Case Study Practical Drills | 案例分析实战演练

📚 Case Study Practical Drills | 案例分析实战演练

In KS3 Edexcel Advanced Mathematics, applying concepts to real-world scenarios is key to deepening understanding. This article presents a series of practical case studies that integrate algebra, geometry, statistics, and number skills. By working through these exercises, you will develop problem-solving strategies essential for advanced study.

在KS3爱德思进阶数学中，将概念应用于现实场景是深化理解的关键。本文呈现一系列实践案例分析，融合代数、几何、统计与数字技能。通过完成这些练习，你将培养进阶学习必备的问题解决策略。

1. Case Study 1: Designing a Rectangular Garden | 案例一：设计矩形花园

A gardener has 20 metres of fencing to enclose a rectangular vegetable plot against a long wall. The wall forms one side, 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. The total fencing used is 2x + y = 20, so y = 20 – 2x. The area A = x × y = x(20 – 2x) = 20x – 2x². This is a quadratic expression that opens downwards, so its maximum occurs at the vertex. Complete the square or use symmetry: the maximum is at x = -b/(2a) = -20/(2×(-2)) = 5. Then y = 20 – 2(5) = 10. The maximum area is 5 × 10 = 50 m².

一位园丁有20米长的篱笆，想靠着一面长墙围出一个矩形菜园。墙构成其中一条边，因此只需围三边。设垂直于墙的宽度为x米，平行于墙的长度为y米。所用篱笆总长为2x + y = 20，因此y = 20 – 2x。面积A = x × y = x(20 – 2x) = 20x – 2x²。这是一个开口向下的二次式，最大值出现在顶点。通过配方或利用对称性：最大值在x = -b/(2a) = -20/(2×(-2)) = 5。那么y = 20 – 2(5) = 10。最大面积为5 × 10 = 50 平方米。

In practice, if the gardener also wants a gate in the fence, we must subtract the gate width from the total fencing. Suppose a 1 m wide gate is placed in one of the side widths. The equation becomes (x – 1) + x + y = 20, giving y = 21 – 2x. Recalculate to find the new optimum dimensions.

实际上，如果园丁还想要在篱笆上开一扇门，就必须从总长度中减去门的宽度。假设一扇1米宽的门位于一条侧边上，方程变成(x – 1) + x + y = 20，得到y = 21 – 2x。重新计算以找出新的最优尺寸。

For a fixed budget of £100 for fencing material costing £8 per metre, how many metres can be purchased, and what is the largest possible area? The available length is 100 ÷ 8 = 12.5 m, which leads to 2x + y = 12.5, y = 12.5 – 2x, area = 12.5x – 2x². Find vertex x = 12.5/4 = 3.125 m, area = 19.53125 m².

若篱笆材料每米花费8英镑，预算固定为100英镑，能购买多少米，此时最大面积是多少？可用长度为100 ÷ 8 = 12.5米，从而2x + y = 12.5，y = 12.5 – 2x，面积 = 12.5x – 2x²。求顶点得x = 12.5/4 = 3.125米，面积为19.53125平方米。

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2. Case Study 2: Planning a School Trip | 案例二：规划学校旅行

A school hires a coach that costs £300 and seats up to 50 students. Museum entry tickets cost £5 per student. The total cost C for n students is C = 300 + 5n. To find the cost per student c, divide by n: c = 300/n + 5. This is a rational function that decreases as n increases, approaching £5 per student for large n.

一所学校租用一辆大巴，费用为300英镑，最多乘坐50名学生。博物馆门票每人5英镑。对于n名学生，总花费C = 300 + 5n。要计算人均费用c，除以n得：c = 300/n + 5。这是一个有理函数，随着n增大而减小，当n很大时人均费用趋近于5英镑。

If 45 students join the trip, the total cost is 300 + 5 × 45 = £525, giving a cost per student of £11.67. If only 30 students go, the cost is 300 + 150 = £450, and per student that is £15. The difference arises from spreading the fixed coach cost over fewer students.

如果有45名学生参加，总花费为300 + 5 × 45 = 525英镑，人均费用为11.67英镑。若只有30名学生前往，花费为300 + 150 = 450英镑，人均为15英镑。差异在于大巴固定成本分摊的人数不同。

The school treasurer sets a maximum per-student cost of £12. Solve 300/n + 5 <= 12. Subtract 5: 300/n <= 7, so n >= 300/7 ≈ 42.86, thus at least 43 students must attend to meet the target.

学校财务主管设定人均费用上限为12英镑。解不等式 300/n + 5 <= 12。减去5：300/n <= 7，得n >= 300/7 ≈ 42.86，因此至少需要43名学生参加才能达到目标。

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3. Case Study 3: Cookie Recipe Scaling | 案例三：曲奇配方比例缩放

A basic cookie recipe makes 24 cookies and requires 200 g of plain flour, 100 g of sugar, 150 g of butter, and 2 large eggs. To make 60 cookies for a bake sale, all ingredients must be scaled by the factor 60/24 = 2.5. Multiply each quantity by 2.5: flour = 200 × 2.5 = 500 g, sugar = 250 g, butter = 375 g, eggs = 5.

一份基础曲奇配方可制作24块曲奇，所需材料为200克普通面粉、100克糖、150克黄油和2个大鸡蛋。要为义卖活动制作60块曲奇，所有原料必须按比例因子60/24 = 2.5进行缩放。每种用量乘以2.5：面粉 = 200 × 2.5 = 500克，糖 = 250克，黄油 = 375克，鸡蛋 = 5个。

What if you only have 3 eggs left? Then the limiting factor is eggs: the recipe for 24 needs 2 eggs, so with 3 eggs you can make 24 × (3/2) = 36 cookies. Adjust other ingredients to the factor 3/2 = 1.5: flour 300 g, sugar 150 g, butter 225 g.

如果只剩下3个鸡蛋怎么办？那么限制因素就是鸡蛋：24块的配方需要2个鸡蛋，因此用3个鸡蛋你可以制作24 × (3/2) = 36块曲奇。其他原料按因子3/2 = 1.5调整：面粉300克，糖150克，黄油225克。

Scaling recipes involves proportional reasoning. For mixed units, convert all to the same unit first. For example, 0.25 kg of butter equals 250 g, making scaling clearer.

配方缩放涉及比例推理。对于混合单位，先全部转换为同一单位。例如，0.25千克黄油等于250克，这样缩放更清晰。

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4. Case Study 4: Mobile Phone Data Plan | 案例四：手机数据套餐

Two mobile phone deals are compared. Plan A charges a £10 monthly line rental plus 5p per minute of calls. Plan B charges £20 monthly line rental plus 2p per minute. Write expressions for the monthly cost: Cₐ = 10 + 0.05m, C_b = 20 + 0.02m, where m is the number of minutes used.

比较两种手机套餐。套餐A月租10英镑，通话每分钟5便士。套餐B月租20英镑，通话每分钟2便士。写出月费用表达式：Cₐ = 10 + 0.05m，C_b = 20 + 0.02m，其中m为使用分钟数。

To find the break-even point, set Cₐ = C_b: 10 + 0.05m = 20 + 0.02m. Subtract 0.02m and 10: 0.03m = 10, so m = 10/0.03 ≈ 333.33 minutes. For usage above 334 minutes, Plan B is cheaper; below 334 minutes, Plan A is better.

要找出盈亏平衡点，令Cₐ = C_b：10 + 0.05m = 20 + 0.02m。两边减0.02m和10：0.03m = 10，得m = 10/0.03 ≈ 333.33分钟。当月使用分钟数超过334分钟时，套餐B更便宜；低于334分钟时，套餐A更优。

Graphing both cost functions shows two straight lines with different slopes. The intersection point confirms the algebraic solution. This linear model assumes no free minutes or additional data charges; modifying it makes a stronger real-life case.

绘制两个费用函数图像，得到两条不同斜率的直线，交点确认了代数解。该线性模型假设没有免费分钟或额外流量费；对其进行修改可得出更真实的实际案例。

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5. Case Study 5: Water Tank Volume and Filling Time | 案例五：水箱体积与注水时间

A cylindrical water tank has a radius of 0.5 m and a height of 1.2 m. The volume V of a cylinder is given by V = πr²h. Substitute: V = π × (0.5)² × 1.2 = π × 0.25 × 1.2 = 0.3π m³. Using π ≈ 3.142, V ≈ 0.9426 m³.

一个圆柱形水箱半径为0.5米，高1.2米。圆柱体积公式为V = πr²h。代入得：V = π × (0.5)² × 1.2 = π × 0.25 × 1.2 = 0.3π 立方米。取π ≈ 3.142，V ≈ 0.9426 立方米。

Since 1 m³ = 1000 litres, the tank holds approximately 942.6 litres. Water flows into the tank at a rate of 5 litres per minute. To fill the tank, the time required is 942.6 ÷ 5 = 188.52 minutes, which is about 3 hours and 8.5 minutes.

因为1立方米 = 1000升，水箱约可容纳942.6升。水以每分钟5升的速率注入水箱。充满水箱所需时间为942.6 ÷ 5 = 188.52分钟，约3小时8.5分钟。

If the tank is only 80% filled, volume of water = 0.8 × 942.6 = 754.08 litres, time = 754.08 / 5 = 150.816 minutes. Converting between cubic metres and litres and accounting for rate changes are valuable unit conversion skills.

如果水箱只装80%的水，水体积 = 0.8 × 942.6 = 754.08升，时间 = 754.08 / 5 = 150.816分钟。在立方米和升之间转换，并考虑速率变化，是重要的单位转换技能。

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6. Case Study 6: Survey Analysis | 案例六：调查分析

A survey of 50 students on their favourite fruit gave these results: apples (15), bananas (20), oranges (10), grapes (5). Find the relative frequencies: apples 15/50 = 0.3 (30%), bananas 0.4, oranges 0.2, grapes 0.1. To draw a pie chart, multiply each fraction by 360°: apples 108°, bananas 144°, oranges 72°, grapes 36°. The sector angles sum to 360°.

一项关于50名学生最喜欢水果的调查得出以下结果：苹果（15人），香蕉（20人），橙子（10人），葡萄（5人）。求相对频率：苹果15/50 = 0.3（30%），香蕉0.4，橙子0.2，葡萄0.1。要绘制饼图，将每个分数乘以360°：苹果108°，香蕉144°，橙子72°，葡萄36°。各扇区角度总和为360°。

The mode is the fruit with the highest frequency, which is bananas. There is no single ‘average’ categorical variable like mean, but we can interpret the modal category as the most popular.

众数是出现频率最高的水果，即香蕉。像平均数之类的数值无法用于分类变量，但我们可以将众数类别解读为最受欢迎的水果。

If another survey collected numerical data—say, number of texts sent per day by 10 students: 20, 35, 22, 40, 35, 38, 25, 30, 35, 28. The mean is (20+35+22+40+35+38+25+30+35+28)/10 = 308/10 = 30.8. Order the data: 20,22,25,28,30,35,35,35,38,40. Median = (30+35)/2 = 32.5. Mode = 35. The range is 40 – 20 = 20.

假如另一项调查收集了数值型数据——例如10名学生每天发送短信数：20, 35, 22, 40, 35, 38, 25, 30, 35, 28。平均值为(20+35+22+40+35+38+25+30+35+28)/10 = 308/10 = 30.8。排序：20,22,25,28,30,35,35,35,38,40。中位数 = (30+35)/2 = 32.5。众数 = 35。极差为40 – 20 = 20。

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7. Case Study 7: Budgeting for a Party | 案例七：派对预算

Ali is planning a birthday party with a total budget of £200. Venue hire costs £60, a DJ costs £45, and food costs £7 per guest. How many guests can he invite? Write an inequality: 60 + 45 + 7g ≤ 200, where g is the number of guests. Simplify: 105 + 7g ≤ 200, so 7g ≤ 95, g ≤ 13.57. Therefore, he can invite at most 13 guests if whole numbers are required.

阿里正在计划一场生日派对，总预算为200英镑。场地租用费60英镑，DJ费用45英镑，每位客人餐饮费7英镑。他能邀请多少位客人？写出不等式：60 + 45 + 7g ≤ 200，其中g为客人数量。化简得：105 + 7g ≤ 200，因此7g ≤ 95，g ≤ 13.57。因此，若要求整数，他最多可邀请13位客人。

If he finds a 10% discount on the DJ, the DJ cost becomes £40.50. Then inequality: 60 + 40.5 + 7g ≤ 200 → 100.5 + 7g ≤ 200 → 7g ≤ 99.5 → g ≤ 14.21, so 14 guests can be invited, increasing headcount by 1.

如果他获得DJ费用九折优惠，DJ费用变为40.50英镑。那么不等式：60 + 40.5 + 7g ≤ 200 → 100.5 + 7g ≤ 200 → 7g ≤ 99.5 → g ≤ 14.21，因此可邀请14位客人，增加了1个名额。

He also considers a different venue costing £80 with no DJ option, food £6.50 per guest. Compare total costs for 15 guests: Venue A with DJ = 60+45+105=£210 (over budget); Venue B = 80 + 6.5×15 = 80+97.5=£177.50, which fits the budget.

他还考虑另一个场地，费用80英镑且不请DJ，餐饮每位6.50英镑。比较15位客人的总费用：场地A含DJ = 60+45+105=210英镑（超预算）；场地B = 80 + 6.5×15 = 80+97.5=177.50英镑，符合预算。

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8. Case Study 8: Crime Scene Investigation (Algebra) | 案例八：犯罪现场调查（代数应用）

Forensic scientists often use foot length to estimate a suspect’s height. One simplified model says: Height (H cm) = 2.5 × Foot length (F cm) + 30. If a footprint at a crime scene measures 28 cm, the estimated height is H = 2.5 × 28 + 30 = 70 + 30 = 100 cm, which is unrealistic for an adult. A more realistic formula might be H = 3.2F + 65. Test with F = 28 cm: H = 3.2 × 28 + 65 = 89.6 + 65 = 154.6 cm, about 1.55 m, plausible for a teenager or shorter adult.

法医学家常用足长来估算嫌疑人身高。一个简化的模型为：身高（H 厘米）= 2.5 × 足长（F 厘米）+ 30。如果犯罪现场的一枚鞋印长28厘米，估算身高为 H = 2.5 × 28 + 30 = 70 + 30 = 100 厘米，这对成年人来说不太现实。一个更真实的公式可能是 H = 3.2F + 65。以 F = 28 厘米测试：H = 3.2 × 28 + 65 = 89.6 + 65 = 154.6 厘米，约1.55米，对青少年或较矮的成年人而言是合理的。

Given a suspect’s height of 175 cm, use the formula F = (H – 65) / 3.2 to predict his foot length: F = (175 – 65) / 3.2 = 110 / 3.2 = 34.375 cm, a size around UK 5-6. Detectives can compare this with evidence to narrow down suspects.

已知某嫌疑人身高175厘米，用公式 F = (H – 65) / 3.2 预测其足长：F = (175 – 65) / 3.2 = 110 / 3.2 = 34.375 厘米，大致对应英国尺码5-6号。侦探可以将其与证据比对以缩小嫌疑人范围。

Algebra also helps in analysing blood splatter angles or time of death using Newton’s law of cooling, introducing exponential equations at a basic conceptual level.

代数还有助于分析血迹溅射角度或利用牛顿冷却定律推断死亡时间，从而在基础概念层面引入指数方程。

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9. Case Study 9: Fibonacci and Patterns in Nature | 案例九：斐波那契与自然界中的模式

The Fibonacci sequence starts 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,… where each term is the sum of the two preceding ones: Fₙ = Fₙ₋₁ + Fₙ₋₂. This pattern emerges in the arrangement of leaves, the spirals of a pinecone, and the number of petals on many flowers. In a sunflower, the number of clockwise and counterclockwise spirals are often consecutive Fibonacci numbers, such as 34 and 55.

斐波那契数列以 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,… 开头，每一项是前两项之和：Fₙ = Fₙ₋₁ + Fₙ₋₂。这一模式出现在叶序、松果的螺旋以及许多花朵的花瓣数目中。在向日葵中，顺时针和逆时针螺旋的数量往往是连续的斐波那契数，比如34和55。

Calculate the ratio of consecutive Fibonacci terms: 3/2=1.5, 5/3≈1.667, 8/5=1.6, 13/8=1.625, 21/13≈1.615, 34/21≈1.619, 55/34≈1.618. This ratio slowly approaches the Golden Ratio φ = (1+√5)/2 ≈ 1.6180339887… The Golden Ratio appears in art, architecture, and natural beauty.

计算相邻斐波那契项的比值：3/2=1.5，5/3≈1.667，8/5=1.6，13/8=1.625，21/13≈1.615，34/21≈1.619，55/34≈1.618。该比值逐渐逼近黄金比例 φ = (1+√5)/2 ≈ 1.6180339887… 黄金比例出现在艺术、建筑和自然美学中。

Use the recurrence to find the 12th term: start 1,1,2,3,5,8,13,21,34,55,89,144. So F₁₂ = 144. Pattern recognition helps build algebraic thinking and links to sequences in advanced study.

利用递推关系求出第12项：从1,1,2,3,5,8,13,21,34,55,89,144。因此F₁₂ = 144。模式识别有助于建立代数思维，并与进阶学习中的数列联系起来。

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10. Case Study 10: Bridge Design – Forces and Triangles | 案例十：桥梁设计：力与三角形

Engineers use triangles in truss bridges because triangles are rigid shapes that resist deformation. Consider an equilateral triangle truss with side length 4 m. Its height can be found by splitting it into two right-angled triangles: height h = √(4² – 2²) = √(16 – 4) = √12 = 2√3 ≈ 3.464 m. The area is (base × height)/2 = (4 × 3.464)/2 = 6.928 m².

工程师在桁架桥中使用三角形，因为三角形是刚性形状，能抵抗变形。考虑一个边长为4米的等边三角形桁架。通过将其分割成两个直角三角形可求高：高 h = √(4² – 2²) = √(16 – 4) = √12 = 2√3 ≈ 3.464 米。面积为(底 × 高)/2 = (4 × 3.464)/2 = 6.928 平方米。

In a right-angled support triangle, if the vertical leg is 3 m and horizontal leg is 4 m, the hypotenuse is √(3² + 4²) = √25 = 5 m, a classic 3-4-5 Pythagorean triple. This relation is vital for checking square corners in construction.

在一个直角三角形支撑结构中，如果竖直直角边为3米，水平直角边为4米，斜边为√(3² + 4²) = √25 = 5米，即经典的3-4-5勾股数。这一关系对于检查建筑施工中的直角至关重要。

Scaling the triangle by a factor of 2.5, the sides become 7.5 m, 10 m, and 12.5 m. The angles remain the same, demonstrating similarity. Similar triangles allow engineers to model smaller prototypes and scale up safely.

将三角形按2.5的比例缩放后，各边变为7.5米、10米和12.5米。角度保持不变，展示了相似性。相似三角形使工程师能够制作较小的模型并安全地放大实际结构。

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