📚 Year 9 SQA Advanced Mathematics: Case Study Practical Drills | Year 9 SQA 进阶数学：案例分析实战演练
In Senior Phase mathematics, applying concepts to real‑world scenarios is just as important as mastering algebraic manipulation. This article presents a series of case study drills designed for Year 9 SQA Advanced Mathematics learners, each targeting a specific skill area. Work through every scenario, translate the situation into mathematical language, and then reflect on whether your solution makes sense in context.
在高年级数学中，将概念应用到现实情境与掌握代数运算同样重要。本文为 Year 9 SQA 进阶数学学生设计了一系列案例分析实战演练，每个案例针对一个特定的技能领域。请逐一完成情境，把情况转化为数学语言，然后反思你的解在情境中是否合理。
1. Linear Equations – Comparing Mobile Plans | 线性方程 – 手机套餐比较
Mia is choosing between two monthly mobile phone contracts. Plan A charges a flat fee of £12 plus £0.05 per minute of calls. Plan B charges £7 per month and £0.08 per minute. Mia typically uses 400 minutes of calls each month, but she wants to know the exact number of minutes for which both plans cost the same.
Mia 正在比较两种手机月套餐。套餐 A 每月固定费用 12 英镑，通话每分钟 0.05 英镑。套餐 B 每月 7 英镑，每分钟 0.08 英镑。Mia 通常每月通话 400 分钟，但她想知道两种套餐费用完全相同时的通话分钟数。
Let the number of minutes be m. Cost for Plan A: 12 + 0.05m. Cost for Plan B: 7 + 0.08m. Set them equal: 12 + 0.05m = 7 + 0.08m.
设通话分钟数为 m。套餐 A 费用：12 + 0.05m。套餐 B 费用：7 + 0.08m。令两者相等：12 + 0.05m = 7 + 0.08m。
Subtract 0.05m from both sides: 12 = 7 + 0.03m. Then subtract 7: 5 = 0.03m. Divide by 0.03: m = 5 ÷ 0.03 ≈ 166.67 minutes. So the break‑even point is about 167 minutes. At 400 minutes, Plan A costs £12 + 0.05 × 400 = £32, while Plan B costs £7 + 0.08 × 400 = £39. For high usage, Plan A is cheaper.
两边减去 0.05m：12 = 7 + 0.03m。再减 7：5 = 0.03m。除以 0.03：m = 5 ÷ 0.03 ≈ 166.67 分钟。因此盈亏平衡点约为 167 分钟。在 400 分钟时，套餐 A 费用为 12 + 0.05 × 400 = 32 英镑，套餐 B 为 7 + 0.08 × 400 = 39 英镑。对于高用量用户，套餐 A 更便宜。
2. Pythagoras’ Theorem – Ladder Safety | 勾股定理 – 梯子安全
A 5‑metre ladder is leaning against a vertical wall. Safety regulations require the foot of the ladder to be at least 1.5 metres from the wall to avoid slipping. When the foot is exactly 1.5 metres away, how high up the wall does the ladder reach? Is the reach still above the 4‑metre mark needed to clean a window?
一架 5 米长的梯子斜靠在竖直墙壁上。安全规定要求梯脚离墙至少 1.5 米以防滑倒。当梯脚正好 1.5 米时，梯子顶端能达到墙上的高度是多少？这个高度是否仍高于清洁窗户所需的 4 米标记？
Let the height up the wall be h metres. By Pythagoras: h² + 1.5² = 5². So h² = 25 − 2.25 = 22.75. Taking the square root gives h = √22 r reaches above 4 metres, so the window can be cleaned safely as long as the foot stays at 1.5 m.
设墙上高度为 h 米。由勾股定理：h² + 1.5² = 5²。因此 h² = 25 − 2.25 = 22.75。开平方得 h = √22.75 ≈ 4.77 米（保留两位小数）。是的，梯子超过 4 米，只要梯脚保持在 1.5 米处，窗户可以安全清洁。
3. Inequalities – Exam Grade Boundaries | 不等式 – 考试分数边界
In a maths exam, a student’s final mark m (out of 80) is converted to a grade. To earn a ‘B’, the mark must be between 60% and 75% inclusive. Express this condition as an inequality and find the range of possible marks. Additionally, the teacher offers a bonus of 5 marks for a project. After adding the bonus, what marks out of 80 originally guarantee a final grade of B?
在一次数学考试中，学生的最终分数 m（满分 80）转换为等级。要获得 ‘B’，分数必须在 60% 到 75% 之间（含边界）。将此条件表示为不等式，并求出可能的分数范围。此外，老师为一个项目提供 5 分加分。加上加分后，原始卷面多少分能保证最终等级为 B？
60% of 80 is 48, 75% of 80 is 60. So the inequality is 48 ≤ m ≤ 60. With a 5‑mark bonus, the final mark is m + 5. We require 48 ≤ m + 5 ≤ 60. Subtract 5 throughout: 43 ≤ m ≤ 55. So a raw mark between 43 and 55 (inclusive) will guarantee a B after the bonus.
80 的 60% 是 48，75% 是 60。因此不等式为 48 ≤ m ≤ 60。加上 5 分加分后，最终分数为 m + 5。要求 48 ≤ m + 5 ≤ 60。整体减去 5：43 ≤ m ≤ 55。因此原始分数在 43 到 55 之间（含边界）可保证加分后获得 B。
4. Quadratic Functions – Projectile Height | 二次函数 – 抛射体高度
A firework is launched from ground level. Its height h (in metres) at time t seconds is given by h = 30t − 5t². Determine when the firework hits the ground again and the maximum height it reaches. Use the symmetry of the parabola to check your maximum.
一枚烟花从地面发射。其高度 h（米）与时间 t（秒）的关系为 h = 30t − 5t²。确定烟花何时落回地面以及达到的最大高度。利用抛物线的对称性检验最大值。
Ground level means h = 0: 30t − 5t² = 0 → 5t(6 − t) = 0. So t = 0 or t 6. The firework lands after 6 seconds. The axis of symmetry is halfway between the roots: t = 3. Maximum height is h(3) = 30×3 − 5×3² = 90 − 45 = 45 metres. The vertex is (3, 45).
地面即 h = 0：30t − 5t² = 0 → 5t(6 − t) = 0。得 t = 0 或 t = 6。烟花在 6 秒后落地。对称轴位于两根中点：t = 3。最大高度为 h(3) = 30×3 − 5×3² = 90 − 45 = 45 米。顶点为 (3, 45)。
5. Statistics – Misleading Averages in Salaries | 统计 – 工资中的误导性平均数
A small tech company has 10 employees with annual salaries: £22k, £24k, £25k, £26k, £27k, £28k, £30k, £32k, £35k, and the CEO earns £150k. The company claims “the average salary is £39.9k”. Which measure of central tendency are they using, and why is it misleading? Calculate the mean, median and decide which better represents the typical salary.
一家小型科技公司有 10 名员工，年薪（千英镑）为：22, 24, 25, 26, 27, 28, 30, 32, 35，而 CEO 赚 15 万。公司宣称“平均工资为 3.99 万英镑”。他们使用的是哪种集中趋势度量，为什么会产生误导？计算平均数和中位数，并判断哪一个更能代表典型工资。
0 = 399. Mean = 399 ÷ 10 = 39.9k. This is the arithmetic mean, heavily skewed by the CEO’s £ orting the mean without context is misleading because one extreme value inflates the average.
所有工资总和 = 22+24+25+26+27+28+30+32+35+150 = 399。平均数 = 399 ÷ 10 = 39.9k。这是算术平均数，被 CEO 的 150k 严重拉高。顺序排列：22,24,25,26,27,28,30,32,35,150。中位数 = (27+28)/2 = 27.5k。中位数更接近普通员工收入。在没有背景的情况下报告平均数具有误导性，因为一个极端值抬高了平均值。
6. Probability – Designing a Fair Game | 概率 – 设计公平游戏
James and Laura play a game with two fair six‑sided dice. James wins if the sum of the dice is 6, 7 or 8. Laura wins if the sum is 2, 3, 4, 10, 11 or 12. If any other sum appears, the round is a draw. By constructing a sample space or using symmetry, determine whether the game is fair. Calculate the probability that a round results in a win for each player.
James 和 Laura 用两个公平的六面骰子玩游戏。如果点数和为 6、7 或 8，James 赢；如果和为 2、3、4、10、11 或 12，Laura 赢。如果出现其他和，该局平局。通过构建样本空间或使用对称性，判断游戏是否公平。计算每局每人获胜的概率。
12 (1) → total 12 ways. Remaining sum 5 (4 ways) and sum 9 (4 ways) are draws. P(James wins) = 16/36 = 4/9 ≈ 0.444; P(Laura wins) = 12/36 = 1/3 ≈ 0.333. The game is not fair; James has a higher chance of winning.
总结果数：36。James 获胜的和：和为 6（5 种）、7（6 种）、8（5 种）→ 共 16 种。Laura 获胜的和：2（1 种）、3（2 种）、4（3 种）、10（3 种）、11（2 种）、12（1 种）→ 共 12 种。其余和 5（4 种）和 9（4 种）为平局。P(James 赢) = 16/36 = 4/9 ≈ 0.444；P(Laura 赢) = 12/36 = 1/3 ≈ 0.333。游戏不公平，James 获胜机会更高。
7. Sequences – Saving for a Bike | 数列 – 存钱买自行车
Emma saves money each week in an arithmetic sequence. In week 1 she saves £5, week 2 she saves £8, week 3 £11, and so on. She continues until the week she saves exactly £50. How many weeks does this take, and what is the total amount saved by then? Use the formula for the nth term and the sum of an arithmetic series.
Emma 每周的存款构成一个等差数列。第 1 周存 5 英镑，第 2 周存 8 英镑，第 3 周存 11 英镑，依此类推。她一直存到某周正好存 50 英镑为止。这需要多少周？到那时总共存了多少钱？使用第 n 项公式和等差数列求和公式。
First term a = 5, common difference d = 3. The nth term Uₙ = 5 + (n−1)×3. Set 5 + 3(n−1) = 50 → 3(n−1) = 45 → n−1 = 15 → n = 16 weeks. Total saved Sₙ = n/2 × (first + last) = 16/2 × (5 + 50) = 8 × 55 = £440. So after 16 weeks, Emma has saved £440.
首项 a = 5，公差 d = 3。第 n 项 Uₙ = 5 + (n−1)×3。令 5 + 3(n−1) = 50 → 3(n−1) = 45 → n−1 = 15 → n = 16 周。总存款 Sₙ = n/2 × (首项 + 末项) = 16/2 × (5 + 50) = 8 × 55 = 440 英镑。所以 16 周后，Emma 共存了 440 英镑。
8. Distance–Time Graphs – Interpreting Journeys | 距离–时间图 – 解读行程
The graph below describes a cyclist’s journey from home. (A sketch would show: flat section 0–20 min, distance stays 0 km; steady increase 20–50 min, reaching 12 km; flat 50–70 min; then a straight line back to 0 km at 90 min.) Without drawing, answer: During which intervals is the cyclist stationary? When is speed greatest? Calculate the speed on the return leg in km/h.
下图描述了一位骑行者从家出发的行程。（草图显示：0–20 分钟平直线段，距离保持 0 公里；20–50 分钟稳步上升，达到 12 公里；50–70 分钟持平；然后直线回到 0 公里，到达 90 分钟。）不画图回答：骑行者在哪段时间静止？哪段速度最快？计算返程速度（千米/小时）。
Stationary sections are when distance does not change: 0–20 min (still at home) and 50– reatest speed corresponds to the steepest slope. The outward leg rises 12 km in 30 min (20–50 min), speed = 12 km ÷ 0.5 h = 24 km/h. The return leg covers 12 km in 20 min (70–90 min), speed = 12 km ÷ (20/60) h = 36 km/h. So the return leg is faster.
静止段为距离不变时：0–20 分钟（仍在家）和 50–70 分钟（休息）。最大速度对应最陡的坡度。去程在 30 分钟内（20–50 分钟）上升 12 公里，速度 = 12 km ÷ 0.5 h = 24 km/h。返程在 20 分钟内（70–90 分钟）行驶 12 公里，速度 = 12 km ÷ (20/60) h = 36 km/h。因此返程更快。
9. Simultaneous Equations – Mixing Alloys | 联立方程 – 合金混合
A metallurgist needs 500 grams of an alloy containing 40% copper. She has two stocks: Alloy X with 30% copper and Alloy Y with 60% copper. How many grams of each must she mix? Set up two equations and solve.
一位冶金学家需要 500 克含铜 40% 的合金。她有两种库存：合金 X 含铜 30%，合金 Y 含铜 60%。她必须混合多少克各种合金？建立两个方程并求解。
Let x = grams of Alloy X, y = grams of Alloy Y. Total mass: x + y = 500. Copper mass: 0.30x + 0.60y = 0.40 × 500 = 200. Solve: from first, y = 500 − x. Substitute: 0.30x + 0.60(500 − x) = 200 → 0.30x + 300 − 0.60x = 200 → −0.30x = −100 → x = 100/0.30 ≈ 333.33 g. Then y ≈ 166.67 g. So about 333 g of X and 167 g of Y are needed.
设 x = 合金 X 的克数，y = 合金 Y 的克数。总质量：x + y = 500。铜质量：0.30x + 0.60y = 0.40 × 500 = 200。求解：由第一式得 y = 500 − x。代入：0.30x + 0.60(500 − x) = 200 → 0.30x + 300 − 0.60x = 200 → −0.30x = −100 → x = 100/0.30 ≈ 333.33 g。则 y ≈ 166.67 g。因此需要约 333 克合金 X 和 167 克合金 Y。
10. Trigonometry – Height of a Tower | 三角学 – 塔的高度
a possible error of ±1°, find the range of possible heights. Use tangent and discuss the impact of measurement error.
从塔底水平距离 50 米的点测得塔顶仰角为 38°。计算塔的高度。如果角度测量可能有 ±1° 的误差，求出可能的高度范围。使用正切并讨论测量误差的影响。
Height h satisfies tan 38° = h / 50 → h = 50 × tan 38°. Using a calculator: tan 38° ≈ 0.7813 → h ≈ 39.07 m. With angle 37°, h = 50 × tan 37° ≈ 50 × 0.7536 = 37.68 m. With angle 39°, h he true height likely lies between 37.7 m and 40.5 m. A small angular error leads to a noticeable difference in height, showing the importance of precise measurement.
高度 h 满足 tan 38° = h / 50 → h = 50 × tan 38°。用计算器：tan 38° ≈ 0.7813 → h ≈ 39.07 米。角度为 37° 时，h = 50 × tan 37° ≈ 50 × 0.7536 = 37.68 米。角度为 39° 时，h = 50 × tan 39° ≈ 50 × 0.8098 = 40.49 米。因此真实高度可能在 37.7 米到 40.5 米之间。小的角度误差导致高度明显差异，说明精确测量的重要性。
11. Area and Perimeter – Composite Shape Design | 面积与周长 – 复合图形设计
A park has a rectangular lawn of length 30 m and width 20 m, with a semicircular flower bed added on one of the shorter sides. Find the total area of the park and the length of fencing required to enclose it (excluding the boundary between the lawn and flower bed). Give your answers in terms of π and as decimals.
一个公园有一块长 30 米、宽 20 米的长方形草坪，在一条短边上附加一个半圆形花坛。求公园的总面积以及围栏所需长度（不包括草坪与花坛之间的边界）。用 π 和小数形式给出答案。
m². Perimeter to fence: rectangle has three sides (two long, one short) plus the curved part of semicircle. Length = 30 + 30 + 20 + π × 10 = 80 + 10π ≈ 111.42 m. The shared line (width 20 m) is not fenced.
长方形面积 = 30 × 20 = 600 平方米。半圆直径 = 20 米，半径 = 10 米。半圆面积 = ½ × π × 10² = 50π ≈ 157.08 平方米。总面积 = 600 + 50π ≈ 757.08 平方米。围栏周长：长方形有三边（两条长边、一条短边）加上半圆的弧长。长度 = 30 + 30 + 20 + π × 10 = 80 + 10π ≈ 111.42 米。共用的边界（宽 20 米）不设围栏。
12. Bivariate Data – Correlation vs. Causation | 双变量数据 – 相关与因果
A study records the number of ice‑ nds. The scatter plot shows a strong positive correlation (r = 0.92). A journalist writes, “Eating ice cream increases the risk of drowning.” ion is flawed. Identify a lurking variable and discuss how the data should be interpreted.
一项研究记录了 12 个周末海滩冰淇淋销量与溺水事件数。散点图显示强正相关（r = 0.92）。一位记者写道：“吃冰淇淋会增加溺水风险。”解释为什么这个结论有缺陷。找出一个潜在变量并讨论应如何解读数据。
Correlation does not imply causation. Both ice‑cream sales and drownings are linked to warmer weather and more people visiting the beach. The lurking variable is temperature or number of beachgoers. When it is hot, more people buy ice cream and more people swim, which naturally raises the risk of drowning. A more accurate statement: higher temperatures are associated with both increased ice‑cream sales and drowning incidents.
相关不代表因果。冰淇淋销量和溺水都与天气变暖及海滩游客增多有关。潜在变量是气温或海滩人数。天气炎热时，更多人买冰淇淋，也有更多人游泳，这自然增加了溺水风险。更准确的说法是：高温与冰淇淋销量增加和溺水事件增多都有关联。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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