📚 Year 8 CCEA Maths: Case Study Practical Exercises | CCEA 八年级数学：案例分析实战演练
Mathematics is at its most powerful when used to solve real-world problems. This article presents a series of case studies and practice exercises designed specifically for Year 8 pupils following the CCEA curriculum. Each scenario invites you to apply number skills, algebra, geometry, and statistics in contexts such as baking, travel, gardening, and school life. Work through the examples carefully, checking your reasoning and methods. The bilingual layout supports learners studying in English, Chinese, or both, reinforcing understanding through clear paired explanations.
数学在解决现实问题时最为强大。本文为遵循 CCEA 课程的八年级学生设计了一系列案例研究和实践练习。每个情景都邀请你在烘焙、旅行、园艺和学校生活等背景中运用数字技能、代数、几何和统计知识。请仔细完成示例，检查你的推理和方法。中英双语的布局支持学习英语、中文或两者的学生，通过清晰的配对解释加深理解。
1. Charity Bake Sale – Cost, Profit and Percentage | 慈善烘焙义卖——成本、利润与百分比
A group of Year 8 pupils baked 240 cupcakes for a charity sale. The ingredients cost £18.60 in total. They sold each cupcake for 35p. All cupcakes were sold. What was the total money taken from sales? How much profit did they make? Express the profit as a percentage of the cost price, correct to one decimal place.
一群八年级学生为慈善义卖烤了 240 个杯形蛋糕。原料总共花费 £18.60。每个杯形蛋糕售价 35 便士。所有蛋糕都已售出。销售总收入是多少？他们赚了多少利润？将利润表示为成本价的百分比，精确到一位小数。
First, find the total sales in pence: 240 × 35 = 8400 pence. Convert to pounds: £84.00. Profit = total sales − cost = £84.00 − £18.60 = £. To find the profit percentage, use (profit ÷ cost) × 100 = (65.40 ÷ 18.60) × 100 ≈ 351.6%. Always check division carefully.
首先，计算总销售额（便士）：240 × 35 = 8400 便士。转换为英镑：£84.00。利润 = 总销售额 − 成本 = £84.00 − £18.60 = £65.40。求利润百分比，使用 (利润 ÷ 成本) × 100 = (65.40 ÷ 18.60) × 100 ≈ 351.6%。务必仔细检查除法。
Key skill: converting pence to pounds, applying the percentage profit formula.
关键技能：便士与英镑的换算，应用利润百分比公式。
2. Travel Planning – Distance, Speed and Time | 旅行规划——距离、速度与时间
A family plans a trip from Belfast to Dublin, a distance of 168 km. They leave at 09:15 and arrive at 11:45. Calculate the average speed for the journey in km/h. On the return trip, traffic slows them to an average speed of 56 km/h. How long will the return journey take in hours and minutes?
一个家庭计划从贝尔法斯特到达柏林旅行，距离 168 公里。他们在 09:15 出发，11:45 到达。计算旅程的平均速度（公里/小时）。在返回途中，交通使他们平均速度降至 56 公里/小时。返回旅程需要多少时间（小时和分钟）？
tance ÷ speed = 168 ÷ 56 = 3 hours exactly. So the return journey takes 3 hours 0 minutes.
所用时间 = 11:45 − 09:15 = 2 小时 30 分钟 = 2.5 小时。平均速度 = 距离 ÷ 时间 = 168 ÷ 2.5 = 67.2 km/h。返回旅程，时间 = 距离 ÷ 速度 = 168 ÷ 56 = 3 小时整。因此返回旅程需要 3 小时 0 分钟。
e speed = total distance ÷ total time. Convert minutes to decimal hours by dividing by 60.
记住：平均速度 = 总距离 ÷ 总时间。将分钟转换为小时的小数形式需除以 60。
3. Garden Design – Perimeter and Area | 花园设计——周长与面积
A rectangular lawn measures 12.5 m by 8.4 m. The gardener wants to put a fence around the lawn and also lay turf over the whole area. Fencing costs £5.60 per metre. Turf costs £4.20 per square metre. Work out the total cost of the fence and the turf separately, then find the combined cost.
一块矩形草坪尺寸为 12.5 米 × 8.4 米。园丁想在草坪周围建围栏，并在整个区域铺上草皮。围栏每米费用 £5.60。草皮每平方米费用 £4.20。分别计算围栏和草皮的总费用，然后求合计费用。
Perimeter = 2 × (12.5 + 8.4) = 2 × 20.9 = 41.8 m. Cost of fencing = 41.8 × 5.60 = £234.08. Area = 12.5 × 8.4 = 105 m². Cost of turf = 105 × 4.20 = £441.00. Total cost = £234.08 + £441.00 = £675.08.
周长 = 2 × (12.5 + 8.4) = 2 × 20.9 = 41.8 米。围栏费用 = 41.8 × 5.60 = £234.08。面积 = 12.5 × 8.4 = 105 平方米。草皮费用 = 105 × 4.20 = £441.00。总费用 = £234.08 + £441.00 = £675.08。
Check units: perimeter in metres, area in square metres. Multiply carefully with decimals.
检查单位：周长以米为单位，面积以平方米为单位。小数乘法要仔细。
4. Test Score Analysis – Mean, Median and Mode | 考试成绩分析——平均数、中位数和众数
A class of 20 Year 8 pupils received the following marks out of 30 in a maths test: 18, 22, 25, 19, 24, 22, 20, 21, 23, 22, 26, 18, 27, 22, 21, 28, 22, 19, 24, 20. Calculate the mean score to one decimal place. Find the median and the mode. Which average best represents the data? Explain your reasoning.
一个 20 名八年级学生的班级在数学测验中获得了以下分数（满分 30）：18, 22, 25, 19, 24, 22, 20, 21, 23, 22, 26, 18, 27, 21, 28, 22, 19, 24, 20。计算平均分数（精确到一位小数）。求中位数和众数。哪一个平均数最能代表这组数据？解释你的理由。
put the numbers in order: 18,18,19,19,20,20,21,21,22,22,22,22,22,23,24,24,25,26,27,28. Median = average of 10th and 11th values = (22 + 22) ÷ 2 = 22. Mode = 22 (appears 5 times). The mean is 22.2, median 22, mode 22 – if there are extreme scores, but here the mean works well as there are no strong outliers.
所有分数之和 = 443。平均数 = 443 ÷ 20 = 22.15 → 22.2（一位小数）。求中位数时，将数字排序：中位数是第 10 和第 11 个值的平均数 = (22 + 22) ÷ 2 = 22。众数 = 22（出现 5 次）。平均数是 22.2，中位数 22，众数 22——三者接近。如果有极端值，中位数或众数可能更好，但这里没有明显异常值，所以平均数也适用。
Always order data for median. Count frequencies carefully for mode.
求中位数时一定要排序。统计众数频数要仔细。
5. Shopping Discount – Percentage and Final Price | 购物折扣——百分比与最终价格
A video game has a normal price of £48. In a sale, all items are reduced by 15%. The shop also offers an extra 5% off the already reduced price for loyalty card holders. What is the final price if someone uses a loyalty card? Give your answer to the nearest penny.
一款电子游戏正常售价 £48。在促销中，所有商品降价 15%。商店还为会员卡持有者提供在已打折价格的基础上再减 5% 的额外优惠。如果某人使用会员卡，最终价格是多少？结果保留到便士。
First discount: 15% of £48 = 0.15 × 48 = £7.20. Reduced price = £48 − £7.20 = £40.80. Extra 5% on reduced price: 5% of £40.80 = 0.05 × 40.80 = £2.04. Final price = £40.80 − £2.04 = £38.76.
首次折扣：£48 的 15% = 0.15 × 48 = £7.20。折后价 = £48 − £7.20 = £40.80。折后价再减 5%：£40.80 的 5% = 0.05 × 40.80 = £2.04。最终价格 = £40.80 − £2.04 = £38.76。
Note: successive percentages are not simply added; the second discount is taken on the reduced amount.
注意：连续百分比不能直接相加；第二次折扣是基于折后金额计算的。
6. Recipe Scaling – Ratio and Proportion | 食谱缩放——比与比例
A recipe for 6 people requires 200 g of flour, 80 g of sugar, and 3 eggs. How much flour, sugar and how many eggs are needed for 15 people? Give flour and sugar in grams. Later, if only 5 eggs are available, what is the maximum number of people you can serve, keeping the ratios the same?
一份 6 人食谱需要 200 克面粉、80 克糖和 3 个鸡蛋。15 人需要多少面粉、糖和鸡蛋？面粉和糖以克为单位。之后，如果只有 5 个鸡蛋，保持比例不变，最多可供多少人食用？
Sugar: 80 × 2.5 = 200 g. Eggs: 3 × 2.5 = 7.5, but you cannot use half an egg in a mixture – 6 people, simplified to 1 egg : 2 people. With 5 eggs, maximum people = 5 × 2 = 10 people.
比例因子 = 15 ÷ 6 = 2.5。面粉：200 × 2.5 = 500 g。糖：80 × 2.5 = 200 g。鸡蛋：3 × 2.5 = 7.5，实际不能使用半个鸡蛋，但数学上保留 7.5。第二部分，鸡蛋与人数比例是 3 个鸡蛋 : 6 人，化简为 1 个鸡蛋 : 2 人。有 5 个鸡蛋，最多可供 5 × 2 = 10 人。
Use the unitary method: divide by the given number to find value for 1, then multiply.
使用单位法：除以给定数量求单一量，再乘以所需数量。
7. Savings Plan – Using Algebra and Sequences | 储蓄计划——运用代数与数列
Emma saves £5 in week 1, £7 in week 2, £9 in week 3, and so on, increasing her saving by £2 each week. Using the nth term formula for an arithmetic sequence, find how much she saves in week 12. In which week will she first save more than £30? How much has she saved in total after 12 weeks?
艾玛第 1 周存 £5，第 2 周存 £7，第 3 周存 £9，依此类推，每周增加 £2。使用等差数列的第 n 项公式，求她在第 12 周存多少钱。在哪一周她的储蓄首次超过 £30？12 周后她总共存了多少钱？
d = 2. nth term: T(n) = 5 + (n − 1) × 2 = 2n + 3. For week 12, T(12) = 2×12 + 3 = 24 + 3 = £27. To exceed £30, solve 2n + 3 > 30 → 2n > 27 → n > 13.5, so in week 14 she saves 2×14+3 = £31, first over £ ter 12 weeks: sum of arithmetic series = n/2 × (first + last) = 12/2 × (5 + 27) = 6 × 32 = £192.
序列为 5, 7, 9, ... 公差 d = 2。第 n 项：T(n) = 5 + (n − 1) × 2 = 2n + 3。第 12 周：T(12) = 2×12 + 3 = £27。超过 £30，解 2n + 3 > 30 → n > 13.5，所以在第 14 周她存了 £31，首次超过 £30。12 周后总数：等差数列求和 = n/2 × (首项 + 末项) = 12/2 × (5 + 27) = 6 × 32 = £192。
Creating a rule with n helps answer 'when' questions easily. Sum formula: S = n/2 (a + l).
用 n 建立规则可以轻松回答“何时”的问题。求和公式：S = n/2 (a + l)。
8. Angle Puzzle – Straight Lines and Triangles | 角度谜题——直线与三角形
In a diagram (not shown, but described), three straight lines meet at a point. One angle is 47°. Another angle is 3x + 5, and the third angle is 2x − 10. All three angles lie on a straight line. Form an equation and solve for x. Then calculate the size of each of the three angles. Afterwards, one of these angles is used as an interior angle of an isosceles triangle. If the other two angles are equal, what are they?
在一图中（无图，但描述如下），三条直线交于一点。一个角为 47°。另一个角为 3x + 5，第三个角为 2x − 10。这三个角都在一条直线上。列出方程并解出 x。然后计算每个角的大小。随后，其中一个角被用作等腰三角形的一个内角。如果另外两个角相等，它们是多少度？
+ 5 = 82.8 + 5 = 87.8°, 2×27.6 − 10 = 55.2 − 10 = 45.2°. Check sum: 47 + 87.8 + 45.2 = 180. For the isosceles triangle, if we take the 87.8° angle as the 'different' ex = 86°; many possibilities depending on choice. We assume using the largest angle 87.8° as the apex.
直线上的角之和为 180°。因此：47 + (3x + 5) + (2x − 10) = 180。化简：42 + 5x = 180 → 5x = 138 → x = 27.6。各角为：47°，87.8°，45.2°。验证和为 180°。对于等腰三角形，如果取 87.8° 为顶角，则两个相等底角分别为 (180 − 87.8) ÷ 2 = 46.1°。也可以取其他角为底角，有多种可能。我们假定用最大的角 87.8° 作为顶角。
Always use the 'sum of angles on a straight line = 180°' and 'angles in a triangle = 180°'. Solve equations step by step.
始终使用“直线上的角之和 = 180°”以及“三角形内角和 = 180°”。逐步解方程。
9. Volume of a Cuboid – Packing Boxes | 长方体体积——包装箱
Small toy boxes each measure 5 cm by 4 cm by 3 cm. They are packed into a larger carton measuring 60 cm by 40 cm by 30 cm. Assuming the boxes fit perfectly with no wasted space, how many toy boxes can be packed into one large carton? If 1500 toy boxes need to be shipped, how many cartons are required?
小型玩具盒每个尺寸为 5 厘米 × 4 厘米 × 3 厘米。它们被装入一个 60 厘米 × 40 厘米 × 30 厘米的大纸箱中。假设完美填充无空隙，一个大纸箱能装多少个玩具盒？如果需要运送 1500 个玩具盒，需要多少个大纸箱？
. For 1500 toy boxes, number of cartons = 1500 ÷ 1200 = 1.25, so 2 cartons are needed (can't have a fraction of a carton).
一个玩具盒体积 = 5 × 4 × 3 = 60 cm³。纸箱体积 = 60 × 40 × 30 = 72000 cm³。每箱装多少个 = 72000 ÷ 60 = 1200 个。1500 个玩具盒需要 1500 ÷ 1200 = 1.25 个，因此需要 2 个纸箱（不能有分数个纸箱）。
Volume = length × width × height. Round up when real objects require whole cartons.
体积 = 长 × 宽 × 高。实际装箱需要整箱时，要向上取整。
10. Interpreting a Pie Chart – School Clubs | 饼图解读——学校俱乐部
A pie chart represents the favourite after-school clubs of 360 Year 8 pupils. The football section has an angle of 120°, drama 80°, art 50°, computing 70°, and the rest 'other'. How many pupils chose each club? If the 'other' category is 40°, what angle does it occupy and how many pupils does that represent? Check your totals add to 360° and 360 pupils.
一张饼图表示 360 名八年级学生最喜欢的课后俱乐部。足球部分的圆心角为 120°，戏剧 80°，美术 50°，计算机 70°，其余为“其他”。各俱乐部有多少学生选择？如果“其他”类别占 40°，它代表的圆心角是多少，代表多少学生？检查角度总和为 360°，学生总数为 360。
80 pupils. Art: 50 pupils. Computing: 70 pupils. The sum so far = 120+80+50+70 = 320°, so 'other' angle = 360 − 320 = 40°, which equals 40 pupils. All totals match.
学生总数 360，总度数 360°，因此 1° 代表 1 名学生。足球：(120/360)×360 = 120 名学生。戏剧：80 名。美术：50 名。计算机：70 名。目前总和 = 320°，所以“其他”角度 = 40°，代表 40 名学生。所有总数吻合。
In a pie chart, the fraction = angle/360. Multiply by total frequency.
在饼图中，比例 = 角度/360。乘以总频数。
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