Year 13 CCEA Computer Science Mock Paper Analysis | Year 13 CCEA 计算机单元测试模拟卷解析

📚 Year 13 CCEA Computer Science Mock Paper Analysis | Year 13 CCEA 计算机单元测试模拟卷解析

This comprehensive mock paper analysis revisits the core themes of the CCEA AS Computer Science specification, designed to help Year 13 students consolidate their understanding and exam technique. Each question mirrors the style of unit tests, covering data representation, Boolean algebra, processor architecture, networking, databases, algorithms, and software development. Detailed solutions are explained step by step, highlighting key revision points alongside subject-specific terminology.

这份全面的模拟卷解析重新梳理了CCEA AS计算机科学课程的核心主题，旨在帮助Year 13学生巩固理解与应试技巧。每道题目仿照单元测试风格，涵盖数据表示、布尔代数、处理器架构、网络、数据库、算法和软件开发等领域。解答逐步解析，结合学科术语突出关键复习要点。

1. Data Representation: Binary to Hexadecimal Conversion | 数据表示：二进制转十六进制

Question: Convert the binary number 11011010₂ into hexadecimal. Show all grouping steps and briefly explain why hexadecimal is useful in computing.

题目：将二进制数 11011010₂ 转换为十六进制。展示所有分组步骤，并简要说明十六进制在计算中的优势。

Solution: Begin by grouping the binary digits into nibbles (sets of four) from the right. The number 11011010 already has eight bits, so we write 1101 1010. For the leftmost nibble, 1101, evaluate using place values: 1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8+4+0+1 = 13, which corresponds to D in hex. The right nibble 1010 gives: 1×2³ + 0×2² + 1×2¹ + 0×2⁰ = 8+0+2+0 = 10, which is A. Combining gives DA₁₆. Hexadecimal is useful because it shortens long binary strings, reducing the risk of human error when reading or writing memory addresses and machine code.

解答：首先将二进制位从右向左分组为半字节（每组四位）。数字11011010已经是8位，因此写成1101 1010。左半字节1101按权展开：1×2³ + 1×2² + 0×2¹ + 1×2⁰ = 8+4+0+1 = 13，对应十六进制数字D。右半字节1010：1×2³ + 0×2² + 1×2¹ + 0×2⁰ = 8+0+2+0 = 10，即A。组合得到 DA₁₆。十六进制之所以有用，是因为它缩短了冗长的二进制串，在读写内存地址和机器码时能减少人为错误。

Key point: Always pad the leftmost group with leading zeros if it contains fewer than four bits after grouping from the right.

关键点：如果从右分组后最左组不足四位，务必用前导零补齐。

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2. Two’s Complement: Negative Number Representation | 补码：负数表示

Question: Represent the decimal number -25 in 8-bit two’s complement. Show the steps of conversion and verify your answer by converting back to decimal.

题目：用8位二进制补码表示十进制数 -25。展示转换步骤，并通过转回十进制来验证。

Solution: First write the positive magnitude, 25, in 8-bit binary: 00011001₂. Invert all bits (one’s complement): 11100110₂. Add 1 to obtain the two’s complement: 11100111₂. Verification: a negative two’s complement number can be checked by converting it as if it were unsigned and subtracting 256. 11100111₂ = 128+64+32+4+2+1 = 231. 231 – 256 = -25, which confirms the representation. Alternatively, invert and add 1 again: bit inverse of 11100111₂ is 00011000₂, plus 1 gives 00011001₂, which is 25.

解答：首先写出正数25的8位二进制形式：00011001₂。将所有位取反（反码）：11100110₂。加1得到补码：11100111₂。验证方法：将该补码按无符号数计算为231，然后用231 – 256 = -25，确认正确。也可以再次取反加1：11100111₂取反得00011000₂，加1得00011001₂，即25，同样证明表示正确。

Important concept: Two’s complement is the standard method for representing signed integers because it avoids a duplicate zero and allows subtraction to be performed by addition.

重要概念：二进制补码是表示有符号整数的标准方法，因为它避免了双零问题，并且允许用加法实现减法运算。

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3. Boolean Algebra Simplification | 布尔代数化简

Question: Using the laws of Boolean algebra, simplify the expression A·(A + B) and then verify the result with a truth table for the case A=1, B=0.

题目：使用布尔代数定律化简表达式 A·(A + B)，并通过在 A=1, B=0 情况下的真值表验证结果。

Solution: Apply the distributive law: A·(A + B) = A·A + A·B. Since A·A = A (idempotent law), this becomes A + A·B. Factor A: A·(1 + B). In Boolean algebra, 1 + B = 1 (annulment), so the expression simplifies to A. Verification: for A=1, B=0, the original expression A·(A + B) = 1·(1+0) = 1·1 = 1. The simplified expression is A = 1. Hence the simplification is correct.

解答：应用分配律：A·(A + B) = A·A + A·B。由幂等律 A·A = A，得 A + A·B。提取公因子 A：A·(1 + B)。布尔代数中 1 + B = 1（湮灭律），因此化简为 A。验证：A=1, B=0 时，原式 A·(A + B) = 1·(1+0)=1·1=1；化简结果 A=1，一致，正确。

Revision note: Be familiar with fundamental Boolean laws—identity, idempotent, annulment, complement, and distributive—as they frequently appear in paper questions.

复习建议：熟练掌握基本布尔定律——同一律、幂等律、湮灭律、互补律和分配律，这些常出现在考题中。

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4. Logic Gate Circuit to Boolean Expression | 逻辑门电路转布尔表达式

Question: A logic circuit consists of a 2-input AND gate receiving A and B, whose output is connected to one input of a 2-input OR gate. The other input of the OR gate is connected to a NOT gate that takes A as input. Derive the Boolean expression for the output Q and draw the truth table.

题目：某逻辑电路包含一个二输入与门，输入端为 A 和 B，其输出连接到一个二输入或门的一个输入端；或门的另一个输入端连接一个非门，该非门以 A 为输入。推导输出 Q 的布尔表达式并画出真值表。

Solution: The AND gate produces A·B. The NOT gate produces A’. The OR gate combines these: Q = (A·B) + A’. Truth table: when A=0, B=0, A·B=0, A’=1, Q=1; A=0, B=1, A·B=0, A’=1, Q=1; A=1, B=0, A·B=0, A’=0, Q=0; A=1, B=1, A·B=1, A’=0, Q=1. Thus the output is HIGH unless A is 1 and B is 0.

解答：与门输出为 A·B，非门输出为 A’，或门组合后 Q = (A·B) + A’。真值表：A=0, B=0 时，A·B=0，A’=1，Q=1；A=0, B=1 时，A·B=0，A’=1，Q=1；A=1, B=0 时，A·B=0，A’=0，Q=0；A=1, B=1 时，A·B=1，A’=0，Q=1。因此除了 A=1 且 B=0 的情况外，输出均为高电平。

Tip for exams: When deriving expressions, label intermediate wires; when completing truth tables, evaluate all input combinations systematically.

应试技巧：推导表达式时标注中间连线；填写真值表时系统地评估所有输入组合。

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5. Processor Components and the Fetch-Decode-Execute Cycle | 处理器组件与取指-解码-执行周期

Question: Describe the roles of the MAR, MDR, and CIR in the fetch-decode-execute cycle. Illustrate with the steps that occur when the CPU fetches an instruction from memory.

题目：描述 MAR、MDR 和 CIR 在取指-解码-执行周期中的作用，并用 CPU 从内存中取指令的步骤加以说明。

Solution: The Memory Address Register (MAR) holds the address of the memory location to be read from or written to. The Memory Data Register (MDR) stores the data or instruction that has been fetched from memory or is to be written. The Current Instruction Register (CIR) holds the current instruction being decoded and executed. During the fetch stage: the PC value is copied into the MAR; a read signal is sent; the instruction is placed onto the data bus and transferred to the MDR; the instruction is then copied into the CIR; finally the PC is incremented.

解答：内存地址寄存器（MAR）保存待读取或写入的内存地址。内存数据寄存器（MDR）存储从内存中取出的数据或指令，或将要写入的数据。当前指令寄存器（CIR）保存正在被

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