📚 Year 9 OCR Maths: Case Study Practice | Year 9 OCR 数学：案例分析实战演练
Case studies are a powerful way to prepare for OCR maths assessments because they mirror the real-world problems you will face in exams and in everyday life. By working through practical scenarios, you learn to apply skills such as percentages, ratio, algebra, and geometry in a meaningful context.
案例学习是备考 OCR 数学考试的有效方法，因为它们反映了你在考试与日常生活中会遇到的真实问题。通过处理实际情境，你将学会在有意义的环境中应用百分比、比、代数和几何等技能。
Below are ten carefully designed case studies covering key Year 9 topics. Work through each problem step by step, then try similar variations on your own.
以下精心设计了十个覆盖九年级核心知识的案例。请逐步完成每个问题，然后自行尝试类似的变式练习。
1. Shopping Discounts | 购物折扣
A shop offers a 20% discount on a jacket originally priced at £45. How much do you pay after the discount?
一件夹克原价 45 英镑，商店打八折。打完折后你需要付多少钱？
Step 1: Find 20% of £45. 20% means 20/100 = 0.2, so multiply: 0.2 × 45 = 9.
第一步：计算 45 英镑的 20%。20% 即 20/100 = 0.2，因此 0.2 × 45 = 9。
Step 2: Subtract the discount from the original price: 45 − 9 = 36.
第二步：从原价中减去折扣额：45 − 9 = 36。
Alternative method: Since 100% − 20% = 80%, you pay 80% of £45, so 0.8 × 45 = 36.
另一种方法：100% − 20% = 80%，即支付原价的 80%，0.8 × 45 = 36。
Final price: £36.
最终价格：36 英镑。
2. Recipe Ratios | 食谱中的比
A pancake recipe uses 200 g of flour for 4 people. How much flour is needed for 10 people? Give your answer in grams.
一份煎饼食谱为 4 人份使用 200 克面粉。10 人份需要多少面粉？答案以克为单位。
Identify the ratio: 200 g : 4 people. Divide by 4 to find the amount for 1 person: 200 ÷ 4 = 50 g per person.
确定比：200 克 : 4 人。除以 4 求出一人份量：200 ÷ 4 = 每人 50 克。
Multiply by 10 to scale up: 50 × 10 = 500 g.
乘以 10 放大：50 × 10 = 500 克。
You can also set up a proportion: (200/4) = (x/10), cross-multiply: 4x = 2000, so x = 500.
也可以列比例式：(200/4) = (x/10)，交叉相乘得 4x = 2000，得 x = 500。
For 10 people, you need 500 g of flour.
10 人份需要 500 克面粉。
3. Travel: Speed, Distance and Time | 行程：速度、距离与时间
A cyclist travels 45 km at a constant speed of 15 km/h. How long does the journey take? Express your answer in hours and minutes.
一位自行车手以 15 公里/时的恒定速度骑行 45 公里。行程需要多长时间？用小时和分钟表示答案。
Time = Distance ÷ Speed
时间 = 距离 ÷ 速度
Substitute the values: Time = 45 ÷ 15 = 3 hours.
代入数值：时间 = 45 ÷ 15 = 3 小时。
Since the answer is exactly 3 hours, there are no minutes to add.
由于答案恰好是 3 小时，没有额外的分钟数。
If the speed were 12 km/h, then Time = 45 ÷ 12 = 3.75 hours. 0.75 of an hour is 0.75 × 60 = 45 minutes, so 3 hours 45 minutes.
如果速度为 12 公里/时，则时间 = 45 ÷ 12 = 3.75 小时。0.75 小时为 0.75 × 60 = 45 分钟，即为 3 小时 45 分钟。
4. Mobile Phone Plans | 手机套餐
A mobile company charges a fixed monthly fee of £10 plus £0.05 per minute of calls. Write an algebraic expression for the total monthly cost and find the cost for 120 minutes.
一家移动公司每月收取固定费用 10 英镑，外加每分钟通话费 0.05 英镑。写出每月总费用的代数表达式，并计算通话 120 分钟的费用。
Let m = number of minutes used. Total cost = 10 + 0.05m.
设 m = 使用分钟数。总费用 = 10 + 0.05m。
For m = 120, Cost = 10 + 0.05 × 120 = 10 + 6 = £16.
当 m = 120 时，费用 = 10 + 0.05 × 120 = 10 + 6 = 16 英镑。
If the cost is £25, we can solve: 10 + 0.05m = 25 → 0.05m = 15 → m = 15 ÷ 0.05 = 300 minutes.
如果费用为 25 英镑，则可解方程：10 + 0.05m = 25 ⇨ 0.05m = 15 ⇨ m = 15 ÷ 0.05 = 300 分钟。
This linear model helps compare different plans.
这个线性模型有助于比较不同套餐。
5. Garden Design: Perimeter and Area | 花园设计：周长与面积
A rectangular garden is 8.5 m long and 4.2 m wide. Calculate the perimeter and the area. If a path 0.5 m wide is built inside the garden all around, find the area of the path.
一个长方形花园长 8.5 米，宽 4.2 米。计算周长和面积。如果在花园内部四周修建一条宽 0.5 米的小路，求小路的面积。
Perimeter: 2 × (length + width) = 2 × (8.5 + 4.2) = 2 × 12.7 = 25.4 m.
周长：2 × (长 + 宽) = 2 × (8.5 + 4.2) = 2 × 12.7 = 25.4 米。
Area of whole garden: 8.5 × 4.2 = 35.7 m².
整个花园面积：8.5 × 4.2 = 35.7 平方米。
After the path, the inner rectangle has length 8.5 − 2×0.5 = 8.5 − 1 = 7.5 m, width 4.2 − 1 = 3.2 m.
修建小路后，内部矩形长 8.5 − 2×0.5 = 7.5 米，宽 4.2 − 1 = 3.2 米。
Inner area: 7.5 × 3.2 = 24 m².
内部面积：7.5 × 3.2 = 24 平方米。
Path area: total area − inner area = 35.7 − 24 = 11.7 m².
小路面积：总面积 − 内部面积 = 35.7 − 24 = 11.7 平方米。
6. School Survey Statistics | 学校调查统计
Students recorded the number of books read in a month: 3, 5, 2, 7, 5, 4, 5, 3, 6. Find the mean, median, mode, and range.
学生记录了一个月阅读的书籍数量：3, 5, 2, 7, 5, 4, 5, 3, 6。求平均数、中位数、众数和极差。
Order the data: 2, 3, 3, 4, 5, 5, 5, 6, 7.
数据排序：2, 3, 3, 4, 5, 5, 5, 6, 7。
Mean: sum = 2+3+3+4+5+5+5+6+7 = 40. Number of values = 9, mean = 40 ÷ 9 ≈ 4.44.
平均数：总和 = 2+3+3+4+5+5+5+6+7 = 40。数据个数 9，平均数 = 40 ÷ 9 ≈ 4.44。
Median (middle value): the 5th value is 5.
中位数（中间值）：第 5 个值为 5。
Mode (most frequent): 5 appears three times, so mode = 5.
众数（出现最频繁）：5 出现三次，众数 = 5。
Range: 7 − 2 = 5.
极差：7 − 2 = 5。
These statistics summarise reading habits effectively.
这些统计量有效概括了阅读习惯。
7. Currency Exchange | 货币兑换
The exchange rate is £1 = $1.28. Convert £350 into dollars. Later, the rate changes to £1 = $1.32 – is this better or worse for the traveller?
汇率为 1 英镑 = 1.28 美元。将 350 英镑兑换为美元。之后汇率变为 1 英镑 = 1.32 美元——这对旅行者更有利还是更不利？
Conversion: 350 × 1.28 = 448 dollars.
兑换：350 × 1.28 = 448 美元。
When the rate rises to 1.32, each pound buys more dollars. For the same £350, you would get 350 × 1.32 = 462 dollars, so the traveller gets more foreign currency – it is better.
当汇率升至 1.32 时，每英镑可兑换更多美元。同样 350 英镑可换得 350 × 1.32 = 462 美元，因此旅行者能获得更多外币——更有利。
If you were converting back from dollars to pounds, a higher exchange rate means fewer pounds received, which is worse.
如果是从美元换回英镑，更高的汇率意味着收到的英镑更少，这是不利的。
8. Profit and Loss | 利润与亏损
A shop buys a bicycle for £120 and sells it for £168. Calculate the profit and the percentage profit on the cost price.
一家商店以 120 英镑购入一辆自行车，并以 168 英镑售出。计算利润以及基于成本价的利润率。
Profit = selling price − cost price = 168 − 120 = £48.
利润 = 售价 − 成本价 = 168 − 120 = 48 英镑。
Percentage profit = (profit ÷ cost price) × 100 = (48 ÷ 120) × 100 = 40%.
利润率 = (利润 ÷ 成本价) × 100 = (48 ÷ 120) × 100 = 40%。
If the bicycle was sold for £90 instead, the loss would be 120 − 90 = £30, and the percentage loss = (30 ÷ 120) × 100 = 25%.
如果自行车以 90 英镑售出，则亏损为 120 − 90 = 30 英镑，亏损率 = (30 ÷ 120) × 100 = 25%。
9. Mixing Solutions | 混合问题
A chemist mixes 300 ml of a 10% salt solution with 200 ml of a 25% salt solution. Find the concentration of the final mixture.
一位化学家将 300 毫升 10% 的盐溶液与 200 毫升 25% 的盐溶液混合。求最终混合物的浓度。
Salt in first solution: 10% of 300 ml = 0.10 × 300 = 30 ml.
第一种溶液中的盐：300 毫升的 10% = 0.10 × 300 = 30 毫升。
Salt in second solution: 25% of 200 ml = 0.25 × 200 = 50 ml.
第二种溶液中的盐：200 毫升的 25% = 0.25 × 200 = 50 毫升。
Total salt = 30 + 50 = 80 ml. Total volume = 300 + 200 = 500 ml.
总盐量 = 30 + 50 = 80 毫升。总体积 = 300 + 200 = 500 毫升。
Concentration = (80 ÷ 500) × 100 = 16%
浓度 = (80 ÷ 500) × 100 = 16%
The mixture is a 16% salt solution.
混合物为 16% 的盐溶液。
10. Pythagoras in Real Life | 现实生活中的勾股定理
A ladder 5.2 m long leans against a wall. The foot of the ladder is 1.8 m from the wall. How high up the wall does the ladder reach?
一架长 5.2 米的梯子斜靠在墙上。梯脚距墙 1.8 米。梯子顶端离地面多高？
Use Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse (ladder length).
使用勾股定理：a² + b² = c²，其中 c 为斜边（梯长）。
Let the height be h. Then h² + 1.8² = 5.2².
设高度为 h，则 h² + 1.8² = 5.2²。
Calculate squares: 1.8² = 3.24, 5.2² = 27.04.
计算平方：1.8² = 3.24，5.2² = 27.04。
So h² = 27.04 − 3.24 = 23.8. Therefore h = √23.8 ≈ 4.88 m (to 2 decimal places).
因此 h² = 27.04 − 3.24 = 23.8。所以 h = √23.8 ≈ 4.88 米（保留两位小数）。
Always check if the triangle is right-angled; here the wall and ground are perpendicular.
务必检查三角形是否为直角三角形；此处墙与地面垂直。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导