📚 Year 10 SQA Maths: Case Study Practical Exercises | Year 10 SQA 数学：案例分析实战演练
This article presents a series of real‑world case studies designed to sharpen your problem‑solving skills for the SQA National 5 Mathematics course. Each case study links multiple topics, including algebra, geometry, trigonometry, statistics and finance, and shows you how to break down a context‑rich question into manageable steps. You will gain confidence by working through structured examples that mirror the style of SQA exam papers.
本文通过一系列真实情境的案例分析，帮助你强化 SQA National 5 数学课程的问题解决能力。每个案例都融合了代数、几何、三角学、统计和金融等多个主题，并展示如何将一个信息丰富的题目分解为可操作的步骤。你将通过演练与 SQA 真题风格一致的示例来增强信心。
1. Optimising a Rectangular Enclosure | 矩形围栏的最优设计
A gardener wants to build a rectangular vegetable plot against a long brick wall. She has 20 metres of flexible edging to form the three other sides. Let the width perpendicular to the wall be x metres, and the length parallel to the wall be L metres. The edging constraint gives L + 2x = 20, so L = 20 − 2x.
一位园丁想靠着一堵长砖墙围出一个矩形菜地。她手上有 20 米柔性围边用于另外三边。设垂直于墙的宽度为 x 米，平行于墙的长度为 L 米。围边限制条件为 L + 2x = 20，因此 L = 20 − 2x。
The area A of the plot is given by A = x(20 − 2x). Expanding gives the quadratic expression A = −2x² + 20x. To find the maximum area we locate the vertex of this parabola. x = −b/(2a) = −20/(2 × (−2)) = 5. The corresponding length is L = 20 − 2(5) = 10 m.
菜地面积 A 由 A = x(20 − 2x) 给出。展开得到二次表达式 A = −2x² + 20x。为求最大面积，我们找到抛物线的顶点。x = −b/(2a) = −20/(2 × (−2)) = 5。对应的长度为 L = 20 − 2(5) = 10 米。
The maximum area is A = 5 × 10 = 50 m². Always check that lengths stay positive: for x = 5 both dimensions are valid. This optimisation uses knowledge of quadratic functions and is a classic exam scenario.
最大面积为 A = 5 × 10 = 50 平方米。务必检查边长保持正数：当 x = 5 时两边尺寸均合理。这一优化问题运用了二次函数的知识，是经典的考试情境。
2. Interpreting a Histogram of Waiting Times | 等候时间直方图的解读
A café recorded the waiting times (in minutes) of 40 customers. The grouped frequency table is: 0−5 min: 6 customers; 5−10 min: 14; 10−15 min: 12; 15−25 min: 8. Histogram bars should use frequency density , 5, 5 and 10 minutes. Frequency densities: 6/5 = 1.2, 14/5 = 2.8, 12/5 = 2.4, 8/10 = 0.8.
一家咖啡馆记录了 40 位顾客的等候时间（分钟）。分组频数表为：0−5 分钟：6 人；5−10 分钟：14 人；10−15 分钟：12 人；15−25 分钟：8 人。直方图的条形应使用 频率密度，因为最后一组的组距是其他组的两倍。组距分别为 5，5，5 和 10 分钟。频率密度为：6/5 = 1.2，14/5 = 2.8，12/5 = 2.4，8/10 = 0.8。
To estimate the mean waiting time, use midpoints: 2.5, 7.5, 12.5 and 20 minutes. Sum of f × x )+(14×7.5)+(12×12.5)+(8×20) = 15 + 105 + 150 + 160 = 430. Mean ≈ 430 ÷ 40 = 10.75 minutes.
估算平均等候时间使用组中值：2.5，7.5，12.5 和 20 分钟。f × x 的总和 = (6×2.5)+(14×7.5)+(12×12.5)+(8×20) = 15 + 105 + 150 + 160 = 430。均值 ≈ 430 ÷ 40 = 10.75 分钟。
The quartiles can be located from a cumulative frequency diagram. Drawing a smooth curve through cumulative frequencies (6, 20, 32, 40) at upper class boundaries (5, 10, 15, 25) allows you to read off Q₁, Q₂ (median) and Q₃. This links data handling with graphical accuracy.
四分位数可以从累积频数图确定。通过在各组的上限 (5, 10, 15, 25) 处标出累积频数 (6, 20, 32, 40) 并连成光滑曲线，即可读取 Q₁、Q₂（中位数）和 Q₃。这要求数据处理与作图准确性相结合。
3. Height of a Tower Using an Angle of Elevation | 利用仰角求塔高
An engineer stands 30 metres from the base of a cell tower. She measures the angle of elevation to the top as 58°. Her eye level is 1.6 m above the ground. The trigonometric relation is tan 58° = h / 30, where h is the height of the tower above her eye level.
一位工程师站在距通信塔底部 30 米处，测得塔顶的仰角为 58°。她的视线高度距地面 1.6 米。三角函数关系为 tan 58° = h / 30，其中 h 是塔身超出视线的高度。
From a calculator, tan 58° ≈ 1.6003. So h = 30 × 1.6003 ≈ 48.0 m. The total tower height is 48.0 + 1.6 = 49.6 m. Rounding to three significant figures gives 49.6 m. Always convert the angle to degrees and check your calculator mode.
由计算器得 tan 58° ≈ 1.6003。因此 h = 30 × 1.6003 ≈ 48.0 米。塔的总高度为 48.0 + 1.6 = 49.6 米。四舍五入到三位有效数字为 49.6 m。务必确保角度单位为度并检查计算器模式。
If the angle were measured from a different distance, the same method applies. This case study reinforces right‑angled trigonometry, rounding and the practical adjustment for the observer's height.
若从不同距离测量角度，方法相同。本案例强化了直角三角形三角学、四舍五入以及对观察者高度的实际修正。
4. Simultaneous Equations from a Shopping Trip | 购物中的方程组
Amir buys 3 notebooks and 2 pens for £5.80. Chloe buys 5 notebooks and 3 pens for £9.70. Let n be the price of one notebook (in pence) and p the price of one pen. The equations are: 3n + 2p = 580 and 5n + 3p = 970.
Amir 买了 3 本笔记本和 2 支钢笔花费 5.80 英镑。Chloe 买了 5 本笔记本和 3 支钢笔花费 9.70 英镑。设 n 为一本笔记本的价格（以便士计），p 为一支钢笔的价格。方程为：3n + 2p = 580 和 5n + 3p = 970。
Solve by elimination: multiply the first equation by 3, the second by 2: 9n + 6p = 1740 and 10n + 6p = 1940. Subtract to get n = 200. Substitute back into the first equation: 3(200) + 2p = 580 → 600 + 2p = 580 → 2p = −20 → p = −10, which suggests a mistake. Re‑examine: properly scaling to eliminate p, multiply first by 3 and second by 2: 9n+6p=1740, 10n+6p=1940, subtract → n = 200 00)+2p=580 → 600+2p=580 → 2p = -20 → p = -10. Negative price indicates impossible scenario — e pen price unrealistic. In reality, check the numbers. Possibly the original amounts were £5.90 and £ c. This situation demonstrates the need to verify the feasibility of solutions in context.
用消元法求解：将第一个方程乘以 3，第二个方程乘以 2：9n + 6p = 1740，10n + 6p = 1940。相减得 n = 200。代回第一个方程：3(200) + 2p = 580 → 600 + 2p = 580 → 2p = −20 → p = −10，得出钢笔价格为负值，这虽然数学上成立但不现实。此案例提醒我们在实际情境中必须检验解的合理性。
A corrected typical problem: 3 notebooks and 2 pens cost £5.30; 5 notebooks and 3 pens cost £8.50. Then 3n+2p=530, 5n+3p=850. Solving gives n=£0.90, p=£1.30. This shows the method of creating and solving linear equations from a written description.
一个典型的修正问题：3 本笔记本和 2 支钢笔花费 5.30 英镑；5 本笔记本和 3 支钢笔花费 8.50 英镑。于是 3n+2p=530，5n+3p=850。解得 n=0.90 英镑，p=1.30 英镑。这展示了从文字描述中建立和求解线性方程组的方法。
5. Navigation and Pythagoras’ Theorem | 航海与毕达哥拉斯定理
A ferry leaves port and sails 24 km due north, then turns and sails 10 km due east. How far is the ferry from the port in a straight line? Using Pythagoras: distance² = 24² + 10² → distance² = 576 + 100 = 676. The straight‑line distance is √676 = 26 km.
一艘渡轮离港后向北航行 24 公里，然后转向正东航行 10 公里。渡轮离港口的直线距离有多远？运用毕达哥拉斯定理：距离² = 24² + 10² → 距离² = 576 + 100 = 676。直线距离为 √676 = 26 公里。
If the ferry had sailed 24 km north and then 10 km east‑northeast (bearing 067.5°), you would need the cosine rule. In National 5, the focus is on right‑angled triangles and straightforward bearings. Make sure to draw a clear diagram and label the right angle.
若渡轮向北航行 24 公里后转向东‑北‑东方向（方位角 067.5°）再航行 10 公里，则需使用余弦定理。在 National 5 课程中，重点是直角三角形和简单的方位角。务必画出清晰的示意图并标出直角。
This case study also connects to vectors: the displacement vector is (10, 24) and its magnitude is found exactly with Pythagoras. Be ready to express bearings and distances accurately.
本案例也与向量紧密相关：位移向量为 (10, 24)，其大小可通过毕达哥拉斯定理准确求得。要准备好精确表达方位角和距离。
6. Compound Interest for a Savings Goal | 为储蓄目标计算的复利
Megan deposits £800 in an account that pays 3.2% annual compound interest. She wants to know how much she will have after 5 years. The multiplier is 1 + 0.032 = 1.032. After 5 years: A = 800 × (1.032)⁵.
Megan 在一个年复利 3.2% 的账户中存入 800 英镑。她想知道 5 年后将有多少钱。增长因子为 1 + 0.032 = 1.032。5 年后：A = 800 × (1.032)⁵。
065024, then (1.032)⁴ ≈ 1.134, and multiply once more by 1.032 to get about 1.170. Finally 800 × 1.170 ≈ £936. A more precise calculator value gives £936.73 (to the nearest penny). Always use the power key and round only at the final answer.
逐步计算：(1.032)² ≈ 1.065024，然后 (1.032)⁴ ≈ 1.134，再乘以 1.032 得约 1.170。最终 800 × 1.170 ≈ 936 英镑。用计算器更精确的值为 936.73 英镑（精确到分）。务必使用乘方键并只在最后答案处舍入。
If interest were paid half‑yearly, the rate per half‑year would be 1.6% and the number of periods would be 10. This variation tests your understanding of compounding frequency. In exams, watch for phrases like 'paid monthly' or 'each quarter'.
若利息每半年支付一次，半年利率为 1.6%，期数为 10。这一变化可以考察你对复利频率的理解。考试中要注意 “按月支付” 或 “每季度” 等表述。
7. Vector Journeys and Resultant Displacement | 向量行程与合位移
A drone flies from a control point O. Its first leg is represented by vector u = (12, 5) km, meaning 12 km east and 5 km north. The second leg is v = (−4, 9) km. The resultant displacement is u + v = (12+(−4), 5+9) = (8, 14) km.
一架无人机从控制点 O 起飞。第一段航程用向量 u = (12, 5) 公里表示，即向东 12 公里、向北 5 公里。第二段为 v = (−4, 9) 公里。合位移为 u + v = (12+(−4), 5+9) = (8, 14) 公里。
The distance from O is the magnitude: √8² + 14² = √64 + 196 = √260. Simplify: √260 = √4×65 = 2√65 km, approximately 16.1 km. The bearing from O is found using tan θ = 14/8, giving θ ≈ 60.3° from the north, so the bearing is 060.3°.
离 O 点的距离为其模：√8² + 14² = √64 + 196 = √260。化简：√260 = √4×65 = 2√65 公里，约 16.1 公里。从 O 点出发的方位角由 tan θ = 14/8 求得，θ ≈ 60.3°（自北偏东），故方位角为 060.3°。
Expressing the vector as a column and adding components is a core National 5 skill. Practise also finding a vector from one point to another given coordinates.
将向量表示为列向量并相加各分量是 National 5 的核心技能。还要练习根据坐标求两点之间的向量。
8. Straight Line Equations from a Taxi Fare | 出租车计费的直线方程
A taxi company charges a fixed pickup fee of £2.50 plus £1.40 per kilometre travelled. Let C be the total cost in pounds and d the distance in km. The equation is C = 1.40d + 2.50, which is in the form y = mx + c. Here the gradient m = 1.40 and the y‑intercept c = 2.50.
某出租车公司收取固定的起步费 2.50 英镑，并每公里加收 1.40 英镑。设 C 为总费用（英镑），d 为行驶距离（公里）。方程为 C = 1.40d + 2.50，形式为 y = mx + c。这里斜率 m = 1.40，y 轴截距 c = 2.50。
Calculate the cost for a 12 km journey: C = 1.40 × 12 + 2.50 = 16.80 + 2.50 = £19.30. If a customer paid £26.30, the distance travelled is found by solving 26.30 = 1.40d + 2.50 → 23.80 = 1.40d → d = 17 km.
计算 12 公里行程的费用：C = 1.40 × 12 + 2.50 = 16.80 + 2.50 = 19.30 英镑。若一位顾客支付了 26.30 英镑，行驶距离通过解 26.30 = 1.40d + 2.50 → 23.80 = 1.40d → d = 17 公里 求得。
Sketch the graph, label the axes 'Distance (km)' and 'Cost (£)', and check that the gradient represents the rate per km. This linear model is a practical way to link equations with real data.
画出草图，坐标轴标注 “距离 (km)” 和 “费用 (£)”，并验证斜率代表每公里收费。这一线性模型是将方程与实际数据紧密联系的实用方法。
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