📚 Case Study Practical Exercises for Year 9 OCR Advanced Mathematics | Year 9 OCR 进阶数学：案例分析实战演练
In Year 9 OCR Advanced Mathematics, applying concepts to real-world scenarios deepens understanding and sharpens problem-solving skills. These case studies cover optimisation, probability, sequences, trigonometry, and more, each designed to mirror the kind of contextual problems you will meet. Work through them step by step, using clear mathematical reasoning. This bilingual guide provides English explanations paired with Chinese translations, helping you master both the language of mathematics and international exam techniques.
在 Year 9 OCR 进阶数学中，将概念应用于实际问题能加深理解并强化解题能力。这些案例分析涵盖优化、概率、数列、三角学等内容，模拟了你在考试中会遇到的场景题。请一步步推演，运用清晰的数学推理。这份双语指南提供英文解释与对应的中文翻译，帮助你掌握数学语言和国际考试技巧。
1. Maximising the Volume of an Open Box | 无盖盒子的最大体积
We start with a 20 cm by 20 cm square sheet of card. Identical squares of side length x cm are cut from each corner, and the sides are folded up to form an open box. The challenge is to find the value of x that produces the maximum possible volume.
我们从一张 20 cm × 20 cm 的正方形卡纸开始。从每个角切去边长为 x cm 的相同正方形，再将四边折起形成一个无盖盒子。目标是找出使容积最大的 x 值。
The base of the box becomes a square of side (20 – 2x) cm, and the height is x cm. The volume V in cm³ is therefore V = x(20 – 2x)².
盒底成为边长为 (20 – 2x) cm 的正方形，高为 x cm。因此容积 V（立方厘米）为 V = x(20 – 2x)²。
Since x must be between 0 and 10, we build a table of values:
由于 x 必须在 0 到 10 之间，我们建立一个数值表：
| x (cm) | Width = 20 – 2x (cm) | Volume V = x(20 – 2x)² (cm³) |
|---|---|---|
| 1 | 18 | 1 × 18² = 324 |
| 2 | 16 | 2 × 16² = 512 |
| 3 | 14 | 3 × 14² = 588 |
| 4 | 12 | 4 × 12² = 576 |
| 3.3 | 13.4 | 3.3 × 13.4² ≈ 592.5 |
| 3.33 | 13.34 | 3.33 × 13.34² ≈ 592.6 |
× (40/3)² = 16000/27 ≈ 592.6 cm³. A graph of V against x confirms the peak at this value.
最大容积出现在 x = 3.33 cm 附近，精确值为 10/3 cm。此时 V = (10/3) × (40/3)² = 16000/27 ≈ 592.6 cm³。画出 V 随 x 变化的图像可确认此处为最高点。
2. Designing a Fair Game: Probability and Expected Value | 设计公平游戏：概率与期望值
A game uses a spinner divided into three sectors: red occupies 1/2 of the circle, blue 1/3, and yellow 1/6. A player pays £1 to spin. If it lands on red, they win £0.50; on blue, they win £1.00; on yellow, they win £2.00. Is the game fair? If not, adjust the yellow prize to make the expected profit for the organiser equal to zero.
一个游戏使用一个转盘，分为三个扇形：红色占 1/2，蓝色占 1/3，黄色占 1/6。玩家支付 £1 转一次。停在红色赢 £0.50，蓝色赢 £1.00，黄色赢 £2.00。此游戏是否公平？如不公平，调整黄色奖金使得组织者的期望利润为零。
1/3)×1.00 + (1/6)×2.00 = 0.25 + 0.333... + 0.333... = 0.9167 (or 11/12). Since the cost is £1, the player's expected loss is 1 – 11/12 = 1/12, so the game favours the organiser.
首先计算玩家的期望奖金：E(win) = (1/2)×0.50 + (1/3)×1.00 + (1/6)×2.00 = 0.25 + 0.333... + 0.333... = 0.9167 (即 11/12)。由于成本为 £1，玩家期望损失为 1 – 11/12 = 1/12，因此游戏对组织者有利。
To make the game fair, let the new yellow prize be £ (win) = 1: (1/2)×0.50 + (1/3)×1.00 + (1/6)×W = 1. Simplify: 0.25 + 1/3 + W/6 = 1 → W/6 = 1 – (1/4 + 1/3) = 1 – 7/12 = 5/12. Hence W = 6 × 5/12 = 2.5. So the yellow prize should be £2.50 for a zero expected profit.
为使游戏公平，设新的黄色奖金为 £W。令 E(win) = 1：(1/2)×0.50 + (1/3)×1.00 + (1/6)×W = 1。化简得：0.25 + 1/3 + W/6 = 1 → W/6 = 1 – (1/4 + 1/3) = 1 – 7/12 = 5/12。因此 W = 6 × 5/12 = 2.5。所以黄色奖金需设为 £2.50 以获得零期望利润。
3. Modelling the Path of a Projectile | 抛射体轨迹建模
, the horizontal and vertical displacements after t seconds are x = 20 cos45° t and y = 20 sin45° t – ½ × 10 × t². Since cos45° = sin45° = √2/2 ≈ 0.7071, these simplify to x = 10√2 t and y = 10√2 t – 5t².
一个球从地面以初速度 20 m s⁻¹、与水平面成 45° 角踢出。忽略空气阻力并取 g = 10 m s⁻²，t 秒后的水平位移和垂直位移分别为 x = 20 cos45° t 和 y = 20 sin45° t – ½ × 10 × t²。由于 cos45° = sin45° = √2/2 ≈ 0.7071，可简化为 x = 10√2 t，y = 10√2 t – 5t²。
Eliminating t yields the trajectory equation: from x = 10√2 t we have t = x/(10√2). Substituting into y gives y = 10√2 (x/(10√2)) – 5 [x/(10√2)]² = x – 5x²/(200) = x – x²/40. Thus the path is a parabola: y = x – x²/40.
消去 t 得到轨迹方程：由 x = 10√2 t 得 t = x/(10√2)。代入 y 得到 y = 10√2 (x/(10√2)) – 5 [x/(10√2)]² = x – 5x²/(200) = x – x²/40。因此路径是一条抛物线：y = x – x²/40。
y = –(1/40)x² + x
Completing the square: y = –1/40 (x² – 40x) = –1/40[(x – 20)² – 400] = –1/40 (x – ht is 10 m, reached when x = 20 m. The total horizontal range is found by setting y = 0: x – x²/40 = 0 → x(1 – x/40) = 0, so x = 0 or x = 40 m. The ball travels 40 m horizontally.
配方法：y = –1/40 (x² – 40x) = –1/40[(x – 20)² – 400] = –1/40 (x – 20)² + 10。最大高度为 10 m，在 x = 20 m 处达到。总水平射程通过设 y = 0 求得：x – x²/40 = 0 → x(1 – x/40) = 0，故 x = 0 或 x = 40 m。球水平飞行 40 m。
4. Comparing Savings Plans: Sequences and Series | 比较储蓄计划：数列与级数
Plan A involves depositing £10 in week 1, £12 in week 2, £14 in week 3, and so on, increasing by £2 each week. Plan B uses a compound growth account where a single deposit of £200 earns 1% interest per week. Find the total saved in each plan after 20 weeks.
方案 A：第1周存 £10，第2周存 £12，第3周存 £14，依此类推，每周增加 £2。方案 B：一次性存入 £200 到一个每周复利1%的账户。求20周后每种方案的总储蓄额。
with first term a = 10 and common difference d = 2. The n-th week deposit is aₙ = 10 + (n – 1)×2. Total after 20 weeks is the sum S₂₀ = (20/2)[2×10 + (20 – 1)×2] = 10[20 + 38] = 580. Thus Plan A yields £580.
方案 A 是一个等差数列，首项 a = 10，公差 d = 2。第 n 周存款为 aₙ = 10 + (n – 1)×2。20周后的总额为 S₂₀ = (20/2)[2×10 + (20 – 1)×2] = 10[20 + 38] = 580。因此方案 A 得到 £580。
this is 200 × 1.01²⁰. Using a calculator, 1.01²⁰ ≈ 1.22019, so the total is approximately £ tal is much lower because only the initial sum compounds; Plan A adds new money each week.
方案 B 遵循等比增长：n 周后金额 = 200 × (1.01)ⁿ。20 周后为 200 × 1.01²⁰。用计算器得 1.01²⁰ ≈ 1.22019，总额约为 £244.04。尽管有利息，但总额低得多，因为只有初始本金在复利；方案 A 每周增加新存款。
5. Maximising Enclosed Area with a Fixed Length of Fence | 固定篱笆长度下的最大围栏面积
A farmer has 120 m of fencing and wants to fence a rectangular area against a long straight wall (no fence needed on that side). Let the two sides perpendicular to the wall have length x metres, and the side parallel to the wall have length (120 – 2x) metres. Find the dimensions that maximise the area.
一农夫有 120 m 长的篱笆，想靠着一道长直墙围出一块矩形区域（靠墙一边不需要篱笆）。设垂直于墙的两边长度为 x 米，平行于墙的一边长度为 (120 – 2x) 米。求能使面积最大的尺寸。
The area A = x(120 – 2x) = –2x² + 120x. This is a quadratic opening downwards. Complete the square: –2(x² – 60x) = –2[(x – 30)² – 900] = –2(x – 30)² + 1800.
面积 A = x(120 – 2x) = –2x² + 120x。这是一个开口向下的二次函数。配方法：–2(x² – 60x) = –2[(x – 30)² – 900] = –2(x – 30)² + 1800。
A = –2(x – 30)² + 1800
The maximum area of 1800 m² occurs when x = 30 m. Then the parallel side is 120 – 60 = 60 m. So the optimal rectangle is 30 m by 60 m.
最大面积 1800 m² 在 x = 30 m 时取得。此时平行边长度为 120 – 60 = 60 m。因此最优矩形尺寸为 30 m × 60 m。
6. Mixing Alloys: Simultaneous Equations | 混合合金：联立方程组
A metallurgist has two copper-zinc alloys. Alloy 1 contains 60% copper and 40% zinc. Alloy 2 contains 45% copper and 55% zinc. How many kilograms of each must be melted together to obtain 50 kg of an alloy that is 52.5% copper?
一位冶金学家有两种铜锌合金。合金 1 含 60% 铜和 40% 锌。合金 2 含 45% 铜和 55% 锌。需要熔化多少公斤的每种合金，才能得到 50 kg 含铜 52.5% 的新合金？
Alloy 2. Then total mass: x + y = 50. The copper mass: 0.60x + 0.45y = 0.525 × 50 = 26.25.
设 x 为合金 1 的质量（kg），y 为合金 2 的质量。总质量方程：x + y = 50。铜的质量方程：0.60x + 0.45y = 0.525 × 50 = 26.25。
ide by 15: 4x + 3y = 175. From x + y = 50, multiply by 3: 3x + 3y = 150. Subtract: (4x+3y) – (3x+3y) = 175 – 150 → x = 25. Then y = 25. So we need 25 kg of each alloy.
将铜方程乘以 100 消去小数：60x + 45y = 2625。除以 15 得：4x + 3y = 175。由 x + y = 50 乘以 3 得：3x + 3y = 150。相减：(4x+3y) – (3x+3y) = 175 – 150 → x = 25。则 y = 25。因此每种合金各需 25 kg。
7. The Golden Ratio through a Quadratic Equation | 通过二次方程探究黄金比例
he longer part to the shorter. If the longer part is φ and the shorter is 1, then (φ + 1) / φ = φ / 1. This gives φ + 1 = φ², or φ² – φ – 1 = 0.
黄金比例 φ 常定义为将一条线段分成两段，使得整体与较长部分的比值等于较长部分与较短部分的比值。若较长部分为 φ，较短部分为 1，则有 (φ + 1) / φ = φ / 1。得到 φ + 1 = φ²，即 φ² – φ – 1 = 0。
φ² – φ – 1 = 0
Solve using the quadratic formula: φ = [1 ± √1 + 4] / 2 = (1 ± √5 180339887. This number appears in art, architecture, and nature, such as in the spiral of a nautilus shell or the dimensions of the Parthenon.
用求根公式解：φ = [1 ± √1 + 4] / 2 = (1 ± √5)/2。取正根得到 φ ≈ 1.6180339887。这个数字出现在艺术、建筑和自然界中，例如鹦鹉螺的螺旋或帕特农神庙的尺寸比例。
The reciprocal of φ is 1/φ ≈ 0.618, which equals φ – makes the golden ratio a fascinating case study linking algebra, geometry, and aesthetics.
φ 的倒数为 1/φ ≈ 0.618，恰好等于 φ – 1。这一性质使黄金比例成为联系代数、几何与美学的迷人案例。
8. Navigation Using Trigonometry and the Cosine Rule | 应用三角学与余弦定理的导航问题
A ship sails from port P on a bearing of 060° for 12 km to point Q. Then it changes course to a bearing of 150° and sails 9 km to point R. Find the distance PR and the bearing from P to R.
一艘船从港口 P 出发，沿 060° 方位角航行 12 km 到达点 Q。然后改变航向至 150° 方位角，再航行 9 km 到达点 R。求距离 PR 以及从 P 到 R 的方位角。
The angle at Q can be found from the bearings. The initial bearing 060° means the path PQ makes 60° with north. The new bearing 150° means QR makes 150° with
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