📚 Year 11 WJEC Maths: Case Study Practical Drill | 威尔斯联合考试委员会11年级数学：案例分析实战演练
In the WJEC GCSE Mathematics specification, applying mathematical reasoning to real‑world scenarios is a core skill. This article presents a series of case study‑style questions that blend number, algebra, geometry and statistics, mirroring the style of exam paper problem‑solving sections. Each scenario is broken down with fully worked reasoning, helping you see how to translate a written problem into a structured mathematical solution.
在 WJEC 中等教育普通证书数学考试说明中，将数学推理应用于现实情境是一项核心能力。本文通过一系列案例分析式的问题，融合了数、代数、几何和统计，模拟考试卷中问题解决部分的风格。每个情境都配有完整的推导过程，帮助你学会如何将文字问题转化为条理清晰的数学解答。
1. Budgeting a School Prom | 学校毕业舞会预算
A student committee is organising a Year 11 prom. The venue costs £720 to hire, the DJ charges £180, and decorations come to £95. They plan to sell tickets at £14 each. They want to know the minimum number of tickets they must sell to cover all costs and also raise at least £250 for a charity donation.
学生委员会正在组织 11 年级毕业舞会。场地租金为 720 英镑，DJ 收费 180 英镑，装饰花费 95 英镑。他们计划以每张 14 英镑的价格出售门票。他们想知道至少售出多少张门票才能覆盖所有成本，并为慈善捐款筹集至少 250 英镑。
First, let the number of tickets sold be n. Total cost C is the sum of the three fixed expenses: C = 720 + 180 + 95 = £995. The total income from tickets is 14n. To cover costs and have a surplus of at least £ nce tickets must be whole numbers, they need to sell at least 89 tickets. Check: 89 × 14 = £1246, which gives 1246 − 995 = £251 for charity.
首先，设售出门票数为 n。总成本 C 是三项固定支出的总和：C = 720 + 180 + 95 = 995 英镑。门票总收入为 14n。为了覆盖成本并有至少 250 英镑的盈余，不等式为：14n ≥ 995 + 250 → 14n ≥ 1245。两边除以 14 得 n ≥ 88.928……由于门票必须是整数，他们至少需要售出 89 张票。检验：89 × 14 = 1246 英镑，扣除成本后为慈善留下 251 英镑。
2. Mobile Phone Contract Comparison | 手机合约比较
A parent is choosing between two mobile phone deals for their child. Deal A has a monthly fee of £12 with 500 free texts and then 8p per extra text. Deal B costs £18 per month with unlimited texts. The child sends about 1100 texts per month. Calculate the monthly cost for each deal with this usage and decide which is cheaper.
一位家长在两种手机套餐之间为孩子选择。套餐 A 月费 12 英镑，包含 500 条免费短信，超出部分每条 8 便士。套餐 B 月费 18 英镑，短信不限量。孩子每月大约发送 1100 条短信。计算在该使用量下每种套餐的月费用，并判断哪一种更便宜。
For Deal A: included texts = 500, so extra texts = 1100 − 500 = 600. Extra cost = 600 × £0.08 = £48. Total monthly cost = £12 + £48 = £60. For Deal B: total monthly cost = £18. Clearly Deal B is significantly cheaper by £42 per month. However, if usage drops to 600 texts, Deal A would cost £12 + 100 × 0.08 = £20, which is still slightly more than £18. A linear equation could model the cost: Cₐ = 12 + 0.08(t − 500) for t > 500.
对于套餐 A：包含短信 500 条，因此额外短信 = 1100 − 500 = 600 条。额外费用 = 600 × 0.08 英镑 = 48 英镑。每月总费用 = 12 + 48 = 60 英镑。对于套餐 B：每月总费用 = 18 英镑。显然套餐 B 每月便宜 42 英镑。但如果使用量降至 600 条，套餐 A 费用为 12 + 100 × 0.08 = 20 英镑，仍略高于 18 英镑。成本可用线性方程建模：当 t > 500 时，Cₐ = 12 + 0.08(t − 500)。
3. Designing a Rectangular Garden with a Fixed Perimeter | 用固定周长设计矩形花园
A gardener has 36 metres of fencing to enclose a rectangular vegetable plot against an existing wall. The wall forms one of the longer sides. Express the area of the plot in terms of the width and find the dimensions that maximise the area.
一位园丁有 36 米长的围栏，想靠着一堵已有的墙壁围出一个矩形菜地。墙壁充当其中一条长边。用宽度表示地块的面积，并求出使面积最大化的尺寸。
Let the width perpendicular to the wall be x ther length). The area A is: A = x(36 − 2x) = 36x − 2x². This is a quadratic with a negative coefficient of x², so the maximum occurs at the vertex. Using symmetry or differentiation, the x-coordinate of the vertex is −b/(2a) with a = −2, b = 36, giving x = −36/(2 × −2) = 9. So width = 9 m, length = 36 − 18 = 18 m. Maximum area = 9 × 18 = 162 m².
设垂直于墙壁的宽度为 x 米。那么平行于墙壁的边长为 (36 − 2x) 米，因为围栏需用于两个宽度和仅一个长度（墙壁充当另一长度）。面积 A 为：A = x(36 − 2x) = 36x − 2x²。这是二次函数，x² 系数为负，最大值出现在顶点。利用对称性或求导，顶点 x 坐标为 −b/(2a)，其中 a = −2，b = 36，得 x = −36/(2 × −2) = 9。因此宽度 = 9 米，长度 = 36 − 18 = 18 米。最大面积 = 9 × 18 = 162 平方米。
4. Interpreting a Travel Graph | 解读行程图
A distance‑time graph shows a car journey. From 0 to 30 minutes the car travels at a steady speed covering 25 miles. Then it stops for 15 minutes. Finally it continues at 50 mph until the total distance reaches 60 miles. Calculate the average speed for the whole journey.
一张距离‑时间图显示了一段汽车行程。从 0 到 30 分钟，汽车以稳定速度行驶了 25 英里。然后停车 15 分钟。最后以 50 英里/小时的速度继续行驶，直到总距离达到 60 英里。计算整段行程的平均速度。
speed = 50 mph, so time for this leg = 35/50 = 0.7 hours = 42 minutes. Total journey time = 0.5 h + 0.25 h (stop) + 0.7 h = 1.45 hours. Average speed = total distance / total time = 60 miles / 1.45 hours ≈ 41.38 mph. The graph would consist of a sloped line, then a horizontal segment, then a steeper sloped line if the scale were different, but here the final speed equals the first, so slopes are equal.
第一段：时间 = 0.5 小时，距离 = 25 英里，速度 = 25/0.5 = 50 英里/小时（一致）。停车后，剩余距离 = 60 − 25 = 35 英里。第三段速度 = 50 英里/小时，因此该段时间 = 35/50 = 0.7 小时 = 42 分钟。总行程时间 = 0.5 小时 + 0.25 小时（停车）+ 0.7 小时 = 1.45 小时。平均速度 = 总距离 / 总时间 = 60 英里 / 1.45 小时 ≈ 41.38 英里/小时。图表将包含一段斜线，接着一段水平线段，最后一段斜线（如果比例不同会更陡，但此处最终速度与第一段相等，因此斜率相同）。
5. Probability in a Raffle | 抽奖活动中的概率
At a school fete, a raffle has 200 tickets sold. There are three prizes: a first prize of a tablet, a second prize of a voucher, and a third prize of a box of chocolates. You buy 5 tickets. What is the probability that you win at least one prize? (Assume tickets are drawn without replacement and each ticket can win at most one prize.)
在学校游园会上，一场抽奖活动售出 200 张彩票。共有三个奖项：头等奖平板电脑、二等奖代金券、三等奖一盒巧克力。你买了 5 张票。你至少赢得一个奖项的概率是多少？（假设彩票不重复抽取，且每张彩票最多中一个奖。）
easier to calculate the probability of winning no prize and subtract from 1. Number of non‑ning tickets = 200 − 3 = 197. The probability that all 5 of your tickets are among the non‑ 96 × 195 × 194 × 193) / (200 × 199 × 198 × 197 × 196). The 197 and 196 cancel, leaving (195 × 194 × 193) / (200 × 199 × 198). Compute numerator: 195 × 194 = 37830, ×193 = 37830 × 193 = 37830 × 200 − 37830 × 7 = 7,566,000 − 264,810 = 7,301,190. Denominator: 200 × 199 = 39800, ×198 = 39800 × 198 = 39800 × 200 − 39800 × 2 = 7,960,000 − 79,600 = 7,880,400. So P(none) ≈ 0.9265. Thus P(at least one) ≈ 1 − 0.9265 = 0.0735, about 7.35%.
计算未中奖的概率并从 1 中减去更为简便。非中奖彩票数量 = 200 − 3 = 197。你的 5 张票全部属于非中奖彩票的概率为：P（无一中奖）= (197/200) × (196/199) × (195/198) × (194/197) × (193/196)。化简乘积，许多项可约去：可写作 (197 × 196 × 195 × 194 × 193) / (200 × 199 × 198 × 197 × 196)。197 和 196 约去，剩下 (195 × 194 × 193) / (200 × 199 × 198)。计算分子：195 × 194 = 37830，乘以 193 = 37830 × 193 = 37830 × 200 − 37830 × 7 = 7,566,000 − 264,810 = 7,301,190。分母：200 × 199 = 39800，乘以 198 = 39800 × 198 = 39800 × 200 − 39800 × 2 = 7,960,000 − 79,600 = 7,880,400。因此 P（无一中奖）≈ 0.9265。故 P（至少一个）≈ 1 − 0.9265 = 0.0735，约 7.35%。
6. Ratio and Proportion in a Recipe | 食谱中的比与比例
A recipe for 8 people requires 300 g of flour, 200 g of sugar and 4 eggs. You need to adapt the quantities for 10 people. Also, if you only have 5 eggs, what is the maximum number of people you can serve while keeping the ingredients in proportion?
一份供 8 人食用的食谱需要 300 克面粉、200 克糖和 4 个鸡蛋。你需要将用量调整为 10 人份。此外，如果你只有 5 个鸡蛋，在保持配料比例的前提下，最多可供多少人食用？
cond part, the limiting factor is the eggs. The ratio requires 4 eggs per 8 people, i.e. 0.5 eggs per person. With 5 eggs available, the number of people = 5 / 0.5 = 10 people. So you can exactly serve 10 people with 5 eggs if you have the other ingredients scaled accordingly. Note: The recipe scaling shows that 10 people need 5 eggs, so 5 eggs suffice for 10 people.
要调整为 10 人份，将每种材料乘以因数 10/8 = 1.25。面粉：300 × 1.25 = 375 克；糖：200 × 1.25 = 250 克；鸡蛋：4 × 1.25 = 5 个。对于第二个问题，限制因素是鸡蛋。该比例要求每 8 人用 4 个鸡蛋，即每人 0.5 个鸡蛋。现有 5 个鸡蛋，可服务人数 = 5 / 0.5 = 10 人。因此，如果其他材料也按比例准备，5 个鸡蛋正好可供 10 人食用。注：食谱缩放显示 10 人需要 5 个鸡蛋，因此 5 个鸡蛋足够 10 人份。
7. Exchange Rates and Commission | 汇率与手续费
A family travels from Wales to the Eurozone. The exchange rate is £1 = €1.15. A bureau de change charges a fixed commission of £3.50 plus 2% of the sterling amount exchanged. The family wants to obtain at least €800 after all charges. How many pounds must they exchange?
一个家庭从威尔士前往欧元区旅行。汇率为 £1 = €1.15。一家外币兑换所收取固定手续费 3.50 英镑，外加兑换英镑金额的 2% 作为佣金。该家庭希望扣除所有费用后至少得到 800 欧元。他们必须兑换多少英镑？
Let the sterling amount exchanged be x pounds. The gross euros before commission would be 1.15x, but the commission is £ ive pounds available for currency conversion is (x − 3.50 − 0.02x) = 0.98x − 3.50. Then the euros received = 1.15 × (0.98x − 3.50). We need this to be ≥ 800: 1.15(0.98x − 3.50) ≥ 800. Divide by 1.15: 0.98x − 3.50 ≥ 695.652... (since 800/1.15 ≈ 695.652). Then 0.98x ≥ 699.152..., so x ≥ 699.152... / 0.98 ≈ 713.42. Thus they must exchange at least £713.42, in practice £714 if only whole pounds are accepted.
设兑换的英镑金额为 x 英镑。未扣除手续费前的毛欧元为 1.15x，但手续费为 3.50 英镑 + 0.02x，这实际上是以英镑扣除的，或者我们可以考虑净得欧元：他们另付手续费，因此可用于兑换的英镑金额为 (x − 3.50 − 0.02x) = 0.98x − 3.50。然后得到的欧元 = 1.15 × (0.98x − 3.50)。我们需要此值 ≥ 800：1.15(0.98x − 3.50) ≥ 800。除以 1.15：0.98x − 3.50 ≥ 695.652……（因 800/1.15 ≈ 695.652）。然后 0.98x ≥ 699.152……，故 x ≥ 699.152…… / 0.98 ≈ 713.42。因此他们必须兑换至少 713.42 英镑，实际中如果只接受整数英镑，则为 714 英镑。
8. Surface Area and Paint Coverage | 表面积与油漆覆盖量
A cylindrical water tank has a diameter of 1.8 m and a height of 2.4 m. It is open at the top. You need to paint the exterior curved surface and the base. One litre of paint covers 8 m². How many litres of paint are required for two coats?
一个圆柱形水箱的直径为 1.8 米，高为 2.4 米，顶部敞开。你需要给外部曲面和底部上漆。每升油漆可覆盖 8 平方米。刷两层需要多少升油漆？
0.81)π = 5.13π ≈ 16.113 m². For two coats, area = 2 × 5.13π = 10.26π ≈ 32.226 m². Litres needed = total area / coverage per litre = 10.26π / 8 = 1.2825π ≈ 4.03 litres. Therefore, you would need to buy a 5‑litre can. This shows how geometry combines with practical rounding up for purchasing.
半径 r = 0.9 米。曲面面积 = 2πrh = 2 × π × 0.9 × 2.4 = 4.32π 平方米。底部面积 = πr² = π × 0.9² = 0.81π 平方米。需涂漆的外部总面积（一层）= (4.32 + 0.81)π = 5.13π ≈ 16.113 平方米。涂两层时，面积 = 2 × 5.13π = 10.26π ≈ 32.226 平方米。所需升数 = 总面积 / 每升覆盖量 = 10.26π / 8 = 1.2825π ≈ 4.03 升。因此，你需要购买一桶 5 升的油漆。这表明几何如何与实际购买中向上取整相结合。
9. Statistical Analysis of Test Scores | 测验成绩的统计分析
highest score = 50. Identify any outliers using the interquartile range (IQR) rule, and comment on the skewness of the distribution.
一个 25 名学生的班级参加了数学测验。他们满分为 50 分的成绩摘要如下：中位数 = 38，下四分位数 = 29，上四分位数 = 44，最低分 = 15，最高分 = 50。使用四分位距 (IQR) 法则识别任何异常值，并评论分布的偏斜情况。
5 and the highest 50 is less than 66.5, there are no outliers according to the 1.5×IQR rule. To assess skew, compare the distances: Q₂ − Q₁ = 38 − 29 = 9, Q₃ − Q₂ = 44 − 38 = 6. The median is closer to the upper quartile, suggesting a slight negative skew (more concentration of scores at the higher end). Also, the range is 35, while the IQR is 15, indicating moderate spread.
IQR = Q₃ − Q₁ = 44 − 29 = 15。下界限 = Q₁ − 1.5 × IQR = 29 − 22.5 = 6.5。上界限 = Q₃ + 1.5 × IQR = 44 + 22.5 = 66.5。由于最低分 15 大于 6.5 且最高分 50 小于 66.5，根据 1.5×IQR 法则没有异常值。要评估偏斜，比较距离：Q₂ − Q₁ = 38 − 29 = 9，Q₃ − Q₂ = 44 − 38 = 6。中位数更靠近上四分位数，表明存在轻微的负偏斜（分数更集中在较高端）。此外，极差为 35，而 IQR 为 15，表明中等离散度。
10. Using Trigonometry to Measure a Building Height | 利用三角学测量建筑物高度
From a point 40 metres from the base of a building, the angle of elevation to the top is 34°. A second point is located on the same horizontal line, 15 metres closer to the building. Find the angle of elevation from this second point, and hence verify consistency.
从距离建筑物底部 40 米的一点，测得顶部仰角为 34°。在同一水平线上，另有一点比第一点更接近建筑物 15 米。求从这第二点测得的仰角，并由此验证一致性。
Let the height of the building be h cond point to the base is 40 − 15 = 25 m. The angle of elevation θ from this point satisfies tan θ = h / 25 ≈ 26.98 / 25 = 1.0792. Then θ = tan⁻¹(1.0792) ≈ 47.1°. This makes sense because as you get closer, the angle increases. The case demonstrates how trigonometry and inverse functions solve practical height problems.
设建筑物的高度为 h 米。从第一点：tan 34° = h / 40，因此 h = 40 tan 34°。使用计算器，tan 34° ≈ 0.6745，故 h ≈ 40 × 0.6745 = 26.98 米。第二点到建筑物底部的距离为 40 − 15 = 25 米。从该点测得的仰角 θ 满足 tan θ = h / 25 ≈ 26.98 / 25 = 1.0792。因此 θ = tan⁻¹(1.0792) ≈ 47.1°。这很合理，因为距离越近角度越大。此案例展示了三角学与反函数如何解决实际高度问题。
11. Compound Interest on Savings | 储蓄的复利计算
A parent invests £2000 in a savings account paying 2.4% interest per annum, compounded annually. They add no further deposits. Calculate the value after 7 years. Also determine how many whole years it takes for the investment to double.
一位家长将 2000 英镑投资于一个储蓄账户，年利率 2.4%，按年复利。他们不再额外存款。计算 7 年后的价值。并确定投资翻倍所需的整年数。
576)² ≈ 1.0995, then 1.024⁷ = 1.024⁴ × 1.024² × 1.024 ≈ 1.0995 × 1.048576 × 1.024 ≈ 1.1812. Thus A ≈ 2000 × 1.1812 = £ 1.024) / ln(10) using base 10: ln(1.024) ≈ 0.0237, ln(10) ≈ 2.3026, so log₁₀(1.024) ≈ 0.0103). n ≈ 0.3010 / 0.0103 ≈ 29.2 years. So it takes 30 whole years to double. This shows the power but also the patience required with low‑interest compounding.
复利公式为 A = P(1 + r/100)ⁿ。此处 P = 2000，r = 2.4，n = 7。因此 A = 2000 × (1.024)⁷。计算 1.024⁷：1.024² = 1.048576，1.024⁴ = (1.048576)² ≈ 1.0995，然后 1.024⁷ = 1.024⁴ × 1.024² × 1.024 ≈ 1.0995 × 1.048576 × 1.024 ≈ 1.1812。因此 A ≈ 2000 × 1.1812 = 2362.40 英镑。对于翻倍，求解 4000 = 2000 × (1.024)ⁿ → 2 = 1.024ⁿ。取对数：n = log(2) / log(1.024)。log(2) ≈ 0.3010，log(1.024) ≈ 0.0103（由于以 10 为底：ln(1.024) ≈ 0.0237，ln(10) ≈ 2.3026，因此 log₁₀(1.024) ≈ 0.0103）。n ≈ 0.3010 / 0.0103 ≈ 29.2 年。因此需要 30 个整年才能翻倍。这展示了低利率复利的威力，但也体现了所需的耐心。
12. Bearing and Map Scales | 方位角与地图比例尺
On a map with a scale of 1:25000, two villages are 7.2 cm apart. The bearing from A to B is 125°. What is the actual straight‑line distance in kilometres, and explain the bearing meaning in a real‑world navigation context.
在一幅比例尺为 1:25000 的地图上，两个村庄相距 7.2 厘米。从 A 到 B 的方位角为 125°。实际直线距离是多少公里？并解释方位角在现实导航中的含义。
. A bearing of 125° is measured clockwise from north (0°). So from A, you would face north, rotate 125° clockwise (which is between east 90° and south 180°), roughly southeast. This means B lies in a southeast direction from A. Bearings are used in navigation to describe direction precisely without ambiguity.
地图比例尺 1:25000 意味着图上 1 厘米代表实际 25000 厘米。25000 厘米 = 250 米 = 0.25 公里。因此，实际距离 = 7.2 × 0.25 = 1.8 公里。125° 的方位角是从正北（0°）顺时针测量的。因此从 A 点，面向北方，顺时针旋转 125°（介于东 90° 和南 180° 之间），大致为东南方向。这意味着 B 位于 A 的东南方向。方位角用于导航中精确且无歧义地描述方向。
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