📚 Case Study Practice for KS3 Cambridge Advanced Mathematics | KS3 剑桥进阶数学案例分析实战演练
In KS3 Cambridge Advanced Mathematics, applying your knowledge to real-world scenarios is an essential skill. This article presents ten case studies that require you to use algebra, geometry, statistics, ratio, and probability to solve practical problems. Each case study is designed to strengthen your problem-solving abilities and prepare you for Checkpoint assessments.
在 KS3 剑桥进阶数学中，将所学知识应用于现实情境是一项关键技能。本文提供了十个案例分析，要求你运用代数、几何、统计、比率和概率来解决实际问题。每个案例都旨在提升你的问题解决能力，为 Checkpoint 评估做好准备。
1. Case Study: Designing a Garden – Using Algebra | 案例1: 设计花园——运用代数
A gardener needs to fence a rectangular garden. The length is 3 metres longer than the width, and the total length of fencing available is 26 metres.
一位园丁需要为矩形花园围上栅栏。花园的长度比宽度多3米，现有栅栏总长为26米。
Let the width be w metres. Then the length is w + 3 metres.
设宽为 w 米，则长为 w + 3 米。
The perimeter of a rectangle is given by P = 2(length + width).
矩形周长公式为 P = 2 × (长 + 宽)。
Substituting the values: 2(w + w + 3) = 26.
代入数值：2(w + w + 3) = 26。
2(2w + 3) = 26
4w + 6 = 26
4w = 20
w = 5
So the width is 5 m and the length is 5 + 3 = 8 m. The area of the garden is length × width = 8 × 5 = 40 m².
因此，宽为5米，长为 5 + 3 = 8 米。花园的面积为 长 × 宽 = 8 × 5 = 40 平方米。
If each square metre of flowerbeds costs £5 to plant, the total planting cost is 40 × 5 = £200.
如果每平方米花坛的种植成本为5英镑，则总种植成本为 40 × 5 = 200 英镑。
2. Case Study: Planning a Road Trip – Distance, Speed, Time | 案例2: 规划公路旅行——距离、速度、时间
A family drives 240 km to the seaside at an average speed of 80 km/h. How long does the journey take?
一家人以平均速度 80 公里/小时驾车 240 公里前往海边。行程需要多长时间？
Using the formula time = distance ÷ speed:
使用公式 时间 = 距离 ÷ 速度：
time = 240 ÷ 80 = 3 hours
时间 = 240 ÷ 80 = 3 小时
On the return journey, roadworks reduce their speed by 25%. The new speed is 80 × 0.75 = 60 km/h.
返程时，因道路施工，速度降低25%。新的速度为 80 × 0.75 = 60 公里/小时。
The time for the return journey is 240 ÷ 60 = 4 hours.
返程所需时间为 240 ÷ 60 = 4 小时。
The delay caused by roadworks is 4 − 3 = 1 hour.
因道路施工延误的时间为 4 − 3 = 1 小时。
3. Case Study: School Canteen Survey – Data Handling | 案例3: 学校食堂调查——数据处理
150 students voted for their favourite lunch option. The results were: Pizza – 45, Sandwich – 30, Salad – 15, Pasta – 60.
150 名学生投票选出他们最喜欢的午餐。结果为：披萨 45 票、三明治 30 票、沙拉 15 票、意大利面 60 票。
To draw a pie chart, calculate the angle for each category. The total frequency is 45 + 30 + 15 + 60 = 150.
为了绘制饼图，需要计算每个类别的角度。总频数为 45 + 30 + 15 + 60 = 150。
Pizza: fraction = 45/150 = 3/10, angle = 3/10 × 360° = 108°.
披萨：比例 = 45/150 = 3/10，角度 = 3/10 × 360° = 108°。
Sandwich: 30/150 = 1/5, angle = 1/5 × 360° = 72°.
三明治：30/150 = 1/5，角度 = 1/5 × 360° = 72°。
Salad: 15/150 = 1/10, angle = 36°.
沙拉：15/150 = 1/10，角度 = 36°。
Pasta: 60/150 = 2/5, angle = 2/5 × 360° = 144°.
意大利面：60/150 = 2/5，角度 = 2/5 × 360° = 144°。
Check: 108° + 72° + 36° + 144° = 360°. The most popular choice is Pasta with 40% of the votes.
验证：108° + 72° + 36° + 144° = 360°。最受欢迎的是意大利面，获得了40%的选票。
4. Case Study: Scaling a Recipe – Ratio and Proportion | 案例4: 按比例调整食谱——比与比例
A biscuit recipe requires 400 g flour, 250 g sugar, and 200 g butter to make 16 biscuits. How much of each ingredient is needed to make 24 biscuits?
一份饼干食谱需要 400 克面粉、250 克糖和 200 克黄油，可制作 16 块饼干。要制作 24 块饼干，每种原料各需多少？
The scaling factor is 24 ÷ 16 = 1.5.
比例因子为 24 ÷ 16 = 1.5。
Flour needed: 400 × 1.5 = 600 g.
所需面粉：400 × 1.5 = 600 克。
Sugar needed: 250 × 1.5 = 375 g.
所需糖：250 × 1.5 = 375 克。
Butter needed: 200 × 1.5 = 300 g.
所需黄油：200 × 1.5 = 300 克。
Later, you discover only 200 g of sugar is left. What is the maximum number of biscuits you can make with the available sugar?
随后你发现只剩下 200 克糖。利用现有的糖，最多能制作多少块饼干？
Original recipe: 250 g sugar → 16 biscuits. So 1 g sugar makes 16 ÷ 250 = 0.064 biscuits.
原食谱：250 克糖可制作 16 块，因此每克糖可制作 16 ÷ 250 = 0.064 块。
With 200 g sugar, biscuits possible = 200 × 0.064 = 12.8. You can make 12 whole biscuits (assuming other ingredients are sufficient).
有 200 克糖时，可制作饼干数 = 200 × 0.064 = 12.8。可以制作 12 块饼干（假设其他原料充足）。
5. Case Study: Tile Patterns – Linear Sequences | 案例5: 瓷砖图案——线性数列
A tiler arranges square tiles to form a pattern. Pattern 1 uses 5 tiles, Pattern 2 uses 8 tiles, and Pattern 3 uses 11 tiles.
一位瓷砖工用方形瓷砖排列图案。图案1用了5块，图案2用了8块，图案3用了11块。
The number of tiles increases by 3 each time, so the nth term rule is T(n) = 3n + 2.
瓷砖数量每次增加3块，所以第 n 项的规则是 T(n) = 3n + 2。
Test: for n=1, T=3(1)+2=5; n=2, T=8; n=3, T=11. Correct.
验证：当 n=1 时，T=3×1+2=5；n=2 时，T=8；n=3 时，T=11。正确。
How many tiles are needed for Pattern 15? T(15) = 3×15 + 2 = 47 tiles.
第15个图案需要多少块瓷砖？T(15) = 3×15 + 2 = 47 块。
If the tiler has 100 tiles, what is the highest pattern number that can be completed? Solve 3n + 2 ≤ 100.
如果瓷砖工有100块瓷砖，最多能完成到第几个图案？解不等式 3n + 2 ≤ 100。
3n ≤ 98
n ≤ 32.666…
The greatest whole number is 32. Pattern 32 requires 3×32+2 = 98 tiles.
最大的整数为32。第32个图案需要 3×32+2 = 98 块瓷砖。
6. Case Study: Buying in Bulk – Best Buy Problems | 案例6: 批量购买——最优选购问题
A shop sells ice cream in two sizes: a small 500 ml tub for £2.20 and a large 1.2 litre tub for £5.00. Which size gives better value for money?
一家商店出售两种规格的冰淇淋：小盒 500 毫升，售价 £2.20；大盒 1.2 升，售价 £5.00。哪种更划算？
Convert to a common unit (litres). Small tub: 500 ml = 0.5 litres. Price per litre = £2.20 ÷ 0.5 = £4.40/L.
转换为统一单位（升）。小盒：500 毫升 = 0.5 升。每升价格 = 2.20 ÷ 0.5 = 4.40 英镑/升。
Large tub: 1.2 L, price per litre = £5.00 ÷ 1.2 ≈ £4.167/L.
大盒：1.2 升，每升价格 = 5.00 ÷ 1.2 ≈ 4.167 英镑/升。
The large tub offers better value per litre. However, the shop offers a 15% discount on the large tub. The discounted price is £5.00 × 0.85 = £4.25.
大盒每升价格更划算。然而商店对大盒提供 15% 的折扣。折扣后价格为 5.00 × 0.85 = 4.25 英镑。
New price per litre = £4.25 ÷ 1.2 ≈ £3.542/L. This is significantly better than the small tub.
新的每升价格 = 4.25 ÷ 1.2 ≈ 3.542 英镑/升。这明显比小盒划算。
7. Case Study: Saving Pocket Money – Percentages and Simple Interest | 案例7: 存储零花钱——百分比与单利
Alex receives £40 pocket money each month. She decides to save 25% of it every month to buy a video game console costing £150.
亚历克斯每月获得 £40 零花钱。她决定每月存下其中的 25%，以便购买一台售价 £150 的游戏机。
Monthly saving = 25% of £40 = 0.25 × 40 = £10.
每月存款 = £40 的 25% = 0.25 × 40 = £10。
If no interest is earned, how many months must she save? 150 ÷ 10 = 15 months.
如果不计利息，她需要存多少个月？150 ÷ 10 = 15 个月。
Instead, Alex puts the money into a savings account that pays 5% simple interest per year on the total amount saved. After one year (12 months), she has saved 12 × £10 = £120.
相反，亚历克斯将钱存入一个储蓄账户，该账户每年按存款总额支付 5% 的单利。一年（12个月）后，她共存了 12 × 10 = 120 英镑。
Simple interest earned = 5% of £120 = 0.05 × 120 = £6.
获得的单利 = £120 的 5% = 0.05 × 120 = 6 英镑。
Total after one year = £120 + £6 = £126. She still needs £150 − £126 = £24. She must save for 3 more months (since she cannot buy the console earlier).
一年后总额为 120 + 6 = 126 英镑。她仍然需要 150 − 126 = 24 英镑。她必须再存 3 个月（因为她无法提前购买游戏机）。
8. Case Study: Packaging Design – Volume and Surface Area | 案例8: 包装设计——体积与表面积
A company designs two rectangular boxes for packaging. Box A measures 20 cm by 15 cm by 10 cm. Box B measures 25 cm by 12 cm by 8 cm.
一家公司设计了两种矩形包装盒。盒子A尺寸为 20 厘米 × 15 厘米 × 10 厘米。盒子B尺寸为 25 厘米 × 12 厘米 × 8 厘米。
Volume of Box A = 20 × 15 × 10 = 3000 cm³.
盒子A的体积 = 20 × 15 × 10 = 3000 立方厘米。
Volume of Box B = 25 × 12 × 8 = 2400 cm³. Box A holds more volume.
盒子B的体积 = 25 × 12 × 8 = 2400 立方厘米。盒子A容量更大。
Surface area of Box A = 2(20×15 + 20×10 + 15×10) = 2(300 + 200 + 150) = 2 × 650 = 1300 cm².
盒子A的表面积 = 2(20×15 + 20×10 + 15×10) = 2(300 + 200 + 150) = 2 × 650 = 1300 平方厘米。
Surface area of Box B = 2(25×12 + 25×8 + 12×8) = 2(300 + 200 + 96) = 2 × 596 = 1192 cm².
盒子B的表面积 = 2(25×12 + 25×8 + 12×8) = 2(300 + 200 + 96) = 2 × 596 = 1192 平方厘米。
Although Box A has a larger volume, Box B uses less material (smaller surface area) per unit volume. The company must decide based on cost and customer needs.
尽管盒子A体积更大，但盒子B的单位体积所用材料更少（表面积更小）。公司必须根据成本和客户需求做出决定。
9. Case Study: Fair Game? – Probability | 案例9: 公平的游戏？——概率
A game uses a spinner divided into 5 equal sectors: 3 blue and 2 red. If the spinner lands on blue, you win £1. If it lands on red, you lose £2.
一个游戏使用一个分成5个相等扇区的转盘：3个蓝色，2个红色。如果转盘停在蓝色区域，你赢 1 英镑。如果停在红色区域，你输 2 英镑。
Probability of blue = 3/5; probability of red = 2/5.
蓝色的概率 = 3/5Published by TutorHao | KS3 进阶数学 Revision Series | aleveler.com
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