📚 Year 10 WJEC Mathematics: Case Study Practical Exercises | Year 10 WJEC 数学：案例分析实战演练
Real-life problems bring mathematics to life. In the WJEC Year 10 course, you are expected to apply number, algebra, geometry, statistics and probability to solve genuine challenges. This article presents eight case studies that mirror the style of exam questions and coursework tasks, helping you build confidence through guided practice. Each case study includes a scenario, key calculations and reasoning steps, with English and Chinese explanations side by side.
现实生活问题让数学变得生动。在 WJEC 十年级课程中，你需要运用数、代数、几何、统计和概率知识解决真实的挑战。本文提供八个案例分析，模拟考试题和课程作业的题型，通过引导式练习帮助你建立信心。每个案例包含情境、关键计算和推理步骤，并配有中英文对照讲解。
1. Personal Finance: Budgeting and Savings | 个人理财：预算与储蓄
A Year 10 student earns £96 per month from a weekend job. She plans to buy a laptop priced at £480. She decides to save 30% of her monthly income and keep the rest for other expenses. How many months will she need to reach her goal? If she also receives a one-off birthday gift of £60, how does this shorten the saving period?
一名十年级学生通过周末兼职每月赚取 96 英镑。她计划购买一台标价 480 英镑的笔记本电脑，决定将月收入的 30% 存起来，其余留作他用。她需要多少个月才能实现目标？如果她还收到一笔 60 英镑的生日礼金，储蓄时间会如何缩短？
Monthly saving amount = 30% of £96 = 0.30 × £96 = £28.80. Without extra money, months needed = £480 ÷ £28.80 = 16.67, so 17 full months (since she must save in whole months). With the £60 gift, the remaining amount is £480 − £60 = £420. Then £420 ÷ £28.80 = 14.58, so 15 months. The gift reduces the time by 2 months.
每月储蓄额 = 96 英镑的 30% = 0.30 × 96 = 28.80 英镑。在没有额外资金时，所需月数 = 480 ÷ 28.80 = 16.67，即需要 17 个完整的月份（因为按月取整储蓄）。有了 60 英镑礼金后，剩余金额 = 480 − 60 = 420 英镑。此时 420 ÷ 28.80 = 14.58，需要 15 个月。礼金节省了 2 个月时间。
Simple interest can also be considered. If she deposits savings into an account paying 4% per annum simple interest, the interest earned after one year on £28.80 monthly deposits would be calculated on the average balance, but here we focus on the essential percentage and division skills tested at Year 10 level.
还可以考虑单利。如果她把储蓄存入年利率 4% 的单利账户，一年后每月存入 28.80 英镑所产生的利息可通过平均余额计算，但此处我们聚焦于十年级需要掌握的百分比和除法核心技能。
2. Travel Planning: Speed, Distance and Currency | 旅行规划：速度、距离与货币兑换
ter a 45-minute rest stop, they continue the journey. What is their estimated arrival time? The family also needs to exchange £500 into euros for a trip to France. The exchange rate is £1 = €1.14, and the bureau charges a 2.5% commission on the sterling amount. How many euros will they receive?
一个家庭从卡迪夫开车前往伦敦，距离为 210 公里。他们 09:15 出发，平均车速保持 70 公里/小时。途中休息 45 分钟后继续行驶。预计到达时间是什么？该家庭还需要为法国之旅将 500 英镑兑换成欧元。汇率为 1 英镑 = 1.14 欧元，兑换所按英镑金额收取 2.5% 的手续费。他们将获得多少欧元？
Departure at 09:15 plus 3 h 45 min gives arrival at 13:00. For currency: commission = 2.5% of £500 = £12.50, so amount exchanged = £500 − £12.50 = £487.50. Euros received = £487.50 × 1.14 = €555.75.
行驶时间 = 距离 ÷ 速度 = 210 ÷ 70 = 3 小时。加上休息的总时间 = 3 小时 + 45 分钟 = 3 小时 45 分钟。09:15 出发加 3 小时 45 分钟，到达时间为 13:00。货币计算：手续费 = 500 英镑的 2.5% = 12.50 英镑，因此实际兑换金额 = 500 − 12.50 = 487.50 英镑。获得欧元 = 487.50 × 1.14 = 555.75 欧元。
3. Home Improvement: Area and Paint Costs | 家庭装修：面积与油漆成本
A bedroom wall measures 3.6 m in length and 2.4 m in height. It has a rectangular window of dimensions 1.2 m by 0.9 m and a door of 2.0 m by 0.8 m. One tin of paint covers 10 m² and costs £14.99. How many tins are needed to apply two coats on the wall (excluding window and door)? What is the total paint cost?
一面卧室墙壁长 3.6 米，高 2.4 米。墙上有一扇 1.2 米 × 0.9 米的矩形窗户和一扇 2.0 米 × 0.8 米的门。一罐油漆可涂刷 10 平方米，售价 14.99 英镑。如果要在墙面（不含门窗）刷两层漆，需要多少罐油漆？总油漆费用是多少？
l area to cover = 2 × 5.96 = 11.92 m². Tins needed = ceil(11.92 ÷ 10) = 2 tins. Cost = 2 × £14.99 = £29.98.
墙壁面积 = 3.6 × 2.4 = 8.64 平方米。窗户面积 = 1.2 × 0.9 = 1.08 平方米。门面积 = 2.0 × 0.8 = 1.60 平方米。净面积 = 8.64 − 1.08 − 1.60 = 5.96 平方米。两层漆的总涂刷面积 = 2 × 5.96 = 11.92 平方米。所需罐数 = 向上取整 (11.92 ÷ 10) = 2 罐。费用 = 2 × 14.99 = 29.98 英镑。
4. Data Analysis: School Survey on Exercise | 数据分析：学校运动调查
A school surveyed 60 Year 10 students about the number of hours they exercise per week. The results are summarised: 0–2 hours: 8 students; 2–4 hours: 22; 4–6 hours: 18; 6–8 hours: 10; 8–10 hours: 2. Estimate the mean number of hours and identify the modal class. What percentage of students exercise more than 6 hours per week?
某学校对 60 名十年级学生每周锻炼的小时数进行了调查。结果汇总如下：0–2 小时：8 人；2–4 小时：22 人；4–6 小时：18 人；6–8 小时：10 人；8–10 小时：2 人。请估算平均小时数，并找出众数所在组。每周锻炼超过 6 小时的学生占百分之多少？
18 + 7×10 + 9×2 = 8 + 66 + 90 + 70 + 18 = 252. Mean ≈ 252 ÷ 60 = 4.2 hours. Modal class: 2–4 hours (frequency 22). Percentage exercising >6 hours = (10+2)/60 × 100% = 12/60 × 100% = 20%.
估算平均值时使用组中值：1, 3, 5, 7, 9。（组中值 × 频数）之和 = 1×8 + 3×22 + 5×18 + 7×10 + 9×2 = 8 + 66 + 90 + 70 + 18 = 252。平均 ≈ 252 ÷ 60 = 4.2 小时。众数所在组：2–4 小时（频数 22）。锻炼超过 6 小时的百分比 = (10+2)/60 × 100% = 12/60 × 100% = 20%。
5. Recipe Scaling: Ratios and Proportions | 食谱调整：比与比例
A cake recipe for 8 people requires 300 g of flour, 200 g of sugar, 4 eggs and 150 ml of milk. You need to make the same cake for 12 people. How much of each ingredient is required? Express the ratio of flour to sugar in its simplest form for the original recipe, and check whether this ratio remains the same after scaling.
一份供 8 人食用的蛋糕食谱需要 300 克面粉、200 克糖、4 个鸡蛋和 150 毫升牛奶。你需要为 12 人制作同样的蛋糕。每种原料需要多少？请将原食谱中面粉与糖的比化为最简形式，并检验调整后该比例是否保持不变。
ur:sugar = 300:200 = 3:2. After scaling, 450:300 = 3:2 as well, so the ratio is preserved.
比例因子 = 12 ÷ 8 = 1.5。面粉：300 × 1.5 = 450 克；糖：200 × 1.5 = 300 克；鸡蛋：4 × 1.5 = 6 个；牛奶：150 × 1.5 = 225 毫升。原食谱面粉与糖之比 = 300:200 = 3:2。调整后 450:300 = 3:2，因此比例保持不变。
6. Sports Performance: Interpreting Distance-Time Graphs | 运动表现：解读距离-时间图
Jamie goes for a cycle ride. The distance-time graph shows: from 0 to 10 minutes, distance increases steadily from 0 km to 4 km; from 10 to 15 minutes, the line is horizontal; from 15 to 25 minutes, distance increases from 4 km to 10 km; then from 25 to 30 minutes, distance increases from 10 km to 14 km. Calculate Jamie's speed in km/h for each moving segment and identify when he was resting.
杰米骑车出行。距离-时间图显示：0 至 10 分钟，距离从 0 公里匀速增加至 4 公里；10 至 15 分钟，线段水平；15 至 25 分钟，距离从 4 公里增加至 10 公里；然后 25 至 30 分钟，距离从 10 公里增加至 14 公里。请计算杰米在每段运动中的速度（km/h），并指出他何时在休息。
Segment 1 (0–10 min): 4 km in 10 min = 4 km ÷ (10/60) h = 4 × 6 = 24 km/h. Segment 2 (10–15 min): horizontal line means zero distance change, so rest. Segment 3 (15–25 min): distance increase 6 km in 10 min = 6 ÷ (10/60) = 36 km/h. Segment 4 (25–30 min): 4 km in 5 min = 4 ÷ (5/60) = 48 km/h.
第一段（0–10 分钟）：10 分钟行驶 4 公里，速度 = 4 ÷ (10/60) = 4 × 6 = 24 km/h。第二段（10–15 分钟）：水平线段表示距离不变，所以休息。第三段（15–25 分钟）：10 分钟行驶 6 公里，速度 = 6 ÷ (10/60) = 36 km/h。第四段（25–30 分钟）：5 分钟行驶 4 公里，速度 = 4 ÷ (5/60) = 48 km/h。
7. Business Maths: Profit, Loss and Equations | 商业数学：盈亏与方程
A shop buys hoodies at £22 each and sells them at £35. Write an expression for the profit made when N hoodies are sold. In a sale, the price is reduced by 15%, and the shop still makes a profit of £5 per hoodie. Determine the new selling price and the profit if 50 hoodies are sold at the sale price. Also, set up and solve an equation to find how many hoodies must be sold to earn a total profit of £600 at the original price.
一家商店以每件 22 英镑的价格购入卫衣，以 35 英镑出售。写出销售 N 件卫衣所获利润的表达式。在促销活动中，价格下调 15%，商店每件仍获利 5 英镑。求新的销售价格，以及按促销价销售 50 件时的利润。此外，建立并求解方程，计算按原价需销售多少件卫衣才能获得 600 英镑的总利润。
Profit per hoodie at original price = £35 − £22 = £13. Expression for N hoodies: 13N. Sale price = £35 × (1 − 0.15) = £35 × 0.85 = £29.75. Profit at sale price = £29.75 − £22 = £7.75 (but the problem states £5 profit, so we use given: new profit is £5). Total profit for 50 = 50 × £5 = £250. For £600 target at original profit: 13N = 600 → N = 600 ÷ 13 ≈ 46.15, so at least 47 hoodies.
原价每件利润 = 35 − 22 = 13 英镑。销售 N 件的利润表达式：13N。促销价 = 35 × (1 − 0.15) = 35 × 0.85 = 29.75 英镑。促销利润 = 29.75 − 22 = 7.75 英镑（但题目明确利润为 5 英镑，我们按题意取新利润 5 英镑）。销售 50 件的总利润 = 50 × 5 = 250 英镑。要达成 600 英镑目标利润：13N = 600 → N = 600 ÷ 13 ≈ 46.15，因此至少需要销售 47 件。
8. Map Scales and Similarity | 地图比例尺与相似形
A map uses a scale of 1:25000. Two villages are 8.4 cm apart on the map. Calculate the actual distance in kilometres. A rectangular park on the map measures 5 cm by 3 cm. Find the real area of the park in square kilometres. Another map of the same park has a scale of 1:50000. Determine the park's dimensions on this second map and explain the relationship between the area scale factors.
某地图使用 1:25000 的比例尺。两座村庄在地图上相距 8.4 厘米。计算实际距离（公里）。地图上一个矩形公园的尺寸为 5 厘米 × 3 厘米。求公园的实际面积（平方公里）。同一公园的另一幅地图使用 1:50000 的比例尺。求公园在第二幅地图上的尺寸，并解释面积比例因子之间的关系。
Real area = 1.25 × 0.75 = 0.9375 km². For the 1:50000 map, scale factor relative to real life is 1/50000, so dimensions: length = 125000 cm ÷ 50000 = 2.5 cm; width = 75000 cm ÷ 50000 = 1.5 cm. Area scale factor = (1/50000)² = 1/2500000000, whereas for 1:25000 it is 1/625000000. The area on the second map is (1/2)² = 1/4 of the area on the first map.
实际距离 = 8.4 cm × 25000 = 210000 cm = 2100 m = 2.1 km。公园实际长 = 5 cm × 25000 = 125000 cm = 1.25 km；实际宽 = 3 cm × 25000 = 75000 cm = 0.75 km。实际面积 = 1.25 × 0.75 = 0.9375 km²。对于 1:50000 的地图，相对于实际的比例因子为 1/50000，因此尺寸：长 = 125000 cm ÷ 50000 = 2.5 cm；宽 = 75000 cm ÷ 50000 = 1.5 cm。面积比例因子 = (1/50000)² = 1/2500000000，而 1:25000 的面积因子为 1/625000000。第二幅地图上的面积是第一幅地图面积的 (1/2)² = 1/4。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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