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

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

This walkthrough takes you through a mock unit test designed for Year 11 AQA Computer Science, unpacking key answers and showing the reasoning behind each question. You’ll encounter topics from data representation, algorithms, programming, and computer systems, all aligned with the AQA specification. Use this guide to identify common mistakes, reinforce your understanding, and build confidence for the real exam.

本文逐题解析一套为 Year 11 AQA 计算机科学设计的单元测试模拟卷，拆解关键答案并展示每道题的推理过程。内容覆盖数据表示、算法、编程和计算机系统等主题，完全匹配 AQA 考纲。通过这份指南，你可以识别常见错误、巩固知识，并为正式考试建立信心。

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1. Binary to Hexadecimal Conversion | 二进制与十六进制转换

Question: Convert the binary number 1011 1100₂ into hexadecimal. Show your working and explain why hexadecimal is often used by programmers.

题目：将二进制数 1011 1100₂ 转换为十六进制。展示计算过程，并解释程序员为何经常使用十六进制。

Split the binary string into nibbles (groups of 4 bits) starting from the right: 1011 and 1100. No zero-padding is needed here because both groups already contain 4 bits. Use the conversion table: 1011₂ = 11 in denary = B in hex; 1100₂ = 12 in denary = C in hex. Combining the nibbles gives the hexadecimal value BC₁₆.

从右边开始将二进制串每4位分为一组（半字节）：1011 和 1100。此处无需补零，因为两组都已有4位。根据转换表：1011₂ 对应十进制 11，即十六进制的 B；1100₂ 对应十进制 12，即十六进制的 C。组合后得到十六进制值 BC₁₆。

Hexadecimal is a compact way to represent binary. One hex digit stands for exactly 4 bits, so a byte can be written as two hex characters. This helps developers read memory dumps, colour codes and error messages far more easily than long binary strings.

十六进制是一种紧凑的二进制表示法。一个十六进制数字恰好表示4个二进制位，因此一个字节可用两个十六进制字符书写。这让开发者在查看内存转储、颜色码和错误信息时，远比阅读长串二进制便捷得多。

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2. Lossy vs Lossless Compression | 有损与无损压缩

Question: A student says, “I always use lossy compression because it makes files much smaller.” Criticise this claim by giving one advantage and one disadvantage of lossy compression compared with lossless compression.

题目：一名学生说：“我总是使用有损压缩，因为它能把文件变得小得多。” 通过与无损压缩对比，指出有损压缩的一个优点和一个缺点，从而批评这一说法。

The advantage is that lossy compression drastically reduces file size by permanently discarding some data, which saves storage space and speeds up transmission over a network. This is ideal for streaming music or video where a perfect copy is not essential.

优点在于有损压缩通过永久丢弃部分数据大幅缩减文件体积，既节省存储空间又能加速网络传输。这对于流媒体音乐或视频等无需完美副本的场景非常理想。

The main disadvantage is that the discarded data cannot be recovered, so the quality of the reconstructed file is lower than the original. If you repeatedly edit and save a lossy file, quality degrades further. In contrast, lossless compression retains all original data, making it suitable for text documents, software and archival purposes.

主要缺点是被丢弃的数据无法恢复，因此重建文件的质量低于原始文件。如果反复编辑并保存有损文件，质量会进一步下降。相反，无损压缩保留全部原始数据，适用于文本文档、软件和存档用途。

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3. Logic Gate Circuit Analysis | 逻辑门电路分析

Question: A circuit takes two inputs, A and B, feeds them into an AND gate, and then passes the output through a NOT gate. Draw a truth table for the final output Q.

题目：一电路有两个输入 A 和 B，将它们接入一个 AND 门，再将其输出通过 NOT 门。画出最终输出 Q 的真值表。

A	B	A AND B	Q (NOT)
0	0	0	1
0	1	0	1
1	0	0	1
1	1	1	0

The truth table shows that Q is 1 for all input combinations except when both A and B are 1. The Boolean expression for the circuit is Q = ¬(A ∧ B), which is simply A NAND B. NAND gates are very common in digital electronics because they are functionally complete.

真值表显示除 A 与 B 均为 1 的情况外，Q 恒为 1。该电路的布尔表达式为 Q = ¬(A ∧ B)，即 A NAND B。NAND 门在数字电路中被广泛使用，因为它是功能完备门。

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4. Tracing a Pseudocode Algorithm | 伪代码算法追踪

Question: Study the pseudocode below. Complete a trace table showing the values of a and b after each iteration of the loop.
a ← 3
b ← 1
FOR i ← 1 TO 3
b ← b + a
a ← a × 2
ENDFOR

题目：研究下列伪代码。完成追踪表，展示每次循环后 a 与 b 的值。
a ← 3
b ← 1
FOR i ← 1 TO 3
b ← b + a
a ← a × 2
ENDFOR

i (loop counter)	a	b
–	3	1
1	6	4
2	12	10
3	24	22

The loop runs three times. In each iteration, b is first increased by the current value of a, then a is multiplied by 2. After the final iteration, a holds 24 and b holds 22. Tracing exercises like this strengthen your ability to simulate code in your head—a vital exam skill.

循环执行三次。每次迭代中，b 先加上当前 a 值，然后 a 自乘 2。最终迭代后 a 为 24，b 为 22。此类追踪练习可强化大脑模拟代码的能力，是重要的应试技巧。

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5. Identifying Programming Errors | 识别编程错误

Question: The following Python function is meant to return ‘Hot’ if the temperature is above 30, otherwise ‘Cold’. Identify and explain two mistakes.
1 def check_temp(t):
2 if t > 30
3 return ‘Hot’
4 else
5 return ‘Cold’

题目：以下 Python 函数旨在温度高于 30 时返回 ‘Hot’，否则返回 ‘Cold’。找出并解释两处错误。
1 def check_temp(t):
2 if t > 30
3 return ‘Hot’
4 else
5 return ‘Cold’

The first error is a missing colon at the end of the ‘if’ statement (line 2). In Python, a colon is required to start the code block. The second error is a missing colon after the ‘else’ keyword (line 4). Both are syntax errors that will cause the interpreter to raise an error before executing the code.

第一个错误是 ‘if’ 语句末尾（第2行）缺少冒号。Python 要求用冒号开始代码块。第二个错误是 ‘else’ 关键字后缺少冒号（第4行）。两处均为语法错误，会导致解释器在执行代码前抛出异常。

A logically correct version would be:
def check_temp(t):
if t > 30:
return ‘Hot’
else:
return ‘Cold’
Always check for missing colons and indentation when debugging Python—these are among the most common mistakes for GCSE candidates.

逻辑正确的版本应为：
def check_temp(t):
if t > 30:
return ‘Hot’
else:
return ‘Cold’
调试 Python 时务必检查缺失的冒号及缩进——这是 GCSE 考生最常见的失误之一。

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6. Linear Search vs Binary Search | 线性搜索与二分搜索

Question: Explain why binary search is faster than linear search on large, sorted lists. State one situation where linear search would be the better choice.

题目：解释为何在大型有序列表中二分搜索比线性搜索更快，并指出一种线性搜索更优的情形。

Binary search works by repeatedly dividing the search interval in half, comparing the middle element with the target. This eliminates half of the remaining items each step, giving a worst-case time complexity of O(log n). Linear search checks every element one by one, yielding O(n). For a list of one million items, binary search takes at most about 20 comparisons, while linear search may need one million.

二分搜索通过反复将搜索区间一分为二并比较中间元素来工作，每次可剔除剩余项的一半，最坏时间复杂度为 O(log n)。线性搜索逐一检查每个元素，复杂度为 O(n)。对于一百万个元素的列表，二分搜索最多约需20次比较，而线性搜索可能需要一百万次。

Linear search is the better choice when the list is unsorted, because binary search requires a sorted list to guarantee correctness. It is also simpler to code and may be faster for very small datasets where the overhead of sorting outweighs the benefit.

当列表无序时线性搜索更优，因为二分搜索要求列表有序才能保证正确。此外线性搜索编程更简单，对于极小数据集，排序带来的开销可能超过二分搜索的优势。

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7. Data Structures – 2D Arrays | 数据结构 —— 二维数组

Question: A game board is stored in a 2D array ‘board’. The board has 5 rows and 6 columns, and both indices start at 0. Write the index expression to access the cell in the fourth row and the second column. Then explain what board[3][0] refers to.

题目：游戏棋盘存储在一个二维数组 ‘board’ 中。该棋盘有5行6列，两个索引均从0开始。写出访问第4行第2列单元格的索引表达式。然后说明 board[3][0] 指向什么。

Since counting starts at 0, the fourth row is index 3, and the second column is index 1. The expression is board[3][1] (or board[3,1] in some pseudocode conventions). board[3][0] refers to the cell in the fourth row (index 3) and the first column (index 0).

由于计数从0开始，第4行索引为3，第2列索引为1。表达式为 board[3][1]（某些伪代码规范中写作 board[3,1]）。board[3][0] 指向第四行（索引3）第一列（索引0）的单元格。

Two-dimensional arrays are very common in board games, spreadsheets and pixel images. Remembering that index numbering usually starts at 0 prevents off-by-one errors in the exam.

二维数组在棋类游戏、电子表格和像素图像中极为常见。牢记索引通常从0开始编号，可避免考试中的 off-by-one（错位）错误。

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8. CPU Components and the FDE Cycle | CPU 组件与取指-解码-执行周期

Question: Describe the roles of the Program Counter (PC) and Memory Data Register (MDR) during the Fetch-Decode-Execute cycle. Explain how the PC changes after a jump instruction.

题目：描述程序计数器（PC）和内存数据寄存器（MDR）在取指-解码-执行周期中的作用。解释遇到跳转指令后 PC 如何变化。

During the fetch stage, the PC holds the address of the next instruction to be processed. This address is copied to the Memory Address Register (MAR), and the instruction is fetched from RAM into the M

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