📚 Year 10 OCR Mathematics: Case Study Practice | Year 10 OCR 数学：案例分析实战演练
In the OCR GCSE (9-1) Mathematics course, case study questions test your ability to apply mathematical skills to real-life situations. These multi-step problems blend number, algebra, geometry and statistics into a single scenario. Mastering them is essential for achieving a high grade in Year 10 and beyond.
在OCR GCSE(9-1)数学课程中，案例分析题考查你将数学技能应用于现实情境的能力。这类多步骤问题将数、代数、几何和统计融合在同一个场景中。掌握它们对于在Year 10及以后取得高分至关重要。
1. Understanding the Structure of Case Study Questions | 理解案例分析题的结构
Case study questions usually begin with a block of text that describes a situation, such as planning a party or calculating travel costs. You need to extract numerical information, identify the operations required and often work with units, percentages or ratio.
案例分析题通常以一段描述情境的文字开篇，比如策划派对或计算旅行花费。你需要提取数字信息，识别所需的运算，并经常处理单位、百分比或比例。
The mark scheme rewards clear working, correct use of formulas and sensible rounding. Always read the whole stem before starting, and highlight key figures. Look for hidden links — for example, a map scale might feed into an area calculation.
评分方案奖励清晰的解题过程、正确的公式使用和合理的四舍五入。开始前务必通读全题，并标出关键数字。寻找隐含联系——例如，地图比例尺可能输入到面积计算中。
2. Case Study 1: Home Budget Planning | 案例一：家庭预算规划
Scenario: A family has a monthly take-home income of £2,840. They spend 30% on rent, 15% on food, and 10% on transport. The rest is split equally between savings and entertainment. A sudden car repair costs £220. How much can they still save this month?
场景：一个家庭每月税后收入为2840英镑。他们将收入的30%用于房租，15%用于食品，10%用于交通。剩余部分平均分配给储蓄和娱乐。突然的汽车修理费为220英镑。他们本月还能储蓄多少钱？
First, calculate the fixed expenses: rent = 0.30 × 2840 = £852, food = 0.15 × 2840 = £426, transport = 0.10 × 2840 = £284. Total fixed = 852 + 426 + 284 = £1,562. Remaining = 2840 – 1562 = £1,278.
首先，计算固定支出：房租=0.30×2840=852英镑，食品=0.15×2840=426英镑，交通=0.10×2840=284英镑。固定总和=852+426+284=1562英镑。剩余=2840–1562=1278英镑。
Half of the remainder is for savings: 1278 ÷ 2 = £639. After the repair, they have 639 – 220 = £419 left for savings. However, the repair may come from the entertainment half; the problem says “still save”, implying the repair is taken from the final savings amount. So they save £419.
剩余的一半用于储蓄：1278÷2=639英镑。修理后，他们还剩下639–220=419英镑用于储蓄。然而，修理费可能来自娱乐部分；题目说“还能储蓄”，暗示修理费从最终储蓄额中扣除。因此他们储蓄419英镑。
3. Case Study 2: Travel Distance and Speed | 案例二：旅行距离与速度
Situation: A student cycles to school at an average speed of 12 km/h for the first 4 km, then walks the remaining 1.2 km at 4.8 km/h. Calculate the total journey time and the overall average speed in m/s.
情境：一名学生以12 km/h的平均速度骑行前4公里，然后以4.8 km/h步行剩余的1.2公里。计算整个行程的时间和整体平均速度（以米/秒为单位）。
Time for cycling: distance / speed = 4 / 12 = 1/3 hour = 20 minutes. Time for walking: 1.2 / 4.8 = 0.25 hour = 15 minutes. Total time = 20 + 15 = 35 minutes = 35 × 60 = 2100 seconds.
骑行时间：距离/速度=4/12=1/3小时=20分钟。步行时间：1.2/4.8=0.25小时=15分钟。总时间=20+15=35分钟=35×60=2100秒。
Total distance = 4 + 1.2 = 5.2 km = 5200 m. Average speed = total distance / total time = 5200 / 2100 ≈ 2.476 m/s. Rounded to 1 decimal place: 2.5 m/s. Always convert units consistently.
总距离=4+1.2=5.2公里=5200米。平均速度=总距离/总时间=5200/2100≈2.476米/秒。四舍五入到一位小数：2.5米/秒。务必保持单位一致。
4. Case Study 3: Garden Design and Area | 案例三：花园设计与面积
A rectangular garden measures 15 m by 8 m. A circular pond of radius 1.5 m sits in the centre. The rest is covered by grass. Grass seed costs £3.20 per square metre, plus a fixed delivery charge of £12. Find the total cost to cover the garden.
一个矩形花园长15米、宽8米。中央有一个半径1.5米的圆形池塘。其余部分覆盖草皮。草籽每平方米3.20英镑，固定运费12英镑。求覆盖花园的总费用。
Area of rectangle: 15 × 8 = 120 m². Area of pond: π × r² = π × 1.5² = π × 2.25 ≈ 7.0686 m² (use π = 3.14159). Grass area = 120 – 7.0686 = 112.9314 m².
矩形面积：15×8=120平方米。池塘面积：π×r²=π×1.5²=π×2.25≈7.0686平方米（使用π=3.14159）。草皮面积=120–7.0686=112.9314平方米。
Cost of seed = 112.9314 × 3.20 = £361.38048, round to £361.38. Add delivery: 361.38 + 12 = £373.38. So total cost = £373.38.
草籽费用=112.9314×3.20=361.38048英镑，取整为361.38英镑。加上运费：361.38+12=373.38英镑。因此总费用为373.38英镑。
5. Case Study 4: Data Statistics and Chart Analysis | 案例四：数据统计与图表分析
A survey asked 40 students about their favourite fruit. The results: Apple (14), Banana (10), Orange (8), Grapes (6), Other (2). Draw a pie chart and work out the percentage for each sector. What fraction of students chose Apple or Banana?
一项调查询问了40名学生最喜欢的水果。结果：苹果(14)、香蕉(10)、橘子(8)、葡萄(6)、其他(2)。绘制饼图并计算每一扇区的百分比。选择苹果或香蕉的学生占几分之几？
Total frequency = 40. Percentage for Apple: (14/40) × 100 = 35%. Banana: (10/40) × 100 = 25%. Orange: 20%, Grapes: 15%, Other: 5%. Pie chart angles: Apple = 0.35 × 360° = 126°, Banana = 90°, Orange = 72°, Grapes = 54°, Other = 18°.
总频数=40。苹果百分比：(14/40)×100=35%。香蕉：(10/40)×100=25%。橘子：20%，葡萄：15%，其他：5%。饼图角度：苹果=0.35×360°=126°，香蕉=90°，橘子=72°，葡萄=54°，其他=18°。
The fraction who chose Apple or Banana = (14 + 10) / 40 = 24/40 = 3/5. Simplify fractions to gain marks. In OCR exams you may need to interpret such a chart and answer further questions about the mode or median.
选择苹果或香蕉的比例=(14+10)/40=24/40=3/5。化简分数以获得分数。在OCR考试中，你可能需要解读这样的图表并回答关于众数或中位数的问题。
6. Case Study 5: Ratio and Discount Shopping | 案例五：比例与折扣购物
A shop offers a discount: “Buy 3 items and get 25% off the cheapest one.” Three friends buy items costing £28, £22, and £16. They split the total cost in the ratio of the original prices. How much does each pay?
一家商店提供折扣：”购买3件商品，最便宜的一件享受25%的折扣。” 三个朋友购买了价格分别为28英镑、22英镑和16英镑的商品。他们按原价比例分摊总费用。每人应付多少？
Discount applies to the cheapest item (16). Discount = 25% of 16 = 0.25 × 16 = £4. So the item costs 16 – 4 = £12. Total bill = 28 + 22 + 12 = £62.
折扣适用于最便宜的商品(16)。折扣额=16的25%=0.25×16=4英镑。所以该商品支付12英镑。总账单=28+22+12=62英镑。
Total original sum = 28 + 22 + 16 = 66. Ratio of shares: 28:22:16 simplifies to 14:11:8. Sum of ratio parts = 14+11+8=33. First friend pays (14/33) × 62 ≈ 26.30, second (11/33) × 62 = 20.67, third (8/33)×62 ≈ 15.03. Check: 26.30+20.67+15.03=62.
原价总和=28+22+16=66。分摊比例：28:22:16简化为14:11:8。比例份数总和=14+11+8=33。第一个朋友支付(14/33)×62≈26.30，第二个(11/33)×62=20.67，第三个(8/33)×62≈15.03。核对：26.30+20.67+15.03=62。
7. Case Study 6: Savings and Compound Interest | 案例六：储蓄与复利
Maya invests £800 in a savings account that pays 3.5% compound interest per annum. No withdrawals are made. Calculate the balance after 3 years. Compare with simple interest at the same rate.
玛雅将800英镑存入一个年利率3.5%的复利储蓄账户。没有提款。计算3年后的余额。并与相同利率下的单利进行比较。
Compound interest formula:
A = P(1 + r/100)ⁿ
where P=800, r=3.5, n=3. A = 800 × (1 + 0.035)³ = 800 × (1.035)³. (1.035)³ = 1.035 × 1.035 × 1.035 ≈ 1.108717. So A ≈ 800 × 1.108717 = £886.97 (nearest penny).
复利公式：
A = P(1 + r/100)ⁿ
其中P=800，r=3.5，n=3。A=800×(1+0.035)³=800×(1.035)³。(1.035)³=1.035×1.035×1.035≈1.108717。所以A≈800×1.108717=886.97英镑（精确到便士）。
Simple interest: Interest per year = 800 × 0.035 = £28. Total interest for 3 years = 3 × 28 = £84. Balance = 800 + 84 = £884. So compound interest earns an extra £2.97 over 3 years.
单利：每年利息=800×0.035=28英镑。3年总利息=3×28=84英镑。余额=800+84=884英镑。因此，复利在3年后多赚了2.97英镑。
8. Common Mistakes and Checking Techniques | 常见错误与检查技巧
Many students lose marks by misreading units (e.g. minutes vs hours) or forgetting to convert before using formulas. Always underline the units in the question and change everything to a consistent system — metres, seconds, pounds, etc.
许多学生因误读单位（例如分钟与小时）或在套用公式前忘记换算而失分。务必在题目中标出单位，并将所有数据转换为一致的系统——如米、秒、英镑等。
Using unrounded values in intermediate steps gives more accurate final answers. However, you should only round at the very end. A rough estimate before calculating can catch errors. For example, 112.9 × 3.2 should be roughly 360, so £373.38 is reasonable.
在中间步骤使用未舍入的数值能得到更精确的最终答案。但你应该只在最后一步舍入。计算前先估算一下可以捕获错误。例如，112.9×3.2大约为360，所以373.38英镑是合理的。
In ratio problems, always check that your parts add up to the total. In percentage questions, verify whether the percentage is “of” or “increase”. Reading the question twice before writing anything helps avoid misunderstandings.
在比例问题中，务必检查各部分总和等于总数。在百分比问题中，确认该百分比是“占…的”还是“增加了…”。动笔前把题目读两遍有助于避免误解。
9. Exam Preparation Tips for OCR Case Studies | OCR案例分析备考建议
Practice with past OCR Foundation and Higher papers. Note that annual case study questions often involve a mixture of topics — algebra might appear within a geometry context. Focus on showing every step of your working.
练习过去的OCR基础级和更高级别试卷。注意每年的案例分析题往往混合多个主题——代数可能会出现在几何情境中。注重展示你的每一步解答过程。
Create a “toolkit” of formulas and conversion factors: area, volume, speed, density, percentage change, compound interest. Memorise them and understand when to use each. Flash cards are very effective.
建立一个公式和换算因子的“工具箱”：面积、体积、速度、密度、百分比变化、复利。记住它们并理解何时使用。闪卡非常有效。
Under timed conditions, allocate roughly one minute per mark. If stuck, move on and return later. For 5-mark case study questions, plan your approach for 30 seconds before writing — list steps: extract data, convert units, set up equation, solve, interpret answer.
在计时条件下，大约每分钟完成1分的题目。如果卡住了，先跳过，回头再解决。对于5分的案例分析题，先花30秒构思方案再动笔——列出步骤：提取数据、换算单位、建立方程、求解、解释答案。
10. Summary and Moving Forward | 总结与展望
Real-world case studies are an excellent way to see how mathematics connects to everyday life. By working through budgeting, travel, design, statistics, ratios and interest, you build transferable problem-solving skills. Keep practising, and you will find that these questions become manageable and even enjoyable.
现实世界的案例分析是理解数学如何与日常生活相连接的绝佳方式。通过处理预算、旅行、设计、统计、比例和利息问题，你培养了可迁移的解决问题能力。坚持练习，你会发现这些题目变得得心应手，甚至充满乐趣。
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