📚 Cross-Disciplinary Problem Solving for Year 9 CIE Further Mathematics | CIE 九年级进阶数学跨学科解题训练
In Year 9 CIE Further Mathematics, modelling real-world situations from physics, economics, biology and chemistry helps consolidate algebraic, geometric and probabilistic thinking. This article presents ten integrated problem-solving exercises that connect mathematical techniques to cross-disciplinary contexts, reinforcing both skills and reasoning.
在九年级 CIE 进阶数学中，将物理、经济、生物和化学中的真实情境建立数学模型，有助于巩固代数、几何与概率思维。本文提供十个跨学科的综合解题练习，把数学技巧与多学科背景联系起来，强化技能和推理能力。
1. Motion and Quadratic Equations | 运动与二次方程
A ball is launched upwards with an initial speed of 20 m/s. Its height h metres above the ground after t seconds is given by h = 20t – 5t². Find when it hits the ground and its maximum height.
一个小球以 20 米/秒的初速度向上发射，t 秒后距地面的高度 h 米满足 h = 20t – 5t²。求小球何时落地及其最大高度。
Set h = 0: 20t – 5t² = 0 → 5t(4 – t) = 0, so t = 0 (launch) or t = 4 seconds (landing). The maximum height occurs at the vertex of the quadratic. For y = ax² + bx + c, the axis of symmetry is t = -b/(2a). Here a = -5, b = 20, so t = -20/(2 × -5) = 2 s. Substitute t = 2 into h: h = 20×2 – 5×2² = 40 – 20 = 20 m.
令 h = 0：20t – 5t² = 0 → 5t(4 – t) = 0，得 t = 0（发射时刻）或 t = 4 秒（落地）。最大高度在二次函数的顶点处取得。对于 y = ax² + bx + c，对称轴为 t = -b/(2a)。这里 a = -5，b = 20，所以 t = -20/(2 × -5) = 2 秒。将 t = 2 代入 h：h = 20×2 – 5×2² = 40 – 20 = 20 米。
2. Revenue and Cost Functions | 收益与成本函数
A company produces x units of a gadget. The total cost is C(x) = 1000 + 15x (in dollars), and the selling price per unit is 45 – 0.1x. Write the revenue function R(x) and the profit function P(x). Find the number of units that maximise profit.
某公司生产 x 件小器具，总成本为 C(x) = 1000 + 15x（美元），每件售价为 45 – 0.1x。写出收益函数 R(x) 和利润函数 P(x)，并求使利润最大的产量。
Revenue: R(x) = x × (45 – 0.1x) = 45x – 0.1x². Profit P(x) = R(x) – C(x) = (45x – 0.1x²) – (1000 + 15x) = -0.1x² + 30x – 1000. This is a quadratic with a = -0.1, b = 30. Maximum at x = -b/(2a) = -30/(2 × -0.1) = 150 units. Profit at 150 units: P(150) = -0.1×150² + 30×150 – 1000 = -2250 + 4500 – 1000 = 1250 dollars.
收益：R(x) = x × (45 – 0.1x) = 45x – 0.1x²。利润 P(x) = R(x) – C(x) = (45x – 0.1x²) – (1000 + 15x) = -0.1x² + 30x – 1000。此为二次函数，a = -0.1，b = 30。最大值在 x = -b/(2a) = -30/(2 × -0.1) = 150 件。此时利润 P(150) = -0.1×150² + 30×150 – 1000 = -2250 + 4500 – 1000 = 1250 美元。
3. Exponential Growth in Biology | 生物学中的指数增长
A bacteria culture contains 500 cells at 9 a.m. The population doubles every 3 hours. Write an exponential function P(t) for the number of cells after t hours. Calculate the expected population at 6 p.m. on the same day.
一个细菌培养在上午 9 时有 500 个细胞，每 3 小时数量翻倍。写出 t 小时后的细胞数指数函数 P(t)，并计算当天下午 6 时的预期数量。
The exponential model is P(t) = 500 × 2^(t/3), where t is in hours. From 9 a.m. to 6 p.m. is 9 hours. Substitute t = 9: P(9) = 500 × 2^(9/3) = 500 × 2³ = 500 × 8 = 4000 cells.
指数模型为 P(t) = 500 × 2^(t/3)，t 以小时为单位。从上午 9 时到下午 6 时共 9 小时。代入 t = 9：P(9) = 500 × 2^(9/3) = 500 × 2³ = 500 × 8 = 4000 个细胞。
4. Trigonometry in Surveying | 测量中的三角学
A surveyor measures the angle of elevation to the top of a tower as 35° from a point 50 metres from its base. Determine the height of the tower. Then find the angle of elevation from a point 25 metres closer to the base.
测量员在距塔底 50 米处测得塔顶的仰角为 35°，求塔的高度。然后求在向塔底靠近 25 米后的位置，仰角是多少。
Let h be the height. Using tangent: tan 35° = h / 50 → h = 50 tan 35° ≈ 50 × 0.7002 = 35.01 m. When 25 m closer, distance = 25 m. New angle θ satisfies tan θ = h / 25 ≈ 35.01 / 25 = 1.4004 → θ ≈ tan⁻¹(1.4004) ≈ 54.5°.
设高度为 h。由正切关系：tan 35° = h / 50 → h = 50 tan 35° ≈ 50 × 0.7002 = 35.01 m。靠近 25 米后，距离为 25 米。新仰角 θ 满足 tan θ = h / 25 ≈ 35.01 / 25 = 1.4004 → θ ≈ tan⁻¹(1.4004) ≈ 54.5°。
5. Chemical Mixtures and Ratios | 化学混合物与比例
A chemist mixes a 30% acid solution with a 10% acid solution to obtain 400 mL of a 15% acid solution. Determine how many millilitres of each original solution are required.
一位化学家用 30% 的酸溶液和 10% 的酸溶液混合，得到 400 毫升 15% 的酸溶液。求每种原溶液各需多少毫升。
Let x mL of 30% solution and y mL of 10% solution. Total volume: x + y = 400. Acid content: 0.30x + 0.10y = 0.15 × 400 = 60. Multiply the second equation by 10: 3x + y = 600. Subtract the first: (3x + y) – (x + y) = 600 – 400 → 2x = 200 → x = 100 mL. Then y = 300 mL.
设需要 30% 溶液 x 毫升，10% 溶液 y 毫升。总体积：x + y = 400。酸含量：0.30x + 0.10y = 0.15 × 400 = 60。将第二式乘以 10：3x + y = 600。减去第一式：(3x + y) – (x + y) = 600 – 400 → 2x = 200 → x = 100 毫升，y = 300 毫升。
6. Energy and Work-Rate Problems | 能量与工作效率问题
A pump can fill a tank in 6 hours. A second pump can fill it in 4 hours. How long will it take to fill the tank if both pumps work together? The tank holds 2400 litres of water; determine the combined flow rate.
一台水泵 6 小时可注满一箱水，另一台 4 小时可注满。两泵同时工作，注满水箱需要多长时间？水箱容量为 2400 升，求组合流量。
Rate of first pump: 1/6 tank per hour; second pump: 1/4 tank per hour. Combined rate: 1/6 + 1/4 = 2/12 + 3/12 = 5/12 tank per hour. Time to fill = 1 ÷ (5/12) = 12/5 = 2.4 hours. Flow rates: first = 2400/6 = 400 L/h; second = 2400/4 = 600 L/h; combined = 1000 L/h.
第一台水泵的效率为每小时 1/6 箱，第二台为 1/4 箱。组合效率：1/6 + 1/4 = 2/12 + 3/12 = 5/12 箱每小时。所需时间 = 1 ÷ (5/12) = 12/5 = 2.4 小时。流量：第一台 2400/6 = 400 升/时；第二台 2400/4 = 600 升/时；合计 1000 升/时。
7. Geometric Optimization | 几何最优化
A farmer has 120 metres of fencing to enclose a rectangular field next to a straight river, requiring no fence along the river. Express the area A in terms of width x, and find the dimensions that maximise the area.
一位农民有 120 米长的围栏，用来在一条笔直河流旁围出一个矩形场地，靠河一侧不用围栏。用宽 x 表示面积 A，并求使面积最大的尺寸。
Let the width perpendicular to the river be x metres. The side parallel to the river has length (120 – 2x). Area A = x(120 – 2x) = 120x – 2x². Maximise the quadratic: x = -b/(2a) where a = -2, b = 120 → x = -120/(2 × -2) = 30 m. Then length = 120 – 2×30 = 60 m. Maximum area = 30 × 60 = 1800 m².
设与河流垂直的宽为 x 米，则平行河岸的边长 (120 – 2x) 米。面积 A = x(120 – 2x) = 120x – 2x²。最大化该二次函数：x = -b/(2a)，其中 a = -2，b = 120，得 x = -120/(2 × -2) = 30 米。长为 120 – 60 = 60 米，最大面积 1800 平方米。
8. Probability in Games | 游戏中的概率
A game involves rolling two fair dice. You win if the sum is at least 10 or if the two numbers are equal. Calculate the probability of winning in a single roll. Express your answer as a fraction in simplest form.
一个游戏需要掷两个均匀的骰子，若点数之和至少为 10 或两个数字相同则获胜。求单次掷骰获胜的概率，用最简分数表示。
Total outcomes = 6 × 6 = 36. Favorable for sum ≥ 10: (4,6),(5,5),(6,4),(5,6),(6,5),(6,6) — that is 6 outcomes. Favorable for equal numbers (doubles): (1,1)…(6,6) — 6 outcomes. Intersection (both conditions): (5,5) and (6,6) are in both sets, so |A ∪ B| = 6 + 6 – 2 = 10. Probability = 10/36 = 5/18.
总结果数 = 6 × 6 = 36。点数之和 ≥ 10 的有：(4,6),(5,5),(6,4),(5,6),(6,5),(6,6) 共 6 种。数字相同（对子）：(1,1)…(6,6) 共 6 种。两条件的交集：(5,5) 和 (6,6) 属两者，所以 |A ∪ B| = 6 + 6 – 2 = 10。概率 = 10/36 = 5/18。
9. Inequalities and Resource Allocation | 不等式与资源分配
A workshop produces chairs (x) and tables (y). Each chair requires 2 hours of labour and 3 units of wood; each table requires 5 hours of labour and 4 units of wood. There are at most 40 labour hours and 36 units of wood available. Write the inequalities and identify which of the points (5,6) or (8,4) is feasible.
某车间生产椅子 (x) 和桌子 (y)。每把椅子需 2 小时人工和 3 单位木材；每张桌子需 5 小时人工和 4 单位木材。可用人工最多 40 小时，木材最多 36 单位。写出不等式组，并判断 (5,6) 和 (8,4) 哪个可行。
Labour: 2x + 5y ≤ 40; Wood: 3x + 4y ≤ 36; x ≥ 0, y ≥ 0. For (5,6): 2×5 + 5×6 = 10+30=40 ≤ 40; 3×5 + 4×6 = 15+24=39 > 36, not feasible. For (8,4): 2×8 + 5×4 = 16+20=36 ≤ 40; 3×8 + 4×4 = 24+16=40 > 36, also not feasible. Both violate wood constraint; to find feasible integer points would require checking vertices.
人工：2x + 5y ≤ 40；木材：3x + 4y ≤ 36；x ≥ 0, y ≥ 0。(5,6)：2×5 + 5×6 = 40 ≤ 40；3×5 + 4×6 = 39 > 36，不可行。(8,4)：2×8 + 5×4 = 36 ≤ 40；3×8 + 4×4 = 40 > 36，也不可行。木材约束均不满足，可行整点需检查顶点。
10. Data Analysis and Linear Models | 数据分析与线性模型
The table shows the temperature T (°C) of a chemical solution at time t minutes after heating begins. Use the points (2, 45) and (8, 69) to model T as a linear function of t. Then predict the temperature after 12 minutes and interpret the gradient.
| t (min) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| T (°C) | 45 | 53 | 61 | 69 |
表格显示加热开始后 t 分钟化学溶液的温度 T (°C)。用点 (2, 45) 和 (8, 69) 将 T 建模为 t 的线性函数，预测 12 分钟时的温度并解释斜率的含义。
Gradient m = (69 – 45)/(8 – 2) = 24/6 = 4 °C/min. Linear equation: T = 4t + c. Using (2,45): 45 = 4×2 + c → c = 37. So T = 4t + 37. At t = 12, T = 4×12 + 37 = 48 + 37 = 85 °C. The gradient of 4 °C/min means the temperature rises by 4 degrees Celsius each minute under constant heating.
斜率 m = (69 – 45)/(8 – 2) = 24/6 = 4 °C/分钟。线性方程：T = 4t + c。代入 (2,45)：45 = 4×2 + c → c = 37，故 T = 4t + 37。t = 12 时，T = 4×12 + 37 = 85 °C。斜率 4 °C/分钟 表示在持续加热下每分钟温度升高 4 摄氏度。
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