📚 Year 8 AQA Maths: Case Study Workout | 案例分析实战演练
Mathematics is not just about numbers on a page — it is a tool for solving real-world problems. In Year 8, students encounter increasingly complex scenarios where they must apply their understanding of fractions, ratios, algebra, geometry, and statistics. This article presents a series of case study exercises that mirror the style of AQA assessment tasks, helping learners build confidence through practice. Each case study encourages logical thinking, step-by-step working, and clear communication of mathematical reasoning.
数学不仅仅是纸面上的数字——它是解决现实世界问题的工具。在八年级，学生将遇到越来越复杂的场景，需要运用分数、比例、代数、几何和统计等知识。本文提供一系列案例分析练习，模拟AQA评估任务的风格，帮助学生通过练习建立信心。每个案例鼓励逻辑思维、逐步推导，并清晰表达数学推理过程。
1. Fractions in Cooking | 分数在烹饪中的应用
A recipe requires 3/4 cup of sugar to make 12 cookies. Emily wants to make 30 cookies for a bake sale. She has 2 1/2 cups of sugar. Does she have enough sugar? Show your working.
一份食谱需要 3/4 杯糖制作 12 块饼干。艾米丽想为烘焙义卖制作 30 块饼干。她有 2 1/2 杯糖。她的糖够吗？请展示计算过程。
ount of sugar needed for one cookie: (3/4) ÷ 12 = 3/4 × 1/12 = 3/48 = 1/16 cup per cookie.
首先，求出一块饼干所需的糖量：(3/4) ÷ 12 = 3/4 × 1/12 = 3/48 = 1/16 杯每块。
ugh sugar. The difference is 2 1/2 - 1 7/8 = 5/2 - 15/8 = 20/8 - 15/8 = 5/8 cup left over.
30 块饼干所需糖量：30 × (1/16) = 30/16 = 15/8 = 1 7/8 杯。由于 1 7/8 小于 2 1/2，她有足够的糖。差值为 2 1/2 - 1 7/8 = 5/2 - 15/8 = 20/8 - 15/8 = 5/8 杯剩余。
2. Ratio and Map Scales | 比例与地图比例尺
A map has a scale of 1 : 25 000. Two towns are 8 cm apart on the map. Calculate the actual distance in kilometres. Further, a cyclist travels at a steady speed of 15 km/h. How long will it take to cycle between the two towns?
一张地图的比例尺为 1 : 25 000。地图上两个城镇相距 8 厘米。计算实际距离（以公里为单位）。此外，一名自行车手以 15 公里/小时的匀速骑行。骑行通过这两个城镇需要多长时间？
cm. Convert to metres: 200 000 cm ÷ 100 = 2 000 m. Convert to km: 2 000 m ÷ 1 000 = 2 km.
实际距离 = 8 厘米 × 25 000 = 200 000 厘米。转换为米：200 000 厘米 ÷ 100 = 2 000 米。转换为公里：2 000 米 ÷ 1 000 = 2 公里。
= 2/15 hour. Convert to minutes: (2/15) × 60 = 8 minutes. So, the journey takes 8 minutes.
时间 = 距离 ÷ 速度 = 2 公里 ÷ 15 公里/小时 = 2/15 小时。转换为分钟：(2/15) × 60 = 8 分钟。因此，行程需要 8 分钟。
3. Solving Equations in Context | 情境中的方程求解
Alex buys 3 identical notebooks and a pen for £4.10. The pen costs £0.80. Write an equation and find the cost of one notebook.
亚历克斯买了 3 本相同的笔记本和一支笔，共花费 4.10 英镑。笔的价格是 0.80 英镑。列方程并求一本笔记本的价格。
Let the cost of one notebook be £ 3x + 0.80 = 4.10. Subtract 0.80: 3x = 3.30. Divide by 3: x = 1.10. So, each notebook costs £1.10.
设一本笔记本的价格为 £x。方程：3x + 0.80 = 4.10。两边减 0.80：3x = 3.30。除以 3：x = 1.10。因此，每本笔记本价格为 1.10 英镑。
4. Angles in a Triangle and Quadrilateral | 三角形和四边形的角度
A triangle has angles of 45° and 72°. Find the third angle. Then, a quadrilateral has angles of 90°, 105°, and 110°. Find the fourth angle.
一个三角形有两个角分别为 45° 和 72°。求第三个角。接着，一个四边形有三个角分别为 90°、105° 和 110°。求第四个角。
Sum of angles in a triangle = 180°. Third angle = 180° - (45° + 72°) = 180° - 117° = 63°.
三角形内角和为 180°。第三个角 = 180° - (45° + 72°) = 180° - 117° = 63°。
in a quadrilateral = 360°. Fourth angle = 360° - (90° + 105° + 110°) = 360° - 305° = 55°.
四边形内角和为 360°。第四个角 = 360° - (90° + 105° + 110°) = 360° - 305° = 55°。
5. Percentages and Sales | 百分比与折扣
A jacket originally priced at £48 is reduced by 15% in a sale. Work out the sale price. Additionally, a further 10% discount is applied at the checkout for loyalty card holders. Find the final price.
一件原价 48 英镑的夹克在打折期间降价 15%。计算打折后的价格。此外，结账时持会员卡可再享受 10% 折扣。求最终价格。
15% of £48 = 0.15 × 48 = £7.20. Sale price = £48 - £7.20 = £40.80.
48 英镑的 15% = 0.15 × 48 = 7.20 英镑。打折价 = 48 - 7.20 = 40.80 英镑。
Extra 10% on sale price: 10% of £40.80 = 0.10 × 40.80 = £4.08. Final price = £40.80 - £4.08 = £36.72. Alternatively, multiply by 0.9: £40.80 × 0.9 = £36.72.
打折价再享 10% 折扣：40.80 的 10% = 0.10 × 40.80 = 4.
Published by TutorHao | Year 8 Mathematics Revision Series | aleveler.com
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