📚 Case Study Practice for Year 10 CCEA Maths | Year 10 CCEA 数学：案例分析实战演练
In Year 10 CCEA Mathematics, applying mathematical concepts to real-world scenarios is not only a key assessment objective but also the best way to understand why maths matters. This article guides you through a series of case studies designed to build your problem-solving skills in algebra, geometry, statistics, and more. Each case study mimics the style of CCEA exam questions, encouraging you to reason, calculate, and communicate your thinking clearly.
在 Year 10 CCEA 数学课程中，将数学概念应用于实际问题不仅是评估的核心目标，也是理解数学为何重要的最佳途径。本文通过一系列案例研究，帮助你提升代数、几何、统计等方面的解题能力。每个案例都模拟 CCEA 考试题型，促使你进行推理、计算并清晰表达思路。
1. Planning a School Event Budget | 规划学校活动预算
A school is organising a charity fun day. The organising committee has £500 to spend. They plan to hire a bouncy castle for £80, buy snacks for £2.50 per person, and set aside money for prizes. If they expect 120 people to attend, how much money can be allocated to prizes so the total does not exceed the budget? Represent the spending using an inequality and solve it.
一所学校计划举办慈善欢乐日。组委会拥有 500 英镑预算。他们打算以 80 英镑租用充气城堡，按每人 2.50 英镑购买零食，并预留奖金款项。如果预计有 120 人参加，在不超预算的情况下，最多能为奖金分配多少钱？用不等式表示支出并求解。
Let x be the amount spent on prizes. The total cost is 80 + 2.5 × 120 + x. This must be at most 500:
80 + 300 + x ≤ 500 → 380 + x ≤ 500 → x ≤ 120. So they can spend up to £120 on prizes.
设 x 为奖金支出。总花费为 80 + 2.5 × 120 + x，不得超过 500：
80 + 300 + x ≤ 500 → 380 + x ≤ 500 → x ≤ 120。因此奖金支出最多 120 英镑。
2. Designing a Rectangular Garden | 设计矩形花园
A gardener wants to enclose a rectangular vegetable patch using 30 metres of fencing. One side is against a wall, so no fencing is needed there. Find the maximum area that can be enclosed and the dimensions that achieve it. This is a classic optimisation problem leading to a quadratic expression for area.
一位园丁想用 30 米长的围栏围起一个矩形菜地，其中一侧靠墙，无需围栏。求出能围成的最大面积以及此时的长宽尺寸。这是一个经典的优化问题，会引出面积的二次表达式。
Let the side parallel to the wall be y metres, and each perpendicular side be x metres. Then 2x + y = 30 → y = 30 − 2x. The area A = xy = x(30 − 2x) = 30x − 2x². This is a quadratic with a maximum at x = −30/(2×(−2)) = 7.5 m. Then y = 30 − 15 = 15 m. Maximum area = 7.5 × 15 = 112.5 m².
设平行于墙的边长为 y 米，两条垂直边长为 x 米。则 2x + y = 30 → y = 30 − 2x。面积 A = xy = x(30 − 2x) = 30x − 2x²。这是开口向下的二次函数，最大值在 x = −30/(2×(−2)) = 7.5 m 处。此时 y = 15 m，最大面积为 112.5 m²。
3. Data-Driven Survey on Mobile Phone Usage | 数据驱动的手机使用调查
A group of Year 10 students surveyed how many hours per day their classmates spend on social media. The results from 20 respondents were: 2, 3, 1, 4, 2, 3, 5, 2, 1, 0, 4, 3, 2, 2, 6, 1, 3, 4, 2, 3. Construct a frequency table, calculate the mean, median, and mode, and decide which average best represents the data. Draw a bar chart to visualise the distribution.
一群 Year 10 学生调查了同学们每天花在社交媒体上的小时数。20 名受访者的结果如下：2, 3, 1, 4, 2, 3, 5, 2, 1, 0, 4, 3, 2, 2, 6, 1, 3, 4, 2, 3。构建频率表，计算平均数、中位数和众数，并判断哪个平均值最能代表数据。绘制条形图展示分布。
First, organise the data into a frequency table:
首先将数据整理成频率表：
| Hours (h) | Frequency |
|---|---|
| 0 | 1 |
| 1 | 3 |
| 2 | 6 |
| 3 | 5 |
| 4 | 3 |
| 5 | 1 |
| 6 | 1 |
Mean = (0×1 + 1×3 + 2×6 + 3×5 + 4×3 + 5×1 + 6×1) ÷ 20 = (0+3+12+15+12+5+6)/20 = 53/20 = 2.65 hours. Median is the average of the 10th and 11th values in order: both are 2 and 3? Let’s order the data: ten values are 0|1,1,1|2,2,2,2,2,2| The 10th and 11th values are 2 and 3, so median = 2.5. Mode = 2 (most frequent). The median and mode are slightly lower than the mean due to a few high values; median may be the most representative.
平均数 = (0×1 + 1×3 + 2×6 + 3×5 + 4×3 + 5×1 + 6×1) ÷ 20 = (0+3+12+15+12+5+6)/20 = 53/20 = 2.65 小时。中位数：排序后第 10 和 11 个值为 2 和 3，中位数 = 2.5。众数 = 2。由于少数高值拉高了平均数，中位数可能最具代表性。
4. Travel Planning: Speed, Distance, and Time | 旅行规划：速度、距离与时间
A family drives from Belfast to Dublin, a distance of 165 km. They travel at an average speed of 60 km/h on the motorway for the first 120 km, but then roadworks reduce their speed to 40 km/h for the remaining distance. Calculate the total journey time. If they need to arrive by 14:00, what is the latest departure time?
一个家庭从贝尔法斯特驾车前往都柏林，全程 165 公里。他们在高速公路上以平均 60 km/h 行驶了前 120 公里，但因道路施工，剩余路段速度降至 40 km/h。计算总行程时间。若需在 14:00 到达，最晚何时出发？
Time for first part: 120 ÷ 60 = 2 hours. Remaining distance: 165 − 120 = 45 km; time = 45 ÷ 40 = 1.125 hours = 1 hour 7.5 minutes (1 h 7 min 30 s). Total time = 3 h 7.5 min. To arrive by 14:00, subtract 3 h 7.5 min → departure by 10:52:30, so practically by 10:52. Use formula time = distance ÷ speed and ensure units are consistent.
第一部分时间：120 ÷ 60 = 2 小时。剩余距离：165 − 120 = 45 km；时间 = 45 ÷ 40 = 1.125 小时 = 1 小时 7.5 分钟。总时间 = 3 小时 7.5 分钟。要在 14:00 前到达，需在 10:52:30 前出发，实际可定为 10:52。使用公式 时间 = 距离 ÷ 速度，注意单位统一。
5. Comparing Mobile Phone Plans | 比较手机套餐
A store offers two monthly plans. Plan A: £10 fixed charge plus £0.05 per minute of calls. Plan B: £15 fixed charge plus £0.03 per minute. Determine for how many minutes the two plans cost the same. Represent both plans as linear equations and graph them to visually confirm the break-even point. Recommend which plan is cheaper for a customer who talks 400 minutes per month.
一家商店提供两种月套餐。A 套餐：固定 10 英镑，通话每分钟 0.05 英镑。B 套餐：固定 15 英镑，通话每分钟 0.03 英镑。求出使用多少分钟时两种套餐费用相同。将两个套餐表示为线性方程并绘制图形，直观确认盈亏平衡点。建议每月通话 400 分钟的客户选择哪个套餐更省钱。
Let m = minutes. Cost A = 10 + 0.05m; Cost B = 15 + 0.03m. Set equal: 10 + 0.05m = 15 + 0.03m → 0.02m = 5 → m = 250 minutes. For m = 400, A = 10 + 20 = £30; B = 15 + 12 = £27. Plan B is cheaper. Graph both lines on the same axes; the intersection at (250, 22.5) shows the break-even point.
设 m 为分钟数。A 费用 = 10 + 0.05m；B 费用 = 15 + 0.03m。令其相等：10 + 0.05m = 15 + 0.03m → 0.02m = 5 → m = 250 分钟。当 m = 400 时，A = 10 + 20 = 30 英镑；B = 15 + 12 = 27 英镑，B 更便宜。将两条线画在同一坐标轴上，交点 (250, 22.5) 即为盈亏平衡点。
6. Measuring the Height of a Flagpole | 测量旗杆高度
During a school field trip, students need to estimate the height of a flagpole using a clinometer and trigonometry. They stand 20 metres away from the pole and measure the angle of elevation to the top as 38°. Calculate the height of the flagpole, assuming the clinometer is held at a height of 1.5 m above the ground. Use right-angled triangle trigonometry.
在学校实地考察中，学生需要使用测斜仪和三角学估算旗杆的高度。他们站在离旗杆 20 米远的地方，测得顶部仰角为 38°。假设测斜仪离地高度为 1.5 米，计算旗杆高度。使用直角三角形三角比。
In the right triangle, tan(38°) = opposite/adjacent = (h − 1.5) / 20, where h is the flagpole’s total height. So h − 1.5 = 20 × tan(38°). Using tan(38°) ≈ 0.7813 (from calculator), we get h − 1.5 ≈ 15.626 → h ≈ 17.13 m (to 2 d.p.). Always consider the instrument height.
在直角三角形中，tan(38°) = 对边/邻边 = (h − 1.5) / 20，其中 h 为旗杆总高度。因此 h − 1.5 = 20 × tan(38°)。用计算器得 tan(38°) ≈ 0.7813，则 h − 1.5 ≈ 15.626 → h ≈ 17.13 米（保留两位小数）。务必考虑仪器高度。
7. Personal Finance: Savings and Compound Interest | 个人理财：储蓄与复利
Ellie deposits £800 into a savings account that earns 2.5% compound interest per annum, paid yearly. She makes no additional deposits or withdrawals. Calculate the balance after 3 years. Then compare this with simple interest at the same rate over the same period. Explain why the difference grows over time.
艾莉将 800 英镑存入年利率为 2.5% 的复利储蓄账户，按年计息。她不进行额外的存取款。计算 3 年后的余额。然后与同等期限、同等利率的单利进行比较。解释为何差异随时间增大。
Compound amount after 3 years: 800 × (1 + 0.025)³ = 800 × (1.025)³ ≈ 800 × 1.07689 = £861.51. Simple interest: 800 + 800 × 0.025 × 3 = 800 + 60 = £860. The difference is £1.51 after 3 years. Over longer periods, compound interest earns ‘interest on interest’, so the gap widens significantly.
3 年后的复利本息和：800 × (1 + 0.025)³ = 800 × (1.025)³ ≈ 800 × 1.07689 = 861.51 英镑。单利：800 + 800 × 0.025 × 3 = 800 + 60 = 860 英镑。3 年后差额 1.51 英镑。期限更长时，复利会产生“利滚利”效应，差距会显著扩大。
8. Probability in a Raffle Draw | 抽奖活动中的概率
At the school fete, a raffle drum contains 200 tickets: 5 win a large prize, 20 win a small prize, and the rest win nothing. If you buy one ticket, what is the probability you win a prize? If you buy two tickets, what is the probability that at least one wins a prize? Assume tickets are not replaced.
在学校游园会上，抽奖箱中有 200 张票：5 张中大奖，20 张中小奖，其余无奖。若你购买一张票，中奖概率是多少？若购买两张票，至少有一张中奖的概率是多少？假设抽票不放回。
P(win with one ticket) = (5+20)/200 = 25/200 = 1/8 = 0.125. For two tickets, P(at least one win) = 1 − P(both lose). P(first loses) = 175/200; P(second loses given first loses) = 174/199. So P(both lose) = (175/200) × (174/199) = (30450)/(39800) ≈ 0.765. Thus P(at least one win) = 1 − 0.765 = 0.235 (23.5%).
一张票中奖概率 = (5+20)/200 = 25/200 = 1/8 = 0.125。两张票的情况：P(至少一张中奖) = 1 − P(两张都无奖)。第一张无奖概率 = 175/200；第二张在首张无奖条件下无奖概率 = 174/199。均无奖 = (175/200) × (174/199) ≈ 0.765。因此至少一中奖概率 = 1 − 0.765 = 0.235 (23.5%)。
9. Transformations and Symmetry in Logo Design | 标志设计中的变换与对称
A graphic designer starts with a basic right-angled triangle with vertices A(1,1), B(5,1), C(1,4). They reflect the triangle in the y‑axis, then rotate the image 90° clockwise about the origin. Determine the final coordinates of the triangle’s vertices. Describe any symmetries in the resulting shape if they also draw the original and the transformed figures on the same grid.
一位平面设计师从一个直角三角形开始，顶点为 A(1,1)、B(5,1)、C(1,4)。他将该三角形关于 y 轴反射，再将图形绕原点顺时针旋转 90°。求最终三角形的顶点坐标。若他将原图和变换后的图形画在同一网格上，描述合成形状的对称性。
Reflection in y‑axis maps (x,y) → (−x,y). So A(1,1) → A'(−1,1); B(5,1) → B'(−5,1); C(1,4) → C'(−1,4). Then rotate 90° clockwise: (x,y) → (y, −x). Apply to A’: (−1,1) → (1,1). B’: (−5,1) → (1,5). C’: (−1,4) → (4,1). Final vertices: (1,1), (1,5), (4,1). The combined picture may show rotational symmetry of order 2 if the original is also drawn.
关于 y 轴反射：(x,y) → (−x,y)。因此 A(1,1) → A'(−1,1)；B(5,1) → B'(−5,1)；C(1,4) → C'(−1,4)。然后顺时针旋转 90°：(x,y) → (y, −x)。应用于 A’：(−1,1) → (1,1)。B’：(−5,1) → (1,5)。C’：(−1,4) → (4,1)。最终坐标为 (1,1)、(1,5)、(4,1)。若同时绘制原图，整体图形可能具有 2 阶旋转对称。
10. Mixed Problem: Planning a House Extension | 混合问题：规划房屋扩建
A homeowner plans to build a rectangular extension 4.2 m by 6.5 m. The local council requires a scale drawing on A4 paper (297 mm × 210 mm) with a scale chosen so the extension fits neatly with at least 2 cm margins on all sides. Determine a suitable scale (e.g., 1:50, 1:100). Then calculate the area of the extension in m² and the cost of tiling the floor if tiles cost £24.50 per m² and 10% extra is allowed for wastage.
一名房主计划建造 4.2 米 × 6.5 米的矩形扩建。当地议会要求在一张 A4 纸（297 毫米 × 210 毫米）上绘制比例图，所选比例应使扩建图充分容纳，且四周至少留有 2 厘米空白。确定合适的比例（如 1:50、1:100）。然后计算扩建面积（平方米）以及铺设地砖的费用，地砖价格为每平方米 24.50 英镑，另加 10% 的损耗余量。
Drawing area available: 297−40 = 257 mm, 210−40 = 170 mm. Convert extension dimensions to mm: 4200 mm, 6500 mm. Try scale 1:50 → 84 mm by 130 mm – fits easily. Scale 1:20 → 210 mm by 325 mm – too large for width. So 1:50 is suitable. Area = 4.2 × 6.5 = 27.3 m². Tiles needed: 27.3 × 1.1 = 30.03 m², cost = 30.03 × 24.50 = £735.735 → £735.74 (rounded).
可用绘图区域：297−40 = 257 mm，210−40 = 170 mm。将扩建尺寸转换为毫米：4200 mm，6500 mm。尝试 1:50 → 84 mm × 130 mm——轻松容下。1:20 → 210 mm × 325 mm——宽度超出。故 1:50 合适。面积 = 4.2 × 6.5 = 27.3 m²。所需瓷砖面积：27.3 × 1.1 = 30.03 m²，费用 = 30.03 × 24.50 = 735.735 英镑，约为 735.74 英镑。
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