Year 10 Edexcel Computer Science: Interdisciplinary Integrated Question Practice

📚 Year 10 Edexcel Computer Science: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练

Interdisciplinary questions in Computer Science test your ability to connect core topics with mathematics, physics, business, and ethics. This comprehensive practice guide presents typical integrated exam-style problems from the Edexcel Year 10 specification, with detailed step-by-step solutions designed to strengthen your analytical thinking and subject links.

计算机科学中的跨学科问题考查你将核心主题与数学、物理、商业和伦理联系起来的能力。这份综合训练指南针对爱德思 Year 10 考纲，提供典型的综合考题，并配有详细的分步解答，旨在强化你的分析思维和学科关联能力。

1. Introduction to Interdisciplinary Questions | 跨学科问题简介

Interdisciplinary questions often combine binary arithmetic with mathematical reasoning, or require you to evaluate network performance using physics formulas. In Edexcel exams, such questions reward not only recall but the ability to apply knowledge across two or more domains. The following sections walk you through key areas where this skill is essential.

跨学科问题通常将二进制运算与数学推理相结合，或者要求运用物理公式评估网络性能。在爱德思考试中，这类问题不仅考查记忆，更看重跨领域应用知识的能力。接下来的小节将带你练习几个核心领域中此项技能的关键题型。

2. Binary and Hexadecimal Conversions with Arithmetic | 二进制与十六进制转换及算术运算

Question 1: Convert the binary number 11010110₂ to its hexadecimal equivalent, then add the hexadecimal value 3A₁₆ and express the result back in binary.

问题1：将二进制数 11010110₂ 转换为等值的十六进制，再将结果与十六进制数 3A₁₆ 相加，把最终结果表示回二进制。

Step-by-step solution: First, split 11010110₂ into nibbles: 1101₂ and 0110₂. 1101₂ = 13₁₀ = D (hex), 0110₂ = 6. So the number is D6₁₆. Second, add D6₁₆ and 3A₁₆: D6 (214₁₀) + 3A (58₁₀) = 272₁₀. Convert 272₁₀ to hex: 272 ÷ 16 = 17 remainder 0 → 11₀, but 17 ÷ 16 = 1 remainder 1 → 110₁₆ (that is 1×16² + 1×16 + 0 = 256+16). Wait, recalculate: 272 / 16 = 17 r 0, 17 / 16 = 1 r 1, so hex digits are 1,1,0 → 110₁₆. Check: 1×256 + 1×16 + 0 = 272. Finally convert 110₁₆ to binary: 1→0001, 1→0001, 0→0000 → 0001 0001 0000₂, or simply 100010000₂. Final answer: 100010000₂.

分步解答：首先将 11010110₂ 按4位分组：1101₂ 和 0110₂。1101₂ = 13₁₀ = D（十六进制），0110₂ = 6。因此该数为 D6₁₆。第二步，计算 D6₁₆ + 3A₁₆：D6（十进制214） + 3A（十进制58） = 272₁₀。将 272₁₀ 转为十六进制：272 ÷ 16 = 17 余 0，17 ÷ 16 = 1 余 1，所以十六进制位依次为 1, 1, 0 → 110₁₆。验证：1×256 + 1×16 + 0 = 272。最后将 110₁₆ 转为二进制：1 → 0001，第二个1 → 0001，0 → 0000，组合为 0001 0001 0000₂，简写为 100010000₂。最终答案：100010000₂。

Question 2: A colour is stored using a 16-bit representation with 5 bits for red, 6 bits for green and 5 bits for blue. Calculate the maximum number of distinct colours and express the answer as a power of 2. Then explain why this uses fewer colours than a 24-bit system.

问题2：一种颜色使用16位表示，其中红色5位、绿色6位、蓝色5位。计算最大可区分的颜色数量，并用2的幂表示。然后解释为何这比24位系统使用的颜色少。

Answer: Total bits = 16, thus maximum colours = 2¹⁶ = 65,536. A 24-bit system gives 2²⁴ = 16,777,216 colours, which is much larger because exponent increases by 8. This links to mathematical indices.

解答：总位数 = 16，因此最大颜色数 = 2¹⁶ = 65,536 种。24位系统提供 2²⁴ = 16,777,216 种颜色，颜色数显著增加，因为指数增大了8。这直接联系到数学中的幂指数概念。

3. Logic Gates and Boolean Algebra | 逻辑门与布尔代数

Interdisciplinary link: Boolean algebra is a branch of algebra where variables take TRUE or FALSE (1 or 0). This directly overlaps with mathematical logic and circuit design in Physics.

跨学科联系：布尔代数是代数的一个分支，变量只能取 TRUE 或 FALSE（1 或 0）。它直接与数学逻辑以及物理中的电路设计重叠。

Question: Simplify the Boolean expression Q = (A ∧ B) ∨ (¬A ∧ B) and represent it using a single logic gate.

问题：化简布尔表达式 Q = (A ∧ B) ∨ (¬A ∧ B)，并用单一逻辑门表示。

Simplification: Factor B: Q = B ∧ (A ∨ ¬A). Since A ∨ ¬A = 1, Q = B ∧ 1 = B. So the expression simplifies to B. The circuit is just a buffer (or a wire). This simplification uses the law of complementarity from Boolean algebra, which reflects mathematical distributive laws.

化简：提取公因子 B：Q = B ∧ (A ∨ ¬A)。因为 A ∨ ¬A = 1，所以 Q = B ∧ 1 = B。因此表达式简化为 B。电路只需一条直通线（或缓冲器）。这个化简运用了布尔代数中的互补律，反映了数学中的分配律。

Truth table practice: Build the truth table for A XOR B and show its equivalence to (A ∨ B) ∧ ¬(A ∧ B).

真值表练习：构建 A XOR B 的真值表，并证明它与 (A ∨ B) ∧ ¬(A ∧ B) 等价。

A	B	A XOR B	(A∨B) ∧ ¬(A∧B)
0	0	0	0
0	1	1	1
1	0	1	1
1	1	0	0

The columns match, confirming the logical equivalence. XOR is crucial in error detection and cryptography.

两列完全一致，证明逻辑等价。XOR 在错误检测和密码学中至关重要。

4. Algorithmic Thinking and Flowcharts | 算法思维与流程图

A flowchart for a temperature control system in a greenhouse combines sensors (Physics) with decision-making. Examine the pseudocode below that reads temperature and turns on a heater if the value is below 5 °C. Identify the mathematical comparison and rewrite using a while loop.

温室温度控制系统的流程图结合了传感器（物理）与决策。请阅读下面的伪代码，它读取温度并在低于5 °C 时打开加热器。识别其中的数学比较，并用 while 循环重新编写。

Original code:

INPUT temp
IF temp < 5 THEN
    OUTPUT 'Heater ON'
ELSE
    OUTPUT 'Heater OFF'
ENDIF

Translation to while loop (monitoring continuously):

WHILE True:
    INPUT temp
    IF temp < 5 THEN
        OUTPUT 'Heater ON'
    ELSE
        OUTPUT 'Heater OFF'
    ENDIF
    WAIT 60 seconds
ENDWHILE

Here the mathematical condition (<) controls the physical actuator. This illustrates how algorithms integrate real-world constraints.

此处数学条件（<）控制物理执行器。这说明了算法如何整合现实世界约束。

Question: Write a flowchart and pseudocode to find the average of three sensor readings. Show how the sum is calculated and divided by 3.

问题：编写流程图和伪代码，求三个传感器读数的平均值。展示如何求和并除以3。

Solution sketch: Start → input r1, r2, r3 → sum = r1+r2+r3 → avg = sum/3 → output avg → End. The division is basic arithmetic.

解答草图：开始 → 输入 r1, r2, r3 → sum = r1+r2+r3 → avg = sum/3 → 输出 avg → 结束。除法属于基本算术。

5. Data Representation: Sound and Images | 数据表示：声音与图像

A sound wave is sampled at 44.1 kHz with 16-bit resolution for 3 minutes. Calculate the file size in megabytes, ignoring compression. This ties into physics (wave sampling) and mathematics (unit conversions).

以 44.1 kHz 的采样率、16 位分辨率录制 3 分钟的声音。计算文件大小（兆字节），忽略压缩。这个问题联系物理（波形采样）和数学（单位换算）。

Calculation: Sampling rate = 44,100 samples per second. Bit depth = 16 bits = 2 bytes per sample. Duration = 3 × 60 = 180 seconds. Data per second = 44,100 × 2 bytes = 88,200 bytes. Total = 88,200 × 180 = 15,876,000 bytes. Convert to MB: 15,876,000 ÷ (1024×1024) ≈ 15.14 MB (or using 1,000,000 approach: 15.876 MB). In exams, follow specification convention; Edexcel often uses 1 MB = 1,000,000 bytes or 1,048,576 bytes. Clarify with teacher. For now, using 1024², it's about 15.14 MB.

计算：采样率 = 44,100 次/秒。位深度 16 位 = 2 字节/采样点。时长 = 3×60 = 180 秒。每秒数据量 = 44,100 × 2 字节 = 88,200 字节。总量 = 88,200 × 180 = 15,876,000 字节。转换为 MB：15,876,000 ÷ (1024×1024) ≈ 15.14 MB（或按 1,000,000 计为 15.876 MB）。考试中遵循考纲约定；爱德思通常使用 1 MB = 1,000,000 字节还是 1,048,576 字节需明确。此处以 1024² 估算约为 15.14 MB。

Image question: A 400×300 pixel image uses a colour depth of 24 bits. Determine the storage needed in kilobytes. Then shrink the image to 200×150 using interpolation. Explain why file size reduces by a factor of 4.

图像问题：一幅 400×300 像素的图像，颜色深度为 24 位。确定所需存储空间（千字节）。然后将图像缩小到 200×150，使用插值法。解释为什么文件大小缩减为原来的四分之一。

Solution: Original pixels = 400×300 = 120,000 px. Bits = 120,000 × 24 = 2,880,000 bits. Bytes = 2,880,000 / 8 = 360,000 bytes. KB = 360,000 / 1024 ≈ 351.56 KB. Resized pixels = 200×150 = 30,000 px. This is one fourth of 120,000, so file size also quarter (ignoring metadata). Mathematically, area scales with product of width and height.

解答：原始像素数 = 400×300 = 120,000 像素。总位 = 120,000 × 24 = 2,880,000 位。字节 = 2,880,000 / 8 = 360,000 字节。KB = 360,000 / 1024 ≈ 351.56 KB。缩放后像素 = 200×150 = 30,000 像素。这是原像素数的四分之一，因此文件大小也变为四分之一（忽略元数据）。数学上，面积随宽和高的乘积缩放。

6. Networking and Data Transmission | 网络与数据传输

A 5 MB file is transmitted over a network with a bandwidth of 100 Mbps. Calculate the minimum transmission time ignoring overheads. Then if the round-trip time (RTT) is 20 ms, estimate the total time with a simple stop-and-wait protocol where an acknowledgement is sent after each packet of 1250 bytes. (Link to physics: signal propagation; maths: unit conversions.)

一个 5 MB 的文件通过带宽为 100 Mbps 的网络发送。计算忽略开销的最小传输时间。若往返时延（RTT）为 20 ms，且使用简单的停止-等待协议，每发送一个 1250 字节的数据包后需等待确认，估算总传输时间。（联系物理：信号传播；数学：单位换算。）

Data size: 5 MB = 5 × 1,000,000 = 5,000,000 bytes (using SI units) or = 5 × 1,048,576 if binary. Use 5,000,000 bytes for simplicity. Bandwidth 100 Mbps = 100,000,000 bits per second = 12,500,000 bytes per second (since 8 bits/byte). Transmission time = 5,000,000 / 12,500,000 = 0.4 seconds. With stop-and-wait: packet size 1250 bytes, number of packets = 5,000,000 / 1250 = 4,000 packets. Each packet requires trans time 1250/12,500,000 = 0.0001 s (0.1 ms), plus RTT 20 ms = 0.02 s for each ack. Total per packet ≈ 0.0201 s. Total time = 4,000 × 0.0201 = 80.4 seconds. This huge increase shows the inefficiency of stop-and-wait.

数据量：5 MB = 5,000,000 字节（采用 SI 单位）。带宽 100 Mbps = 每秒 100,000,000 位 = 12,500,000 字节/秒。传输时间 = 5,000,000 / 12,500,000 = 0.4 秒。停止-等待协议：数据包大小 1250 字节，包数 = 5,000,000 / 1250 = 4,000 个。每个包传输时间 1250/12,500,000 = 0.0001 秒（0.1 ms），加上 RTT 20 ms = 0.02 秒等待确认。每个包总时间 ≈ 0.0201 秒。总时间 = 4,000 × 0.0201 ≈ 80.4 秒。传输时间剧增，说明停止-等待协议效率低下。

7. Database Design and SQL Queries | 数据库设计与 SQL 查询

A school database has tables STUDENT(StudentID, Name, YearGroup) and RESULT(StudentID, Subject, Grade). Write an SQL query to list names of Year 10 students who achieved a Grade 9 in Mathematics. This combines filtering across tables (JOIN) and linking to business data retrieval.

某学校数据库包含表 STUDENT(StudentID, Name, YearGroup) 和 RESULT(StudentID, Subject, Grade)。编写 SQL 查询，列出在数学科目获得 9 级成绩的 Year 10 学生姓名。这结合了跨表筛选（JOIN）和商业数据检索。

SELECT Name
FROM STUDENT
JOIN RESULT ON STUDENT.StudentID = RESULT.StudentID
WHERE YearGroup = 'Year 10' AND Subject = 'Mathematics' AND Grade = '9';

This query uses the mathematical concept of sets – the intersection of conditions. Cartoon analogy: you are finding the overlap of two circles (Year 10 students and Mathematics Grade 9 achievers).

该查询运用了数学中的集合概念——条件的交集。形象地说，你在寻找两个圆（Year 10 学生与数学 9 级获得者）的重叠部分。

Normalization question: Explain how splitting the data into two tables avoids data redundancy, linking to the idea of unique identifiers (keys). This relates to mathematical functions and dependencies.

规范化问题：解释将数据拆分成两个表如何避免数据冗余，并联系唯一标识符（键）的概念。这关联到数学中的函数与依赖关系。

Answer: Without normalization, repeating Student names for each result entry wastes storage. By using StudentID as a foreign key, each name is stored once, mirroring a mathematical mapping from ID to name.

解答：若不规范化，每条成绩记录都要重复学生姓名，浪费存储。通过将 StudentID 作为外键，每个姓名只存储一次，这类似于从 ID 到姓名的数学映射。

8. Ethical, Legal and Environmental Impacts | 道德、法律与环境影响

Computing technologies intersect with social sciences. Consider an AI recruitment tool that shortlists candidates. Discuss an ethical issue and propose a technical safeguard. Relate to bias in data sets, which connects to statistics (sample bias).

计算技术与社会学科交叉。考虑一个用于筛选候选人的 AI 招聘工具。讨论一项道德问题，并提出一项技术保障措施。关联数据集中的偏见，联系统计学中的样本偏差。

Answer: If training data historically favors male applicants, the algorithm may discriminate against women. A technical safeguard is to audit training data for demographic balance and apply re-weighting. Statistically, ensuring the sample is representative prevents skewed outcomes. Legal compliance with the Equality Act 2010 is necessary.

解答：如果训练数据历史上偏向男性申请者，算法可能歧视女性。技术保障措施是审核训练数据的人口平衡性，并施加重新加权。从统计学看，确保样本代表性可防止偏差结果。必须遵守《2010 年平等法案》的法律要求。

Environmental question: Data centres consume huge electricity. A cloud provider claims to use 100% renewable energy. Evaluate the environmental impact considering manufacturing of hardware. Link to physics energy concepts and lifecycle analysis.

环境问题：数据中心消耗大量电力。某云服务商声称使用 100% 可再生能源。评估其环境影响（需考虑硬件制造）。联系物理中的能源概念和生命周期分析。

Evaluation: While operation may be green, manufacturing servers involves mining rare earth metals and emitting CO₂. Embodied carbon should be accounted. This holistic view is a lifecycle assessment, an interdisciplinary approach drawing on environmental science.

评估：尽管运营可能绿色，但制造服务器涉及稀土开采和 CO₂ 排放，应计入隐含碳。这种整体观即生命周期评估，是一种借鉴环境科学的跨学科方法。

9. Programming Constructs with Real-world Applications | 编程结构与现实应用

Write a Python program to simulate a simple savings account with compound interest. Use the formula A = P(1 + r/100)ⁿ. This merges programming with financial mathematics. Provide code and explain how loops handle the exponent n years.

编写一个 Python 程序，模拟复利储蓄账户。使用公式 A = P(1 + r/100)ⁿ。这融合了编程与金融数学。提供代码并解释循环如何处理年数 n 的指数。

P = float(input('Initial deposit: '))
r = float(input('Annual interest rate (%): '))
n = int(input('Number of years: '))
amount = P
for year in range(n):
    amount = amount * (1 + r/100)
print('Final amount:', round(amount, 2))

The loop repeats multiplication n times, effectively computing the power. This demonstrates iteration as repeated multiplication. Could also use amount = P * (1 + r/100) ** n for direct calculation.

循环重复乘 n 次，有效计算了幂。这展示了迭代即重复乘法。也可以使用 amount = P * (1 + r/100) ** n 直接计算。

Another example: Predict the motion of a projectile. Input initial velocity v and angle theta (degrees). Calculate horizontal range. Use formulas from physics (suvat). This tests math library and trigonometric functions.

另一个例子：预测抛射体运动。输入初速度 v 和角度 θ（度）。计算水平射程。使用物理公式（suvat）。这考查数学库及三角函数。

import math
v = float(input('Velocity (m/s): '))
theta_deg = float(input('Angle (degrees): '))
g = 9.81
theta_rad = math.radians(theta_deg)
range_distance = (v**2 * math.sin(2 * theta_rad)) / g
print('Range:', round(range_distance, 2), 'm')

Here, sin(2θ) and unit conversion illustrate deep STEM integration.

此处 sin(2θ) 与单位转换体现了深度的 STEM 融合。

10. Error Detection and Correction | 错误检测与纠正

Even parity is used to detect single-bit errors. Given a 7-bit ASCII byte 1011001, add an even parity bit at the most significant position. Then, if the received byte is 01011001 (with parity), determine whether an error is detected. This connects to modular arithmetic (sum mod 2).

采用偶校验来检测单位错误。给定 7 位 ASCII 字节 1011001，在最高位添加偶校验位。然后，若接收到的字节为 01011001（含校验位），判断是否检测到错误。这联系模运算（总和 mod 2）。

Original data bits: 1+0+1+1+0+0+1 = 4 ones (even). Even parity aims for total even number of 1s. Since already even, parity bit = 0 → transmitted byte 0 1011001 (leading 0). Received byte 01011001 has bits: 0+1+0+1+1+0+0+1 = 4 ones, still even, so no error detected (though a two-bit error could go undetected).

原始数据位：1+0+1+1+0+0+1 = 4 个 1（偶数）。偶校验要求总 1 个数为偶数。因已为偶数，校验位 = 0 → 发送字节 0 1011001。接收字节 01011001 各位之和 = 0+1+0+1+1+0+0+1 = 4，仍为偶数，故未检测到错误（但两位错误可能漏检）。

Checksum: Calculate the checksum of two hexadecimal values A2₁₆ and 7F₁₆. Show addition and 1's complement. Applied maths with number bases.

校验和：计算十六进制值 A2₁₆ 和 7F₁₆ 的校验和。展示加法与反码。应用了不同数制的数学。

A2 + 7F = 0x121 (since A2 = 162, 7F=127, sum=289, 289 decimal = 0x121). For 8-bit checksum, often take lower byte: 0x21. 1's complement = 0xDE. This is used in network protocols.

A2 + 7F = 0x121（A2=162, 7F=127，和 289₁₀ = 0x121）。在 8 位校验和中，取低位字节：0x21，其反码为 0xDE。这用于网络协议中。

11. Integrated Case Study: Smart Home System | 综合案例研究：智能家居系统

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Year 10 Edexcel Computer Science: Interdisciplinary Integrated Question Practice | 跨学科综合题型训练