SQA Year 11 Mathematics: Case Study Practice | SQA Year 11 数学：案例分析实战演练

📚 SQA Year 11 Mathematics: Case Study Practice | SQA Year 11 数学：案例分析实战演练

Case study questions in the SQA National 5 Mathematics exam require you to apply a range of mathematical skills to unfamiliar, real-life contexts. These problems often combine multiple topics and test your ability to model situations, perform accurate calculations and interpret results. This article walks you through three complete case studies covering algebra, trigonometry and statistics. Each is broken down into manageable steps with clear reasoning in English and Chinese, followed by exam tips to help you tackle similar questions with confidence.

SQA National 5 数学考试中的案例分析题要求你将多种数学技能应用于不熟悉的、真实的情境。这类题目通常结合多个主题，考察你建模、精确计算和解读结果的能力。本文将带你完整演练三个案例，涵盖代数、三角学和统计。每个案例都分解为易于管理的步骤，提供清晰的中英文推理，最后还有考试技巧，帮助你自信应对类似题目。

1.1 Modelling: Setting up Variables and Equation | 建模：设定变量与建立方程

A gardener has 40 metres of fencing to enclose a rectangular flower bed against a straight wall. Only three sides need fencing because the wall forms one side. We want to find the dimensions that give the maximum area. Start by introducing a variable: let the width perpendicular to the wall be x metres.

一位园丁有40米篱笆，想靠着一堵直墙围一个矩形花坛。因为墙充当了一边，所以只需要围三边。我们想找出使面积最大的尺寸。首先引入变量：设与墙垂直的宽为 x 米。

Since the fencing runs along two widths and one length, the length parallel to the wall is (40 − 2x) metres. The area A (in square metres) is the product of width and length.

因为篱笆用于两个宽和一个长，平行于墙的长为 (40 − 2x) 米。面积 A（平方米）是宽与长的乘积。

A = x(40 − 2x) = −2x² + 40x

This is a quadratic expression with a negative coefficient of x², so the parabola opens downwards and the vertex will give the maximum area.

这是一个二次表达式，x² 的系数为负，因此抛物线开口向下，顶点将给出最大面积。

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1.2 Completing the Square to Find Maximum | 完全平方求最大值

To locate the vertex, we complete the square for the quadratic −2x² + 40x. First, take out the factor −2 from the x terms:

为了找到顶点，我们对二次式 −2x² + 40x 进行完全平方。首先，从含 x 的项中提取公因子 −2：

A = −2(x² − 20x)

Now, inside the bracket, take half of the coefficient of x (which is −20), giving −10, square it to get 100, and both add and subtract it:

现在，在括号内，取 x 的系数（−20）的一半，得到 −10，平方得 100，然后同时加上并减去它：

A = −2(x² − 20x + 100 − 100) = −2[(x − 10)² − 100]

Expand the outer bracket to obtain vertex form:

展开外层括号得到顶点形式：

A = −2(x − 10)² + 200

x is at (10, 200). This tells us that when x = 10 m, the area reaches a maximum of 200 m².

顶点在 (10, 200)。这告诉我们当 x = 10 米时，面积达到最大值 200 平方米。

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1.3 Interpretation and Verification | 解释与验证

², confirming our algebraic result. From a practical perspective, x must be positive and the length must be non-negative, so 0 < x < 20. Our solution falls within this domain.

宽 x = 10 米时，长变为 40 − 2×10 = 20 米。面积为 10 × 20 = 200 平方米，验证了代数结果。从实际角度看，x 必须为正且长度非负，所以 0 < x < 20。我们的解在这个范围内。

a is 0. The quadratic shape confirms that the maximum occurs at the vertex. This is a common SQA final answer requiring dimensions and the maximum area.

我们还可以检验端点：若 x = 0，面积为 0；若 x = 20，长为 0，面积也为 0。二次函数的形状确认最大值出现在顶点。这是 SQA 常见的最终答案，要求给出尺寸和最大面积。

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2.1 Drawing a Diagram and Given Information | 绘制示意图与已知信息

Two surveyors need to determine the height of a tower. From point A, the angle of elevation to the top (C) is 35°. They walk directly towards the tower for 50 m to point B, where the angle of elevation becomes 50°. We are to find the height of the tower, h metres.

两位测量员需要确定一座塔的高度。从 A 点测得塔顶 C 的仰角为 35°。他们朝塔的方向直走 50 米到达 B 点，此处仰角变为 50°。我们要计算塔高 h 米。

Always start by sketching the right‑ AD = h / tan35°. In right triangle BCD, BD = h / tan50°. The distance AB = AD − BD = 50 m.

始终从画草图开始。设塔底为 D。在直角三角形 ACD 中，AD = h / tan35°。在直角三角形 BCD 中，BD = h / tan50°。距离 AB = AD − BD = 50 米。

ngle CBA = 180° − 50° = 130°, so angle ACB = 180° − 35° − 130° = 15°. The side AB is 50 m.

我们不用直接解正切方程，而是利用三角形 ABC，已知两角及夹边。∠CAB = 35°，∠CBA = 180° − 50° = 130°，故 ∠ACB = 180° − 35° − 130° = 15°。边 AB = 50 米。

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2.2 Applying the Sine Rule | 应用正弦定理

In any triangle, the sides are proportional to the sines of the opposite angles. Applying the sine rule to triangle ABC:

在任意三角形中，边与对角的正弦成比例。对三角形 ABC 应用正弦定理：

AC / sin130° = AB / sin15° ⇒ AC / sin130° = 50 / sin15°

Recall that sin130° = sin(180°−130°) = sin50° ≈ 0.7660. sin15° ≈ 0.2588. Substituting:

注意 sin130° = sin(180°−130°) = sin50° ≈ 0.7660。sin15° ≈ 0.2588。代入：

AC = 50 × sin130° / sin15° ≈ 50 × 0.7660 / 0.2588 ≈ 148.0 m

Now return to right triangle ACD. The height h is opposite the 35° angle, so:

现在回到直角三角形 ACD。高 h 是 35° 角的对边，所以：

sin35° = h / AC ⇒ h = AC × sin35°

This approach avoids simultaneous equations and uses only the sine rule plus right‑triangle trigonometry.

这种方法避免了联立方程，仅使用正弦定理和直角三角形三角比。

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2.3 Calculation and Rounding | 计算与四舍五入

Using a calculator, sin35° ≈ 0.5736. Therefore:

用计算器，sin35° ≈ 0.5736。因此：

h ≈ 148.0 × 0.5736 ≈ 84.9 m (to 3 significant figures)

SQA typically expects rounding to an appropriate degree of accuracy. Since the given data (angles to the nearest degree, distance 50 m) is fairly precise, 3 significant figures is reasonable. Always state the rounding clearly.

SQA 通常要求舍入到适当的精确度。因为给出的数据（角度精确到度，距离50米）比较精确，保留三位有效数字是合理的。总是要清晰地说明舍入。

tan35° ≈ 84.9/0.7002 ≈ 121.3 m, difference 50.1 m, confirming consistency within rounding.

我们可以用正切法验证：BD = h / tan50° ≈ 84.9/1.1918 ≈ 71.2 米，AD = h / tan35° ≈ 84.9/0.7002 ≈ 121.3 米，二者差 50.1 米，在舍入范围内一致。

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3.1 Calculating Measures of Centre and Spread | 计算集中趋势和离散度量

Two classes sat the same test. Their percentage scores are shown below. We will calculate key statistics to compare performances.

两个班级参加了同一测试。他们的百分制成绩如下。我们将计算关键统计量以比较表现。

Class A: 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
Class B: 30, 45, 55, 60, 65, 70, 75, 80, 85, 95

Both data sets have 10 values. We can compute the mean (x̄), median, lower quartile (Q1), upper quartile (Q3), interquartile range (IQR) and standard deviation (s).

两组数据均有 10 个值。我们可以计算平均数（x̄）、中位数、下四分位数（Q1）、上四分位数（Q3）、四分位距（IQR）和标准差（s）。

Published by TutorHao | Year 11 Mathematics Revision Series | aleveler.com

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