📚 Year 10 Edexcel Maths: Case Study Practical Exercises | Year 10 Edexcel 数学：案例分析实战演练
This article presents a selection of real-world case studies that bring Year 10 Edexcel Maths topics to life. Each scenario is designed to help you apply key skills from number, algebra, geometry, statistics and probability to practical situations, just as you would in an exam problem‑solving question. Work through the exercises step by step to strengthen your understanding and build confidence for your GCSE assessments.
本文精选了一系列真实情境的案例分析，将 Year 10 Edexcel 数学知识点融入实际生活。每个案例旨在帮助你将数、代数、几何、统计和概率的核心技能应用于实际问题，正如你在考试中的应用题中遇到的那样。一步一步完成练习，巩固理解，为 GCSE 评估建立信心。
1. Planning a School Trip | 规划学校旅行
A school is organising a day trip for 180 students. The coach company charges a fixed booking fee of £120 plus £8.50 per seat. Teachers travel free and must accompany students in the ratio 1 teacher : 15 students. The school has a budget of £2000 for transport. Determine whether the budget is sufficient and calculate the total cost per student.
一所学校计划为 180 名学生组织一日游。大巴公司收取 120 英镑的固定预订费，再加每个座位 8.50 英镑。教师免费随行，且师生比例必须为 1 名教师对应 15 名学生。学校交通预算为 2000 英镑。判断预算是否足够，并计算生均总费用。
achers. Total seats needed = 180 students + 12 teachers = 192 seats. Variable cost = 192 × £8.50 = £1632. Total cost = £120 + £1632 = £1752. Since £1752 < £2000, the budget is sufficient. Cost per student = £1752 ÷ 180 = £.73 (to 2 d.p.). Importantly, teachers do not pay, so the student cost remains reasonable.
所需教师人数：180 ÷ 15 = 12 名教师。所需座位总数 = 180 名学生 + 12 名教师 = 192 个座位。可变费用 = 192 × 8.50 英镑 = 1632 英镑。总费用 = 120 + 1632 = 1752 英镑。由于 1752 < 2000，预算充足。生均费用 = 1752 ÷ 180 ≈ 9.73 英镑（保留两位小数）。重要的是教师免费，因而学生人均花费合理。
2. Mobile Phone Tariffs | 手机话费套餐比较
Amira is choosing between two monthly mobile plans. Plan A costs £15 per month and includes 500 minutes, with extra minutes at 10p each. Plan B costs £22 with 1000 inclusive minutes, extra minutes at 8p each. Find the number of minutes for which both plans cost the same, and recommend which plan is cheaper if Amira typically uses 750 minutes per month.
Amira 正在两个月度手机套餐之间选择。套餐 A 每月 15 英镑，含 500 分钟，超出部分每分钟 0.10 英镑。套餐 B 每月 22 英镑，含 1000 分钟，超出部分每分钟 0.08 英镑。求两个套餐费用相同时的通话分钟数，并推荐如果 Amira 通常每月使用 750 分钟，哪个套餐更便宜。
: 15 + 0.10(m − 500) = 22 + 0.08(m − 1000) → 15 + 0.10m − 50 = 22 + 0.08m − 80 → 0.10m − 35 = 0.08m − 58 → 0.02m = −23 → m = −1150, which is impossible. So the equal point must be in the range 500 < m ≤ 1000, where Plan B cost is simply £ 15 + 0.10m − 50 = 22 → 0.10m = 57 → m = 570 minutes. For m = 750, Plan A: 15 + 0.10(250) = £40; Plan B: £22. Plan B is far cheaper. Thus, for 750 minutes, Plan B saves £18 per month.
设 m 为使用总分钟数。当 m > 500 时，套餐 A 费用 = 15 + 0.10(m − 500)。当 m > 1000 时，套餐 B 费用 = 22 + 0.08(m − 1000)。令两者相等求平衡点： 0) → 15 + 0.10m − 50 = 22 + 0.08m − 80 → 0.10m − 35 = 0.08m − 58 → 0.02m = −23 → m = −1150，不可能。因此相等点必在 500 < m ≤ 1000 区间内，此时套餐 B 的费用就是 22 英镑。解 15 + 0.10(m − 500) = 22 → 15 + 0.10m − 50 = 22 → 0.10m = 57 → m = 570 分钟。当 m = 750 时，套餐 A：15 + 0.10(250) = 40 英镑；套餐 B：22 英镑。套餐 B 便宜很多。因此，对于 750 分钟，套餐 B 每月节省 18 英镑。
3. Designing a Garden | 花园设计
A rectangular garden measures 12 m by 8 m. A uniform path of width x metres is built around the inside edge, leaving a central lawn. The lawn must cover at least 60% of the total garden area. Form an inequality in x and solve it to find the maximum possible width of the path, correct to one decimal place.
一个矩形花园尺寸为 12 米乘 8 米。围绕内侧边缘修建一条宽度为 x 米的均匀小路，中间留出草坪。草坪面积至少占花园总面积的 60%。建立关于 x 的不等式并求解，找出小路的可能最大宽度，精确到一位小数。
width = 8 − 2x. Lawn area = (12 − 2x)(8 − 2x) = 96 − 40x + 4x². Condition: lawn area ≥ 60% vide by 4: x² − 10x + 9.6 ≥ 0. Solve x² − 10x + 9.6 = 0 using quadratic formula: x = [10 ± √100 − 38.4] / 2 = [10 ± √61.6] / 2. √61 ide the roots. Since x must be less than 4 (half of 8), the feasible interval is x ≤ 1.075… Thus maximum x ≈ 1.1 m (1 d.p.). The path width cannot exceed 1.1 m.
总面积 = 12 × 8 = 96 m²。草坪长 = 12 − 2x，宽 = 8 − 2x。草坪面积 = (12 − 2x)(8 − 2x) = 96 − 40x + 4x²。条件：草坪面积 ≥ 96 的 60% = 57.6 m²。因此 96 − 40x + 4x² ≥ 57.6 → 4x² − 40x + 38.4 ≥ 0 → 除以 4：x² − 10x + 9.6 ≥ 0。解 x² − 10x + 9.6 = 0，用二次公式：x = [10 ± √100 − 38.4] / 2 = [10 ± √61.6] / 2。√61.6 ≈ 7.85。根为 (10 + 7.85)/2 ≈ 8.93；(10 − 7.85)/2 ≈ 1.075。二次式在根之外 ≥ 0。由于 x 必须小于 4（8 的一半），可行区间为 x ≤ 1.075…。因此最大 x ≈ 1.1 m（一位小数）。小路宽度不能超过 1.1 m。
4. Mixing a Fruit Drink | 混合水果饮料
A factory produces a fruit drink by blending pure orange juice with a sugar syrup. The orange juice contains 12% natural sugar, while the syrup is 80% sugar. The final blend must contain exactly 30% sugar and fill a 500‑litre tank. How many litres of orange juice and syrup are needed? Form and solve simultaneous equations.
一家工厂混合纯橙汁与糖浆来生产水果饮料。橙汁含 12% 天然糖，而糖浆含 80% 糖。最终混合液必须含糖 30%，并装满 500 升的罐。需要多少升橙汁和糖浆？建立并求解联立方程组。
− J. Substitute into second: 0.12J + 0.80(500 − J) = 150 → 0.12J + 400 − 0.80J = 150 → −0.68J = −250 → J = 250 / 0.68 = 367.647… ss = 0.12×367.65 + 0.80×132.35 ≈ 44.118 + 105.88 = 150 kg (since 1 litre ≈ 1 kg). Perfect.
设橙汁为 J 升，糖浆为 S 升。总体积：J + S = 500。糖含量：0.12J + 0.80S = 0.30 × 500 = 150。由第一式得 S = 500 − J。代入第二式： + 0.80(500 − J) = 150 → 0.12J + 400 − 0.80J = 150 → −0.68J = −250 → J = 250 / 0.68 ≈ 367.6 升。则 S = 500 − 367.647 ≈ 132.4 升。验证：糖质量 = 0.12×367.65 + 0.80×132.35 ≈ 44.118 + 105.88 = 150 千克（假设 1 升 ≈ 1 千克）。完美。
5. Commute Time Survey | 通勤时间调查
A group of 40 Year 10 students recorded their one‑way commute to school in minutes. The data are summarised in the table. Calculate an estimate of the mean commute time and the modal class. Then construct a cumulative frequency table and draw a cumulative frequency graph to estimate the median and interquartile range.
40 名 Year 10 学生记录了他们上学单程通勤时间（分钟）。数据总结在下表中。计算平均通勤时间的估计值和众数所在组。然后构建累积频数表，并绘制累积频数图以估算中位数和四分位距。
| Time (t minutes) | Frequency |
|---|---|
| 0 ≤ t < 10 | 8 |
| 10 ≤ t < 20 | 14 |
| 20 ≤ t < 30 | 10 |
| 30 ≤ t < 50 | 5 |
| 50 ≤ t < 80 | 3 |
class: 10 ≤ t < 20 (highest frequency 14). Cumulative frequency: 8, 22, 32, 37, 40. Median at 20th value falls in 10−20 class; from graph estimate median ≈ 18 minutes. Q1 at 10th value ≈ 12 min; Q3 at 30th value ≈ 26 min; IQR = 14 min.
平均值估计：组中值 5, 15, 25, 40, 65。fx 总和 = 8×5 + 14×15 + 10×25 + 5×40 + 3×65 = 40 + 210 + 250 + 200 + 195 = 895。平均值 = 895 ÷ 40 = 22.375 分钟。众数所在组：10 ≤ t < 20（频数最高 14）。累积频数：8, 22, 32, 37, 40。中位数位于第 20 个值，落在 10−20 组；从图上估算中位数约为 18 分钟。第一四分位数（第 10 个值）≈ 12 分钟；第三四分位数（第 30 个值）≈ 26 分钟；IQR = 14 分钟。
6. Saving for a Games Console | 为游戏机储蓄
Leo starts saving money in January with £5 and each month he saves £3 more than the previous month. He wants to buy a console that costs £260. How many months will it take for his total savings to reach or exceed the cost? Model this as an arithmetic series and solve.
Leo 从一月开始存钱，第一个月存 5 英镑，之后每月比前一月多存 3 英镑。他想买一台价值 260 英镑的游戏机。他的总储蓄需要多少个月才能达到或超过该价格？将其建模为等差数列并求解。
n(3n + 7)/2 ≥ 260 → 3n² + 7n − 520 ≥ 0. Solve 3n² + 7n − 520 = 0: discriminant = 49 − 4×3×(−520) = 49 + 6240 = 6289. √6289 ≈ 79.3. n = [−7 ± 79.3] / 6. Positive root: (72.3) / 6 ≈ 12.05. So n ≥ 12.05, thus n = 13 months. Check: S₁₂ = 12/2 (3×12+7) = 6 × 43 = £258; S₁₃ = 13/2 (3×13+7) = 6.5 × 46 = £299. It takes 13 months.
每月储蓄构成等差数列：a = 5, d = 3。n 个月后的总金额：Sₙ = n/2 [2a + (n−1)d] = n/2 [10 + 3(n−1)] = n/2 (3n + 7)。需要 Sₙ ≥ 260。即 n(3n + 7)/2 ≥ 260 → 3n² + 7n − 520 ≥ 0。解 3n² + 7n − 520 = 0：判别式 = 49 − 4×3×(−520) = 49 + 6240 = 6289。√6289 ≈ 79.3。n = [−7 ± 79.3] / 6。正根：(72.3) / 6 ≈ 12.05。故 n ≥ 12.05，因此 n = 13 个月。验证：S₁₂ = 12/2 (3×12+7) = 6 × 43 = 258 英镑；S₁₃ = 13/2 (3×13+7) = 6.5 × 46 = 299 英镑。需要 13 个月。
7. Pizza Value Comparison | 披萨性价比比较
A pizzeria offers a small pizza with diameter 9 inches for £6.50 and a large pizza with diameter 12 inches for £9.80. Which pizza gives more area per pound spent? ss is the same, compare the volume of pizza per pound and decide which is the better deal.
一家披萨店售卖直径 9 英寸的小披萨，售价 6.50 英镑；直径 12 英寸的大披萨，售价 9.80 英镑。哪种披萨每英镑获得更多面积？使用 π = 3.14。另外，如果厚度相同，比较每英镑获得的披萨体积，判断哪个更划算。
Small pizza radius = 4.5 in, area = π × 4.5² = 3.14 × 20.25 = 63.585 in². Area per £ = 63.585 ÷ 6.50 ≈ 9.78 in²/£. Large pizza radius = 6 in, area = 3.14 × 36 = 113.04 in². Area per £ = 113.04 ÷ 9.80 ≈ 11.53 in²/£. Large offers more area per pound. Since thickness constant, volume per £ ratio is same as area per £ ratio. Thus large pizza is about 18% better value.
小披萨半径 = 4.5 英寸，面积 = π × 4.5² = 3.14 × 20.25 = 63.585 in²。每英镑面积 = 63.585 ÷ 6.50 ≈ 9.78 in²/£。大披萨半径 = 6 英寸，面积 = 3.14 × 36 = 113.04 in²。每英镑面积 = 113.04 ÷ 9.80 ≈ 11.53 in²/£。大披萨每英镑提供更多面积。由于厚度恒定，每英镑体积之比与面积之比相同。因此大披萨性价比大约高出 18%。
8. Probability Carnival Game | 概率嘉年华游戏
At a fair, a game costs £1 to play. You roll a fair six‑sided die and win £3 if you roll a 6, win £1 if you roll a 5, and nothing otherwise. Calculate the expected gain per game and explain whether the game is profitable for the stallholder in the long run. If the stallholder wants to keep the prize for rolling a 6 unchanged but adjust the prize for rolling a 5 so that the expected gain is −£0.20 (favouring the stall), what should the new prize be?
在一个集市上，一项游戏收费 1 英镑。你投掷一个公平的六面骰子，如果掷出 6 点赢得 3 英镑，掷出 5 点赢得 1 英镑，其他情况无奖。计算每次游戏的期望收益，并说明从长期来看该游戏对摊主是否有利。如果摊主希望保持掷出 6 点的奖金不变，但调整掷出 5 点的奖金，使得期望收益为 −0.20 英镑（对摊主有利），那么新的奖金应为多少？
Let X be net gain (win − cost). P(6) = 1/6, gain = 3 − 1 = £2. P(5) = 1/6, gain = 1 − 1 = £0. P(other) = 4/6, gain = −£1. Expected gain = (1/6)×2 + (1/6)×0 + (4/6)×(−1) = (2 − 4)/6 = −2/6 ≈ −£0.333. Player loses 33p per game on average, so stallholder profits. For new prize £ = (2 + P − 1 − 4)/6 = (P − 3)/6. Set equal to −0.20: (P − 3)/6 = −0.2 → P − 3 = −1.2 → P = £1.80. So the prize for rolling a 5 should be £1.80.
设 X 为净收益（奖金减成本）。P(6) = 1/6，收益 = 3 − 1 = 2 英镑。P(5) = 1/6，收益 = 1 − 1 = 0 英镑。P(其他) = 4/6，收益 = −1 英镑。期望收益 = (1/6)×2 + (1/6)×0 + (4/6)×(−1) = (2 − 4)/6 = −2/6 ≈ −0.333 英镑。玩家平均每局损失 33 便士，因此摊主盈利。若掷出 5 点的新奖金为 P 英镑：期望收益 = (1/6)×2 + (1/6)×(P − 1) + (4/6)×(−1) = (2 + P − 1 − 4)/6 = (P − 3)/6。令其等于 −0.20：(P − 3)/6 = −0.2 → P − 3 = −1.2 → P = 1.80 英镑。因此掷出 5 点的奖金应为 1.80 英镑。
9. Ramp for Wheelchair Access | 轮椅坡道设计
Building regulations state that a wheelchair ramp must have a maximum gradient of 1 : 12. A straight ramp is needed to rise to a doorway that is 0.75 metres above ground level. Calculate the minimum horizontal length and the actual length of the ramp surface. If the available space only allows a horizontal distance of 5 metres, find the maximum rise possible and the resulting gradient as a ratio.
建筑规范规定轮椅坡道的最大坡度为 1 : 12。需建造一直坡道升至离地 0.75 米的门口。计算最小水平长度和坡道表面的实际长度。如果可用空间只允许 5 米的水平距离，求最大可能的提升高度，并将坡度表示为比例。
horizontal = 1 : 12. For rise 0.75 m, horizontal = 0.75 × 12 = 9 m. Ramp surface length = √0.75² + 9² = √0.5625 + 81 = √81 : 5 = 1 : 12 (since 5/0.4167 ≈ 12), still exactly 1 : 12 if rise is set to exactly 5/12 m.
坡度 1 : 12 表示垂直升高 : 水平距离 = 1 : 12。对于 0.75 m 的升高，水平距离 = 0.75 × 12 = 9 m。坡道表面长度 = √0.75² + 9² = √0.5625 + 81 = √81.5625 ≈ 9.03 m。若水平距离限制为 5 m，最大升高 h = 5 / 12 ≈ 0.4167 m ≈ 0.42 m。坡度 = h : 5 = 0.4167 : 5 = 1 : 12（因为 5/0.4167 ≈ 12），如果升高精确设为 5/12 米，坡度仍恰好为 1 : 12。
10. Box Design from a Net | 从展开图设计盒子
A manufacturer wants to create an open‑ in terms of x. Find the volume when x = 3 cm. Determine the maximum possible value of x for the box to exist and sketch the graph of V against x for feasible x values. Use trial values to estimate the x that gives the maximum volume.
一家制造商想从一张 30 cm × 20 cm 的矩形卡纸的四个角各剪去边长为 x cm 的正方形，然后折起侧边，制成无盖盒子。用 x 写出体积 V 的表达式。求 x = 3 cm 时的体积。确定盒子能够存在的 x 的最大可能值，并绘制在可行的 x 值下 V 关于 x 的草图。使用试值估算使体积最大的 x。
). For x = 3: V = 4×27 − 100×9 + 600×3 = 108 − 900 + 1800 = 1008 cm³. Existence requires 30 − 2x > 0 and 20 − 2x > 0 → x < 10 and x < 15, so max possible x is just under 10 cm. To find max volume, test values: if x = 4, V = 4(64) − 100(16) + 600(4) = 256 − 1600 + 2400 = 1056 cm³. x = 4.5 gives V = 4(91.125) − 100(20.25) + 600(4.5) = 364.5 − 2025 + 2700 = 1039.5. Maximum around x ≈ 4.2 cm.
剪去后，长 = 30 − 2x，宽 = 20 − 2x，高 = x。体积 V = x(30 − 2x)(20 − 2x) = 4x(15 − x)(10 − x) = 4x(150 − 25x + x²) = 4x³ − 100x² + 600x（展开）。当 x = 3 时：V = 4×27 − 100×9 + 600×3 = 108 − 900 + 1800 = 1008 cm³。盒子存在需满足 30 − 2x > 0 且 20 − 2x > 0 → x < 10 且 x < 15，因此最大可能 x 略小于 10 cm。为求最大体积，试值：若 x = 4，V = 4(64) − 100(16) + 600(4) = 256 − 1600 + 2400 = 1056 cm³。x = 4.5 时 V = 4(91.125) − 100(20.25) + 600(4.5) = 364.5 − 2025 + 2700 = 1039.5。最大值在 x ≈ 4.2 cm 附近。
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