KS3 OCR Further Maths: Interdisciplinary Problem-Solving Training | KS3 OCR 进阶数学：跨学科综合题型训练

📚 KS3 OCR Further Maths: Interdisciplinary Problem-Solving Training | KS3 OCR 进阶数学：跨学科综合题型训练

In KS3 Further Maths, applying mathematical techniques across different subjects sharpens your analytical thinking. This article provides targeted practice for OCR-style interdisciplinary questions, blending topics like science, geography and economics with core mathematical skills such as ratio, proportion, algebra and statistics. Each section presents a real-world problem and explains the solution step by step.

在 KS3 进阶数学中，将数学技巧应用于不同学科能强化你的分析思维。本文提供针对 OCR 风格的跨学科综合题型训练，融合科学、地理和经济等主题，与比例、代数、统计等核心数学技能相结合。每一节都呈现一个真实世界的问题，并逐步解释解决方案。

1. Ratio and Recipe Scaling (Food Technology) | 比例与食谱调整（食品科技）

A recipe requires 250 g of flour, 120 g of sugar and 80 g of butter to serve 6 people. You need to adjust it for 15 people. Find the mass of each ingredient needed and express the ratio of flour to sugar to butter in simplest form.

一份食谱需要 250 克面粉、120 克糖和 80 克黄油，可供 6 人食用。你需要调整食谱供 15 人食用。求每种食材所需的质量，并用最简形式表示面粉、糖和黄油的比例。

Step 1: Determine the scaling factor. Divide the new number of people by the original: 15 ÷ 6 = 2.5.

步骤1：确定缩放因子。用新人数除以原人数：15 ÷ 6 = 2.5。

Step 2: Multiply each ingredient mass by 2.5. Flour: 250 × 2.5 = 625 g. Sugar: 120 × 2.5 = 300 g. Butter: 80 × 2.5 = 200 g.

步骤2：每种食材质量乘以 2.5。面粉：250 × 2.5 = 625 克。糖：120 × 2.5 = 300 克。黄油：80 × 2.5 = 200 克。

Step 3: Write the original ratio 250 : 120 : 80. Divide each term by their highest common factor, which is 10, to get 25 : 12 : 8. This simplified ratio remains the same for the scaled version because all quantities are multiplied by the same factor.

步骤3：写出原始比例 250 : 120 : 80。每个项除以它们的最大公因数 10，得到 25 : 12 : 8。这个简化比例在缩放后保持不变，因为所有量都乘以了相同的因子。

---

2. Speed, Distance and Time in Physics | 物理中的速度、距离和时间

A car travels at a constant speed of 72 km/h. How far does it travel in 3.5 hours? Express your answer in both kilometres and metres. Then calculate the time required to cover 180 km at the same speed.

一辆汽车以 72 km/h 的恒定速度行驶。它在 3.5 小时内行驶了多远？答案分别用千米和米表示。然后计算以相同速度行驶 180 km 所需的时间。

Step 1: Use the formula d = v × t, where d is distance, v is speed and t is time.

步骤1：使用公式 d = v × t，其中 d 是距离，v 是速度，t 是时间。

Step 2: Substitute v = 72 km/h and t = 3.5 h into the formula. d = 72 × 3.5 = 252 km.

步骤2：将 v = 72 km/h 和 t = 3.5 h 代入公式。d = 72 × 3.5 = 252 km。

Step 3: Convert 252 km to metres: 252 × 1000 = 252 000 m.

步骤3：将 252 km 转换为米：252 × 1000 = 252 000 m。

Step 4: To find time for 180 km, rearrange the formula to t = d ÷ v. t = 180 ÷ 72 = 2.5 hours, or 2 hours 30 minutes.

步骤4：要计算 180 km 所需时间，转换公式为 t = d ÷ v。t = 180 ÷ 72 = 2.5 小时，即 2 小时 30 分钟。

---

3. Map Scales and Real Distance (Geography) | 地图比例尺与实际距离（地理）

A map has a scale of 1 : 50 000. Two villages are 7.4 cm apart on the map. Calculate the actual distance in kilometres. What would be the map distance for a real length of 12 km?

一张地图的比例尺为 1 : 50 000。两个村庄在地图上相距 7.4 cm。计算实际距离，单位为公里。如果实际长度为 12 km，地图上的距离应该是多少？

Step 1: The scale means 1 cm on map represents 50 000 cm in reality. For 7.4 cm, real distance = 7.4 × 50 000 = 370 000 cm.

步骤1：该比例尺意味着地图上的 1 cm 代表实际 50 000 cm。因此 7.4 cm 的实际距离 = 7.4 × 50 000 = 370 000 cm。

Step 2: Convert centimetres to kilometres. Since 1 km = 100 000 cm, divide by 100 000: 370 000 ÷ 100 000 = 3.7 km.

步骤2：将厘米转换为公里。由于 1 km = 100 000 cm，除以 100 000：370 000 ÷ 100 000 = 3.7 km。

Step 3: To find map distance for 12 km, first convert 12 km to cm: 12 × 100 000 = 1 200 000 cm.

步骤3：要求 12 km 在地图上的距离，先将 12 km 转换为 cm：12 × 100 000 = 1 200 000 cm。

Step 4: Divide the real distance in cm by the scale factor: 1 200 000 ÷ 50 000 = 24 cm. So the map distance is 24 cm.

步骤4：用实际距离（cm）除以比例尺因子：1 200 000 ÷ 50 000 = 24 cm。因此地图距离为 24 cm。

---

4. Osmosis and Cell Percentage Change (Biology) | 渗透与细胞体积百分比变化（生物）

When placed in a dilute solution, a plant cell’s volume increases from 4 500 µm³ to 5 400 µm³. Calculate the percentage increase in volume. If the cell then returns to its original volume, express this decrease as a percentage of the enlarged volume.

当放入稀溶液中，一个植物细胞的体积从 4 500 µm³ 增加到 5 400 µm³。计算体积增加的百分比。如果细胞随后恢复到原来的体积，用相对于增大后体积的百分比表示这种减少。

Step 1: Find the difference in volume: 5 400 – 4 500 = 900 µm³.

步骤1：计算体积差：5 400 – 4 500 = 900 µm³。

Step 2: Percentage increase = (difference ÷ original) × 100 = (900 ÷ 4 500) × 100 = 0.2 × 100 = 20%.

步骤2：增加百分比 = (差值 ÷ 原始值) × 100 = (900 ÷ 4 500) × 100 = 0.2 × 100 = 20%。

Step 3: For the decrease back to 4 500 µm³ from 5 400 µm³, the decrease is 900 µm³. Percentage decrease = (900 ÷ 5 400) × 100 = 0.1666… × 100 ≈ 16.7% (to 3 s.f.).

步骤3：对于从 5 400 µm³ 减回到 4 500 µm³，减少量是 900 µm³。减少百分比 = (900 ÷ 5 400) × 100 = 0.1666… × 100 ≈ 16.7%（保留三位有效数字）。

---

5. Experimental Error and Statistics (Science) | 实验误差与统计（科学）

A student measures the boiling point of water six times and obtains these values in °C: 100.1, 99.8, 100.0, 99.9, 100.3, 99.7. Calculate the mean boiling point. Then find the range and comment on the precision of the measurements.

一名学生六次测量水的沸点，得到以下摄氏度数值：100.1, 99.8, 100.0, 99.9, 100.3, 99.7。计算平均沸点。然后求出极差，并评价测量的精密度。

Step 1: Add all readings: 100.1 + 99.8 + 100.0 + 99.9 + 100.3 + 99.7 = 599.8.

步骤1：将所有读数相加：100.1 + 99.8 + 100.0 + 99.9 + 100.3 + 99.7 = 599.8。

Step 2: Mean = total ÷ number of readings = 599.8 ÷ 6 ≈ 99.9667, rounded to 100.0 °C (to 1 d.p.).

步骤2：平均值 = 总和 ÷ 读数个数 = 599.8 ÷ 6 ≈ 99.9667，四舍五入为 100.0 °C（保留一位小数）。

Step 3: Range = highest – lowest. Highest = 100.3, lowest = 99.7, so range = 0.6 °C.

步骤3：极差 = 最大值 – 最小值。最大值 = 100.3，最小值 = 99.7，因此极差 = 0.6 °C。

Step 4: A small range indicates good precision, but the mean close to 100.0 °C also suggests high accuracy if we accept the true boiling point as 100 °C.

步骤4：极差小说明精密度良好，而平均值接近 100.0 °C 也表明，如果我们将真实沸点视为 100 °C，则准确度很高。

---

6. Simple Interest and Budgeting (Economics) | 单利与预算（经济）

You save £400 in a bank account that offers 3% simple interest per year. How much interest is earned after 2 years, and what is the total amount? If you want to buy a £450 item, will the total savings be enough?

你将 400 英镑存入一个年利率为 3% 的单利银行账户。2 年后可获得多少利息，总金额为多少？如果你想买一件 450 英镑的物品，总储蓄够吗？

Step 1: Simple interest I = P × r × t, where P = principal (£400), r = rate (0.03), t = time (2 years). I = 400 × 0.03 × 2 = £24.

步骤1：单利 I = P × r × t，其中 P = 本金（400 英镑），r = 利率（0.03），t = 时间（2 年）。I = 400 × 0.03 × 2 = 24 英镑。

Step 2: Total amount A = P + I = 400 + 24 = £424.

步骤2：总金额 A = P + I = 400 + 24 = 424 英镑。

Step 3: Compare £424 with the cost £450. £424 < £450, so the savings are not sufficient. You would need an extra £26.

步骤3：比较 424 英镑和 450 英镑。424 < 450，因此储蓄不够。你还需额外 26 英镑。

---

7. Area, Volume and Material Costs (Design & Technology) | 面积、体积与材料成本（设计与技术）

A rectangular storage box has internal dimensions: length 80 cm, width 50 cm and height 30 cm. Calculate its volume in cm³ and convert to litres. The material for the base costs £0.05 per cm² of area. Find the cost of the base material.

一个长方体储物盒的内部尺寸为：长 80 cm，宽 50 cm，高 30 cm。计算它的体积（cm³）并换算成升。底部材料每平方厘米成本为 0.05 英镑。求底部材料的成本。

Step 1: Volume = length × width × height = 80 × 50 × 30 = 120 000 cm³.

步骤1：体积 = 长 × 宽 × 高 = 80 × 50 × 30 = 120 000 cm³。

Step 2: 1 litre = 1 000 cm³, so volume in litres = 120 000 ÷ 1 000 = 120 L.

步骤2：1 升 = 1 000 cm³，因此体积以升为单位 = 120 000 ÷ 1 000 = 120 L。

Step 3: Base area = length × width = 80 × 50 = 4 000 cm².

步骤3：底部面积 = 长 × 宽 = 80 × 50 = 4 000 cm²。

Step 4: Cost = area × unit cost = 4 000 × 0.05 = £200.

步骤4：成本 = 面积 × 单位成本 = 4 000 × 0.05 = 200 英镑。

---

8. Percentage Change and Population (Geography) | 百分比变化与人口（地理）

A town’s population was 24 000 in 2015 and increased to 27 600 by 2023. Calculate the percentage increase over this period. If the same percentage increase occurs over the next 8 years, what will the population be in 2031?

一个城镇的人口在 2015 年为 24 000，到 2023 年增长到 27 600。计算这段时期内的增长百分比。如果未来 8 年发生相同的百分比增长，2031 年的人口会是多少？

Step 1: Population change = 27 600 – 24 000 = 3 600.

步骤1：人口变化 = 27 600 – 24 000 = 3 600。

Step 2: Percentage increase = (3 600 ÷ 24 000) × 100 = 0.15 × 100 = 15%.

步骤2：增长百分比 = (3 600 ÷ 24 000) × 100 = 0.15 × 100 = 15%。

Step 3: For the next 8 years, apply the same 15% increase to the 2023 population: 27 600 × 1.15 = 31 740. So the projected population is 31 740.

步骤3：对于未来 8 年，对 2023 年的人口应用同样的 15% 增长率：27 600 × 1.15 = 31 740。因此预计人口为 31 740。

---

9. Interpreting Scientific Graphs | 解读科学图表

In a chemistry experiment, the temperature of a solution is recorded every minute. The graph shows a linear increase from 20 °C at 0 min to 50 °C at 10 min. Find the rate of temperature increase per minute. Write an equation linking temperature T (°C) and time t (min). Use the equation to predict the temperature at 15 min.

在一个化学实验中，每分钟记录溶液的温度。图表显示从 0 分钟时的 20 °C 线性增加到 10 分钟时的 50 °C。求每分钟温度升高的速率。写出联系温度 T (°C) 与时间 t (min) 的方程。用该方程预测 15 分钟时的温度。

Step 1: Rate of increase = change in T / change in t = (50 – 20) / (10 – 0) = 30 ÷ 10 = 3 °C per minute.

步骤1：升高速率 = T 的变化量 / t 的变化量 = (50 – 20) / (10 – 0) = 30 ÷ 10 = 3 °C/分钟。

Step 2: The linear equation is T = starting temperature + rate × time, so T = 20 + 3t.

步骤2：线性方程为 T = 起始温度 + 速率 × 时间，因此 T = 20 + 3t。

Step 3: For t = 15 min, T = 20 + 3 × 15 = 20 + 45 = 65 °C.

步骤3：当 t = 15 分钟时，T = 20 + 3 × 15 = 20 + 45 = 65 °C。

---

10. Algebraic Formulae in Physics (Newton’s Second Law) | 物理中的代数公式（牛顿第二定律）

Newton’s second law states that F = m × a, where F is force in newtons (N), m is mass in kg, and a is acceleration in m/s². A cart of mass 12 kg accelerates at 2.5 m/s². Find the force required. Then, if the same force is applied to a 15 kg cart, what acceleration results?

牛顿第二定律指出 F = m × a，其中 F 是力（牛顿），m 是质量（kg），a 是加速度（m/s²）。一辆质量为 12 kg 的小车以 2.5 m/s² 加速。求需要的力。然后，如果同样的力施加到一辆 15 kg 的小车上，会产生多大的加速度？

Step 1: Substitute m = 12 and a = 2.5 into F = m × a: F = 12 × 2.5 = 30 N.

步骤1：将 m = 12 和 a = 2.5 代入 F = m × a：F = 12 × 2.5 = 30 N。

Step 2: Rearrange the formula to make a the subject: a = F ÷ m. With F = 30 and m = 15, a = 30 ÷ 15 = 2 m/s².

步骤2：重新排列公式，使 a 成为主体：a = F ÷ m。代入 F = 30 和 m = 15，a = 30 ÷ 15 = 2 m/s²。

Step 3: Notice that for the same force, a larger mass results in a smaller acceleration, consistent with inverse proportion.

步骤3：注意，同样大小的力，质量越大加速度越小，符合反比关系。

---

Published by TutorHao | Further Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）