📚 Year 10 CCEA Further Mathematics: Case Study Practical Exercises | Year 10 CCEA 进阶数学：案例分析实战演练
In Year 10 CCEA Further Mathematics, students move beyond routine calculations to apply concepts in meaningful contexts. This article presents a series of case studies designed to sharpen problem-solving skills and deepen understanding of advanced topics such as quadratics, trigonometry, vectors, probability, financial mathematics, linear programming, sequences, coordinate geometry, statistics, algebraic fractions, matrices and functions. Each section pairs a realistic scenario with step-by-step reasoning, helping you see how mathematics models and solves real-world challenges.
在 Year 10 CCEA 进阶数学课程中，学生不再只是进行常规计算，而是要在有意义的情境中应用概念。本文通过一系列案例分析，旨在锻炼解决问题的能力，加深对二次方程、三角学、向量、概率、金融数学、线性规划、数列、坐标几何、统计学、代数分式、矩阵及函数等进阶主题的理解。每个部分都将真实场景与逐步推理相结合，帮助你体会数学如何建模并解决现实难题。
1. Solving Quadratic Equations in Real Life | 现实生活中的二次方程求解
ses produced in hundreds. To break even, the business needs to find the production levels where P = 0, meaning they earn exactly enough to cover costs.
一家小企业生产定制手机壳。其月利润 P（单位：英镑）可用模型 P = -2x² + 120x - 800 表示，其中 x 是以百为单位的生产数量。为了达到盈亏平衡，企业需要找到使 P = 0 的生产水平，即收入刚好覆盖成本。
dividing through by -2, giving x² - 60x + 400 = 0. Using the quadratic formula x = [ -b ± √b² - 4ac pute the discriminant: b² - 4ac = 3600 - 1600 = 2000. Thus x = [60 ± √2000] / 2 = [60 ± 20√5] / 2 = 30 ± 10√5. Approximating √5 52.36. So the company breaks even when producing about 764 cases or 5236 cases per month.
令二次方程 -2x² + 120x - 800 = 0，除以 -2 化简得 x² - 60x + 400 = 0。应用求根公式 x = [ -b ± √b² - 4ac ] / 2a，其中 a = 1, b = -60, c = 400，计算判别式：b² - 4ac = 3600 - 1600 = 2000。因此 x = [60 ± √2000] / 2 = [60 ± 20√5] / 2 = 30 ± 10√5。近似计算 √5 ≈ 2.236，两解为 x ≈ 7.64 和 x ≈ 52.36。所以企业每月生产约764个或5236个手机壳时可达到盈亏平衡。
0) - 800 = -1800 + 3600 - 800 = 1000 pounds. The business should aim to produce 3000 cases for the highest profit.
利润最大值出现在顶点处。由于二次函数开口向下，最大值在 x = -b/(2a) = 60/2 = 30 处，利润为 P = -2(900) + 120(30) - 800 = 1000 英镑。企业应以生产3000个手机壳为目标，以获得最高利润。
2. Using Trigonometry in Navigation | 三角函数在航海中的应用
A ship sails from a port A on a bearing of 065° for 12 km to point B, then changes course to a bearing of 140° and sails 8 km to point C. We want to find the direct distance and bearing from A to C.
一艘船从港口 A 沿 065° 方位角航行 12 km 到达 B 点，然后转为 140° 方位角航行 8 km 到达 C 点。需求出从 A 到 C 的直线距离及方位角。
or angle at B in triangle ABC is the difference between the two bearings, but we must consider the direction change. The angle between the paths is 140° - 65° = 75°. However, the interior angle of triangle at B is the supplement: the path from A to B goes north-east, and BC goes south-east, so the angle ABC is 180° - 75° = 105°. Then by the cosine rule, AC² = 12² + 8² - 2×12×8×cos 105°. cos 105° = -cos 75° ≈ -0.2588, so AC² = 144 + 64 - 192×(-0.2588) = 208 + 49.69 ≈ 257.69, thus AC ≈ √257.69 ≈ 16.05 km.
首先，将方位角转换为相对于正北的角度。AB 段与正北夹角 65°，BC 段为 140°，三角形 ABC 中在 B 点的内角为两方位角之差，但需考虑方向变化。两段路径夹角为 140° - 65° = 75°。然而，三角形在 B 点的内角是其补角：从 A 到 B 为东北方向，BC 为东南方向，故角 ABC = 180° - 75° = 105°。然后根据余弦定理，AC² = 12² + 8² - 2×12×8×cos 105°。cos 105° = -cos 75° ≈ -0.2588，故 AC² = 144 + 64 - 192×(-0.2588) = 208 + 49.69 ≈ 257.69，因此 AC ≈ √257.69 ≈ 16.05 km。
To find the bearing of C from A, use the sine rule: sin(∠BAC) / 8 = sin 105° / 16.05. sin 105° ≈ 0.9659, so sin(∠BAC) ≈ (8 × 0.9659) / 16.05 ≈ 0.4815, giving ∠BAC ≈ 28.8°. The bearing is then 65° + 28.8° = 93.8°, approximately 094°.
为求从 A 到 C 的方位角，使用正弦定理：sin(∠BAC) / 8 = sin 105° / 16.05。sin 105° ≈ 0.9659，故 sin(∠BAC) ≈ (8 × 0.9659) / 16.05 ≈ 0.4815，得 ∠BAC ≈ 28.8°。则方位角为 65° + 28.8° = 93.8°，约为 094°。
3. Vectors and Forces in Equilibrium | 向量与力的平衡
A lighting fixture is suspended by two cables attached to the ceiling at points 2 m apart. The cables are of length 1.5 m and 2 m, and the fixture weighs 50 N. We can use vectors to find the tension in each cable.
一个灯具由两根分别长1.5 m和2 m的缆绳悬挂在天花板上，两悬挂点相距2 m，灯具重50 N。我们可以用向量求出每根缆绳受到的拉力。
g vector geometry. Let the fixture coordinates be (x, y). Using distances: (x)² + (y-2)² = 1.5² and (x-2)² + (y-2)² = 2². Subtracting the first from the second: (x-2)² - x² = 4 - 2.25 => (x² -4x +4) - x² = 1.75 => -4x +4 = 1.75 => -4x = -2.25 => x = 0.5625. Then (0.5625)² + (y-2)² = 2.25 => (y-2)² = 2.25 - 0.3164 ≈ 1.9336 => y-2 ≈ -1.3905 (negative as below ceiling), so y ≈ 0.6095.
建立坐标系，设天花板两点为 (0, 2) 和 (2, 2)（米）。灯具悬挂在两圆的交点，可用向量几何求解。设灯具坐标为 (x, y)。由距离关系得：(x)² + (y-2)² = 1.5² 及 (x-2)² + (y-2)² = 2²。两式相减：(x-2)² - x² = 4 - 2.25 ⇒ -4x +4 = 1.75 ⇒ x = 0.5625。代入得 (y-2)² = 2.25 - 0.3164 ≈ 1.9336 ⇒ y-2 ≈ -1.3905（因为灯具在天花板下方，取负），故 y ≈ 0.6095。
95) = (1.4375, 1.3905), length 2. The tensions T₁ and T₂ act along these vectors. Their vertical components sum to 50 N. Unit vectors are u = a/1.5 ≈ (-0.375, 0.927) and v = b/2 ≈ (0.71875, 0.69525). Vertical equilibrium: 0.927 T₁ + 0.69525 T₂ = 50. Horizontal equilibrium: -0.375 T₁ + 0.71875 T₂ = 0. From the second, T₁ = (0.71875/0.375) T₂ ≈ 1.9167 T₂. Substitute into vertical: 0.927(1.9167 T₂) + 0.69525 T₂ ≈ 1.777 T₂ + 0.69525 T₂ = 2.47225 T₂ = 50, so T₂ ≈ 20.22 N, T₁ ≈ 38.8 N.
从灯具出发的两根缆绳方向向量：a = (-0.5625, 1.3905)，长度1.5；b = (1.4375, 1.3905)，长度2。拉力 T₁ 和 T₂ 沿这些向量作用。单位向量为 u = a/1.5 ≈ (-0.375, 0.927)，v = b/2 ≈ (0.71875, 0.69525)。垂直平衡：0.927 T₁ + 0.69525 T₂ = 50。水平平衡：-0.375 T₁ + 0.71875 T₂ = 0。由水平式得 T₁ = 1.9167 T₂。代入垂直式得 2.47225 T₂ = 50，故 T₂ ≈ 20.22 N，T₁ ≈ 38.8 N。
4. Probability and Game Strategies | 概率与游戏策略
In a game, a player draws two cards from a standard 52-card deck without replacement. Winning requires drawing at least one ace. Calculate the probability of winning and evaluate whether it is worth paying £1 to play if the prize is £10.
在一个游戏中，玩家从一副标准的52张扑克牌中不放回地抽取两张牌。若至少抽到一张 A 则获胜。计算获胜概率，并评估如果支付1英镑参与、奖金为10英镑是否划算。
P(no aces) = 1128/1326 = 0.8507, so P(at least one ace) = 1 - 0.8507 = 0.1493 (about 14.93%). Expected value: (0.1493 × £10) - £1 = £1.493 - £1 = £0.493. Since the expected gain is positive, it is statistically favourable to play.
从52张中选2张的总方法数为 C(52,2) = 1326。无 A 的方法数：48张非 A 牌中选2张，C(48,2) = 1128。故 P(无 A) = 1128/1326 = 0.8507，因此 P(至少一张 A) = 1 - 0.8507 = 0.1493（约14.93%）。期望值：(0.1493 × £10) - £1 = £1.493 - £1 = £0.493。由于期望收益为正，从统计角度看参与是划算的。
If the prize is changed to £6, expected value drops to (0.1493 × £6) - £1 = £0.8958 - £1 = -£0.1042, making it unfavourable. This shows how probability informs decision-making.
若奖金改为6英镑，期望值降至 (0.1493 × £6) - £1 = -£0.1042，为负值，变得不划算。这体现了概率如何辅助决策。
5. Financial Mathematics: Compound Interest and Loans | 金融数学：复利与贷款
Amelia invests £5000 at an annual interest rate of 4.5% compounded monthly. Find the value of the investment after 8 years. Later, she plans to borrow £10000 for a car, repayable over 3 years at an APR of 6.2% with monthly payments. Calculate the monthly instalment.
Amelia 将5000英镑以年利率4.5%、每月复利的方式投资。求8年后的投资价值。之后，她打算贷款10000英镑买车，分3年偿还，年利率6.2%，按月付款。计算每月应还款额。
Compound interest formula: A = P(1 + r/n)nt. Here P = 5000, r = 0.045, n = 12, t = 8. So A = 5000(1 + 0.045/12)12×8 = 5000(1.00375)96. (1.00375)96 ≈ 1.432, so A ≈ 5000 × 1.432 = £7160. The investment grows to about £7160.
复利公式：A = P(1 + r/n)nt。此处 P = 5000，r = 0.045，n = 12，t = 8。因此 A = 5000(1 + 0.045/12)96 = 5000(1.00375)96。(1.00375)96 ≈ 1.432，故 A ≈ £7160。投资最终约值7160英镑。
For the loan, the present value of annuity formula: P = PMT × [1 - (1 + i)-N] / i, where i = 0.062/12 ≈ 0.0051667, N = 36. Rearranging: 10000 = PMT × [1 - (1.0051667)-36] / 0.0051667. (1.0051667)-36 0.8310, so 1 - 0.8310 = 0.1690. Then PMT = 10000 × 0.0051667 / 0.1690 ≈ 51.667 / 0.1690 ≈ £305.70. Monthly payment is about £305.70.
对于贷款，使用年金现值公式：P = PMT × [1 - (1 + i)-N] / i，其中 i = 0.062/12 ≈ 0.0051667，N = 36。移项得 10000 = PMT × [1 - (1.0051667)-36] / 0.0051667。(1.0051667)-36 ≈ 0.8310，故 1 - 0.8310 = 0.1690。则 PMT = 10000 × 0.0051667 / 0.1690 ≈ £305.70。每月还款约305.70英镑。
6. Linear Programming and Resource Optimisation | 线性规划与资源优化
A workshop produces two types of handcrafted chairs: Classic and Deluxe. Each Classic needs 3 hours of carpentry and 1 hour of finishing, while each Deluxe needs 2 hours of carpentry and 2 hours of finishing. Available per week: 90 carpentry hours and 40 finishing hours. Profit per Classic is £80, per Deluxe £100. Determine the optimal production mix to maximise profit.
一家作坊生产两种手工椅：经典型和豪华型。每把经典款需要3小时木工和1小时表面处理，豪华款需要2小时木工和2小时表面处理。每周可用工时为：木工90小时，表面处理40小时。经典款利润80英镑，豪华款100英镑。确定最大化利润的最佳生产组合。
region: intersections of boundaries. Solving 3x+2y=90 and x+2y=40: subtract to get 2x=50 => x=25, then y=(40-25)/2=7.5. Vertices: (0,0), (0,20) [since x+2y=40 => y=20], (30,0) [3x=90], and (25,7.5). Evaluate P: at (0,20): 80×0+100×20=2000; at (30,0): 80×30=2400; at (25,7.5): 80×25+100×7.5=2000+750=2750. So maximum profit £ : 3×25+16=91>90, not allowed; (24,8): 72+16=88, 24+16=40 ok, profit 1920+800=2720). Best integer solution (24,8) gives £2720. Often integer programming needed, but the vertex gives near-optimal guidance.
设 x 为经典款数量，y 为豪华款数量。约束条件：3x + 2y ≤ 90（木工），x + 2y ≤ 40（表面处理），x ≥ 0，y ≥ 0。利润 P = 80x + 100y。画出可行域：边界交点。解 3x+2y=90 与 x+2y=40 得 x=25，y=7.5。顶点：(0,0), (0,20), (30,0), (25,7.5)。计算 P：(0,20)=2000，(30,0)=2400，(25,7.5)=2750。最大利润2750出现在经典25、豪华7.5，整数解测试 (25,7): 利润2700; (24,8): 利润2720。整数最优为24经典、8豪华，利润2720英镑。
7. Sequences and Series in Design | 数列与设计中的应用
An architect designs a spiral staircase where the step widths form an arithmetic sequence. The first step width is 1.0 m, and each subsequent step is 5 cm wider. There are 20 steps. Find the total length of the handrail needed along the inner edge, and the width of the widest step.
一位建筑师设计的螺旋楼梯，其踏板宽度构成等差数列。第一级宽1.0 m，之后每级增加5 cm。共有20级。求沿内侧所需扶手的总长度，以及最宽一级的宽度。
p inner edges (assuming handrail follows exactly) is the sum S₂₀ = n/2 (a + u₂₀) = 20/2 × (1.0 + 1.95) = 10 × 2.95 = 29.5 m. The handrail length would therefore be 29.5 metres.
等差数列：a = 1.0，d = 0.05，n = 20。第20项（最宽踏步）为 u₂₀ = 1.0 + 19×0.05 = 1.95 m。所有踏步内侧总长度（扶手跟随）为和 S₂₀ = 20/2 × (1.0 + 1.95) = 10 × 2.95 = 29.5 m。因此扶手长度需29.5米。
ratio such that the 20th step is 2.0 m wide, we can solve for r: 1.0 × r¹⁹ = 2.0, so r = 21/19 but the growth is slower; arithmetic gives a predictable linear increase, which is easier for construction.
若设计师希望采用等比数列，首项1.0 m，第20项为2.0 m，可解公比：r¹⁹ = 2.0，r = 21/19 ≈ 1.037。其总和为 a(1 - r²⁰)/(1 - r)，但线性增长的等差数列在施工中更容易控制。
8. Coordinate Geometry and Circle Theorems | 坐标几何与圆的性质
termine whether the path is tangent to the pond, and if not, find the points of intersection.
一个圆形池塘在坐标网格上圆心为 (5, 8)，半径为10米（1单位=1 m）。一条笔直的小路沿直线 y = 2x - 4 修建。判断小路是否与池塘相切，若不相切，求出交点。
=> 5x² - 58x + 169 = 100 => 5x² - 58x + 69 = 0. Discriminant D = (-58)² - 4×5×69 = 3364 - 1380 = 1984. Since D > 0, the line intersects the circle at two points. Solve: x = [58 ± √1984] / 10 = [58 ± √64×31] / 10 = [58 ± 8√31] / 10 = 5.8 ± 0.8√31. √31 308. Intersection points: (10.25, 16.51) and (1.35, -1.31). The path cuts across the pond.
圆的方程：(x - 5)² + (y - 8)² = 100。代入 y = 2x - 4 得 5x² - 58x + 69 = 0。判别式 D = 1984 > 0，直线与圆相交于两点。x = [58 ± √1984] / 10 ≈ 10.254 和 1.346，对应 y 约为 16.508 和 -1.308。因此小路穿过池塘，并非相切。
Distance from centre to line: distance d = |Ax₀ + By₀ + C| / √A² + B² with line 2x - y - 4 = 0. So d = |2×5 - 8 - 4| / √4 + 1 = |10 - 12| / √5 = 2 / √5 4 m. Since d < 10, the line indeed passes through the circle, confirming the intersection.
圆心到直线的距离：d = |2×5 - 8 - 4| / √5 = 2/√5 ≈ 0.894 m，小于半径10，证实相交。
9. Statistics: Interpreting Data and Drawing Conclusions | 统计学：数据解读与结论
A fitness tracker recorded the daily step counts for a student over 30 days. The data has a mean of 8500 steps and a standard deviation of 1200 steps. Assuming a normal distribution, estimate the proportion of days the student exceeded 10000 steps, and find the step count below which the lowest 10% of days fall.
一个健身追踪器记录了一位学生30天的每日步数。数据均值为8500步，标准差为1200步。假设服从正态分布，估计步数超过10000步的天数比例，并求出步数最低的10%的天数对应的步数阈值。
0 = 1500/1200 = 1.25. Using the standard normal table, P(Z > 1.25) ≈ 0.1056, so about 10.6% of days exceed 10000 steps. For the lowest 10% mately -1.2816. So X = μ + zσ = 8500 - 1.2816×1200 ≈ 8500 - 1537.9 = 6962.1 steps. Thus 10% of days are below about 6960 steps.
标准化：z = (X - μ)/σ。X=10000时，z=1.25，P(Z > 1.25) ≈ 0.1056，约10.6%的天数超过10000步。最低10%的z值约为-1.2816，X ≈ 8500 - 1538 = 6962步。即10%的日子步数低于约6960步。
If the student wants to be in the top 5% 00 + 1974 = 10474 steps. These statistical insights can help set realistic activity goals.
若该学生希望进入前5%，z₀.₀₅ = 1.6449，X ≈ 10474步。这些统计推断有助于设定切实可行的运动目标。
10. Algebraic Fractions and Rational Equations | 代数分式与有理方程
₂. A technician needs a combined resistance of 6 ohms. One resistor is 4 ohms more than the other. Find the two resistor values.
在电路中，两个电阻 R₁ 和 R₂ 并联的总电阻 R 满足 1/R = 1/R₁ + 1/R₂。技术员需要总电阻为6欧姆，且其中一个电阻比另一个大4欧姆。求两个电阻值。
) / [x(x+4)] = (2x+4) / [x(x+4)]. So 1/6 = (2x+4) / (x²+4x). Cross-multiply: x² + 4x = 6(2x+4) = 12x + 24. Rearrange: x² + 4x - 12x - 24 = 0 => x² - 8x - 24 = 0. Solve using quadratic formula: x = [8 ± √64 + 96] / 2 = [8 ± √160] / 2 = [8 ± 4√10] / 2 = 4 ± 2√10. √10 n: x ≈ 10.324 ohms, then x+4 ≈ 14.324 ohms. The two resistors are about 10.3 Ω and 14.3 Ω.
设较小电阻为 x 欧姆，则较大电阻为 x+4。方程：1/6 = 1/x + 1/(x+4)。右边通分得 (2x+4)/(x²+4x)。交叉相乘得 x² + 4x = 12x + 24，
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