📚 Year 10 Eduqas Maths: Case Study Practice Drills | Year 10 Eduqas 数学：案例分析实战演练
Real-world mathematics is not just about memorising formulas; it is about applying concepts to solve everyday problems. The Eduqas GCSE Mathematics specification emphasises problem-solving and reasoning, often presenting scenarios that require multi-step thinking. This article provides a series of case study practice drills designed to strengthen your skills in key Year 10 topics, including number, algebra, geometry, statistics, and probability. Each case study mirrors the style of contextual questions you might encounter, and every step is explained in both English and Chinese to support bilingual learners.
现实生活中的数学不仅仅是记忆公式，而是将概念应用于解决日常问题。Eduqas GCSE 数学大纲强调问题解决和推理能力，经常出现需要多步骤思考的情景题。本文提供一系列案例分析实战演练，旨在加强你对于十年级核心主题的技能，包括数、代数、几何、统计和概率。每个案例都模拟了可能遇到的背景题风格，每一步都用中英双语解释，为双语学习者提供支持。
1. Holiday Budget: Managing Money & Percentages | 度假预算：管理金钱与百分比
A family plans a holiday with a total budget of £2400. They spend 35% on flights, 25% on accommodation, and save the rest for food and activities. Calculate the amount spent on flights, accommodation, and the remaining amount. If they later find a discount that reduces the flight cost by 15%, how much money do they save overall?
一个家庭计划度假，总预算为2400英镑。他们花35%在机票上，25%在住宿上，剩下的用于食物和活动。计算机票和住宿的费用，以及剩余的金额。如果他们后来发现一个折扣，可以减少机票费用的15%，他们总共节省了多少钱？
Flight cost = 35% of £2400 = 0.35 × 2400 = £840. Accommodation = 0.25 × 2400 = £600. Remaining = £2400 − £840 − £600 = £960. After discount: new flight cost = £840 × (1 − 0.15) = £840 × 0.85 = £714. Saving = £840 − £714 = £126. Total saving = £126. They could increase their food budget by £126.
机票费用 = 2400的35% = 0.35 × 2400 = 840英镑。住宿 = 0.25 × 2400 = 600英镑。剩余 = 2400 − 840 − 600 = 960英镑。折扣后，新机票费用 = 840 × 0.85 = 714英镑。节省金额 = 840 − 714 = 126英镑。总节省126英镑。他们可以增加食物预算126英镑。
Extension: If the family wants to reduce the total spend by 12% overall, what new budget should they aim for? New budget = £2400 × 0.88 = £2112.
拓展：如果家庭希望总支出减少12%，他们应该设定多少新预算？新预算 = 2400 × 0.88 = 2112英镑。
2. Designing a Rectangular Garden: Area & Perimeter | 设计矩形花园：面积与周长
A gardener has 40 metres of fencing to enclose a rectangular vegetable patch. One side lies along a wall, so only three sides need fencing. Let the width perpendicular to the wall be x metres, and the length parallel to the wall be y metres. Express y in terms of x, and find the maximum area that can be enclosed.
一位园丁有40米长的围栏来围一个矩形菜地。一边靠墙，所以只需要围三面。设垂直于墙的宽度为 x 米，平行于墙的长度为 y 米。用 x 表示 y，并求出可以围成的最大面积。
x − 10)² − 100] = −2(x − 10)² + 200. Maximum area occurs when x = 10, giving A = 200 m². Then y = 40 − 20 = 20 m. So dimensions 10 m by 20 m.
围栏总使用量：2x + y = 40，因此 y = 40 − 2x。面积 A = x × y = x(40 − 2x) = 40x − 2x²。这是一个二次函数。配方法：A = −2(x² − 20x) = −2[(x − 10)² − 100] = −2(x − 10)² + 200。当 x = 10 时取得最大面积 A = 200 平方米。此时 y = 40 − 20 = 20 米。所以尺寸为10米乘20米。
0 + 20 + 10 = 40 m, correct. The garden area is 200 m², enough for several vegetable rows.
检验：所需周长 = 10 + 20 + 10 = 40 米，正确。花园面积200平方米，足够种植多排蔬菜。
3. Mobile Phone Tariffs: Linear Equations & Graphs | 手机资费：线性方程与图像
Two mobile networks offer deals: Network A charges £10 per month plus 5
Published by TutorHao | Year 10 Mathematics Revision Series | aleveler.com
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