Year 11 Edexcel Computer Science: Unit Test Mock Exam Walkthrough | Year 11 Edexcel 计算机：单元测试模拟卷解析

📚 Year 11 Edexcel Computer Science: Unit Test Mock Exam Walkthrough | Year 11 Edexcel 计算机：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Year 11 Edexcel Computer Science, covering key topics such as data representation, algorithms, programming, and computer systems. Each question is explained with the correct answer and common pitfalls.

本文详细解析一份针对 Year 11 Edexcel 计算机科学的单元测试模拟卷，涵盖数据表示、算法、程序设计和计算机系统等关键主题。每道题都给出了正确答案和常见错误分析。

1. Binary Addition | 二进制加法

Question: Add the binary numbers 1011 and 0110. Show your working.

题目：将二进制数 1011 与 0110 相加，并展示运算过程。

the carried 1 = 10, write 0 and carry 1 again. Leftmost column: 1 + 0 + the carried 1 = 10, write 0 and carry 1. Since there is a final carry, it becomes the new leftmost digit. The result is 10001.

首先对齐两个二进制数。从最右列开始：1 + 0 = 1。下一列：1 + 1 = 10，写 0 进 1。第三列：0 + 1 + 进位 1 = 10，写 0 再进 1。最左列：1 + 0 + 进位 1 = 10，写 0 进 1。最终进位成为新的最高位，结果为 10001。

with the carry bit. The decimal check: 1011₂ = 11, 0110₂ = 6, sum = 17₁₀, which is 10001₂.

常见错误是忘记最后一列的进位，得出 10101。务必确保每一列都加上进位。用十进制验证：1011₂ = 11，0110₂ = 6，和为 17₁₀，对应 10001₂。

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2. Logic Gates and Truth Tables | 逻辑门与真值表

Draw the truth table for the logic circuit with inputs A, B and output Q = NOT (A AND B).

题目：画出输入为 A、B，输出为 Q = NOT (A AND B) 的逻辑电路的真值表。

When A=0, B=1: AND = 0, Q = 1. When A=1, B=1: AND = 1, Q = 0. The truth table therefore has three rows where Q=1 and only the last row gives Q=0.

该表达式是一个与非门（NAND）。两个输入共有四种组合。当 A=0, B=0：A AND B = 0，因此 Q = NOT 0 = 1。A=1, B=0：AND = 0，Q = 1。A=0, B=1：AND = 0，Q = 1。A=1, B=1：AND = 1，Q = 0。所以真值表中有三行 Q=1，只有最后一行 Q=0。

Students sometimes confuse NAND with NOR. Remember that NAND is the opposite of AND, so it only outputs 0 when both inputs are 1. Always list all combinations in a systematic order: 00, 01, 10, 11.

学生有时会将 NAND 与 NOR 混淆。记住 NAND 是 AND 的反相，只有当两个输入都为 1 时输出才为 0。务必按顺序列出所有组合：00、01、10、11。

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3. Pseudocode Tracing | 伪代码跟踪

Question: What is the output of this pseudocode?
a ← 5
b ← 2
while a > 0
a ← a - b
output a
endwhile

题目：下列伪代码的输出是什么？
a ← 5
b ← 2
while a > 0
a ← a - b
output a
endwhile

trace the loop step by step. Initially a = 5. First iteration: 5 > 0 is true, a becomes 5 – 2 = 3, output 3. Second iteration: 3 > 0 is true, a = 3 – 2 = 1, output 1. Third iteration: 1 > 0 is true, a = 1 – t -1. Fourth iteration: -1 > 0 is false, so the loop terminates. The outputs are 3, 1, -1.

逐步跟踪循环。最初 a = 5。第一次迭代：5 > 0 为真，a 变为 5 – 2 = 3，输出 3。第二次迭代：3 > 0 为真，a = 3 – 2 = 1，输出 1。第三次迭代：1 > 0 为真，a = 1 – 2 = -1，输出 -1。第四次迭代： -1 > 0 为假，循环结束。输出依次为 3, 1, -1。

A common pitfall is assuming the loop stops once a becomes less than b. In reality the condition is checked only at the start of each iteration, so a negative value is still output before the loop exits.

常见陷阱是认为一旦 a 小于 b 循环就会停止。实际上条件仅在每次循环开始时检查，因此在退出前仍会输出一个负数值。

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4. Image Representation | 图像表示

Question: An image is 200 pixels wide and 300 pixels high, with a colour depth of 24 bits. Calculate the file size in kilobytes (KB).

题目：一幅图像宽 200 像素、高 300 像素，颜色深度为 24 位。计算其文件大小，以千字节（KB）为单位。

000 × 3 = 180,000 bytes. To convert to kilobytes, divide by 1024: 180,000 ÷ 1024 ≈ 175.78 KB.

总像素数 = 宽 × 高 = 200 × 300 = 60,000 像素。每个像素需要 24 位，即 3 字节（因为 1 字节 = 8 位）。总字节数 = 60,000 × 3 = 180,000 字节。转换为千字节，除以 1024：180,000 ÷ 1024 ≈ 175.78 KB。

lso remember that 24-bit colour depth means 8 bits for each of red, green, and blue. The answer can be given to two decimal places.

注意使用正确的换算因子：1 KB = 1024 字节，而非 1000。同时记住 24 位颜色深度意味着红、绿、蓝各占 8 位。答案可保留两位小数。

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5. Data Compression | 数据压缩

Question: Explain the difference between lossy and lossless compression, giving an example of each.

题目：解释有损压缩与无损压缩的区别，并各举一例。

Lossless compression reduces file size without any loss of data. The original file can be perfectly reconstructed. A common example is PNG images for graphics or ZIP for documents. Lossy compression permanently removes some data to reduce file size further. This results in a slight reduction in quality but achieves much smaller files. Examples include JPEG for photographs and MP3 for audio.

无损压缩在减小文件大小的同时不丢失任何数据，原始文件可被完美重建。常见例子如用于图形的 PNG 图像或文档压缩的 ZIP。有损压缩会永久性地移除部分数据以进一步减小文件体积，这会带来少量质量损失，但能获得小得多的文件。例如照片的 JPEG 格式和音频的 MP3 格式。

Lossy is generally preferred for media where perfect reproduction is less critical, whereas lossless is essential for text, programs, and archival purposes. In the exam, it is important to link the example to the explanation, e.g., JPEG reduces detail imperceptible to the human eye.

有损压缩通常适用于对完美再现要求不高的媒体，而无损压缩对于文本、程序和存档至关重要。在考试中，需要将例子与解释结合起来，例如 JPEG 会减少人眼难以察觉的细节。

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6. Network Protocols | 网络协议

Question: Which protocol is used to transfer web pages securely? Explain its role.

题目：哪个协议用于安全传输网页？解释其作用。

The protocol is HTTPS (Hypertext Transfer Protocol Secure). It encrypts the data exchanged between a web browser and a web server using SSL/TLS (Secure Sockets Layer/Transport Layer Security). This encryption ensures that sensitive information such as passwords and credit card numbers cannot be intercepted and read by attackers. It also provides authentication, so the user can trust the identity of the website.

该协议为 HTTPS（安全超文本传输协议）。它使用 SSL/TLS（安全套接层/传输层安全）对浏览器与 Web 服务器之间交换的数据进行加密。这种加密确保密码和信用卡号等敏感信息不会被攻击者截获并读取。它还能提供身份验证，让用户确认网站的真实身份。

A common mistake is writing HTTP instead of HTTPS. HTTP sends data in plain text, making it insecure for sensitive transactions. The padlock icon in the browser indicates an active HTTPS connection.

常见错误是写成 HTTP 而非 HTTPS。HTTP 以明文传输数据，不适合敏感交易。浏览器中的锁形图标表示正在使用 HTTPS 连接。

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7. Ethical and Legal Issues | 伦理与法律问题

Question: Discuss one ethical issue with the use of facial recognition technology.

题目：讨论使用面部识别技术的一个伦理问题。

One major ethical issue is the invasion of privacy. Facial recognition systems can be deployed in public spaces without individuals' consent, enabling mass surveillance. People may be tracked without their knowledge, leading to a loss of anonymity. This raises concerns about how the collected data is stored, used, and who has access to it. Ethically, individuals should have the right to control their biometric data.

一个主要的伦理问题是侵犯隐私。面部识别系统可以在未经个人同意的情况下部署在公共场所，从而实现大规模监控。人们可能在不知情的情况下被追踪，导致匿名性丧失。这引发了关于所收集数据如何存储、使用以及谁能访问的担忧。从伦理上讲，个人应有权控制自己的生物特征数据。

Another issue is bias and fairness. Studies have shown that some facial recognition algorithms are less accurate for certain ethnic groups, which can lead to misidentification and discrimination. In an exam answer, it is sufficient to pick one well-explained ethical point, linking it clearly to the technology.

另一个问题是偏见与公平性。研究表明，某些面部识别算法对特定族群的准确率较低，这可能导致误认和歧视。在考试中，只需选择一个清晰的伦理观点并联系该技术加以充分解释即可。

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8. Identifying Programming Errors | 识别编程错误

Question: Find and explain the errors in the following Python code:
for i in range(5,0):
    print(i)

题目：找出并解释以下 Python 代码中的错误：
for i in range(5,0):
    print(i)

There is a logic error, not a syntax error. The range(5, 0) call creates an empty sequence because the start value 5 is greater than the stop value 0 with a default step of 1. Therefore, the loop body never executes, and nothing is printed. The programmer likely intended to count down from 5 to 1, which requires a negative step: range(5, 0, -1).

这里存在逻辑错误，而非语法错误。调用 range(5, 0) 会生成一个空序列，因为起始值 5 大于终止值 0，且默认步长为 1。因此循环体根本不会执行，没有任何输出。程序员可能想从 5 倒数到 1，这需要使用负步长：range(5, 0, -1)。

Identifying whether an error is syntax, logic, or runtime is crucial. A syntax error would be a missing colon, while this code runs but does not produce the expected output, making it a logic error. Always check the iteration direction and step size.

区分语法错误、逻辑错误和运行时错误至关重要。语法错误例如缺少冒号，而此代码能够运行但没有产生预期输出，属于逻辑错误。务必检查迭代方向和步长大小。

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9. SQL Queries | SQL 查询

Question: Write an SQL statement to select all students with a grade greater than 80 from a table named Students.

题目：编写一条 SQL 语句，从名为 Students 的表中选出所有 grade 大于 80 的学生。

The correct SQL is: SELECT * FROM Students WHERE grade > 80; The * wildcard returns all columns. If only certain fields are required, you could list them, e.g., SELECT name, grade FROM Students WHERE grade > 80. Always terminate with a semicolon, and ensure the condition matches the data type — if grade is numeric, no quotes are needed around 80.

正确的 SQL 语句为：SELECT * FROM Students WHERE grade > 80；使用 * 通配符可返回所有列。如果只需要特定字段，可将其列出，如：SELECT name, grade FROM Students WHERE grade > 80。务必以分号结尾，并确保条件匹配数据类型——若 grade 为数值型，80 两侧不需要引号。

A common mistake is writing WHERE grade > '80', which treats the value as text and may lead to incorrect comparisons. Also, do not confuse the order of clauses; SELECT comes first, followed by FROM, then WHERE.

常见错误是写成 WHERE grade > '80'，这会把值当作文本处理，可能导致错误的比较。同时，不要混淆子句顺序：首先是 SELECT，然后是 FROM，最后是 WHERE。

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10. Sorting Algorithms | 排序算法

Question: Describe how bubble sort works on the list [4, 2, 7, 1] and show the first pass.

题目：描述冒泡排序在列表 [4, 2, 7, 1] 上的工作过程，并展示第一趟排序。

t pass: compare 4 and 2 → swap, list becomes [2, 4, 7, 1]. Compare 4 and 7 → no swap. Compare 7 and 1 → swap, list becomes [2, 4, 1, 7]. The largest element, 7, has 'bubbled' to the end. Subsequent passes will sort the remaining unsorted portion.

冒泡排序通过重复遍历列表，比较相邻元素并在顺序不对时交换它们来工作。第一趟：比较 4 与 2 → 交换，列表变为 [2, 4, 7, 1]。比较 4 与 7 → 不交换。比较 7 与 1 → 交换，列表变为 [2, 4, 1, 7]。此时最大元素 7 已“冒泡”至末尾。后续趟次将对剩余的未排序部分继续排序。

4 and 1 (swap) → [2, 1, 4, 7]. The process repeats until no swaps occur in a pass, indicating the list is sorted.

第一趟结束后，列表并未完全排序，但末尾必定是最大值。第二趟会比较 2 与 4（不交换）、4 与 1（交换）→ [2, 1, 4, 7]。重复此过程，直到某趟没有发生任何交换，表示列表已排序。

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