧元，共住 5 晚。你还计划每天为餐饮和活动预留 50 欧元，共 5 天。
Problem: Work out the total cost of the trip in euros and then convert it back to pounds to see how much you need in sterling. Additionally, if the flight price is reduced by 15% in a sale, what would the new total be in pounds?
问题：计算这次旅行以欧元计的总花费，再将其兑换回英镑，看看你需要多少英镑。另外，如果机票价格在促销中降低 15%，那么以英镑计算的新总花费是多少？
First, find the total in euros. Hotel: €85 × 5 = €425. Daily spending: €50 × 5 = €250. Total in euros (excluding flight) = €425 + €250 = €675. The flight in euros: £180 × 1.14 = €205.20. Total trip amount = €675 + €205.20 = €880.20. Convert to pounds: €880.20 ÷ 1.14 ≈ £772.11. For the sale, flight discount: 15% of £180 = 0.15 × 180 = £27. New flight price = £180 − £27 = £153. In euros: £153 × 1.14 = €174.42. New total: €425 + €250 + €174.42 = €849.42. In pounds: €849.42 ÷ 1.14 ≈ £745.11.
首先，以欧元计算总花费。酒店：€85 × 5 = €425。每日开销：€50 × 5 = €250。不含机票的欧元总数为 €425 + €250 = €675。机票折合欧元：£180 × 1.14 = €205.20。旅行总金额：€675 + €205.20 = €880.20。兑换成英镑：€880.20 ÷ 1.14 ≈ £772.11。促销情况下，机票折扣：180 英镑的 15% = 0.15 × 180 = £27。新机票价格：£180 − £27 = £153。折合欧元：£153 × 1.14 = €174.42。新总花费：€425 + €250 + €174.42 = €849.42。折合英镑：€849.42 ÷ 1.14 ≈ £745.11。
This case study uses multiplication for unit conversion, percentage decrease, and budgeting with multiple components.
这个案例使用了乘除法进行单位换算、百分比减少，以及多项目预算的加法与乘法。
3. Case Study 3: Interpreting Statistical Data – School Canteen Survey | 案例三：解读统计数据——学校食堂调查
Students rated the new canteen menu on a scale from 1 (poor) to 5 (excellent). The results are shown in the frequency table below.
学生们对新食堂菜单进行了评分，评分范围为 1（差）到 5（优秀）分。结果如下面的频数表所示。
| Rating | Frequency |
|---|---|
| 1 | 5 |
| 2 | 4 |
| 3 | 7 |
| 4 | 9 |
| 5 | 15 |
Task: Calculate the mean, median, and mode of the ratings. Also draw a bar chart to represent the data and comment on the overall satisfaction level.
任务：计算评分的平均数、中位数和众数。同时绘制条形图来表示数据，并评价整体的满意度水平。
ighest frequency, which is 5. To calculate the mean, multiply each rating by its frequency and divide by the total number:
学生总人数：5 + 4 + 7 + 9 + 15 = 40。众数是出现次数最多的评分，即 5。计算平均数时，将每个评分乘以其频数，再除以总人数：
Mean = (1×5 + 2×4 + 3×7 + 4×9 + 5×15) ÷ 40
= (5 + 8 + 21 + 36 + 75) ÷ 40 = 145 ÷ 40 = 3.625
h and 21st values are both 4, so the median is 4. A bar chart would show increasing bars, with the highest for rating 5. The mean of 3.625 and median of 4 indicate a positive overall satisfaction, with the most common rating being excellent.
求中位数需要将 40 个数值按大小排列。累计频数为：1 星 5 人，2 星及以下 9 人，3 星及以下 16 人，4 星及以下 25 人，5 星及以下 40 人。第 20 和第 21 个值都落在 4 星，因此中位数为 4。条形图将显示逐渐升高的条形，最高条对应 5 分。平均分 3.625 和中位数 4 表明总体满意度较高，最常见的评价是优秀。
4. Case Study 4: Probability in a Dice Game | 案例四：骰子游戏中的概率
In a game, two fair six-sided dice are rolled and the sum of the numbers is recorded. Players bet on which sum will appear most often.
在一个游戏中，同时掷出两颗均匀的六面骰子，将朝上的点数相加，记录总和。玩家对出现最频繁的总和下注。
Problem: Construct a sample space diagram to show all possible sums. Determine the probability that the sum is 7. Which sum has the highest probability, and why?
问题：构建一个样本空间图，显示所有可能的总和。求出总和为 7 的概率。哪个总和出现的概率最高？为什么？
ample space can be organised in a table. The sum of 7 appears in combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — s 7 because it has the greatest number of combinations. Sums like 2 (only 1,1) and 12 (only 6,6) have a probability of 1/36. This illustrates that not all sums are equally likely even though each individual outcome is.
总共有 6×6 = 36 个等可能的结果。总和范围从 2 到 12。样本空间可以组织成表格。总和 7 出现的组合有：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) —— 共 6 种结果。因此 P(总和=7) = 6/36 = 1/6。最可能出现的总和是 7，因为它拥有的不同组合数最多。而 2（仅有 (1,1)）和 12（仅有 (6,6)）的概率各为 1/36。这说明了尽管每个具体投掷结果是等可能的，但各个总和的概率并不相等。
5. Case Study 5: Speed, Distance and Time – A Rail Journey | 案例五：速度、距离与时间——一次铁路旅行
A train travels from Edinburgh to London, a distance of 640 kilometres, at an average speed of 80 km/h. The journey includes a 30-minute stop at Newcastle.
一列火车从爱丁堡开往伦敦，全程 640 公里，平均时速 80 公里。旅程中包括在纽卡斯尔停靠 30 分钟。
Problem: Calculate the total travel time in hours and minutes. If the train departs at 08:15, what is its arrival time? What would the average speed need to be to reduce the total trip time by 1 hour (excluding the fixed stop)?
问题：计算总共旅行时间（小时和分钟）。如果火车上午 8:15 出发，抵达时间是什么时候？如果希望总行程时间（不包括固定停靠）缩短 1 小时，平均速度需要达到多少？
30 min gives arrival at 16:45. To reduce the moving time by 1 hour, new moving time = 7 hours. The required speed equals distance divided by time: 640 ÷ 7 ≈ 91.43 km/h. Notice how direct proportion is applied: a shorter time requires a higher speed for the same distance.
不停靠的行驶时间：640 ÷ 80 = 8 小时。加上 30 分钟停靠，总行程时间为 8 小时 30 分钟。8:15 出发加上 8 小时 30 分，到达时间为 16:45。要把行驶时间缩短 1 小时，则新的行驶时间为 7 小时。所需速度等于距离除以时间：640 ÷ 7 ≈ 91.43 公里/时。请注意，这里应用了正比关系：相同距离下，时间越短，所需的速度就越高。
6. Case Study 6: Ratio and Proportion in a Recipe | 案例六：食谱中的比与比例
A cookie recipe for 8 servings requires 2 cups of flour, 1 cup of sugar, and 0.5 cups of butter. You need to make enough for 20 people, and you also want to adjust the butter to use exactly 1.2 cups.
一份供 8 人食用的饼干食谱需要 2 杯面粉、1 杯糖和 0.5 杯黄油。你需要制作足够 20 人食用的分量，并且想把黄油用量恰好调整为 1.2 杯。
Problem: Calculate the amount of each ingredient needed for 20 people. Then, if you fix the butter at 1.2 cups, find the scaling factor and the corresponding quantities of flour and sugar.
问题：计算 20 人需要的每种材料的用量。然后，如果把黄油固定为 1.2 杯，求出缩放因子以及相应的面粉和糖的用量。
at 1.2 cups, find the new scaling factor: 1.2 ÷ 0.5 = 2.4. Then flour = 2 × 2.4 = 4.8 cups, sugar = 1 × 2.4 = 2.4 cups. The total servings possible at this factor would be 8 × 2.4 = 19.2, meaning you can serve about 19 people fully.
从 8 人到 20 人的缩放因子：20 ÷ 8 = 2.5。20 人份：面粉 = 2 × 2.5 = 5 杯，糖 = 1 × 2.5 = 2.5 杯，黄油 = 0.5 × 2.5 = 1.25 杯。若将黄油固定为 1.2 杯，求新的缩放因子：1.2 ÷ 0.5 = 2.4。那么面粉 = 2 × 2.4 = 4.8 杯，糖 = 1 × 2.4 = 2.4 杯。在这个缩放因子下，可行的份数为 8 × 2.4 = 19.2，即你可以做出约 19 人份的饼干。
This exercise demonstrates direct proportion and multiplicative reasoning with ratios in recipes.
这个练习展示了食谱中比和比例的直接比例关系以及乘性推理。
7. Case Study 7: Solving Equations in a Business Context | 案例七：商业情境中的方程求解
You run a small business making handmade keyrings. The fixed cost for a market stall is £15. Each keyring costs £2 to make and is sold for £5.
你经营一家制作手工钥匙扣的小生意。市场摊位的固定成本为 15 英镑。每个钥匙扣的制作成本为 2 英镑，售价为 5 英镑。
Problem: How many keyrings must you sell to break even? If you want a profit of at least £30, write and solve an inequality to find the minimum number to sell.
问题：你需要卖出多少个钥匙扣才能达到收支平衡？如果你希望获得至少 30 英镑的利润，写出并求解一个不等式，以找到最少需要卖出的数量。
ost → 5n = 15 + 2n. Subtract 2n from both sides: 3n = 15 → n = 5. For a profit of at least £ 30. Solve: 3n − 15 ≥ 30 → 3n ≥ 45 → n ≥ 15. So you must sell at least 15 keyrings to earn £30 profit or more.
设 n 为卖出的个数。总成本 = 固定成本 + 单件成本 × n = 15 + 2n。总收入 = 5n。收支平衡点：收入 = 成本 → 5n = 15 + 2n。两边减去 2n 得到 3n = 15，解得 n = 5。要获得至少 30 英镑的利润：利润 = 收入 − 成本 = 5n − (15 + 2n) = 3n − 15 ≥ 30。求解：3n − 15 ≥ 30 → 3n ≥ 45 → n ≥ 15。因此，你必须至少卖出 15 个钥匙扣才能获得 30 英镑及以上的利润。
8. Case Study 8: Pythagoras’ Theorem in Construction | 案例八：建筑中的勾股定理
A ladder of length 5 m is leaning against a vertical wall. The base of the ladder is placed 1.5 m away from the wall for safety.
一架 5 米长的梯子斜靠在竖直的墙壁上。为了安全，梯子的底端距离墙壁 1.5 米。
Problem: How high up the wall does the ladder reach? If safety regulations require the base to be at least 1 m from the wall, what is the maximum height the ladder could reach with this constraint?
问题：梯子的顶端能达到墙壁多高的位置？如果安全规定要求底端至少离墙 1 米，那么在这个限制下，梯子能达到的最大高度是多少？
d be h metres. The ladder, wall, and ground form a right-angled triangle. Apply Pythagoras' theorem: h² + (1.5)² = 5² → h² + 2.25 = 25 → h² = 22.75 → h = √22 m. If the base distance d is exactly 1 m, then h² + 1² = 5² → h² + 1 = 25 → h² = 24 → h = √24 ≈ 4.90 m. The maximum safe height with the regulation is approximately 4.90 m.
设达到的高度为 h 米。梯子、墙壁和地面构成一个直角三角形。应用勾股定理：h² + (1.5)² = 5² → h² + 2.25 = 25 → h² = 22.75 → h = √22.75 ≈ 4.77 米。如果底端距离 d 恰好为 1 米，那么 h² + 1² = 5² → h² + 1 = 25 → h² = 24 → h = √24 ≈ 4.90 米。在安全规定下，所能达到的最大高度约为 4.90 米。
9. Case Study 9: Analysing Mobile Phone Plans – Linear Graphs | 案例九：分析手机套餐——线性图
Two mobile phone plans are offered. Plan A charges a fixed monthly fee of £12 and £0.04 per minute of calls. Plan B has no monthly fee but charges £0.10 per minute.
有两个手机套餐可供选择。套餐 A 每月收取固定费用 12 英镑，通话每分钟 0.04 英镑。套餐 B 没有月租费，但每分钟收费 0.10 英镑。
Problem: Write equations for the monthly cost C in pounds for each plan based on minutes used m. Determine the number of minutes for which both plans cost the same, and sketch the graphs on the same axes.
问题：写出每个套餐以分钟数 m 为变量的每月费用 C（英镑）的方程。确定两个套餐费用相同时的通话分钟数，并在同一坐标系中画出草图。
: 12 + 0.04m = 0.10m → 12 = 0.06m → m = 12 ÷ 0.06 = 200 minutes. At m = 200, both cost C =
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