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

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

Mock exams are the best way to identify knowledge gaps before the real IGCSE Computer Science paper. This walkthrough analyses a typical Unit 1–2 test covering data representation, hardware, software, networks and algorithms. Each question is explained with step‑by‑step logic, common pitfalls and examiner tips.

模拟考试是找出知识漏洞的最佳方式。本文解析一份典型的单元1–2测试卷，涵盖数据表示、硬件、软件、网络和算法。每道题都附有分步解析、常见错误和考官提示。

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1. Binary Conversion | 二进制转换

A typical question: Convert the denary number 215 into an 8‑bit binary number, showing all working.

常见题目：将十进制数215转换为8位二进制数，并展示计算过程。

= 1 r 1; 1 ÷ 2 = 0 r 1. Read the remainders from bottom to top to obtain 11010111. As an 8‑bit number, it already has eight bits – no leading zeros required.

不断用2除215并记录余数：215 ÷ 2 = 107 余 1；107 ÷ 2 = 53 余 1；53 ÷ 2 = 26 余 1；26 ÷ 2 = 13 余 0；13 ÷ 2 = 6 余 1；6 ÷ 2 = 3 余 0；3 ÷ 2 = 1 余 1；1 ÷ 2 = 0 余 1。从下往上读取余数得到11010111。作为8位二进制数，它已经有八位，无需补前导零。

Common mistake: writing the remainders in the order they were calculated rather than reversing them. Always remember to take the remainder sequence from the last division upwards.

常见错误：按照计算顺序写下余数，而没有反转顺序。务必记住从最后一次除法开始向上读取余数序列。

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2. Hexadecimal Representation | 十六进制表示

Question: Represent the denary value 254 in hexadecimal. You may use binary as an intermediate step.

题目：用十六进制表示十进制数254。允许使用二进制作为中间步骤。

Method 1 via binary: 254 in 8‑ . Split into nibbles: 1111 1110. Nibble 1111 = F, nibble 1110 = E. Thus hexadecimal is FE.

方法一通过二进制：254的8位二进制为11111110（128+64+32+16+8+4+2）。拆分为半字节：1111 1110。1111 = F，1110 = E，因此十六进制为FE。

remainder 14. Denary 15 is hex F, denary 14 is hex E. Write quotient then remainder to get FE.

方法二直接除法：254 ÷ 16 = 15 余 14。十进制15对应十六进制F，十进制14对应十六进制E。将商和余数依次写出即为FE。

Examiners expect you to show working and label your answer clearly. When converting from binary to hex, always group bits from the right.

考官希望看到计算过程并清楚标注答案。从二进制转换到十六进制时，务必从右开始分组。

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3. Binary Arithmetic and Overflow | 二进制运算与溢出

Question: Add the 8‑bit binary numbers 10101010 and 11001100. State whether an overflow occurs and justify your answer.

题目：将8位二进制数10101010和11001100相加。判断是否发生溢出并说明理由。

10101010 + 11001100 = 1 01110110

Add column by column from the right. The sum generates a carry into the 9th bit. In an 8‑bit register, the leftmost bit (bit 7) represents the sign in two's complement. Here the carry into bit 7 (carry from bit 6) is '1', and the carry out of bit 7 is also '1'. Since the carry in and carry out of the sign bit are the same, no overflow occurs. However, the extra 9th bit is lost, so the result stored is 01110110.

从右向左逐列相加，结果产生向第9位的进位。在8位寄存器中，最左位（第7位）在补码表示中代表符号位。此处进入符号位的进位（来自第6位）是'1'，离开符号位的进位也是'1'。符号位进位输入与输出相同，因此没有溢出。但多出的第9位被丢弃，寄存器中存储的结果为01110110。

Overflow in two's complement occurs only when the carry into the sign bit differs from the carry out of the sign bit. Always check this condition when adding signed binary numbers.

在补码加法中，只有当进入符号位的进位与离开符号位的进位不同时，才会发生溢出。对带符号二进制数做加法时，务必检查这一条件。

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4. Truth Tables and Logic Circuits | 真值表与逻辑电路

e for the logic expression X = NOT (A AND B) OR C. Draw the corresponding circuit diagram.

题目：完善逻辑表达式 X = NOT (A AND B) OR C 的真值表，并画出对应电路图。

A	B	C	A AND B	NOT (A AND B)	X = NOT (A AND B) OR C
0	0	0	0	1	1
0	0	1	0	1	1
0	1	0	0	1	1
0	1	1	0	1	1
1	0	0	0	1	1
1	0	1	0	1	1
1	1	0	1	0	0
1	1	1	1	0	1

To build the circuit, first connect A and B to an AND gate. Feed its output through a NOT gate. Then join the inverted output with input C into an OR gate. The final output is X.

搭建电路时，先将A和B接入与门，其输出经过非门反相，再将该反相输出与输入C一同接入或门，最终输出为X。

When drawing, use standard symbols: a D‑shaped AND gate, a triangle with a bubble for NOT, and a curved‑back OR gate. Labels matter – examiners deduct marks for missing or incorrect gate shapes.

画图时应使用标准符号：D形与门、带小圆圈的三角非门、以及背部弯曲的或门。标注很重要，门形状缺失或不正确会被扣分。

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5. CPU Components and the Fetch-Execute Cycle | CPU部件与取指-执行周期

Question: Describe the roles of the MAR and MDR during the fetch stage of the fetch‑decode‑execute cycle.

题目：描述取指-译码-执行周期的取指阶段中MAR和MDR的作用。

The Memory Address Register (MAR) holds the address of the next instruction to be fetched. The CPU copies the content of the Program Counter (PC) into the MAR at the start of the fetch stage. The MAR then sends this address along the address bus to main memory.

存储器地址寄存器（MAR）保存将要取出的下一条指令的地址。取指阶段开始时，CPU将程序计数器（PC）的内容复制到MAR中，然后MAR通过地址总线将该地址发送到主存。

The Memory Data Register (MDR) acts as a buffer. Once the instruction at the requested address is read from memory, it is placed into the MDR. From there, the instruction is copied to the Current Instruction Register (CIR) for decoding. The MDR is also used when data is written to or read from memory during execution.

存储器数据寄存器（MDR）起到缓冲作用。当从内存中读出位于请求地址的指令时，该指令被放入MDR，然后被复制到当前指令寄存器（CIR）以进行译码。在执行阶段读写数据时，MDR同样会被使用。

Many students confuse MAR and MDR. Use the mnemonic: MAR holds the 'where' (address), MDR holds the 'what' (data or instruction).

很多学生混淆MAR和MDR。可借助记忆法：MAR存放“哪里”（地址），MDR存放“什么”（数据或指令）。

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6. Storage Types and Compression | 存储类型与压缩

Question: A manufacturer decides to use an SSD instead of a magnetic HDD in a new ultrabook. Explain two advantages of this choice.

题目：某制造商决定在新型超极本中使用SSD而非磁介质HDD，请说明这种选择的两个优点。

First, an SSD has no moving parts. This makes it more resistant to physical shock and suitable for portable devices. Second, SSDs offer much faster data access speeds because there are no mechanical delays from spinning platters or moving read/write heads.

第一，SSD没有活动部件，使其更抗物理冲击，适合便携设备。第二，SSD提供快得多的数据访问速度，因为没有旋转盘片和移动读写磁头带来的机械延迟。

Question: Explain lossy compression and give one example of its use.

题目：解释有损压缩并给出一个应用示例。

Lossy compression reduces file size by permanently discarding some of the original data, often data that is less noticeable to humans. An example is the MP3 audio format, which removes sounds outside typical human hearing range to create much smaller files than uncompressed WAV.

有损压缩通过永久丢弃部分原始数据来减小文件大小，通常是人类不易察觉的数据。例子是MP3音频格式，它移除人耳典型听力范围外的声音，生成比无压缩WAV小得多的文件。

A common misconception is that lossy compression simply reduces quality – but it does so by exploiting perceptual limitations, so the change is often invisible or inaudible.

一个常见误解是认为有损压缩只会降低质量——实际上它利用感知限制，因此改变往往不可见或不可闻。

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7. Algorithm Design and Trace Tables | 算法设计与跟踪表

Question: Complete the trace table for the following pseudocode when the user inputs the numbers 3, 9, 2, 7, 4 (in that order).

题目：当用户依次输入数字3、9、2、7、4时，完成以下伪代码的跟踪表。

Pseudocode:

total ← 0
FOR i ← 1 TO 5
    INPUT num
    total ← total + num
NEXT i
OUTPUT total

Trace table setup:

i	num	total	OUTPUT
1	3	3
2	9	12
3	2	14
4	7	21
5	4	25	25

The variable 'total' accumulates the sum as each number is entered. At the end of the loop, i reaches 5 and total becomes 25. The OUTPUT statement then displays 25. Trace tables are essential for understanding how variables change over time; always update each column step by step.

变量'total'在每次输入数字时累加求和。循环结束时i到达5，total变为25。随后OUTPUT语句显示25。跟踪表对于理解变量如何随时间变化至关重要；必须逐步更新每一列。

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8. Bubble Sort Walkthrough | 冒泡排序详解

Question: The array [7, 2, 9, 4] is to be sorted in ascending order using bubble sort. Show the contents of the array after each pass and state the number of comparisons and swaps.

题目：数组[7, 2, 9, 4]需用冒泡排序按升序排列。显示每一遍之后数组的内容，并说明比较和交换次数。

, 7, 9, 4]. Compare 9 and 4 → swap → [2, 7, 4, 9]. End of pass 1. (3 comparisons, 2 swaps)

第一遍：比较7和2 → 交换 → [2, 7, 9, 4]。比较7和9 → 不交换 → [2, 7, 9, 4]。比较9和4 → 交换 → [2, 7, 4, 9]。第一遍结束。（3次比较，2次交换）

7, 9]. The last element is already sorted so only two comparisons. (2 comparisons, 1 swap)

第二遍：比较2和7 → 不交换。比较7和4 → 交换 → [2, 4, 7, 9]。最后一个元素已排好，所以只需两次比较。（2次比较，1次交换）

is sorted. (1 comparison, 0 swaps). A further pass may be performed to confirm no more swaps are needed, but in optimized bubble sort we stop when a pass has zero swaps.

第三遍：比较2和4 → 不交换。数组已排序。（1次比较，0次交换）。虽然可以再执行一遍以确认无需交换，但优化后的冒泡排序在某一遍中无交换时便停止。

. Writing intermediate states neatly earns full marks; missing a pass loses structure marks.

总比较次数 = 6，总交换次数 = 3。整洁地写出中间状态可得满分；遗漏任何一遍会失去结构分。

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9. Networking Devices and IP Addressing | 网络设备与IP地址

Question: Distinguish between a hub and a switch. Also explain why the world is moving from IPv4 to IPv6.

题目：区分集线器与交换机。并解释为何全球正由IPv4向IPv6过渡。

A hub is a simple device that broadcasts incoming data packets to all connected ports, regardless of the destination. This wastes bandwidth and creates security risks. A switch is more intelligent: it learns the MAC addresses of devices connected to each port and forwards data only to the appropriate port, increasing efficiency and security.

集线器是一种简单设备，它将收到的数据包广播到所有连接的端口，不考虑目的地，浪费带宽且带来安全风险。交换机更智能：它会学习每个端口所连设备的MAC地址，并仅将数据转发到相应端口，提高了效率和安全性。

IPv4 uses 32‑bit addresses, allowing about 4.

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