Case Study Workouts in Further Maths | 进阶数学案例分析实战演练

📚 Case Study Workouts in Further Maths | 进阶数学案例分析实战演练

Real-world problem-solving is at the heart of CCEA Further Maths. This article presents a series of practical case studies designed to sharpen your analytical skills and apply concepts such as arithmetic, ratio, algebra, geometry and statistics. Each scenario encourages you to think like a mathematician, break down a situation into manageable steps, and use the most efficient method to reach a solution.

解决现实问题正是 CCEA 进阶数学的核心。本文提供一系列实际案例分析，旨在锻炼你的分析能力，并将算术、比率、代数、几何和统计等概念学以致用。每个情境都鼓励你像数学家一样思考，把情况分解成可操作的小步骤，并用最高效的方法得出答案。

1. Case Study 1: The School Fair Fundraiser | 案例1：学校义卖筹款

The student council organises a school fair. They charge an entry fee of £2.50 per person and expect around 300 visitors. A refreshment stall is predicted to generate a profit of £450 after deducting ingredient costs. However, hiring a bouncy castle costs £200. The council wants to know how much money will be left for the school fund.

学生会策划了一场校园义卖。入场费定为每人 £2.50，预计有 300 名参观者。一个小吃摊位扣除食材成本后预计可获利 £450。但租赁充气城堡需要 £200。学生会想知道最终能为学校基金筹集到多少资金。

First, calculate the income from entry fees:

首先计算门票收入：

Entry income = 300 × £2.50 = £750

Add the refreshment profit to get the total income:

将门票收入加上小吃利润得到总收入：

Total income = £750 + £450 = £1200

Finally, subtract the bouncy castle hire charge to leave the final profit:

最后减去充气城堡租赁费便得到最终利润：

Profit = £1200 − £200 = £1000

The school will gain £1000. To extend the thinking, the council wonders how many visitors would be needed if the entry fee were lowered to £2.00 but they still wanted to reach a profit of £1000. Let n represent the number of visitors. The equation becomes:

学校将获得 £1000。为了进一步思考，学生会想知道如果入场费降至 £2.00 但仍希望获得 £1000 利润，需要多少参观者。设 n 代表参观人数，方程为：

2n + 450 − 200 = 1000

2n = 750 → n = 375

375 visitors would be required, showing how a price change affects demand.

需要 375 名参观者，这揭示了价格变化对需求的影响。

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2. Case Study 2: Designing a New Playground | 案例2：设计新游乐场

A school plans to build a rectangular playground measuring 30 m in length and 20 m in width. The ground needs to be covered with grass turf, and a fence will surround the area. Turf costs £5 per square metre, while fencing costs £8 per metre. The budget officer needs a total cost estimate.

一所学校计划建造一个长 30 米、宽 20 米的长方形游乐场。地面需要铺上草皮，周围要建围栏。草皮每平方米 £5，围栏每米 £8。预算负责人需要估算总费用。

First, work out the perimeter for the fencing:

首先计算围栏所需的周长：

Perimeter = 2 × (30 + 20) = 2 × 50 = 100 m

Then, calculate the fence cost:

然后计算围栏成本：

Fence cost = 100 × £8 = £800

Next, find the area to be turfed:

接着计算铺草皮的面积：

Area = 30 × 20 = 600 m²

The turf cost is:

草皮费用为：

Turf cost = 600 × £5 = £3000

Adding both gives the total estimated cost:

两者相加得出总估算费用：

Total = £800 + £3000 = £3800

If the budget is capped at £3000, the planning committee must reduce either dimensions or material quality. Using algebra they can explore what width would be possible if length remains 30 m and turfing is the priority.

如果预算上限为 £3000，规划委员会就必须缩减尺寸或材料品质。他们可以利用代数，在长度保持 30 米且优先铺草皮的情况下，推算可能的宽度。

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3. Case Study 3: Travel Plans for a School Trip | 案例3：学校旅行计划

A Year 7 group is travelling from Belfast to the Giant’s Causeway, a distance of 120 km. The hired coach travels at an average speed of 60 km/h. The trip organiser needs to know the journey time and the total fuel cost at £0.25 per km. Additionally, the departure time is 8:30 am, so the arrival time must be calculated.

七年级的一个小组从贝尔法斯特前往巨人堤道，路程 120 公里。租用的大巴以平均每小时 60 公里的速度行驶。旅行组织者需要知道行驶时间以及每公里 £0.25 燃料费下的总燃料成本。另外，出发时间是上午 8:30，还需计算到达时间。

Use the formula: Time = Distance ÷ Speed.

使用公式：时间 = 距离 ÷ 速度。

Journey time = 120 ÷ 60 = 2 hours

So the coach will arrive at 10:30 am. Next, calculate the fuel cost:

因此大巴将在上午 10:30 抵达。接着计算燃料成本：

Fuel cost = 120 × £0.25 = £30

If 30 students share the coach hire and fuel equally, the cost per student is:

如果 30 名学生均摊大巴租金和燃料费，每名学生需支付：

Cost per student = £30 ÷ 30 = £1.00

Additional expenses like entry fees can be added later, making this a helpful model for planning real trips.

门票等额外费用可以后期再加，这为规划真实旅行提供了一个实用的模型。

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4. Case Study 4: Baking for Charity | 案例4：慈善烘焙

A group decides to bake cupcakes for a charity sale. A basic recipe makes 12 cupcakes and requires 200 g of flour, 100 g of sugar, 2 eggs and 120 ml of milk. They need to produce 60 cupcakes. Find the scaled quantities and decide how many oven batches are needed if one baking tray holds 24 cupcakes and each batch bakes in 20 minutes.

一个小组决定为慈善义卖烘焙杯子蛋糕。基础配方可制作 12 个杯子蛋糕，需要 200 克面粉、100 克糖、2 个鸡蛋和 120 毫升牛奶。他们需要制作 60 个蛋糕。求出按比例增加的食材用量，并确定如果每个烤盘可放 24 个蛋糕且每批烤制需 20 分钟，需要多少批。

The scale factor from 12 to 60 is:

从 12 到 60 的比例系数为：

Scale factor = 60 ÷ 12 = 5

Multiply each ingredient by 5:

每种食材乘以 5：

• Flour: 200 g × 5 = 1000 g (1 kg)
• Sugar: 100 g × 5 = 500 g
• Eggs: 2 × 5 = 10 eggs
• Milk: 120 ml × 5 = 600 ml

For baking, with 24 cupcakes per batch:

就烘焙而言，每批 24 个蛋糕：

Number of batches = 60 ÷ 24 = 2.5

So they need 3 batches (rounding up). Total baking time = 3 × 20 min = 60 minutes. This exercise combines ratio, multiplication and real-world rounding decisions.

因此需要 3 批（向上取整）。总烘焙时间 = 3 × 20 分钟 = 60 分钟。这个练习综合了比、乘法和现实生活中的取整决策。

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5. Case Study 5: Sport Tournament Statistics | 案例5：体育比赛统计

The school basketball team played five matches. Their scores were: 23, 19, 26, 30 and 17 points. The coach wants to analyse performance using the mean, median and range. He also asks for a simple bar chart to visualise the scores.

学校篮球队进行了五场比赛，得分分别为：23、19、26、30 和 17 分。教练希望通过平均值、中位数和极差来分析表现，并要求画一张简单的条形图来直观显示得分。

First, find the total and the mean:

首先，计算总分和平均值：

Total = 23 + 19 + 26 + 30 + 17 = 115

Mean = 115 ÷ 5 = 23

To find the median, order the scores: 17, 19, 23, 26, 30. The middle value is 23, so the median is 23.

为求中位数，将得分排序：17, 19, 23, 26, 30。中间值是 23，所以中位数是 23。

The range is the difference between the highest and lowest:

极差是最高分与最低分之差：

Range = 30 − 17 = 13

A bar chart would have match numbers on the horizontal axis and points on the vertical axis, showing a clear rise and fall. These statistics help the team see consistency – the mean and median being equal suggests a symmetric distribution of scores.

条形图的横轴为比赛场次，纵轴为得分，可清晰显示起伏。这些统计数据有助于球队了解稳定性——平均值与中位数相等表明得分的分布是对称的。

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6. Case Study 6: Building a Model Bridge | 案例6：建造模型桥

A technology project involves constructing a model bridge to a scale of 1:50. The real bridge is 25 m long and 8 m wide. Students must find the model dimensions in centimetres and purchase wooden sticks for the deck. Each stick is 30 cm long and costs £1.20. If the model deck uses sticks laid side by side lengthwise, calculate how many sticks are needed and their total cost.

一项技术课项目要求按 1:50 的比例搭建一座模型桥。真实桥长 25 米，宽 8 米。学生需计算以厘米为单位的模型尺寸，并购买木条制作桥面。每根木条长 30 厘米，售价 £1.20。如果模型桥面沿长度方向并排铺设木条，计算需要多少根木条以及总花费。

Convert real dimensions to model dimensions using the scale 1:50. Since 1 m = 100 cm, first express real length and width in cm:

使用比例 1:50 将真实尺寸转换为模型尺寸。由于 1 米 = 100 厘米，首先将真实长宽表示为厘米：

Real length = 25 × 100 = 2500 cm, width = 8 × 100 = 800 cm

Divide by the scale factor 50:

除以比例系数 50：

Model length = 2500 ÷ 50 = 50 cm, model width = 800 ÷ 50 = 16 cm

If the deck is made of sticks placed end-to-end along the length, each stick covers 30 cm. Number of sticks along one length = 50 ÷ 30 ≈ 1.67, so 2 sticks are needed to span the length (with a small overhang or cut). However, if the deck needs full coverage across the width using sticks placed side-by-side, we need to think differently. The problem states sticks are laid side by side lengthwise. So there will be multiple rows: each row along the length requires 2 sticks (since 50 cm ÷ 30 cm per stick = 2 sticks with 10 cm left, use another stick). The number of rows across the width is model width divided by stick width. But a stick is likely a thin strip; assume 1 cm wide for simplicity. Then number of rows = 16 ÷ 1 = 16 rows. Total sticks = 16 rows × 2 sticks per row = 32 sticks. Cost:

如果桥面由木条沿长度方向首尾相接铺设，每根木条长 30 厘米。沿单条长度方向需要的木条数 = 50 ÷ 30 ≈ 1.67，所以需 2 根才能覆盖长度（有少量悬挑或需切割）。但若桥面需要用木条并排密铺整个宽度，就要换一种思考方式。题目说明木条沿长度方向并排铺设，因此会有多列：每列沿长度需 2 根木条（因 50 cm ÷ 30 cm/根 ≈ 2，剩余用另一根补齐）。沿宽度方向的列数等于模型宽度除以木条宽度。假设木条宽 1 厘米以简化，列数 = 16 ÷ 1 = 16 列。总木条数 = 16 × 2 = 32 根。成本：

Total cost = 32 × £1.20 = £38.40

This case beautifully integrates scaling conversions and precise material estimation.

这个案例完美地综合了比例换算与精确的材料估算。

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7. Case Study 7: Saving for a New Console | 案例7：为游戏机存钱

Alex wants to buy a games console priced at £280. He already has £40 saved and plans to save £30 per month from his pocket money. His building society account offers simple interest at 3.5% per year, calculated on the total amount saved by the end of the year. Alex wants to know how many months he needs to reach £280, and also how much interest he would earn if he kept saving for exactly one year.

亚历克斯想买一台标价 £280 的游戏机。他已存了 £40，并计划每月从零花钱中省下 £30。他的建房协会账户提供 3.5% 的年单利，按年末余额计算。亚历克斯想知道需要多少个月才能攒够 £280，以及如果他整整存一年，能获得多少利息。

First, find the additional amount needed:

首先，计算还需多少金额：

Amount needed = £280 − £40 = £240

Divide by monthly saving:

除以每月储蓄额：

Months = £240 ÷ £30 = 8 months

So Alex will have enough after 8 months. If he continues to save for a full year (12 months), his total saved principal will be:

因此，8 个月后亚历克斯就能攒够。如果他继续存满一整年（12 个月），总本金将为：

Principal = £40 + (12 × £30) = £40 + £360 = £400

Simple interest for one year at 3.5%:

一年期单利，利率 3.5%：

Interest = 3.5% of £400 = 0.035 × 400 = £14

Total after one year = £414. This demonstrates how simple interest rewards longer saving periods.

一年后总额为 £414。这展示了单利如何奖励更长的储蓄周期。

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8. Case Study 8: Running a Tuck Shop | 案例8：经营小卖部

The school enterprise club sets up a tuck shop. They buy chocolate bars at £0.45 each and sell them for £0.70. The stall has a fixed rental fee of £10 per day. The club wants to calculate the break-even point – the number of bars they must sell so that profit exactly covers the fixed cost. They also want to know how many bars they would need to sell to make a daily profit of £20.

学校创业俱乐部开了一家小卖部。他们以每条 £0.45 的价格购入巧克力棒，并以 £0.70 售出。摊位每天的固定租赁费为 £10。俱乐部想计算盈亏平衡点 —— 即利润刚好覆盖固定成本所需销售的巧克力棒数量。他们还想知道，要赚取 £20 的日利润需要卖出多少条。

The profit made on each bar (contribution) is:

每条巧克力棒的利润（贡献毛益）为：

Contribution = £0.70 − £0.45 = £0.25

At break-even, total contribution equals the fixed cost. Let n be the number of bars:

在盈亏平衡点，总贡献毛益等于固定成本。设 n 为巧克力棒数量：

0.25n = 10 → n = 10 ÷ 0.25 = 40 bars

To achieve a desired profit of £20, the equation becomes:

要获得 £20 的利润，方程为：

0.25n − 10 = 20 → 0.25n = 30 → n = 120 bars

This algebraic approach is fundamental for small business planning and links directly to linear equations studied in further maths.

这种代数方法对于小企业规划至关重要，并与进阶数学中学习的线性方程直接相关。

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9. Case Study 9: Planning a School Garden | 案例9：规划学校花园

The eco committee designs a trapezium-shaped flower bed for the school garden. The parallel sides measure 3 m and 5 m, and the perpendicular height is 2 m. They need to fill the bed with topsoil, which is sold in bags covering 1.5 m² each. Calculate the area, the number of bags required, and the total cost if each bag costs £4.50.

Published by TutorHao | Year 7 进阶数学 Revision Series | aleveler.com

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