Year 12 CCEA Mathematics: Interdisciplinary Problem-Solving Practice

📚 Year 12 CCEA Mathematics: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练

This article presents a collection of interdisciplinary problems designed for Year 12 CCEA Mathematics. Each section integrates core mathematical techniques – from mechanics and calculus to statistics and trigonometry – with real-world contexts such as finance, biology, engineering and the physical sciences. Practising these questions will strengthen your ability to transfer mathematical skills across subjects, a key skill assessed in applied modules like M1 and S1 and in pure topics where modelling is required.

本文为 CCEA 数学 Year 12 学生汇编了一组跨学科综合题。每个小节将力学、微积分、统计和三角学等核心数学技巧与金融、生物、工程和物理科学等真实情境相结合。练习这些题目将强化你跨学科迁移数学技能的能力，这一能力在 M1、S1 等应用模块以及需要建立模型的纯数学专题中都非常关键。

1. Motion and Kinematics | 运动与运动学

A particle is projected vertically upwards from ground level with an initial speed of 14 m s⁻¹. The acceleration due to gravity is 9.8 m s⁻², and air resistance is negligible. By modelling the motion using the SUVAT equations, find the maximum height reached and the time taken to return to the ground.

一质点从地面以 14 m s⁻¹ 的初速度竖直向上抛出。重力加速度为 9.8 m s⁻²，空气阻力忽略不计。利用匀变速直线运动公式建立模型，求能达到的最大高度以及返回地面所需的时间。

At maximum height, the velocity becomes zero. Using v² = u² + 2as with u = 14, v = 0, a = −9.8, we obtain 0 = 196 − 19.6s, so s = 10 m. To find the total time in the air, use s = ut + ½at² for the entire journey from ground to ground, with s = 0: 0 = 14t − 4.9t². Factoring gives t(14 − 4.9t) = 0, hence t = 0 or t = 14/4.9 ≈ 2.857 s. The total time is approximately 2.86 s.

在最大高度处，速度为零。利用 v² = u² + 2as，代入 u = 14，v = 0，a = −9.8，得到 0 = 196 − 19.6s，因此 s = 10 m。求空中总时间时，对从地面到地面的整个过程使用 s = ut + ½at²，且 s = 0：0 = 14t − 4.9t²。因式分解得 t(14 − 4.9t) = 0，故 t = 0 或 t = 14/4.9 ≈ 2.857 s。总时间约为 2.86 秒。

s = ut + ½at² v² = u² + 2as

2. Financial Sequences and Series | 金融数列与级数

An investment of £2000 earns compound interest at an annual rate of 3.6%. Interest is added at the end of each year. Calculate the value of the investment after 8 years, and determine the number of whole years required for the investment to double. This scenario illustrates a geometric progression with common ratio 1.036.

一笔 2000 英镑的投资以 3.6% 的年利率获得复利，每年年底计息。计算 8 年后的投资价值，并确定投资翻倍所需的整年数。这一问题体现了公比为 1.036 的等比数列。

The value after n years is given by A = P × rⁿ, where r = 1 + 0.036 = 1.036. After 8 years, A = 2000 × 1.036⁸ ≈ 2000 × 1.3273 = £2654.60. To double, we require 2000 × 1.036ⁿ = 4000 ⇒ 1.036ⁿ = 2. Taking logs: n = log 2 / log 1.036 ≈ 0.3010 / 0.01536 ≈ 19.6. Thus, 20 whole years are needed.

n 年后的价值为 A = P × rⁿ，其中 r = 1 + 0.036 = 1.036。8 年后，A = 2000 × 1.036⁸ ≈ 2000 × 1.3273 = 2654.60 英镑。翻倍要求 2000 × 1.036ⁿ = 4000 ⇒ 1.036ⁿ = 2。取对数得 n = log 2 / log 1.036 ≈ 0.3010 / 0.01536 ≈ 19.6。因此需要 20 整年。

A = P(1 + r/100)ⁿ n = ln 2 / ln(1 + r/100)

3. Exponential Growth and Decay | 指数增长与衰减

A bacterial culture initially contains 500 cells and grows at a rate proportional to its size. After 3 hours, the population has increased to 1400 cells. Assuming continuous exponential growth, write an equation for the population P(t) and estimate the population after 10 hours. How long does it take for the population to reach 10 000?

一个细菌培养物最初含有 500 个细胞，并以与当前数量成正比的速率增长。3 小时后，种群数量增加到 1400 个细胞。假设连续指数增长，写出种群数量 P(t) 的方程，并估算 10 小时后的数量。种群达到 10 000 需要多长时间？

The model is P(t) = P₀ eᵏᵗ. With P₀ = 500 and P(3) = 1400: 1400 = 500 e³ᵏ ⇒ e³ᵏ = 2.8 ⇒ 3k = ln 2.8 ⇒ k = (ln 2.8)/3 ≈ 1.02962/3 ≈ 0.3432. Thus P(t) = 500 e⁰·³⁴³²ᵗ. After 10 hours, P(10) = 500 e³·⁴³² ≈ 500 × 30.94 ≈ 15 470 cells. For P = 10 000: 10 000 = 500 e⁰·³⁴³²ᵗ ⇒ e⁰·³⁴³²ᵗ = 20 ⇒ t = ln 20 / 0.3432 ≈ 2.9957 / 0.3432 ≈ 8.73 hours.

模型为 P(t) = P₀ eᵏᵗ。由 P₀ = 500 和 P(3) = 1400：1400 = 500 e³ᵏ ⇒ e³ᵏ = 2.8 ⇒ 3k = ln 2.8 ⇒ k = (ln 2.8)/3 ≈ 1.02962/3 ≈ 0.3432。因此 P(t) = 500 e⁰·³⁴³²ᵗ。10 小时后，P(10) = 500 e³·⁴³² ≈ 500 × 30.94 ≈ 15 470 个细胞。当 P = 10 000 时：10 000 = 500 e⁰·³⁴³²ᵗ ⇒ e⁰·³⁴³²ᵗ = 20 ⇒ t = ln 20 / 0.3432 ≈ 2.9957 / 0.3432 ≈ 8.73 小时。

P(t) = P₀ eᵏᵗ k = ln 2 / t_double

4. Probability and Expected Value | 概率与期望值

In a game, a fair six-sided die is rolled. If it shows a 6, the player wins £10; if it shows a 1, the player loses £5; otherwise the player wins £2. Let X be the net gain per game. Construct the probability distribution of X, calculate the expected gain E(X), and explain whether the game is favourable to the player in the long run.

在一个游戏中，抛掷一枚均匀的六面骰子。若出现 6 点，玩家赢得 10 英镑；若出现 1 点，玩家输掉 5 英镑；其他情况玩家赢得 2 英镑。设 X 为每局游戏的净收益。构造 X 的概率分布，计算期望收益 E(X)，并说明从长远看该游戏是否对玩家有利。

X takes values 10, −5 and 2. P(X=10) = 1/6, P(X=−5) = 1/6, P(X=2) = 4/6 = 2/3. The expected value is E(X) = (10)(1/6) + (−5)(1/6) + (2)(4/6) = (10 − 5 + 8)/6 = 13/6 ≈ £2.17. Because E(X) > 0, the player expects to gain around £2.17 per game in the long run; the game is favourable.

X 的取值为 10、−5 和 2。P(X=10) = 1/6，P(X=−5) = 1/6，P(X=2) = 4/6 = 2/3。期望值为 E(X) = (10)(1/6) + (−5)(1/6) + (2)(4/6) = (10 − 5 + 8)/6 = 13/6 ≈ 2.17 英镑。由于 E(X) > 0，玩家在长期内预期每局赚取约 2.17 英镑；该游戏是有利的。

E(X) = Σ xᵢ P(X = xᵢ)

5. Optimisation Using Calculus | 微积分求最优化

An open box is made from a rectangular sheet of cardboard measuring 30 cm by 20 cm by cutting equal squares of side x cm from each corner and folding up the sides. Show that the volume V cm³ is given by V = 4x³ − 100x² + 600x, and find the value of x that maximises the volume. Confirm that this value gives a maximum.

用一个长 30 cm、宽 20 cm 的长方形纸板制作一个无盖盒子，方法是从每个角剪去边长为 x cm 的正方形，然后折起各边。证明体积 V cm³ 可表示为 V = 4x³ − 100x² + 600x，并求使体积最大的 x 值。确认该值给出的是最大值。

After cutting, the length becomes (30 − 2x), width (20 − 2x) and height x. So V = x(30 − 2x)(20 − 2x) = x(600 − 100x + 4x²) = 4x³ − 100x² + 600x. Differentiate: dV/dx = 12x² − 200x + 600. Set to zero: 12x² − 200x + 600 = 0 ⇒ divide by 4: 3x² − 50x + 150 = 0. Use quadratic formula: x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 ≈ (50 ± 26.46)/6. Feasible solution x ≈ (50 − 26.46)/6 ≈ 3.92 cm (since x < 10). Second derivative: d²V/dx² = 24x − 200; at x ≈ 3.92 it is negative, confirming a maximum. Maximum volume ≈ 4(3.92)³ − 100(3.92)² + 600(3.92) ≈ 1056 cm³.

剪裁后，长为 (30 − 2x)，宽为 (20 − 2x)，高为 x。因此 V = x(30 − 2x)(20 − 2x) = x(600 − 100x + 4x²) = 4x³ − 100x² + 600x。求导：dV/dx = 12x² − 200x + 600。令导数为零：12x² − 200x + 600 = 0 ⇒ 除以 4：3x² − 50x + 150 = 0。使用求根公式：x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 ≈ (50 ± 26.46)/6。合理解 x ≈ (50 − 26.46)/6 ≈ 3.92 cm（因为 x < 10）。二阶导数：d²V/dx² = 24x − 200；在 x ≈ 3.92 处为负，确认为最大值。最大体积 ≈ 4(3.92)³ − 100(3.92)² + 600(3.92) ≈ 1056 cm³。

V = x(L − 2x)(W − 2x) dV/dx = 0, d²V/dx² < 0

6. Normal Distribution in Quality Control | 正态分布在质量控制中的应用

A machine produces metal rods with lengths normally distributed, mean 50.0 mm and standard deviation 0.3 mm. Rods are acceptable if their length lies between 49.5 mm and 50.5 mm. A batch of 500 rods is produced. Calculate the probability that a randomly chosen rod is acceptable, and hence predict the number of rods in the batch that are expected to be rejected.

一台机器生产的金属棒长度服从正态分布，均值为 50.0 mm，标准差为 0.3 mm。长度在 49.5 mm 到 50.5 mm 之间的金属棒为合格品。现生产一批 500 根棒。计算随机抽取一根棒合格的概率，并由此预测该批次预期的不合格棒数量。

Let L ~ N(50, 0.3²). For the lower bound: z₁ = (49.5 − 50)/0.3 = −1.666… ≈ −1.67. Upper bound: z₂ = (50.5 − 50)/0.3 = 1.666… ≈ 1.67. Using standard normal tables, P(Z < 1.67) ≈ 0.9525, P(Z < −1.67) ≈ 0.0475. Thus P(acceptable) = 0.9525 − 0.0475 = 0.9050. Expected acceptable rods = 500 × 0.905 = 452.5, so around 453 rods accept. Rejected = 500 − 453 = 47 rods (approximately).

设 L ~ N(50, 0.3²)。下限：z₁ = (49.5 − 50)/0.3 = −1.666… ≈ −1.67。上限：z₂ = (50.5 − 50)/0.3 = 1.666… ≈ 1.67。查标准正态表得 P(Z < 1.67) ≈ 0.9525，P(Z < −1.67) ≈ 0.0475。因此合格概率 = 0.9525 − 0.0475 = 0.9050。预期合格棒数 = 500 × 0.905 = 452.5，大约 453 根合格。不合格数 ≈ 500 − 453 = 47 根。

z = (x − μ) / σ P(acceptable) = Φ(z₂) − Φ(z₁)

7. Trigonometric Waves and Sound | 三角函数波与声音

A sound wave is modelled by the function y = 3 sin(880πt), where y is the displacement in micrometres and t is time in seconds. This represents a pure tone of frequency 440 Hz. Determine the amplitude, period and frequency of the wave. Find the times within the first 0.005 seconds when the displacement is exactly 1.5 µm.

一种声波用函数 y = 3 sin(880πt) 建模，y 是位移，单位为微米，t 是时间，单位为秒。这表示频率为 440 Hz 的纯音。求波的振幅、周期和频率。求在最初 0.005 秒内位移恰好为 1.5 µm 的时刻。

The general form is y = A sin(2πft). Comparing: A = 3 µm, 2πf = 880π ⇒ f = 440 Hz. Period T = 1/f = 1/440 ≈ 0.00227 s. To find when y = 1.5: 3 sin(880πt) = 1.5 ⇒ sin(880πt) = 0.5. The general solutions for sin θ = 0.5 are θ = π/6 + 2πk or θ = 5π/6 + 2πk. Thus 880πt = π/6, 5π/6, … For the first 0.005 s: t = (π/6)/(880π) = 1/(5280) ≈ 0.000189 s; t = (5π/6)/(880π) = 5/(5280) ≈ 0.000947 s; next solution (π/6 + 2π) gives t = (13π/6)/(880π) = 13/5280 ≈ 0.00246 s; (5π/6 + 2π) gives t = 17/5280 ≈ 0.00322 s. All are within 0.005 s.

一般形式为 y = A sin(2πft)。对比可得：A = 3 µm，2πf = 880π ⇒ f = 440 Hz。周期 T = 1/f = 1/440 ≈ 0.00227 s。求 y = 1.5 的时刻：3 sin(880πt) = 1.5 ⇒ sin(880πt) = 0.5。sin θ = 0.5 的通解为 θ = π/6 + 2πk 或 θ = 5π/6 + 2πk。因此 880πt = π/6, 5π/6, … 在最初 0.005 秒内：t = (π/6)/(880π) = 1/5280 ≈ 0.000189 s；t = (5π/6)/(880π) = 5/5280 ≈ 0.000947 s；下一解为 (π/6 + 2π) 给出 t = (13π/6)/(880π) = 13/5280 ≈ 0.00246 s；(5π/6 + 2π) 给出 t = 17/5280 ≈ 0.00322 s。这些均在 0.005 s 内。

y = A sin(2πft) T = 1/f sinθ = sinα ⇒ θ = nπ + (−1)ⁿα

8. Geometry in Construction | 建筑中的几何

A roof truss forms an isosceles triangle with base 8 m and equal sides of 5 m. To add support, a steel strut runs from the apex perpendicular to the base. Calculate the length of the strut and the angle between a sloping side and the base. Hence, if a ceiling is installed at a height of 2 m above the base along the strut, determine how far from the apex the strut must be cut.

一个屋架呈等腰三角形，底边长 8 m，两腰长各 5 m。为增加支撑，从顶点向底边作垂线（即支撑梁）。计算支撑梁的长度以及斜边与底边之间的夹角。如果沿支撑梁在距底边 2 m 高处安装顶棚，求支撑梁应自顶点算起在何处切割。

The strut is the altitude to the base, splitting the base into two 4 m segments. By Pythagoras: strut length = √(5² − 4²) = √9 = 3 m. The angle θ between a sloping side and the base satisfies cosθ = adjacent/hypotenuse = 4/5, so θ = arccos(0.8) ≈ 36.87°. The strut is 3 m long from apex to base. If a ceiling is 2 m above the base, the distance from the apex down the strut is 3 − 2 = 1 m. So the strut must be cut 1 m from the apex.

支撑梁是到达底边的高，它将底边分成两段各 4 m。由勾股定理：梁长 = √(5² − 4²) = √9 = 3 m。斜边与底边的夹角θ满足 cosθ = 邻边/斜边 = 4/5，所以 θ = arccos(0.8) ≈ 36.87°。支撑梁自顶点到底边长 3 m。若天花板在距离底边 2 m 高处，从顶点沿梁向下的距离为 3 − 2 = 1 m。因此梁应在距顶点 1 m 处被切断。

a² + b² = c² cosθ = adjacent / hypotenuse

9. Vectors and Forces | 向量与力

A boat sails on a bearing of 060° for 12 km to a lighthouse and then changes course to a bearing of 150° for 9 km to reach a harbour. Using vector components, express each leg of the journey as a position vector, and find the direct distance and bearing from the starting point to the harbour. Take north and east as positive axes.

一艘船沿方位角 060° 航行 12 km 到达一座灯塔，然后改变航向至方位角 150° 再航行 9 km 到达港口。利用向量分量将各段路程表示为位置向量，并求从起点到港口的直线距离及方位角。以正北、正东为坐标轴。

Represent bearings as angles measured clockwise from north. Leg 1: 060° gives components: east = 12 sin60° = 12 × 0.8660 = 10.392 km, north = 12 cos60° = 6 km. Vector r₁ = (10.392, 6). Leg 2: bearing 150° gives east = 9 sin150° = 9 × 0.5 = 4.5 km, north = 9 cos150° = 9 × (−0.8660) = −7.794 km. Vector r₂ = (4.5, −7.794). Resultant displacement r = r₁ + r₂ = (14.892, −1.794). Distance = √(14.892² + 1.794²) ≈ √(221.78 + 3.22) = √225 = 15.0 km. Bearing: θ = arctan(east/north) = arctan(14.892 / −1.794) gives an angle in the fourth quadrant; the bearing = 90° + arctan(1.794/14.892)? Actually, bearing = 180° − arctan(east/|north|) since north negative, east positive. arctan(14.892/1.794) ≈ arctan(8.30) ≈ 83.1°. So bearing = 180° − 83.1° = 096.9° approximately.

将方位角表示为从正北顺时针测量的角度。第一段：060°，分量：东 = 12 sin60° = 12 × 0.8660 = 10.392 km，北 = 12 cos60° = 6 km。向量 r₁ = (10.392, 6)。第二段：方位角 150°，东 = 9 sin150° = 9 × 0.5 = 4.5 km，北 = 9 cos150° = 9 × (−0.8660) = −7.794 km。向量 r₂ = (4.5, −7.794)。合位移 r = r₁ + r₂ = (14.892, −1.794)。距离 = √(14.892² + 1.794²) ≈ √(221.78 + 3.22) = √225 = 15.0 km。方位角：θ = arctan(东/北) = arctan(14.892 / −1.794)，由于北分量为负、东分量为正，角在第四象限对应方位角：从正北顺时针量测 = 180° − arctan(东/|北|) = 180° − arctan(14.892/1.794) ≈ 180° − 83.1° = 096.9°。

r = (r cosθ, r sinθ) |r| = √(x² + y²) bearing = arctan(east/north)

10. Exponential Decay in Pharmacology | 药理学中的指数衰减

A patient is given a 200 mg dose of a drug. The amount of drug in the bloodstream reduces exponentially with a half-life of 4 hours. Write a function for the amount A(t) remaining after t hours. The drug becomes effective only when at least 30 mg is present. For how long after the dose does a single administration remain effective?

一名患者服用了 200 mg 的药物。血液中的药物量按指数衰减，半衰期为 4 小时。写出 t 小时后剩余药量 A(t) 的函数。该药只有当含量不低于 30 mg 时才有效。问单次服药后药效能维持多久？

Exponential decay model: A(t) = A₀ e⁻ᵏᵗ, with half-life t₁/₂ = ln 2 / k ⇒ k = ln 2 / 4 ≈ 0.1733. So A(t) = 200 e⁻⁰·¹⁷³³ᵗ. For A ≥ 30: 200 e⁻⁰·¹⁷³³ᵗ = 30 ⇒ e⁻⁰·¹⁷³³ᵗ = 0.15 ⇒ −0

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Year 12 CCEA Mathematics: Interdisciplinary Problem-Solving Practice | 跨学科综合题型训练