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

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

This walkthrough dissects a typical Year 11 Eduqas unit test mock paper, covering data representation, logic circuits, algorithms, programming, networking, cybersecurity and ethical computing. Every question is presented with a model solution, step-by-step reasoning and examiner tips to help you refine exam technique.

本文详细解析一份典型的 Year 11 Eduqas 计算机科学单元测试模拟卷，涵盖数据表示、逻辑电路、算法、编程、网络、网络安全与道德计算等考点。每道题均配有标准答案、逐步推导和考官提示，助你优化答题策略。

1. Binary Addition and Overflow | 二进制加法与溢出

Question: Add the two 8-bit binary numbers 01101101₂ and 01011011₂. Show all carries and state whether overflow occurs, justifying your answer.

题目：对两个 8 位二进制数 01101101₂ 和 01011011₂ 做加法。写出所有进位过程，判断是否发生溢出并说明理由。

y 1. Bit 3: 1+1 plus carry 1 = 1 carry 1. Bit 4: 0+0 plus carry 1 = 1 carry 0. Bit 5: 1+1 plus carry 0 = 0 carry 1. Bit 6: 1+0 plus carry 1 = 0 carry 1. Bit 7: 0+0 plus carry 1 = 1. The final result is 11001000₂.

逐步计算：从最低有效位（第0位）开始，1+1=0 进位1。第1位：0+1 加进位1=0 进位1。第2位：1+0 加进位1=0 进位1。第3位：1+1 加进位1=1 进位1。第4位：0+0 加进位1=1 进位0。第5位：1+1 加进位0=0 进位1。第6位：1+0 加进位1=0 进位1。第7位：0+0 加进位1=1。最终结果为 11001000₂。

01101101₂ = 109 and 01011011₂ = 91. Their sum, 200, lies within the 8-bit unsigned range 0– but the Eduqas specification typically assesses unsigned addition unless stated otherwise.

按无符号整数处理，01101101₂ = 109，01011011₂ = 91，和为200，处于8位无符号范围0–255内，因此无溢出。若采用有符号补码，两个正数相加若最高位变为1则表示溢出，但Eduqas规范若无特别说明，通常考察无符号加法。

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

Question: Boolean expression and complete the truth table. Hence find the output when A=1, B=0, C=1.

题目：某逻辑电路有 A、B、C 三个输入。先计算 A AND B，再将该结果与 C 做 OR 运算。写出布尔表达式并完成真值表，然后求出 A=1、B=0、C=1 时的输出。

The Boolean expression is Q = (A ∧ B) ∨ C. The truth table below confirms the output for all combinations.

布尔表达式为 Q = (A ∧ B) ∨ C。下面的真值表验证了所有组合的输出。

A	B	C	A ∧ B	Q = (A ∧ B) ∨ C
0	0	0	0	0
0	0	1	0	1
0	1	0	0	0
0	1	1	0	1
1	0	0	0	0
1	0	1	0	1
1	1	0	1	1
1	1	1	1	1

For the input A=1, B=0, C=1, the intermediate result is 1 ∧ 0 = 0. Then 0 ∨ 1 = 1, so the final output is 1.

对于输入 A=1、B=0、C=1，中间结果 1 ∧ 0 = 0，接着 0 ∨ 1 = 1，因此最终输出为 1。

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3. Representing Characters – ASCII and Unicode | 字符表示 – ASCII 与 Unicode

Question: Explain why the ASCII character set is insufficient for global communication and describe how Unicode overcomes these limitations. Give one concrete example.

题目：解释为什么 ASCII 字符集不适合全球沟通，并说明 Unicode 如何克服这些局限。给出一个具体的例子。

ASCII uses only 7 bits per character, providing 128 code points. This set covers basic Latin letters, digits and control characters, but it cannot represent Cyrillic, Chinese, Arabic or other scripts. Unicode addresses this by using up to 32 bits per character, with the most common form (UTF-8) using variable-length encoding. It currently supports over 143,000 characters, including emojis, mathematical symbols and historic scripts. For instance, the Chinese character '学' cannot be encoded in ASCII; Unicode assigns it the code point U+5B66.

ASCII 每个字符仅用7位，共128个码位，只能涵盖基本拉丁字母、数字和控制符号，无法表示西里尔、中文、阿拉伯文等文字。Unicode 通过每字符最多32位（常用UTF-8采用变长编码）解决了这一问题，目前支持超过143,000个字符，包括表情符号、数学符号和历史文字。例如，中文字符'学'在ASCII中无法编码，而Unicode分配码位U+5B66。

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4. Bubble Sort Trace | 冒泡排序追踪

Question: Apply bubble sort to the list [9, 5, 7, 3, 1]. Show the list after each pass and state the total number of passes and swaps performed.

题目：对列表 [9, 5, 7, 3, 1] 执行冒泡排序。显示每一趟之后的列表，并说明总共进行的趟数和交换次数。

7, 9, 3, 1]; 9 and 3 → swap → [5, 7, 3, 9, 1]; 9 and 1 → swap → [5, 7, 3, 1, 9] (4 swaps).

第1趟：比较9和5 → 交换 → [5, 9, 7, 3, 1]；9和7交换 → [5, 7, 9, 3, 1]；9和3交换 → [5, 7, 3, 9, 1]；9和1交换 → [5, 7, 3, 1, 9] (4次交换)。

no swap; 7,3 swap → [5, 3, 7, 1, 9]; 7,1 swap → [5, 3, 1, 7, 9]; 7,9 no swap → (2 swaps).

第2趟：[5, 7, 3, 1, 9] → 5,7 不交换；7,3 交换 → [5, 3, 7, 1, 9]；7,1 交换 → [5, 3, 1, 7, 9]；7,9 不交换 (2次交换)。

swap → [3, 5, 1, 7, 9]; 5,1 swap → [3, 1, 5, 7, 9]; 5,7 no swap; 7,9 no swap → (2 swaps).

第3趟：[5, 3, 1, 7, 9] → 5,3 交换 → [3, 5, 1, 7, 9]；5,1 交换 → [3, 1, 5, 7, 9]；5,7 不交换；7,9 不交换 (2次交换)。

,7 no swap; 7,9 no swap → (1 swap). List is now sorted. Total passes = 4, total swaps = 9.

第4趟：[3, 1, 5, 7, 9] → 3,1 交换 → [1, 3, 5, 7, 9]; 剩余均无需交换 (1次交换)。列表已有序。总趟数4，总交换次数9。

In the worst case, bubble sort requires n-1 passes and O(n²) comparisons, making it inefficient for large datasets compared to merge sort or quicksort.

最坏情况下，冒泡排序需要 n-1 趟和 O(n²) 次比较，对于大数据集效率远不如归并排序或快速排序。

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5. Python Code Debugging | Python 代码调试

Question: The program below intends to sum the even numbers from 1 to 10. Identify and correct the error, then state the output.

题目：以下程序试图计算1到10之间所有偶数的和。找出并更正错误，然后给出输出结果。

total = 0
for i in range(1, 11):
    if i % 2 = 0:
        total = total + i
print(total)

The bug is the use of the assignment operator = instead of the equality operator == in the condition i % 2 = 0. Python interprets this as an attempt to assign a value to an expression, which causes a syntax error. The correct line should be if i % 2 == 0:. After fixing, the loop adds 2, 4, 6, 8 and 10, giving a final total of 30.

错误是在条件 i % 2 = 0 中使用了赋值运算符 = 而非等号运算符 ==。Python 试图给表达式赋值而导致语法错误。正确写法应为 if i % 2 == 0:。修复后，循环累加 2, 4, 6, 8, 10，最终输出 30。

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6. Network Topologies Comparison | 网络拓扑结构比较

Question: Compare the star and mesh network topologies by stating one advantage of each in a school network environment.

题目：对比星型和网状网络拓扑，针对学校网络环境各指出一个优点。

Topology	Advantage
Star	If a single cable fails, only the connected node is affected; the rest of the network stays operational. It is also easy to add new workstations without disrupting the network.
Mesh	Multiple redundant paths mean there is no single point of failure. If one link breaks, data can be rerouted, offering very high reliability—valuable for critical server rooms.

星型拓扑中，单根线缆故障只影响所连节点，网络其余部分仍可运行，且新增工作站简便，不会中断网络。网状拓扑因为存在多条冗余路径，没有单点故障，一条链路断开后数据可绕行，可靠性极高，适合关键服务器机房。

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7. Cybersecurity Threats and Prevention | 网络安全威胁与预防

Question: Describe a brute-force attack and explain how enforcing a strong password policy helps defend against it.

题目：描述暴力攻击的原理，并解释强制实施强密码策略如何帮助防御这种攻击。

ct password is found. Automated tools can test millions of passwords per second. A strong password policy that requires a minimum length (e.g., 12+ characters), a mix of uppercase, lowercase, digits and symbols drastically increases the number of possible combinations. For instance, an 8-character password drawn from 95 printable ASCII characters has 95⁸ ≈ 6.6 × 10¹⁵ possibilities, making exhaustive search computationally infeasible in a reasonable time. Adding a lockout after repeated failures and using multi-factor authentication further reduce the risk.

暴力攻击是一种试错方法，攻击者系统地尝试所有可能的字符组合，直到找到正确密码。自动化工具每秒可测试数百万个密码。强密码策略要求最小长度（如12+字符）、包含大小写字母、数字和符号，这会极大地增加组合数。例如，从95个可打印字符中选取的8位密码有95⁸ ≈ 6.6 × 10¹⁵ 种可能，使得穷举搜索在合理时间内变得不现实。配合多次失败锁定和多因素认证可进一步降低风险。

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8. Linear Search Pseudocode and Efficiency | 线性搜索伪代码与效率

Question: Write pseudocode for a linear search that finds a target value in an array. Then evaluate its performance when searching a large dataset.

题目：编写线性搜索伪代码，在数组中查找目标值，随后评估其在大数据集上的性能。

PROCEDURE linearSearch(arr, target)
    index ← 0
    WHILE index < LENGTH(arr)
        IF arr[index] = target THEN
            OUTPUT "Found at position", index
            STOP
        ENDIF
        index ← index + 1
    ENDWHILE
    OUTPUT "Not found"
ENDPROCEDURE

The algorithm checks each element sequentially. In the best case the target is at the first position (O(1)); in the worst case it is at the end or absent, requiring n comparisons (O(n)). For a dataset of 10⁶ records, linear search may need up to one million checks, making it unsuitable for large-scale applications. Binary search on a sorted array would achieve O(log n) time, which is far more efficient.

该算法逐个检查元素。最佳情况目标在首位置 (O(1))；最坏情况在末尾或不存在，需 n 次比较 (O(n))。对于包含 10⁶ 条记录的数据集，线性搜索可能需要多达百万次检查，因此不适用于大规模应用。在有序数组上使用二分搜索可实现 O(log n) 时间，效率高得多。

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9. Hexadecimal and Colour Representation | 十六进制与颜色表示

Question: Explain why hexadecimal is preferred over binary for representing RGB colours in HTML. Convert the binary sequence 11111111 00000000 00000000 into its HTML hex colour code.

题目：解释为什么在 HTML 中表示 RGB 颜色时十六进制比二进制更受青睐。将二进制序列 11111111 00000000 00000000 转换为 HTML 十六进制颜色代码。

le. An 8-bit binary value such as 11111111 requires eight digits, while its hex equivalent 'FF' colour in binary would be a string of 24 bits, but in hex it becomes a six-digit code like # 111₂ = FF₁₆ (red), 00000000₂ = 00₁₆ (green), 00000000₂ = 00₁₆ (blue). Concatenating with a '#' gives #FF0000, representing pure red.

十六进制更简洁且易读。一个8位二进制值如11111111需8个数字，而十六进制'FF'只需两位。24位RGB颜色若用二进制需24个数字，而十六进制变为如#FF0000的六位代码，对网页开发者更不易出错。转换：11111111₂ = FF₁₆（红），00000000₂ = 00₁₆（绿

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