Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

In SQA Mathematics, real-life case studies help you apply concepts to practical situations. This article presents a collection of worked case studies covering finance, geometry, trigonometry, statistics, algebra, and vectors. Each case includes a problem statement, step-by-step solution, and key revision notes. Use these exercises to build confidence for your Year 11 assessments.

在 SQA 数学中，实际案例分析帮助你学以致用。本文提供一系列涵盖金融、几何、三角、统计、代数和向量的案例，每题包括题目、逐步解答和重点笔记。通过这些练习，为你 Year 11 的评估积累信心。

1. Case Study 1: Saving with Compound Interest | 案例分析1：复利储蓄

Problem: Alex deposits £800 in a savings account that pays 3.5% annual compound interest. He plans to leave the money untouched for 5 years. How much will be in the account at the end of the period? If he needs £1000, will he have enough?

问题：亚历克斯将800英镑存入一个年利率3.5%的复利储蓄账户。他打算5年内不动用这笔钱。到期时账户里会有多少钱？如果他需要1000英镑，够吗？

We use the compound interest formula A = P(1 + r)ⁿ, where P is the principal, r the decimal interest rate, and n the number of years.

我们使用复利公式 A = P(1 + r)ⁿ，其中 P 为本金，r 为利率（小数形式），n 为年数。

A = 800 × (1 + 0.035)⁵

First calculate (1.035)⁵. 1.035² = 1.071225; 1.035⁴ = (1.071225)² = 1.147523; 1.035⁵ = 1.147523 × 1.035 ≈ 1.187685. Then A = 800 × 1.187685 = 950.148. Rounded to the nearest penny, the final amount is £950.15.

先计算 (1.035)⁵。1.035² = 1.071225; 1.035⁴ = (1.071225)² = 1.147523; 1.035⁵ = 1.147523 × 1.035 ≈ 1.187685。然后 A = 800 × 1.187685 = 950.148。四舍五入到分，最终金额为 950.15 英镑。

Since £950.15 < £1000, Alex will not have enough to reach his target. He would need to either deposit more initially or find a better interest rate.

由于 950.15 英镑 < 1000 英镑，亚历克斯不够达到目标。他需要增加本金或寻找更高利率。

Common mistake: forgetting to convert the percentage to a decimal. Always divide the rate by 100.

常见错误：忘记把百分数转换为小数。务必把利率除以100。

2. Case Study 2: Depreciation of a Car | 案例分析2：汽车折旧

Problem: A new car is bought for £14,500. Its value depreciates by 18% per year. Calculate the car’s value after 4 years. When its value falls below £5000, the owner plans to replace it. In which year does this happen?

问题：一辆新车购买价为14,500英镑，每年折旧18%。计算4年后汽车的价值。当价值降到低于5000英镑时，车主计划换车。这发生在第几年？

Depreciation uses a multiplier of (1 − rate). Here multiplier = 1 − 0.18 = 0.82. After 4 years: V = 14,500 × 0.82⁴.

折旧使用乘数 (1 − 折旧率)。此处乘数 = 1 − 0.18 = 0.82。4年后：V = 14,500 × 0.82⁴。

0.82² = 0.6724; 0.82⁴ = (0.6724)² ≈ 0.4521

V = 14,500 × 0.4521 ≈ 6555.45. After 4 years, the car is worth approximately £6555.45.

V = 14,500 × 0.4521 ≈ 6555.45。4年后汽车价值约6555.45英镑。

To find when it drops below £5000, we test each year: Year 5: 14,500 × 0.82⁵ = 14,500 × 0.4521 × 0.82 ≈ 6555.45 × 0.82 = 5375.47. Year 6: 5375.47 × 0.82 ≈ 4407.89. So the value falls below £5000 during the 6th year.

要找出何时跌破5000英镑，逐年检验：第5年：14,500 × 0.82⁵ = 14,500 × 0.4521 × 0.82 ≈ 6555.45 × 0.82 = 5375.47。第6年：5375.47 × 0.82 ≈ 4407.89。因此价值在第6年跌破5000。

We can also use logarithms: 14,500 × 0.82ⁿ < 5000 ⇒ 0.82ⁿ < 5000/14500 ⇒ 0.82ⁿ < 0.3448. n > log(0.3448)/log(0.82) ≈ 5.33, so n = 6.

也可用对数：14,500 × 0.82ⁿ < 5000 ⇒ 0.82ⁿ < 5000/14500 ⇒ 0.82ⁿ < 0.3448。n > log(0.3448)/log(0.82) ≈ 5.33，故 n = 6。

3. Case Study 3: Similar Triangles in Surveying | 案例分析3：测量中的相似三角形

Problem: A surveyor needs to find the width of a river without crossing it. She places two markers A and B on one bank, 20 m apart. Marker C is directly opposite B on the far bank. Angle ABC = 55°, angle BAC = 80°. Use the sine rule or similar triangles? Actually, she can create a similar triangle on land. She marks a point D on line AB extended so that AD = 40 m. From D, she sights C and measures angle ADC = 28°. Use similarity to calculate BC.

问题：一位测量员需要在不渡河的情况下测出河宽。她在同一侧岸上放置两个标记 A 和 B，相距20米。标记 C 在对岸正对 B。∠ABC = 55°，∠BAC = 80°。她实际上在陆地上构造了相似三角形。她将 AB 延长至 D，使 AD = 40 米。从 D 观测 C，测得 ∠ADC = 28°。用相似性计算 BC。

We notice that triangle ABC and triangle ADC share angle A? Not directly. Instead, check angles: In triangle ABC, angle ACB = 180° − 80° − 55° = 45°. In triangle ADC, angle DAC = 80° (since A is common), angle ADC = 28°, so angle ACD = 180° − 80° − 28° = 72°. These triangles are not similar. The question intends to use the scale factor along AB. Perhaps the surveyor sets D such that DB is a known length and constructs a similar triangle? Let’s correct: She places D on the same bank such that B is between A and D? No. Let’s assume she uses a method of similar triangles: She walks along the bank to a point E such that CE is perpendicular to AB? No.

我们注意到三角形 ABC 和 ADC 并不直接相似。也许测量员使用的方法是作垂线。更实际的案例：为测河宽 BC，她从 B 沿河岸走一段距离到 E，使 BE = 15 m，然后从 E 观测 C，使得视线与河岸夹角等于在 B 处的角度？更好的做法：构造相似三角形，利用比例。假设她测量了从 A 和 B 看 C 的角度，并用正弦定理可直接求解。

We can instead solve directly using the sine rule in triangle ABC. We know AB = 20 m, angle A = 80°, angle B = 55°, so side BC can be found. The problem becomes a straightforward trigonometry case.

我们可直接用正弦定理在三角形 ABC 中求解。已知 AB = 20 m，∠A = 80°，∠B = 55°，可求 BC。这就变成了直接的三角学问题。

Using sine rule: BC / sin(A) = AB / sin(C). We need angle C: 180° − 80° − 55° = 45°. So BC = 20 × sin(80°) / sin(45°).

使用正弦定理：BC / sin(A) = AB / sin(C)。需要角 C：180° − 80° − 55° = 45°。所以 BC = 20 × sin(80°) / sin(45°)。

sin(80°) ≈ 0.9848, sin(45°) ≈ 0.7071

BC = 20 × 0.9848 / 0.7071 ≈ 20 × 1.3926 = 27.85 m. The river is about 27.9 m wide.

BC = 20 × 0.9848 / 0.7071 ≈ 20 × 1.3926 = 27.85 m。河宽约27.9米。

Key point: In many surveying problems, you can create a similar triangle by measuring angles and using proportional lengths, but here the sine rule provides a direct solution.

关键：在许多测量问题中，可以通过测角和比例长度构造相似三角形，但此处正弦定理可直接求解。

4. Case Study 4: Navigation and Bearings | 案例分析4：导航与方位角

Problem: A ship sails from port P on a bearing of 065° for 12 km to point Q. It then changes course to a bearing of 150° and sails 18 km to point R. Find the distance PR and the bearing of R from P.

问题：一艘船从港口 P 以 065° 的方位角航行12公里到达 Q 点。然后改变航向至 150° 方位角，再航行18公里到达 R 点。求 PR 的距离以及从 P 看 R 的方位角。

Sketch the path. Bearing is measured clockwise from north. So PQ: 12 km at 65° east of north. QR: 18 km at 150°, which is 30° east of south (since 180 − 150 = 30). The angle between the two paths is the difference in direction: at Q, the bearing change is 150° − 65° = 85°? Actually, the angle inside the triangle at Q is the angle between the reverse of PQ and QR. The bearing of QP is 65° + 180° = 245°. The angle between QP (245°) and QR (150°) is 245° − 150° = 95°. So triangle PQR has sides PQ = 12, QR = 18, and included angle ∠PQR = 95°. Use cosine rule.

画出路线图。方位角从正北顺时针测量。因此 PQ：12公里，方位065°（东偏北）。QR：18公里，方位150°（南偏东30°，因180−150=30）。两条路径的夹角：在 Q 点，航向变化不是简单的差，要考虑三角形内角。QP 的反向方位是 065°+180°=245°。QP (245°) 与 QR (150°) 的夹角 = 245°−150°=95°。所以三角形 PQR 中，PQ=12，QR=18，夹角 ∠PQR=95°。使用余弦定理。

PR² = 12² + 18² − 2×12×18×cos(95°)

cos(95°) = −cos(85°) ≈ −0.08716. So PR² = 144 + 324 − 432 × (−0.08716) = 468 + 37.65 = 505.65. PR = √505.65 ≈ 22.49 km.

cos(95°) = −cos(85°) ≈ −0.08716。所以 PR² = 144 + 324 − 432 × (−0.08716) = 468 + 37.65 = 505.65。PR = √505.65 ≈ 22.49 公里。

Now find the bearing of R from P. Use sine rule to find angle ∠QPR: sin(∠QPR)/18 = sin(95°)/22.49. sin(95°) ≈ 0.9962. sin(∠QPR) = 18 × 0.9962 / 22.49 ≈ 0.7974. ∠QPR ≈ 52.9°. But this is the angle inside the triangle at P. To get the bearing, we need the angle between north and PR. The line PQ is at 65° from north. The angle ∠QPR is on the eastern side, so the bearing of PR = 65° + 52.9° = 117.9° ≈ 118°.

再求从 P 看 R 的方位角。用正弦定理求 ∠QPR：sin(∠QPR)/18 = sin(95°)/22.49。sin(95°) ≈ 0.9962。sin(∠QPR) = 18 × 0.9962 / 22.49 ≈ 0.7974。∠QPR ≈ 52.9°。这是 P 处的内角。PQ 方位为 65°，∠QPR 在东侧，所以 PR 的方位 = 65° + 52.9° = 117.9° ≈ 118°。

Therefore, PR is about 22.5 km on a bearing of 118°.

因此，PR 距离约22.5公里，方位角118°。

5. Case Study 5: Scatter Graph and Line of Best Fit | 案例分析5：散点图与最佳拟合线

Problem: A student records the temperature (°C) and the number of ice creams sold at a beach kiosk over 8 days. The data: (22, 45), (25, 52), (19, 38), (28, 61), (30, 68), (21, 42), (27, 58), (24, 50). Plot the scatter graph, draw a line of best fit, and estimate the sales when the temperature is 26°C.

问题：一个学生记录了海滩售货亭8天的温度（°C）和冰淇淋销量。数据：(22, 45), (25, 52), (19, 38), (28, 61), (30, 68), (21, 42), (27, 58), (24, 50)。绘制散点图，画出最佳拟合线，并估计温度为26°C时的销量。

First, identify the independent variable (temperature, x-axis) and dependent variable (sales, y-axis). The points show a strong positive correlation.

首先，确定自变量（温度，x轴）和因变量（销量，y轴）。这些点显示出强烈的正相关。

To draw the line of best fit by eye, it should pass through the mean point (x̄, ȳ). Calculate means: x̄ = (22+25+19+28+30+21+27+24)/8 = 196/8 = 24.5. ȳ = (45+52+38+61+68+42+58+50)/8 = 414/8 = 51.75. Plot mean point (24.5, 51.75).

通过目测画出最佳拟合线，它应通过均值点 (x̄, ȳ)。计算均值：x̄ = (22+25+19+28+30+21+27+24)/8 = 196/8 = 24.5。ȳ = (45+52+38+61+68+42+58+50)/8 = 414/8 = 51.75。标出均值点 (24.5, 51.75)。

Choose a second point to determine slope, e.g., (20, 39) and (29, 64). Gradient = (64 − 39) / (29 − 20) = 25 / 9 ≈ 2.78. Equation roughly y = 2.78x + c. Using (24.5, 51.75): 51.75 = 2.78 × 24.5 + c ⇒ c = 51.75 − 68.11 = −16.36. So line equation: sales = 2.78 × temperature − 16.36.

另选一点确定斜率，例如 (20, 39) 和 (29, 64)。斜率 = (64 − 39) / (29 − 20) = 25 / 9 ≈ 2.78。方程约为 y = 2.78x + c。代入 (24.5, 51.75)：51.75 = 2.78 × 24.5 + c ⇒ c = 51.75 − 68.11 = −16.36。故直线方程：销量 = 2.78 × 温度 − 16.36。

For temperature 26°C, predicted sales = 2.78 × 26 − 16.36 = 72.28 − 16.36 = 55.92 ≈ 56 ice creams.

温度26°C时，预测销量 = 2.78 × 26 − 16.36 = 72.28 − 16.36 = 55.92 ≈ 56 个冰淇淋。

Remember: Extrapolation beyond the data range is unreliable; 26°C is within the range so the estimate is reliable.

记住：超出数据范围的外推不可靠；26°C在范围内，故估计可靠。

6. Case Study 6: Quadratic Equation in Projectile Motion | 案例分析6：抛体运动中的二次方程

Problem: A firework rocket is launched from the ground. Its height h (in meters) after t seconds is given by h = −5t² + 40t. Find the time when it hits the ground again and the maximum height reached.

问题：烟花火箭从地面发射。t 秒后高度 h（米）由 h = −5t² + 40t 给出。求它再次落地的时间和达到的最大高度。

When it hits the ground, h = 0. Solve −5t² + 40t = 0 ⇒ −5t(t − 8) = 0. So t = 0 (launch) or t = 8 seconds. The rocket lands after 8 seconds.

落地时 h = 0。解 −5t² + 40t = 0 ⇒ −5t(t − 8) = 0。得 t = 0（发射时）或 t = 8 秒。火箭在8秒后落地。

Maximum height occurs at the vertex. The x-coordinate (t) of the vertex is given by −b/(2a). Here a = −5, b = 40, so t = −40 / (2 × −5) = 40/10 = 4 seconds. Height at t = 4: h = −5(4)² + 40(4) = −80 + 160 = 80 meters.

最大高度出现在顶点处。顶点的 x坐标（t）为 −b/(2a)。此处 a = −5，b = 40，故 t = −40 / (2 × −5) = 40/10 = 4 秒。在 t = 4 时的高度：h = −5(4)² + 40(4) = −80 + 160 = 80 米。

Alternatively, complete the square: h = −5(t² − 8t) = −5[(t − 4)² − 16] = −5(t − 4)² + 80. This confirms maximum 80 m at t = 4.

另一种方法，配方：h = −5(t² − 8t) = −5[(t − 4)² − 16] = −5(t − 4)² + 80。确认 t = 4 时最大高度80米。

7. Case Study 7: Vector Displacement | 案例分析7：向量位移

Problem: A drone flies 200 m on a bearing of 030°. It then turns and flies 150 m on a bearing of 120°. Find the resultant displacement from its starting point, expressed as a column vector and as a distance with bearing.

问题：无人机以 030° 方位飞行200米，然后转向以 120° 方位飞行150米。求从起点的合位移，用列向量和距离与方位角表示。

Resolve each leg into components. Use bearings: from north, eastwards component = length × sin(bearing), northwards component = length × cos(bearing).

将每段分解为分量。利用方位角：从北起算，东分量 = 长度 × sin(方位角)，北分量 = 长度 × cos(方位角)。

First leg: 200 m at 030°: North = 200 × cos30° = 200 × 0.8660 = 173.2, East = 200 × sin30° = 200 × 0.5 = 100.

第一段：200米，030°：北 = 200 × cos30° = 173.2，东 = 200 × sin30° = 100。

Second leg: 150 m at 120°: North = 150 × cos120° = 150 × (−0.5) = −75, East = 150 × sin120° = 150 × 0.8660 = 129.9.

第二段：150米，120°：北 = 150 × cos120° = −75，东 = 150 × sin120° = 129.9。

Total north = 173.2 − 75 = 98.2 m; total east = 100 + 129.9 = 229.9 m. Column vector: (229.9, 98.2)ᵀ or written as (east, north).

总向北 = 173.2 − 75 = 98.2 m；总向东 = 100 + 129.9 = 229.9 m。列向量：(229.9, 98.2)ᵀ。

Resultant distance = √(229.9² + 98.2²) = √(52854.01 + 9643.24) = √62497.25 ≈ 249.99 m ≈ 250 m. Bearing: θ = arctan(east/north) = arctan(229.9/98.2) ≈ arctan(2.341) ≈ 66.9°. So bearing ≈ 067°.

合位移距离 = √(229.9² + 98.2²) ≈ 250 m。方位角：θ = arctan(东/北) = arctan(229.9/98.2) ≈ 66.9°。故方位角 ≈ 067°。

8. Case Study 8: Optimisation with Calculus (for Extension) | 案例分析8：微积分优化（拓展）

Problem: A rectangular pen is to be built against an existing wall, so only three sides need fencing. A total of 40 m of fencing is available. Find the maximum area that can be enclosed.

问题：紧靠一面已有的墙建一个矩形围栏，因此只需围三边。现有40米围栏材料。求能围出的最大面积。

Let the side parallel to the wall be x meters, and each perpendicular side be y meters. Then x + 2y = 40, so x = 40 − 2y. Area A = x × y = (40 − 2y)y = 40y − 2y².

设平行于墙的一边为 x 米，每个垂直边为 y 米。那么 x + 2y = 40，即 x = 40 − 2y。面积 A = x × y = (40 − 2y)y = 40y − 2y²。

Taking derivative: dA/dy = 40 − 4y. Set to zero: 40 − 4y = 0 ⇒ y = 10 m. Then x = 40 − 20 = 20 m. Maximum area = 20 × 10 = 200 m².

求导：dA/dy = 40 − 4y。令导数为零：40 − 4y = 0 ⇒ y = 10 m。则 x = 40 − 20 = 20 m。最大面积 = 20 × 10 = 200 m²。

We can also use quadratic completing the square: A = −2(y² − 20y) = −2[(y − 10)² − 100] = −2(y − 10)² + 200, vertex at y=10 gives maximum 200.

也可用二次函数配方：A = −2(y² − 20y) = −2[(y − 10)² − 100] = −2(y − 10)² + 200，当 y=10 时最大值为200。

9. Common Pitfalls and How to Avoid Them | 常见陷阱与避免方法

Case study questions can trip you up if you misread the context. Always identify the correct mathematical model: is it linear, quadratic, exponential? Check whether you need to multiply by a decimal for percentage change or add a constant. In trigonometry, draw a clear diagram and label all known angles and sides. Use the appropriate rule: sine rule for non‑right triangles with known angle‑opposite side pairs; cosine rule for two sides and the included angle or three sides. For bearings, remember they are measured from north clockwise, and the angle inside a triangle often requires subtracting bearings.

案例分析题若误读情境就容易出错。务必识别正确的数学模型：是线性、二次还是指数？检查是需要乘以百分数变化的小数还是加上常数。在三角学中，画清晰图示并标注所有已知角和边。选择合适定理：当已知一对角和对边时用正弦定理；当已知两边及其夹角或三边时用余弦定理。方位角要记住从北顺时针量度，三角形内角常需通过方位角的加减求得。

In algebra word problems, define variables clearly and write the equation step by step. Check your solution satisfies the original problem, not just the equation. For statistics, the line of best fit should have roughly equal scatter above and below. Use interpolation only within the data range.

在代数应用题中，明确设元并逐步列出方程。检验解是否满足原题情境，而不只是方程。统计中，最佳拟合线上下散点应大致均匀，仅对数据范围内的值进行内插估计。

10. Conclusion: Mastering SQA Case Studies | 结语：掌握 SQA 案例分析

By working through these varied case studies, you have practiced the key skills: financial calculations with multipliers, geometric reasoning with similarity and trigonometry, data interpretation with graphs, algebraic modelling, and vector resolution. The key is to read the problem carefully, translate it into mathematics, solve systematically, and always check your answer in the context.

通过这些多样化的案例分析，你已经练习了关键技能：用乘数进行金融计算、用相似性和三角学进行几何推理、通过图表解读数据、代数建模以及向量分解。关键在于仔细读题，转化为数学语言，系统求解，并始终在情境中检验答案。

Remember that in SQA exams, method marks are awarded for correct steps, not just final answers, so always show your working.

记住，在 SQA 考试中，即使最终答案有误，正确步骤也能得过程分，因此务必展示你的计算过程。

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Case Study Practical Exercises | 案例分析实战演练