📚 Case Study Practice for Year 8 Cambridge Mathematics | Year 8 Cambridge 数学：案例分析实战演练
Case studies in mathematics help you apply number, algebra, geometry, and data skills to real-life situations. In this article, you will work through eight carefully chosen scenarios that mirror the style of Cambridge Year 8 problem-solving tasks. Each case is broken down step by step so you can practise thinking mathematically and communicating your reasoning clearly.
数学中的案例分析帮助你将在数、代数、几何和数据处理等方面的技能应用于现实生活情境。本文将带你演练八个精心挑选的案例，它们反映了剑桥 Year 8 问题解决任务的风格。每个案例都逐步分解，帮助你在练习中学会数学思维并清晰地表达推理过程。
1. Understanding Case Studies in Maths | 理解数学中的案例分析
A case study in mathematics presents a real-world scenario that requires more than one skill to solve. You might need to read a table, calculate percentages, use a formula, and then interpret your answer. The key is to identify the relevant information, choose the correct operations, and check whether your final result makes sense in context.
数学案例分析会给出一个真实世界的情境，解决问题往往需要不止一项技能。你可能需要解读表格、计算百分比、套用公式，然后再解释答案。关键在于识别相关信息、选择正确的运算，并检查最终结果在上下文中是否合理。
Always start by underlining the important numbers and units. Decide what the question is really asking. Then break the problem into smaller steps. Writing down each calculation in a logical order will earn you marks in Cambridge assessments and help avoid careless mistakes.
解题时首先要划出重要的数字和单位，判断题目真正在问什么。然后把问题分解成若干小步骤。将每一步计算按逻辑顺序写下来，不仅能让你在剑桥评估中获得分数，还能避免粗心错误。
Example approach: For a school fete, 240 tickets were printed. ¾ were sold in advance at £2 each, and the rest at the gate for £3 each. Find the total money collected.
示例方法： 学校游园会印了 240 张票。提前售出了 ¾，每张 2 英镑，其余的在门口以每张 3 英镑出售。求收款总额。
1. Advance tickets: ¾ × 240 = 180. Money = 180 × £2 = £360.
1. 预售票数：¾ × 240 = 180。收入 = 180 × 2 英镑 = 360 英镑。
2. Gate tickets: 240 - 180 = 60. Money = 60 × £3 = £180.
2. 门口票数：240 - 180 = 60。收入 = 60 × 3 英镑 = 180 英镑。
3. Total = £360 + £180 = £540.
3. 总计 = 360 + 180 = 540 英镑。
2. Case 1: Shopping Discounts and Best Buys | 案例1：购物折扣与最佳购买
A sports shop has a sale: all items are 30% off marked prices. A pair of trainers originally costs £80, and there is an extra 10% off the sale price for members. Calculate the final price a member pays. Then check whether buying two pairs at full price with a 'buy one get one half price' offer would be cheaper.
一家运动用品商店促销：所有商品按标价打七折。一双运动鞋原价 80 英镑，会员还可在促销价的基础上再打九折。计算会员最终支付的价格。然后判断如果使用“买一第二件半价”的原价优惠买两双是否更便宜。
Sale price: £80 × 0.70 = £56. Member extra 10% off: £56 × 0.90 = £50.40.
促销价：80 × 0.70 = 56 英镑。会员额外九折：56 × 0.90 = 50.40 英镑。
For two pairs: Under member sale, total = 2 × £50.40 = £100.80. With 'buy one get one half price' at original price: first pair £80, second half £40, total = £120. The member sale deal is cheaper by £19.20.
两双鞋的情况：按会员促销价，总价 = 2 × 50.40 = 100.80 英镑。按原价买一第二件半价：第一双 80 英镑，第二双 40 英镑，合计 120 英镑。会员促销便宜了 19.20 英镑。
Cost savings = £120 - £100.80 = £19.20
节省金额 = 120 - 100.80 = 19.20 英镑
3. Case 2: Planning a School Trip Budget | 案例2：规划学校旅行预算
A Year 8 group of 180 students needs coaches for a museum visit. Each coach holds 48 students. The museum entry fee is £3.50 per student, and the total coach hire costs £480. What is the total cost? If the school fund contributes £200, how much must each student pay so that the remaining cost is shared equally?
一个 180 名 Year 8 学生的年级需乘大巴参观博物馆。每辆大巴可容纳 48 名学生。博物馆门票为每名学生 3.50 英镑，大巴租用总费用 480 英镑。总费用是多少？如果学校基金捐助 200 英镑，剩余费用由学生平均分摊，每名学生需要支付多少？
Number of coaches: 180 ÷ 48 = 3.75, so 4 coaches must be hired. Coach cost = £480. Entry fees = 180 × £3.50 = £630. Total cost = £480 + £630 = £1110. After fund: remaining = £1110 - £200 = £910. Cost per student = £910 ÷ 180 ≈ £5.06 (rounded to nearest penny).
大巴数量：180 ÷ 48 = 3.75，因此需要租用 4 辆。大巴费用 = 480 英镑。门票费 = 180 × 3.50 = 630 英镑。总费用 = 480 + 630 = 1110 英镑。基金捐助后剩余：1110 - 200 = 910 英镑。每人分摊 = 910 ÷ 180 ≈ 5.06 英镑（保留到分）。
Notice the use of rounding up for the number of coaches because you cannot hire a fraction of a coach. This is an example of interpreting a decimal result in context.
注意大巴数量向上取整，因为你不能租用部分大巴。这是在具体情境下解读小数结果的一个例子。
4. Case 3: Designing a Rectangular Garden | 案例3：设计矩形花园
A garden is 12 m long and 8 m wide. A path of width 1 m is built along the inside perimeter of the garden. Find the area of the garden that can be planted with flowers. Also calculate the cost of covering the path with gravel if gravel costs £15 per square metre.
一个花园长 12 米、宽 8 米。沿花园内部周边修建一条宽 1 米的小路。求可种植花卉的区域面积。同时计算用砾石铺设小路的费用，假设砾石每平方米 15 英镑。
× 6 = 60 m². Total garden area = 12 × 8 = 96 m². Path area = 96 - 60 = 36 m². Cost = 36 × £15 = £540.
内部矩形尺寸：长 = 12 - 2×1 = 10 米；宽 = 8 - 2×1 = 6 米。种植面积 = 10 × 6 = 60 平方米。花园总面积 = 12 × 8 = 96 平方米。小路面积 = 96 - 60 = 36 平方米。费用 = 36 × 15 = 540 英镑。
6×1, then subtract the four corner squares counted twice. Both approaches must give 36 m².
另一种方法是将小路分割为四个长条：两条尺寸为 12×1，两条尺寸为 6×1，再减去重复计数的四个角部正方形。两种方法的计算结果都应为 36 平方米。
5. Case 4: Interpreting Survey Data | 案例4：解读调查数据
Year 8 students recorded the number of pets owned by 30 families. The results are: 0,1,2,1,0,3,4,1,2,1,0,0,2,1,3,1,0,2,4,1,1,0,3,2,1,0,1,2,1,3. Construct a frequency table and calculate the mean, median and mode. Which measure best represents the typical number of pets? Explain why.
Year 8 学生记录了 30 个家庭拥有的宠物数量。数据如下：（同上）。构建频数表，计算平均数、中位数和众数。哪一个度量最能代表宠物的典型数量？请解释原因。
median, list 30 values in order; the 15th and 16th values are both 1, so median = 1. Mode = 1 (most frequent).
频数：0 只宠物：7 个家庭，1 只：11，2 只：6，3 只：4，4 只：2。平均数 = (0×7+1×11+2×6+3×4+4×2) ÷ 30 = (0+11+12+12+8) ÷ 30 = 43 ÷ 30 ≈ 1.43。找中位数时，将 30 个数据排序；第 15 和第 16 个数均为 1，所以中位数 = 1。众数 = 1。
The median and mode are both 1, while the mean is pulled slightly higher by the few families with 4 pets. Therefore, the median or mode gives a more typical picture of pet ownership in this sample.
中位数和众数均为 1，而平均数因少数拥有 4 只宠物的家庭而被略微拉高。因此，中位数或众数更能体现该样本中宠物拥有量的典型水平。
6. Case 5: Probability in Games | 案例5：游戏中的概率
A fair six-sided dice is rolled twice. What is the probability that the sum of the two numbers is greater than 8? Write your answer as a fraction in its simplest form.
投掷一枚均匀的六面骰子两次。出现两数之和大于 8 的概率是多少？用最简分数表示你的答案。
5,5), (6,4) → 3. Sum 11: (5,6), (6,5) → 2. Sum 12: (6,6) → 1. Total favourable = 4+3+2+1 = 10. Probability = 10/36 = 5/18.
总可能结果 = 6 × 6 = 36。需要和为 >8，即和为 9、10、11、12。计数：和为 9：(3,6), (4,5), (5,4), (6,3) → 4 种。和为 10：(4,6), (5,5), (6,4) → 3 种。和为 11：(5,6), (6,5) → 2 种。和为 12：(6,6) → 1 种。有利结果总数 = 4+3+2+1 = 10。概率 = 10/36 = 5/18。
This case shows how to systematically count outcomes using a table or list. It also reminds you to simplify fractions in probability answers.
这个案例展示了如何用表格或列表系统性地计数。同时提醒你在概率答案中化简分数。
7. Case 6: Scaling a Recipe | 案例6：按比例调整食谱
A recipe for 8 people uses 400 g of flour, 200 g of sugar, and 4 eggs. A chef wants to make a dessert for 14 people and also has only 3 eggs available. How much flour and sugar should she use if she limits the recipe by the number of eggs? Give answers to the nearest gram.
一份为 8 人份的食谱使用面粉 400 克、糖 200 克、鸡蛋 4 个。厨师想要为 14 人制作甜点，但她只有 3 个鸡蛋。如果她按照鸡蛋的数量来限制食谱，面粉和糖各需要用多少克？答案精确到克。
s amount serves 3 × (8÷4) = 6 people. Since 14 people cannot be served with 3 eggs, the scaled recipe is for 6 people only.
原食谱中每个鸡蛋对应：400 ÷ 4 = 100 克面粉；200 ÷ 4 = 50 克糖。使用 3 个鸡蛋时，需要面粉 = 3 × 100 = 300 克；糖 = 3 × 50 = 150 克。这个量可供应 3 × (8÷4) = 6 人，因为 14 人无法用 3 个鸡蛋满足，所以调整后的食谱仅为 6 人份。
The problem highlights proportional reasoning and the idea of a limiting ingredient. In real baking, you must scale all ingredients by the same factor to maintain taste.
这道题强调了比例推理和限制性配料的概念。在实际烘焙中，必须将所有配料按相同倍数调整，以保持口味不变。
8. Case 7: Number Sequences in Nature | 案例7：自然界中的数列
A sunflower head often displays spirals following the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, … Write the rule for generating the next term. If a spiral count gives the 10th term as 55, verify it. Then find the ratio of the 10th term to the 9th term, giving your answer as a decimal to two places.
向日葵花盘通常呈现出遵循斐波那契数列的螺旋线：1, 1, 2, 3, 5, 8, 13, 21, 34, … 请写出生成下一项的规则。假设某个螺旋计数显示第 10 项为 55，请验证。然后求第 10 项与第 9 项的比值，结果用小数表示，保留两位小数。
13, T8=8+13=21, T9=13+21=34, T10=21+34=55. So 55 is correct. Ratio T10/T9 = 55/34 ≈ 1.6176… ≈ 1.62 (to 2 d.p.).
规则：每一项是前两项之和。 , T3=1+1=2, T4=1+2=3, T5=2+3=5, T6=3+5=8, T7=5+8=13, T8=8+13=21, T9=13+21=34, T10=21+34=55。因此 55 正确。比值 T10/T9 = 55/34 ≈ 1.6176… ≈ 1.62（保留两位小数）。
ratio φ ≈ 1.618, a fascinating connection between number sequences and patterns in plants.
这个比值趋近于黄金比例 φ ≈ 1.618，展示了数列与植物图案之间迷人的联系。
9. Case 8: Speed, Distance and Time | 案例8：速度、距离与时间
A cyclist travels from town A to town B at an average speed of 15 km/h and returns at 10 km/h due to a headwind. The total time for the round trip is 5 hours. Form an equation and find the distance between A and B.
一位自行车骑手从 A 镇前往 B 镇的平均速度为 15 公里/小时，由于逆风，返回时的平均速度为 10 公里/小时。往返总时间为 5 小时。列出方程并计算 A、B 两镇之间的距离。
0 (LCM of 15 and 10): 30×(d/15) + 30×(d/10) = 30×5 → 2d + 3d = 150 → 5d = 150 → d = 30 km.
设单程距离为 d 公里。去程时间 = d/15 小时，回程时间 = d/10 小时。d/15 + d/10 = 5。两边同乘 30（15 和 10 的最小公倍数）：30×(d/15)+30×(d/10)=30×5 → 2d+3d=150 → 5d=150 → d=30 公里。
. This is a classic algebraic case involving travel, testing your equation-forming skills.
验证：去程时间 = 30/15 = 2 小时；回程时间 = 30/10 = 3 小时；总时间 = 5 小时。这是一个涉及行程的经典代数案例，考验你建立方程的能力。
10. Key Takeaways for Case Study Success | 案例学习成功的关键
When tackling a mathematical case study, remember these strategies: read the scenario carefully and identify the maths topic; underline key data and units; break the problem into steps; check each calculation before moving on; always interpret your answer in the original context. Use clear headings and show your working, as Cambridge markers reward method marks.
面对数学案例分析时，记住以下策略：仔细阅读情境并识别涉及的数学主题；划出关键数据和单位；将问题拆分成多个步骤；进入下一步前检查每一步计算；始终根据原始情境解释你的答案。使用清晰的标题并展示解题过程，因为剑桥评分会奖励方法分。
Practice with everyday situations—such as comparing mobile phone contracts, planning a sports tournament timetable, or designing a bedroom layout—to become more confident in applying your maths knowledge flexibly.
通过日常情境进行练习——例如比较手机套餐、筹划体育赛事时间表，或设计卧室布局——使你在灵活应用数学知识时更加自信。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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