📚 Year 8 OCR Mathematics: Interdisciplinary Mixed Questions Practice | Year 8 OCR 数学：跨学科综合题型训练
This article brings together real-world problems from physics, geography, biology, chemistry, finance, art, sports, music and design. Each question is written to help you apply your Year 8 mathematics skills—ratios, percentages, algebra, geometry and statistics—in cross-curricular contexts. Work through each section, reading the English first and then the Chinese, to build both your maths and your bilingual understanding.
本文汇集了来自物理、地理、生物、化学、金融、艺术、体育、音乐和设计领域的真实问题。每道题都旨在帮助你运用八年级数学技能——比例、百分比、代数、几何和统计——解决跨学科问题。请逐节练习，先读英文再读中文，同步提升数学能力与双语理解。
1. Speed, Distance, and Time: The Physics of Motion | 速度、距离与时间：运动的物理学
A cyclist travels at a constant speed of 15 km/h for 2 hours and 30 minutes. How far does the cyclist travel? If the cyclist plans to cover the same distance in exactly 2 hours on the next ride, what average speed must they maintain?
一名自行车手以 15 km/h 的恒定速度骑行了 2 小时 30 分钟。他一共骑了多远？如果他计划在下一次骑行中，用恰好 2 小时骑完相同的路程，他必须保持多大的平均速度？
convert 2 hours 30 minutes into hours: 30 minutes = 0.5 hours, so total time = 2.5 hours.
首先，把 2 小时 30 分钟换算成小时：30 分钟 = 0.5 小时，因此总时间 = 2.5 小时。
Distance = Speed × Time
Distance travelled = 15 km/h × 2.5 h = 37.5 km.
骑行距离 = 15 km/h × 2.5 h = 37.5 km。
ime is now 2 hours. Required average speed = Distance ÷ Time = 37.5 km ÷ 2 h = 18.75 km/h.
下一次骑行，距离仍然是 37.5 km，但时间改为 2 小时。所需平均速度 = 距离 ÷ 时间 = 37.5 km ÷ 2 h = 18.75 km/h。
2. Map Scales and Real Distances: Geography in Mathematics | 地图比例尺与实际距离：地理中的数学
On a map with a scale of 1 : 250,000, two landmarks are 8 cm apart. What is the actual distance between them in kilometres? If a third landmark is 15 km away from one of them, how far apart are they on the map?
在一张比例尺为 1:250 000 的地图上，两个地标相距 8 cm。它们之间的实际距离是多少千米？如果第三个地标距离其中一个地标 15 km，它们在地图上的距离是多少？
the map represents 250,000 cm in real life. So 8 cm represents 8 × 250,000 = 2,000,000 cm.
比例尺 1:250 000 表示地图上的 1 cm 代表实际 250 000 cm。因此 8 cm 代表 8 × 250 000 = 2 000 000 cm。
Convert to kilometres: 2,000,000 cm = 20,000 m = 20 km. The actual distance is 20 km.
换算为千米：2 000 000 cm = 20 000 m = 20 km。实际距离为 20 km。
1,500,000 cm. Map distance = actual distance ÷ scale factor = 1,500,000 ÷ 250,000 = 6 cm.
第三个地标的实际距离为 15 km。先换算为 cm：15 km = 15 × 100 000 = 1 500 000 cm。地图距离 = 实际距离 ÷ 比例因子 = 1 500 000 ÷ 250 000 = 6 cm。
3. Percentage Change in Population: Biology and Demographics | 人口百分比变化：生物学与人口统计学
A town's population at the start of 2020 was 12,500. By the end of 2022, it had increased to 13,875. (a) Calculate the percentage increase over these two years. (b) If the population then grows by 5.5% each year from the end of 2022, estimate the population at the end of 2025.
某城镇 2020 年初人口为 12 500 人。到 2022 年底增加到 13 875 人。(a) 计算这两年的增长百分比。(b) 如果从 2022 年底开始，人口每年增长 5.5%，估计 2025 年底的人口数。
(a) Increase = 13,875 - 12,500 = 1,375. Percentage increase = (1,375 ÷ 12,500) × 100% = 11%.
(a) 增长量 = 13 875 - 12 500 = 1 375。增长百分比 = (1 375 ÷ 12 500) × 100% = 11%。
(b) Population at the end of 2022 = 13,875. Annual multiplier = 1 + 5.5% .17424. Estimated population = 13,875 × 1.17424 ≈ 16,292.6, which rounds to 16,293 people.
(b) 2022 年底人口 = 13 875。年增长倍数 = 1 + 5.5% = 1.055。3 年后人口将乘以 1.055 三次。(1.055)³ ≈ 1.17424。估计人口 = 13 875 × 1.17424 ≈ 16 292.6，四舍五入约 16 293 人。
4. Mixing Chemicals: Ratios in Chemistry | 化学混合物：化学中的比例
A chemist needs to mix acid and water in the ratio 2 : 5 by volume. If she uses 300 ml of acid, how much water is needed, and what is the total volume of the mixture? Later she wants to prepare 1.4 litres of the same mixture. How much acid and water should she use?
一位化学家需要按体积比 2 : 5 混合酸和水。如果她用了 300 ml 酸，需要多少水？混合液的总体积是多少？后来她想配制 1.4 升相同混合液，她该用多少酸和水？
÷ 2 = 150 ml. Water = 5 × 150 = 750 ml. Total volume = 300 + 750 = 1,050 ml = 1.05 litres.
酸与水的比例为 2 : 5。每 2 份酸对应 5 份水。若酸为 300 ml，一份为 300 ÷ 2 = 150 ml。水 = 5 × 150 = 750 ml。总体积 = 300 + 750 = 1 050 ml = 1.05 升。
part = 1,400 ÷ 7 = 200 ml. Acid = 2 × 200 = 400 ml. Water = 5 × 200 = 1,000 ml (1 litre).
配制 1.4 升（1 400 ml）时，总份数 = 2 + 5 = 7。一份 = 1 400 ÷ 7 = 200 ml。酸 = 2 × 200 = 400 ml。水 = 5 × 200 = 1 000 ml（1 升）。
5. Electrical Power and Energy: Physics Equations | 电功率与能量：物理方程
A tablet charger delivers a current of 2.5 A at a voltage of 5 V. (a) Calculate the electrical power. (b) If the tablet charges for 3 hours, how much energy does it consume, in watt-hours? Give your answer also in kilowatt-hours.
一台平板电脑充电器输出电流 2.5 A，电压 5 V。(a) 计算电功率。(b) 如果平板充电 3 小时，它消耗了多少能量（以瓦时为单位）？再用千瓦时表示。
(a) Use the power formula:
(a) 使用功率公式：
Power = Current × Voltage (P = I × V)
P = 2.5 A × 5 V = 12.5 W.
功率 = 2.5 A × 5 V = 12.5 W。
(b) Energy = Power × Time. Time = 3 h. Energy = 12.5 W × 3 h = 37.5 Wh.
(b) 能量 = 功率 × 时间。时间 = 3 h。能量 = 12.5 W × 3 h = 37.5 Wh。
To express in kilowatt-hours, divide by 1,000: 37.5 Wh = 0.0375 kWh.
用千瓦时表示，除以 1 000：37.5 Wh = 0.0375 kWh。
6. Simple and Compound Interest: Financial Mathematics | 单利与复利：金融数学
Mia invests £800 in a savings account. Bank A offers 3% simple interest per year. Bank B offers 3% compound interest per year. Calculate the total amount in each account after 3 years.
米娅将 800 英镑存入储蓄账户。银行 A 提供年利率 3% 的单利，银行 B 提供年利率 3% 的复利。计算 3 年后每种账户的总金额。
Simple interest: I = P × r × t. Interest = £800 × 0.03 × 3 = £72. Total = £800 + £72 = £872.
单利：利息 = 本金 × 利率 × 时间。利息 = 800 × 0.03 × 3 = £72。总额 = £800 + £72 = £872。
rest: Amount = P × (1 + r)ᵗ = 800 × (1.03)³. (1.03)³ = 1.092727. Amount = 800 × 1.092727 = £874.18 (to the nearest penny).
复利：总额 = 本金 × (1 + 利率)ᵗ = 800 × (1.03)³。(1.03)³ = 1.092727。总额 = 800 × 1.092727 = £874.18（精确到便士）。
7. The Golden Ratio in Art and Nature | 艺术与自然中的黄金比例
The golden ratio, about 1.618, appears in art, architecture and nature. A famous painting uses a golden rectangle: the ratio of length to width is 1.618 : 1. If the width of the painting is 45 cm, what should the length be? What is the area of the canvas?
黄金比例约为 1.618，出现在艺术、建筑和自然中。一幅名画采用黄金矩形：长与宽之比为 1.618 : 1。如果画作的宽为 45 cm，长应该是多少？画布的面积是多少？
Length = width × 1.618 = 45 cm × 1.618 = 72.81 cm (or approximately 72.8 cm).
长 = 宽 × 1.618 = 45 cm × 1.618 = 72.81 cm（约 72.8 cm）。
Area = length × width = 72.81 cm × 45 cm ≈ 3,276.45 cm².
面积 = 长 × 宽 = 72.81 cm × 45 cm ≈ 3 276.45 cm²。
If another canvas has a length of 100 cm, its golden width would be 100 ÷ 1.618 ≈ 61.8 cm.
如果另一画布长为 100 cm，其黄金比例宽度为 100 ÷ 1.618 ≈ 61.8 cm。
8. Sports Statistics: Mean, Median, and Range | 体育统计：平均数、中位数与极差
A basketball player scored the following points in six games: 18, 24, 12, 30, 18, 22. (a) Find the mean, median, and range of the scores. (b) Which measure best describes the player's typical performance? Explain why.
一名篮球运动员在六场比赛中得分如下：18, 24, 12, 30, 18, 22。(a) 求得分平均数、中位数和极差。(b) 哪个度量最好地描述了该球员的典型表现？解释原因。
he scores: 12, 18, 18, 22, 24, 30. Mean = (12+18+18+22+24+30) ÷ 6 = 124 ÷ 6 ≈ 20.7 points.
将得分排序：12, 18, 18, 22, 24, 30。平均数 = (12+18+18+22+24+30) ÷ 6 = 124 ÷ 6 ≈ 20.7 分。
six numbers, the median is the mean of the 3rd and 4th values: (18 + 22) ÷ 2 = 20 points.
中位数：共六个数，中位数是第 3 和第 4 个数的平均数：(18 + 22) ÷ 2 = 20 分。
Range = 30 - 12 = 18 points.
极差 = 30 - 12 = 18 分。
(b) The median (20) is less affected by the very high score of 30, so it represents a typical performance better than the mean. The range shows consistency; a large range suggests variability.
(b) 中位数
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