Cross-Curricular Integrated Problem-Solving Practice | 跨学科综合题型训练

📚 Cross-Curricular Integrated Problem-Solving Practice | 跨学科综合题型训练

Welcome to a specially designed cross-curricular problem-solving workout for Year 9 OCR Further Mathematics. This article brings together real-world contexts from physics, chemistry, economics, music, computer science and more. Each integrated question challenges you to apply algebraic techniques, proportionality, functions and number skills—exactly as required by the OCR specification. Let's strengthen your mathematical thinking beyond isolated exercises.

欢迎参加为 Year 9 OCR 进阶数学专门设计的跨学科题型训练。本文汇集了物理、化学、经济、音乐、计算机科学等真实情境。每道综合题都要求你运用代数技巧、比例关系、函数和数字技能——完全符合 OCR 考试大纲。让我们超越孤立练习，强化你的数学思维。

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1. Physics – Two Cars Meeting | 物理——两车相遇

Problem: Car A leaves town P travelling at a constant speed of 60 km/h. At the same instant, Car B leaves town Q, located 200 km away, heading directly towards Car A at 80 km/h. Determine the time at which they meet, and the distance from P to the meeting point.

问题：汽车 A 以 60 km/h 的恒定速度离开 P 镇。同一时刻，汽车 B 从 200 km 外的 Q 镇出发，以 80 km/h 的速度径直朝汽车 A 驶去。求它们相遇的时间，以及相遇点距 P 的距离。

Let t represent the time in hours. The distance covered by Car A is d_A = 60t. The distance covered by Car B is d_B = 80t.

设 t 为小时数。汽车 A 行驶的距离为 d_A = 60t。汽车 B 行驶的距离为 d_B = 80t。

When the cars meet, the sum of their distances equals the initial separation:

两车相遇时，各自行驶的距离之和等于初始间隔：

60t + 80t = 200

Combine like terms: 140t = 200. Solve for t: t = 200 ÷ 140 = 10/7 ≈ 1.43 hours (about 1 hour 25.7 minutes).

合并同类项：140t = 200。解出 t：t = 200 ÷ 140 = 10/7 ≈ 1.43 小时（约 1 小时 25.7 分钟）。

To find the distance from P: d_A = 60 × (10/7) = 600/7 ≈ 85.7 km. Thus, they meet approximately 85.7 km from town P.

求距 P 的距离：d_A = 60 × (10/7) = 600/7 ≈ 85.7 km。因此，它们在距 P 镇约 85.7 km 处相遇。

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2. Economics – Profit Maximisation | 经济学——利润最大化

A company finds that its weekly profit P (in pounds) from selling x units of a product is modelled by the quadratic function P = -2x² + 120x - 800. How many units should be sold to achieve maximum profit, and what is that maximum profit?

某公司发现其每周利润 P（英镑）与销售量 x 之间的关系可用二次函数 P = -2x² + 120x - 800 表示。要达到最大利润，应销售多少件产品？最大利润是多少？

Since the coefficient of x² is negative, the parabola opens downward, and the vertex gives the maximum. The x-coordinate of the vertex is x = -b / (2a) with a = -2, b = 120.

由于 x² 系数为负，抛物线开口向下，顶点对应最大值。顶点的 x 坐标为 x = -b / (2a)，其中 a = -2，b = 120。

x = -120 ÷ (2 × -2) = -120 ÷ -4 = 30

Substitute x = 30 into the profit function: P = -2(30)² + 120(30) - 800 = -1800 + 3600 - 800 = 1000.

将 x = 30 代入利润函数：P = -2(30)² + 120(30) - 800 = -1800 + 3600 - 800 = 1000。

The maximum weekly profit is £1000, achieved when 30 units are sold.

最大周利润为 £1000，在销售 30 件产品时实现。

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3. Chemistry – Mixture Concentration | 化学——混合浓度

A chemist mixes x mL of a 20% saline solution with 50 mL of a 10% saline solution. The resulting mixture has a concentration of 14%. Find the value of x.

一位化学家将 x 毫升浓度为 20% 的盐水与 50 毫升浓度为 10% 的盐水混合。得到的混合物浓度为 14%。求 x 的值。

Amount of salt from the first solution: 0.20x. Amount of salt from the second: 0.10 × 50 = 5. Total salt in the mixture: 0.20x + 5.

来自第一种溶液的盐量：0.20x。来自第二种溶液的盐量：0.10 × 50 = 5。混合物中盐的总量：0.20x + 5。

The total volume is (x + 50) mL, and its concentration is 14%, so we set up the equation:

总体积为 (x + 50) 毫升，其浓度为 14%，因此建立方程：

0.20x + 5 = 0.14(x + 50)

Expand the right-hand side: 0.20x + 5 = 0.14x + 7. Subtract 0.14x from both sides: 0.06x + 5 = 7.

展开右边：0.20x + 5 = 0.14x + 7。两边减去 0.14x：0.06x + 5 = 7。

Subtract 5: 0.06x = 2. Finally, x = 2 ÷ 0.06 = 200 ÷ 6 = 100/3 ≈ 33.3 mL.

减去 5：0.06x = 2。最后，x = 2 ÷ 0.06 = 200 ÷ 6 = 100/3 ≈ 33.3 毫升。

The chemist must use 100/3 mL of the 20% solution.

化学家必须使用 100/3 毫升的 20% 溶液。

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4. Biology – Bacterial Growth | 生物——细菌繁殖

A single bacterial cell divides every 20 minutes. How long will it take for the population to reach 1024 cells, assuming no cells die and perfect doubling?

一个细菌细胞每 20 分钟分裂一次。假设没有细胞死亡且倍增完美，需要多久才能使细胞数达到 1024 个？

The population after t minutes follows the model N = N₀ × 2^t/20, where N₀ = 1. We need N = 1024.

t 分钟后的种群数量遵循模型 N = N₀ × 2^t/20，其中 N₀ = 1。我们需要 N = 1024。

2^t/20 = 1024

Recognise that 1024 = 2¹⁰. Therefore, 2^t/20 = 2¹⁰. Equating exponents: t/20 = 10.

注意到 1024 = 2¹⁰。因此，2^t/20 = 2¹⁰。令指数相等：t/20 = 10。

Multiply both sides by 20: t = 200 minutes, which is 3 hours and 20 minutes.

两边乘以 20：t = 200 分钟，即 3 小时 20 分钟。

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5. Geography – Map Scale | 地理——地图比例尺

3.2 cm by 2.5 cm. Calculate the actual area of the field in hectares. (1 hectare = 10000 m²)

在一张比例尺为 1:50000 的地图上，一块矩形田地长 3.2 厘米，宽 2.5 厘米。计算该田地的实际面积（公顷）。 (1 公顷 = 10000 m²)

gths. 1 cm on the map represents 50000 cm in reality, which is 500 m (since 100 cm = 1 m).

首先，将图上长度换算为实际长度。地图上 1 厘米代表实际 50000 厘米，即 500 米（因为 100 cm = 1 m）。

Real length: 3.2 × 500 = 1600 m. Real width: 2.5 × 500 = 1250 m.

实际长度：3.2 × 500 = 1600 m。实际宽度：2.5 × 500 = 1250 m。

Actual area = 1600 m × 1250 m = 2,000,000 m². To convert to hectares, divide by 10000: 2,000,000 ÷ 10000 = 200 hectares.

实际面积 = 1600 m × 1250 m = 2,000,000 m²。换算为公顷，除以 10000：2,000,000 ÷ 10000 = 200 公顷。

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6. Physics – Lever Balance | 物理——杠杆平衡

A seesaw is perfectly balanced. On the left side, a 20 N weight sits 0.4 m from the pivot. On

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