📚 Case Study Practice for OCR Year 11 Maths | 案例分析实战演练
OCR Year 11 Maths case study questions bring together multiple topics in a single real‑world scenario. This guide walks you through practical examples so you can break down complex problems, identify the relevant maths, and present clear solutions.
OCR Year 11 数学的案例分析题将多个知识点融合在一个真实场景中。本文通过实战演练带你拆解复杂问题，识别相关的数学内容，并给出清晰的解答。
1. Understanding the Case Study Format | 理解案例分析题结构
Every case study begins with a block of text, data tables, or diagrams. Your first job is to highlight the quantities given, the unknowns, and any assumptions. Underline key numbers and units.
每个案例分析题都从一段文字、数据表格或示意图开始。你首先要标出已知量、未知量和任何假设条件，用下划线标出关键数字及单位。
Then separate the problem into smaller parts – often labelled (a), (b), (c). Each part usually builds on the previous one, so accuracy early on is essential.
然后把问题拆成小问——通常标有 (a)、(b)、(c)。每一问往往建立在前一问的基础上，因此早期的准确性至关重要。
Decide whether you need to use algebra, geometry, statistics or a combination. OCR examiners expect you to show method steps, not just final answers.
判断需要用到代数、几何、统计还是它们的组合。OCR 考官期望看到方法步骤，而不只是最终答案。
2. Financial Algebra: Compound Interest Savings | 金融代数：复利储蓄
Case study: Maya deposits £2000 into a savings account that pays 3.5% interest compounded annually. She wants to know how much the investment will be worth after 6 years, and how many whole years it takes to exceed £2500.
案例：Maya 把 2000 英镑存入年利率 3.5%、每年复利一次的储蓄账户。她想知道 6 年后的投资价值，以及需要多少个整年才能超过 2500 英镑。
For compound interest, use A = P(1 + r/100)n. Here P = 2000, r = 3.5, n = 6. Substitute: A = 2000 × (1.035)6. Evaluate stepwise: (1.035)6 ≈ 1.229255, so A ≈ £2458.51.
对于复利，使用 A = P(1 + r/100)n。此处 P = 2000，r = 3.5，n = 6。代入：A = 2000 × (1.035)6。逐步计算：(1.035)6 ≈ 1.229255，所以 A ≈ 2458.51 英镑。
To find when A > 2500, solve 2000 × (1.035)n > 2500 ⇒ (1.035)n Use trial or logs: n=6 gives 1.229 (too low), n=7 gives 1.272. So 7 full years are needed.
要找出 A > 2500 的时间，解 2000 × (1.035)n > 2500 ⇒ (1.035)n > 1.25。用尝试或对数：n=6 得 1.229（太低），n=7 得 1.272。因此需要 7 整年。
3. Geometry & Measurement: Designing a Garden | 几何与测量：花园设计
A rectangular lawn measures 12 m by 8 m. A path of uniform width x is built around the inside edge, reducing the lawn area to 60 m². Find x.
一块长方形草坪尺寸为 12 m × 8 m。沿内侧边缘修建一条宽度为 x 的均匀小路，使草坪面积减少到 60 m²。求 x。
(12 − 2x) by (8 − 2x). Area equation: (12 − 2x)(8 − 2x) = 60. Expand: 96 − 40x + 4x² = 60 ⇒ 4x² − 40x + 36 = 0 ⇒ x² − 10x + 9 = 0.
新草坪尺寸为 (12 − 2x) × (8 − 2x)。面积方程：(12 − 2x)(8 − 2x) = 60。展开：96 − 40x + 4x² = 60 ⇒ 4x² − 40x + 36 = 0 ⇒ x² − 10x + 9 = 0。
9) = 0. x = 1 or x = 9. x = 9 is impossible (would remove the lawn), so path width = 1 m.
因式分解：(x − 1)(x − 9) = 0。x = 1 或 x = 9。x = 9 不符合实际（会把草坪修没了），因此小路宽度 = 1 m。
4. Statistics & Probability: Interpreting Survey Data | 统计与概率：解读调查数据
A company surveys 200 customers about their satisfaction. Results: 120 are satisfied, 50 are neutral, 30 are dissatisfied. If three customers are chosen at random, find the probability that exactly two are satisfied.
一家公司调查了 200 名顾客的满意度。结果：120 人满意，50 人中立，30 人不满意。如果随机选取三位顾客，求恰好有两位满意的概率。
be found using combinations or the hypergeometric approach. First, the probability of SSS' 119/199) × (80/198). Then multiply by 3 (the number of arrangements: S S N, S N S, N S S).
这是从有限总体中无放回抽样。可通过组合或超几何方法求恰好两个满意和一个非满意的概率。先算特定顺序 SSS' 的概率：(120/200) × (119/199) × (80/198)。再乘以 3（排列数：S S N, S N S, N S S）。
ct ≈0.6×0.59799×0.40404≈0.1449. Times 3 ≈ 0.4347. The probability is about 0.435 (3 s.f.).
计算：120/200=0.6，119/199≈0.59799，80/198≈0.40404。乘积 ≈0.6×0.59799×0.40404≈0.1449。乘以 3 ≈ 0.4347。概率约为 0.435（三位有效数字）。
5. Ratio & Proportion: Scaling Recipes | 比例推理：调整食谱配方
A cake recipe for 8 people uses 300 g flour, 200 g sugar, and 4 eggs. How much of each ingredient is needed for 14 people? Give quantities in whole numbers of eggs and appropriate units.
一份供 8 人食用的蛋糕食谱使用 300 g 面粉、200 g 糖和 4 个鸡蛋。为 14 人准备需要每种原料多少？给出整数的鸡蛋数量和合适的单位。
.75 = 525 g. Sugar: 200 × 1.75 = 350 g. Eggs: 4 × 1.75 = 7 eggs. All quantities are exact.
比例因子 = 14/8 = 1.75。面粉：300 × 1.75 = 525 g。糖：200 × 1.75 = 350 g。鸡蛋：4 × 1.75 = 7 个。所有分量均为精确值。
Always check the reasonableness: larger number of people needs larger amounts, and eggs must be integer. The recipe scales perfectly without leftovers.
务必检查合理性：人数增加则用量增加，鸡蛋必为整数。该食谱可按比例完美缩放，无剩余。
6. Linear Equations: Mobile Phone Plans | 线性方程：手机套餐比较
Plan A costs £15 per month plus 5p per minute. Plan B costs £20 per month plus 3p per minute. Determine the number of minutes for which the two plans cost the same.
套餐 A 月租 15 英镑，每分钟 5 便士。套餐 B 月租 20 英镑，每分钟 3 便士。求使两种套餐花费相同的通话分钟数。
Convert pence to pounds: 5p = £0.05, 3p = £0.03. Cost functions: CA = 15 + 0.05m, CB = 20 + 0.03m. Set equal: 15 + 0.05m = 20 + 0.03m ⇒ 0.02m = 5 ⇒ m = 250 minutes.
将便士转换为英镑：5p = £0.05，3p = £0.03。费用函数：CA = 15 + 0.05m，CB = 20 + 0.03m。令相等：15 + 0.05m = 20 + 0.03m ⇒ 0.02m = 5 ⇒ m = 250 分钟。
Interpretation: At exactly 250 minutes, both plans cost £27.50. For usage below 250 minutes, Plan A is cheaper; above 250 minutes, Plan B becomes cheaper.
解释：恰好 250 分钟时，两款套餐均花费 £27.50。使用量低于 250 分钟时，套餐 A 更便宜；超过 250 分钟则套餐 B 更划算。
7. Quadratic Modelling: Projectile Motion | 二次函数建模：抛体运动
+ 1, where t is time in seconds. Find the maximum height and the time it takes to reach the ground again.
烟花离地高度 h（米）由 h = −5t² + 40t + 1 给出，t 为时间（秒）。求最大高度和再次落地所需的时间。
The maximum occurs at vertex: t = −b/(2a) = −40/(2 × −5) = 4 seconds. Substitute: hmax = −5(4)² + 40×4 + 1 = −80 + 160 + 1 = 81 m.
最大值出现在顶点处：t = −b/(2a) = −40/(2 × −5) = 4 秒。代入：hmax = −5(4)² + 40×4 + 1 = −80 + 160 + 1 = 81 米。
For hitting the ground, set h = 0: −5t² + 40t + 1 = 0 ⇒ 5t² − 40t − 1 = 0. Use quadratic formula: t = [40 ± √1600 + 20 40.249. Positive root: (40 + 40.249)/10 ≈ 8.025 s. The firework lands after about 8.03 s.
求落地时间，设 h = 0：−5t² + 40t + 1 = 0 ⇒ 5t² − 40t − 1 = 0。用求根公式：t = [40 ± √1600 + 20] / 10 = [40 ± √1620] / 10。√1620 ≈ 40.249。正根：(40 + 40.249)/10 ≈ 8.025 秒。烟花大约在 8.03 秒后落地。
8. Trigonometry: Calculating the Height of a Tower | 三角学：计算塔高
From a point 50 m away from the base of a tower, the angle of elevation to the top is 38°. A second point, 20 m closer, has an elevation of 52°. Use trigonometry to verify the tower's height.
从塔底外 50 米处测得塔顶仰角为 38°。再靠近 20 米处，仰角为 52°。用三角学验证塔高。
From the first point: tan 38° = h / 50 ⇒ an 38° ≈ 50 × 0.7813 = 39.065 m. From the second point (distance = 30 m): tan 52° = h / 30 ⇒ ° ≈ 30 × 1.2799 = 38.397 m. Slight variation due to rounding, but average height ≈ 38.7 m.
从第一个点：tan 38° = h / 50 ⇒ h = 50 tan 38° ≈ 50 × 0.7813 = 39.065 米。从第二个点（距离 30 m）：tan 52° = h / 30 ⇒ h = 30 tan 52° ≈ 30 × 1.2799 = 38.397 米。因四舍五入有微小差异，平均高度约 38.7 米。
In case studies, cross-checking with two methods improves reliability. Always state which measurement you trust more and why.
在案例分析中，用两种方法交叉验证可提高可靠性。务必说明你更信赖哪次测量及其原因。
9. Graphs & Charts: Analysing Sales Trends | 图表分析：销售趋势
A company's quarterly sales (£1000s) were: Q1: 24, Q2: 33, Q3: 30, Q4: 41. Plot a line graph and describe the trend. Calculate the percentage increase from Q1 to Q4.
某公司的季度销售额（千英镑）为：Q1: 24，Q2: 33，Q3: 30，Q4: 41。画出折线图并描述趋势。计算从 Q1 到 Q4 的增长百分比。
Trend: Sales rose from 24 to 33 in Q2, dipped slightly to 30 in Q3, then recovered strongly to 41 in Q4. Overall upward trend with seasonal dip.
趋势：销售从 24 上升到 Q2 的 33，Q3 小幅下降至 30，然后在 Q4 强劲回升至 41。整体呈上升趋势，伴有季节性回落。
Percentage increase = ((41 − 24) / 24) × 100% = (17/24)×100% ≈ 70.8% (1 d.p.). Showing working with formula gains full marks.
增长百分比 = ((41 − 24) / 24) × 100% = (17/24)×100% ≈ 70.8%（保留一位小数）。写出公式及计算过程才能拿到满分。
10. Bounds & Error Intervals: Manufacturing Tolerances | 误差界限：制造公差
A metal rod is cut to a nominal length of 2.5 m, measured to the nearest 0.1 m. Calculate the upper and lower bounds of its length, and the maximum possible area of a square frame made from four such rods.
一根金属棒标称长度为 2.5 m，测量精确到 0.1 m。计算其长度的上界和下界，以及由四根这样的棒制成的正方形框架的最大可能面积。
th directly. Maximum area uses the upper bound: side = 2.55 m, area = (2.55)² = 6.5025 m².
精确到 0.1 m，误差区间为 ±0.05 m。下界 = 2.45 m，上界 = 2.55 m。正方形的边长直接使用棒的长度。最大面积取上界：边长 = 2.55 m，面积 = (2.55)² = 6.5025 m²。
State clearly: 'Maximum possible area = 6.5025 m²' in reality, joints might require deductions, but the mathematical bound is as calculated.
明确表述：“最大可能面积 = 6.5025 m²”，并说明实际情况下接口可能需要扣除，但数学上的界限如上计算。
11. Advanced Problem: Multi-step Investigation | 综合提升：多步骤调查
A cinema sells adult tickets for £9.50 and child tickets for £5.80. On a Saturday, 320 tickets were sold, generating £2614. The manager wants to estimate next week's mix if total tickets remain constant but revenue target rises by 15%.
一家电影院成人票 £9.50，儿童票 £5.80。某个周六售出 320 张票，总收入 £2614。经理想预测下一周若总票数不变但收入目标提高 15% 时的票种结构。
9.5a + 5.8c = 2614. Solve: from first, c = 320 − a. Substitute: 9.5a + 5.8(320 − a) = 2614 ⇒ 9.5a + 1856 − 5.8a = 2614 ⇒ 3.7a = 758 ⇒ a = 205, c = 115.
设 a 为成人票数，c 为儿童票数。方程：a + c = 320，9.5a + 5.8c = 2614。代入求解：c = 320 − a，9.5a + 5.8(320 − a) = 2614 ⇒ 9.5a + 1856 − 5.8a = 2614 ⇒ 3.7a = 758 ⇒ a = 205，c = 115。
New revenue target = £2614 × 1.15 = £ 320, solve 9.5a + 5.8c = 3006.1 and a + c = 320. This gives 3.7a = 3006.1 − 1856 = 1150.1 ⇒ .2. Not integer, so adjust: 311 adults, 9 children gives 9.5×311 + 5.8×9 = 2954.5 + 52.2 = £3006.70, slightly over target. The mix must shift heavily towards adult tickets.
新收入目标 = £2614 × 1.15 = £3006.10。保持总票数 320，解 9.5a + 5.8c = 3006.1 和 a + c = 320。得 3.7a = 3006.1 − 1856 = 1150.1 ⇒ a ≈ 310.8，c ≈ 9.2。非整数，调整：311 成人、9 儿童，收入为 9.5×311 + 5.8×9 = 2954.5 + 52.2 = £3006.70，略微超标。票种结构必须大幅向成人票倾斜。
Interpretation: The cinema needs to attract many more adults, perhaps through special screenings. This illustrates how case studies can lead to business decisions.
解读：影院需要大幅增加成人观众，或许通过特别放映等。这体现了案例分析如何导向商业决策。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图