Year 10 AQA Computer Science: In-Depth Analysis of Past Paper Questions | AQA 计算机历年真题深度解析

📚 Year 10 AQA Computer Science: In-Depth Analysis of Past Paper Questions | AQA 计算机历年真题深度解析

Mastering AQA GCSE Computer Science (8525) requires more than just memorising facts; it demands the ability to apply knowledge to unfamiliar problems. Past paper questions offer the best insight into the examiners’ expectations. This article analyses typical questions across key topics, highlighting common pitfalls, efficient methods, and precise answers. By studying these model solutions, you will sharpen your computational thinking and build the confidence needed to excel in your exams.

掌握 AQA GCSE 计算机科学 (8525) 不仅仅需要记忆知识，更需要将知识应用到陌生问题中的能力。历年真题是了解考官期望的最佳途径。本文深入解析了各关键主题中的典型问题，突出常见错误、高效方法和精准答案。通过学习这些模型解答，你将锻炼计算思维，建立应考信心。

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

Question: Convert the binary number 10111010 to hexadecimal. Show your working. (2 marks)

题目：将二进制数 10111010 转换为十六进制。展示计算过程。（2分）

First, split the binary number into groups of four bits starting from the right: 1011 1010. The left group 1011 equals 8+2+1 = 11, which is B in hex. The right group 1010 equals 8+2 = 10, which is A in hex. So the answer is BA. A common mistake is to group from the left or forget to pad leading zeros; always group from the right.

首先，从右侧开始将二进制数分成四位一组：1011 1010。左组 1011 等于 8+2+1 = 11，即十六进制的 B。右组 1010 等于 8+2 = 10，即十六进制的 A。所以答案为 BA。常见错误是从左侧分组或忘记补前导零；务必从右侧分组。

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

Question: Add the 8-bit binary numbers 01101101 and 01011001. Does overflow occur? Explain. (3 marks)

题目：将8位二进制数 01101101 与 01011001 相加。是否发生溢出？请解释。（3分）

Perform column addition: 0+0=0, 1+1=0 carry 1, etc. The sum is 11000110. Since both numbers are positive (MSB=0) but the result has MSB=1, an overflow has occurred. Overflow happens when the result exceeds the range representable by 8 bits in two’s complement (here it appears positive overflow). In unsigned binary, this would be a carry out, but in two’s complement it is an error flag.

进行逐位加法：0+0=0，1+1=0 进1，等等。和为 11000110。由于两个数均为正数（最高位=0）而结果最高位=1，发生了溢出。当结果超出8位二进制补码可表示的范围时就会溢出。无符号二进制可能产生进位，但在补码中这是一个错误标志。

3. Boolean Logic and Truth Tables | 布尔逻辑与真值表

Question: Draw the truth table for the expression (A OR B) AND NOT C. (4 marks)

题目：画出表达式 (A OR B) AND NOT C 的真值表。（4分）

List all 8 combinations of A, B, C (000 to 111). Evaluate A OR B first, then NOT C, finally AND. For example, A=0,B=1,C=0: (0 OR 1)=1, NOT 0=1, 1 AND 1=1. Examiners expect clearly labelled columns and correct 0/1 outputs. A missing combination or inverted logic is a common error.

列出A、B、C的8种组合（000到111）。先计算 A OR B，再计算 NOT C，最后计算 AND。例如 A=0,B=1,C=0：(0 OR 1)=1，NOT 0=1，1 AND 1=1。考官期望清晰的列标题和正确的0/1输出。遗漏组合或逻辑反转是常见错误。

4. Representing Images – Bitmap vs Vector | 图像表示 – 位图与矢量图

Question: An image is 200 pixels wide and 300 pixels high, using a colour depth of 24 bits. Calculate the file size in kilobytes. State an appropriate unit. (3 marks)

题目：一幅图像宽200像素、高300像素，色深24位。计算文件大小（千字节）。说明合适的单位。（3分）

Total pixels = 200 x 300 = 60,000 pixels. Bits = 60,000 x 24 = 1,440,000 bits. Bytes = 1,440,000 / 8 = 180,000 bytes. Kilobytes = 180,000 / 1024 = 175.78125 KB. Accept ‘175.8 KB’ or ‘176 KB’ if rounded. Remember to divide by 1024, not 1000, as this is a binary system. Leaving the answer in bits or using MB incorrectly loses marks.

总像素 = 200 x 300 = 60,000 像素。位数 = 60,000 x 24 = 1,440,000 位。字节数 = 1,440,000 / 8 = 180,000 字节。千字节数 = 180,000 / 1024 = 175.78125 KB。可接受 ‘175.8 KB’ 或四舍五入的 ‘176 KB’。注意除以1024而非1000，因为是二进制系统。以位为单位或误用 MB 会丢分。

5. Representing Sound – Sampling Rate and Bit Depth | 声音表示 – 采样率与位深度

Question: A music track is recorded in stereo for 3 minutes with a sampling rate of 44.1 kHz and 16-bit resolution. Calculate the uncompressed file size in megabytes. (3 marks)

题目：一段立体声音乐录制3分钟，采样率44.1 kHz，16位分辨率。计算未压缩文件大小（兆字节）。（3分）

Duration in seconds = 3 x 60 = 180 s. Bits per second = 44,100 x 16 x 2 (stereo) = 1,411,200 bps. Total bits = 1,411,200 x 180 = 254,016,000 bits. Bytes = 254,016,000 / 8 = 31,752,000 B. MB = 31,752,000 / (1024²) approx 30.28 MB. Many students forget to multiply by 2 for stereo, leading to half the correct size.

时长秒数 = 3 x 60 = 180 秒。每秒位数 = 44,100 x 16 x 2 (立体声) = 1,411,200 bps。总位数 = 1,411,200 x 180 = 254,016,000 位。字节 = 254,016,000 / 8 = 31,752,000 B。MB = 31,752,000 / (1024²) 约30.28 MB。许多学生忘记乘以2（立体声），导致大小仅为正确值的一半。

6. CPU Components and the Fetch-Execute Cycle | CPU组件与取指–执行周期

Question: Describe the role of the Program Counter (PC) in the fetch-execute cycle. (2 marks)

题目：描述程序计数器（PC）在取指–执行周期中的作用。（2分）

The PC holds the memory address of the next instruction to be fetched. During the fetch stage, the address in the PC is copied to the Memory Address Register (MAR), then the PC is incremented to point to the subsequent instruction. Without this increment, the same instruction would be fetched repeatedly. Do not confuse PC with MAR or MDR.

PC 保存下一条要取指令的内存地址。在取指阶段，PC 中的地址被复制到内存地址寄存器（MAR），然后 PC 递增以指向后续指令。没有递增，相同指令会被反复取出。不要混淆 PC 与 MAR 或 MDR。

7. Primary and Secondary Storage Comparison | 主存储与辅助存储对比

Question: Compare RAM and hard disk drive (HDD) in terms of volatility, speed and typical capacity. (4 marks)

题目：从易失性、速度和典型容量方面比较 RAM 和硬盘驱动器（HDD）。（4分）

RAM is volatile (data lost when power off), while HDD is non-volatile. RAM access speed is measured in nanoseconds, much faster than HDD (milliseconds). Typical RAM capacity in a personal computer is 8-16 GB, whereas HDD can store several terabytes. Confusing ROM with RAM or assuming SSD has same characteristics as HDD are common errors.

RAM 易失（断电数据丢失），而 HDD 非易失。RAM 访问速度以纳秒计，远快于 HDD（毫秒）。个人计算机中 RAM 典型容量为 8-16 GB，而 HDD 可存储数 TB。混淆 ROM 与 RAM，或认为固态硬盘与 HDD 特性相同，是常见错误。

8. Network Protocols and the TCP/IP Model | 网络协议与TCP/IP模型

Question: A student types a URL into a web browser. Which protocol resolves the domain name to an IP address? Explain the process. (3 marks)

题目：学生在浏览器中输入URL。哪个协议将域名解析为IP地址？解释该过程。（3分）

The Domain Name System (DNS) protocol resolves the domain name. The browser sends a DNS query to a DNS server; if the server does not have the record, it forwards the query until an authoritative server returns the IP address. DNS uses UDP on port 53. Relying on ‘HTTP does this’ is a misunderstanding – HTTP is for web page transfer after resolution.

域名系统（DNS）协议解析域名。浏览器向 DNS 服务器发送查询；若服务器无记录，则转发查询直到权威服务器返回 IP 地址。DNS 使用 UDP，端口53。认为“HTTP 负责解析”是错误的——HTTP 是在解析后传输网页。

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

Question: A list of 2000 names is stored alphabetically. State the algorithm more efficient for searching, and calculate the maximum number of comparisons needed. (3 marks)

题目：一个包含2000个姓名的列表按字母顺序存储。说明哪种搜索算法更高效，并计算所需的最大比较次数。（3分）

Binary search is more efficient on a sorted list. The maximum number of comparisons for 2000 items is approximately log₂(2000) rounded up. log₂(2048)=11, so log₂(2000) is between 10 and 11, thus a maximum of 11 comparisons. Linear search would need up to 2000. Students often incorrectly apply binary search to unsorted data.

二分搜索对有序列表更高效。2000个项目的最大比较次数约为 log₂(2000) 向上取整。log₂(2048)=11，故 log₂(2000) 介于10和11之间，因此最大为11次比较。线性搜索可能需要多达2000次。学生经常错误地对无序数据应用二分搜索。

10. Bubble Sort and Insertion Sort | 冒泡排序与插入排序

Question: Show the passes of a bubble sort on the list [4, 2, 7, 1]. How many comparisons and swaps are performed in total? (4 marks)

题目：对列表 [4, 2, 7, 1] 进行冒泡排序，展示各趟过程。总共进行了多少次比较和交换？（4分）

Pass 1: compare 4&2 swap → [2,4,7,1]; 4&7 no swap; 7&1 swap → [2,4,1,7]. Pass 2: 2&4 no swap; 4&1 swap → [2,1,4,7]; 4&7 no swap. Pass 3: 2&1 swap → [1,2,4,7]. Total comparisons = 3+3+3=9 (or 9 in a naive implementation). Swaps = 1+1+1=3. An optimised bubble sort would stop early if no swaps occur. Mark schemes credit the correct sequence.

第1趟：比较4&2 交换→ [2,4,7,1]；4&7不交换；7&1交换→ [2,4,1,7]。第2趟：2&4不交换；4&1交换→ [2,1,4,7]；4&7不交换。第3趟：2&1交换→ [1,2,4,7]。总比较次数 = 3+3+3=9（或简单实现为9）。交换次数 = 1+1+1=3。优化的冒泡排序若无交换则提前终止。评分方案认可正确顺序。

11. Programming – Variables, Constants, and Data Types | 编程 – 变量、常量与数据类型

Question: A program stores the value of VAT as 20%. Explain why a constant should be used instead of a variable, and state an appropriate data type. (2 marks)

题目：程序将增值税值存储为20%。解释为什么应使用常量而非变量，并说明合适的数据类型。（2分）

A constant should be used because the VAT rate does not change during execution; using a constant prevents accidental modification and makes the program more readable. The suitable data type is float/real as it holds a decimal percentage (0.20). Using integer would lose precision. Constants are declared with ‘const’ in many languages.

应使用常量，因为增值税率在执行期间不变；使用常量可防止意外修改并使程序更具可读性。合适的数据类型是浮点数/实数，因为它保存小数百分比 (0.20)。使用整数会丢失精度。常量在许多语言中用 ‘const’ 声明。

12. Defensive Design and Testing | 防御性设计与测试

Question: A user is prompted to enter their age. Write a validation rule that uses a range check. Give an example of erroneous and boundary test data. (4 marks)

题目：提示用户输入年龄。编写一个使用范围检查的验证规则。给出错误测试数据和边界测试数据的例子。（4分）

Validation rule: age must be an integer between 0 and 120 inclusive. Erroneous data: -5 or 200 (should be rejected). Boundary data: 0 and 120 (valid, on the edges), and 121 (just outside, invalid). Normal test data like 25 should also be accepted. This tests the robustness of input handling. Both black-box and white-box testing can be applied.

验证规则：年龄必须是0到120之间的整数（含）。错误数据：-5 或 200（应拒绝）。边界数据：0 和 120（有效，在边缘），以及 121（刚超出，无效）。正常测试数据如25应被接受。这测试了输入处理的稳健性。黑盒测试和白盒测试均可应用。

Published by TutorHao | Computer Science Revision Series | aleveler.com

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