📚 Year 7 CIE Computer Science: Interdisciplinary Question Practice | Year 7 CIE 计算机：跨学科综合题型训练
In the CIE Year 7 Computer Science syllabus, cross-curricular links play a vital role. Students are often expected to apply computing concepts in mathematical, scientific, and real-world contexts. This article provides a rich set of integrated practice questions that blend binary arithmetic, logic, flowcharts, data handling, algorithms, and more, helping you build deeper understanding and exam confidence.
在 CIE 七年级计算机科学课程中，跨学科联系起着至关重要的作用。学生经常需要将计算概念应用于数学、科学和现实世界情境。本文提供了一套丰富的综合训练题，融合了二进制运算、逻辑、流程图、数据处理、算法等内容，帮助你加深理解并提升考试信心。
1. Binary Numbers and Maths | 二进制与数学
Binary numbers are the foundation of all computer data. Converting between binary and decimal, and performing binary addition, directly links to place value and arithmetic skills from Mathematics.
二进制数是所有计算机数据的基础。二进制与十进制之间的转换以及二进制加法运算，直接关联到数学中的位值和算术技能。
Problem: Add the binary numbers 1011₂ and 1101₂. Then verify your answer by converting everything to decimal.
题目：将二进制数 1011₂ 与 1101₂ 相加，然后通过转换为十进制进行验证。
Solution: Start from the rightmost bit: 1+1 = 0 carry 1; next column: 1+0+carry1 = 0 carry 1; next: 0+1+carry1 = 0 carry 1; leftmost: 1+1+carry1 = 1 carry 1. The result is 11000₂. Check: 1011₂ = 11₁₀, 1101₂ = 13₁₀. 11+13 = 24₁₀ = 11000₂. Correct!
解答：从最右边位开始：1+1 = 0 进 1；下一列：1+0+进位1 = 0 进 1；再下一列：0+1+进位1 = 0 进 1；最左列：1+1+进位1 = 1 进 1。结果为 11000₂。验证：1011₂ = 11₁₀，1101₂ = 13₁₀。11+13 = 24₁₀ = 11000₂。正确！
2. Logic Gates and Boolean Algebra | 逻辑门与布尔代数
Logic gates implement Boolean functions and are closely tied to logical reasoning in Mathematics. Understanding truth tables helps in designing circuits and solving logic puzzles.
逻辑门实现布尔函数，与数学中的逻辑推理紧密相连。理解真值表有助于设计电路和解决逻辑谜题。
Problem: For the logic circuit with inputs A=1, B=0 going into an AND gate, and that output fed into a NOT gate, find the final output Q and write the Boolean expression.
题目：对于输入 A=1, B=0 进入与门，输出再进入非门的逻辑电路，找出最终输出 Q 并写出布尔表达式。
Solution: AND output = A AND B = 1 AND 0 = 0. NOT (0) = 1. So Q = 1. Expression: Q = NOT (A AND B).
解答：与门输出 = A AND B = 1 AND 0 = 0。非门输出 = NOT(0) = 1。所以 Q = 1。表达式：Q = NOT (A AND B)。
Now build a truth table for all four input combinations: (0,0)→0→1, (0,1)→0→1, (1,0)→0→1, (1,1)→1→0. This is the NAND operation.
现在为所有四种输入组合建立真值表：(0,0)→0→1, (0,1)→0→1, (1,0)→0→1, (1,1)→1→0。这是一个与非运算。
3. Flowcharts and Problem Solving | 流程图与问题解决
Flowcharts are used to plan algorithms visually. Designing a flowchart to compute the area of a circle combines geometry with algorithmic thinking.
流程图用于可视化地规划算法。设计计算圆面积的流程图结合了几何与算法思维。
Problem: Draw a flowchart that asks for the radius r, then calculates and outputs the area using the formula Area = π × r². Use 3.14 for π.
题目：绘制一个流程图，要求输入半径 r，然后使用公式 面积 = π × r² 计算并输出面积，π 取 3.14。
Solution: The flowchart would have Start → Input r → Set Area = 3.14 × r × r → Output Area → Stop. If r is negative, we could include a decision: “Is r > 0?” with error handling.
解答：流程图应有：开始 → 输入 r → 设置 Area = 3.14 × r × r → 输出 Area → 停止。如果 r 为负数，可以加入判断：“r > 0 ?”，并处理错误。
4. Spreadsheets and Data Analysis | 电子表格与数据分析
Spreadsheets are powerful tools for data analysis. Students must understand cell references, formulas, and functions such as SUM, AVERAGE, MAX, MIN, which mirror statistical operations in Mathematics.
电子表格是强大的数据分析工具。学生必须理解单元格引用、公式以及 SUM、AVERAGE、MAX、MIN 等函数，这些与数学中的统计运算相对应。
| A | Score |
| 1 | 78 |
| 2 | 85 |
| 3 | 92 |
| 4 | 67 |
Problem: Using the table, write a formula to find the average score in cell B6, and another to find the highest score in B7.
题目：使用上表，在单元格 B6 中写出求平均分的公式，在 B7 中写出求最高分的公式。
Solution: B6: =AVERAGE(B1:B4) returns 80.5. B7: =MAX(B1:B4) returns 92. These demonstrate basic statistical functions.
解答：B6: =AVERAGE(B1:B4) 返回 80.5。B7: =MAX(B1:B4) 返回 92。这些展示了基本的统计函数。
5. Algorithms in Science | 科学中的算法
Algorithms can model scientific processes. For example, converting Celsius to Fahrenheit is a simple sequence that links computing with Physics.
算法可以模拟科学过程。例如，将摄氏度转换为华氏度是一个简单的序列，将计算与物理联系起来。
Problem: Write an algorithm in pseudocode that takes a temperature in Celsius and converts it to Fahrenheit using the formula F = (C × 9/5) + 32. The program should output the result.
题目：用伪代码写一个算法，输入摄氏温度，使用公式 F = (C × 9/5) + 32 将其转换为华氏度并输出结果。
Solution:
START
INPUT C
F = (C × 9/5) + 32
OUTPUT F
END
Test with C=0, F should be 32; C=100 gives F=212.
测试：C=0，F 应为 32；C=100 得到 F=212。
6. Data Representation and Measurements | 数据表示与测量单位
Representing images, sound, and text requires understanding bits, bytes, and measurement units. This intersects with Mathematics in calculating file sizes and using units like KB, MB.
表示图像、声音和文本需要理解位、字节和计量单位。这涉及数学中计算文件大小和使用 KB、MB 等单位。
Problem: A 200×200 pixel image uses 3 bytes per pixel for colour. Calculate the file size in kilobytes (KB), given that 1 KB = 1024 bytes.
题目：一幅 200×200 像素的图像，每个像素使用 3 字节表示颜色。如果 1 KB = 1024 字节，计算其文件大小（KB）。
Solution: Total pixels = 200 × 200 = 40000. Total bytes = 40000 × 3 = 120000 bytes. Size in KB = 120000 ÷ 1024 ≈ 117.19 KB.
解答：总像素数 = 200 × 200 = 40000。总字节数 = 40000 × 3 = 120000 字节。KB 大小 = 120000 ÷ 1024 ≈ 117.19 KB。
7. Cybersecurity and Encryption | 网络安全与加密
Encryption methods like the Caesar cipher use modular arithmetic, linking computing to number theory. Understanding simple ciphers builds awareness of data security.
像凯撒密码这样的加密方法使用了模运算，将计算与数论联系起来。理解简单密码有助于建立数据安全意识。
Problem: Encrypt the word “HELLO” with a Caesar cipher of shift 3. Then decrypt the ciphertext “KHOOR” with shift 3. Use A=0, B=1, …, Z=25.
题目：用移位 3 的凯撒密码加密单词 “HELLO”。然后用移位 3 解密密文 “KHOOR”。使用 A=0, B=1, …, Z=25 的规则。
Solution: H(7)+3=10→K, E(4)+3=7→H, L(11)+3=14→O, L→O, O(14)+3=17→R → “KHOOR”. Decrypting “KHOOR”: K(10)-3=7→H, H(7)-3=4→E, O(14)-3=11→L, O→L, R(17)-3=14→O → “HELLO”.
解答：H(7)+3=10→K, E(4)+3=7→H, L(11)+3=14→O, L→O, O(14)+3=17→R → “KHOOR”。解密 “KHOOR”: K(10)-3=7→H, H(7)-3=4→E, O(14)-3=11→L, O→L, R(17)-3=14→O → “HELLO”。
8. Programming and Geometry | 编程与几何
Programming graphics often involves drawing shapes using angles and side lengths, directly applying concepts from Geometry. Turtle graphics in Scratch or Python are classic examples.
编程图形常常需要使用角度和边长来绘制形状，直接应用几何概念。Scratch 或 Python 中的海龟绘图就是典型例子。
Problem: Describe a program to draw a square of side 100 units. Include the number of repetitions and the turning angle.
题目：描述一个绘制边长为 100 单位的正方形的程序。说明重复次数和转向角度。
Solution: Repeat 4 times: move forward 100, turn right 90°. This uses the fact that interior angles sum to 360° and a square has four equal angles of 90°.
解答：重复 4 次：前进 100，右转 90°。这利用了内角和为 360° 且正方形四个角均为 90° 的事实。
9. Networks and Communication | 网络与通信
Data transmission speed calculations combine knowledge of units and rates, akin to time-distance problems in Mathematics.
数据传输速率的计算结合了单位和速率的知识，类似于数学中的时间-距离问题。
Problem: A 5 MB file is being downloaded at a speed of 2 Mbps (megabits per second). How long does it take? (1 byte = 8 bits, 1 MB ≈ 1,000,000 bytes for simplicity).
题目：一个 5 MB 的文件正以 2 Mbps（兆比特每秒）的速度下载。需要多长时间？（为简化，1 字节 = 8 比特，1 MB ≈ 1,000,000 字节）。
Solution: File size in bits = 5 × 1,000,000 × 8 = 40,000,000 bits. Speed = 2,000,000 bits per second. Time = 40,000,000 ÷ 2,000,000 = 20 seconds.
解答：文件大小（比特）= 5 × 1,000,000 × 8 = 40,000,000 比特。速率 = 2,000,000 比特/秒。时间 = 40,000,000 ÷ 2,000,000 = 20 秒。
10. Database Queries and Data Handling | 数据库查询与数据处理
Simple database queries using logical operators (AND, OR) are perfect for integrating Boolean logic with data management, a key skill in Computer Science.
使用逻辑运算符（AND、OR）的简单数据库查询是将布尔逻辑与数据管理相结合的绝佳方式，是计算机科学中的关键技能。
Problem: A ‘Students’ table has fields: Name, Age, House. Write a query to find all students who are in ‘Blue’ house AND age > 12. Explain your logic.
题目：一个 “Students” 表包含字段：姓名、年龄、学院。编写查询以找出所有学院为 ‘Blue’ 且年龄大于 12 的学生。解释你的逻辑。
Solution: SELECT * FROM Students WHERE House = ‘Blue’ AND Age > 12; The AND operator ensures both conditions are true simultaneously.
解答：SELECT * FROM Students WHERE House = ‘Blue’ AND Age > 12; AND 运算符确保两个条件同时为真。
11. Computational Thinking in Real-Life Scenarios | 现实生活中的计算思维
Computational thinking involves decomposition, pattern recognition, abstraction, and algorithm design. Applying these to organize a school event demonstrates cross-curricular value.
计算思维包括分解、模式识别、抽象和算法设计。将这些应用于组织学校活动，展示了跨学科价值。
Problem: You need to plan a school sports day. Break down the problem, identify patterns (e.g., timetabling events), and write an algorithm for the schedule.
题目：你需要策划一次学校运动会。分解问题，识别模式（如活动时间表），并编写一个日程算法。
Solution: Decompose into: venues, participants, events, timing. Pattern: each event takes 15 min + 5 min transition. Algorithm: 1. List events and participants. 2. Allocate time slots avoiding clashes. 3. Output timetable. Abstract away from specific names.
解答：分解为：场地、参与者、项目、时间安排。模式：每个项目用时 15 分钟加 5 分钟转场。算法：1. 列出项目和参与者。2. 分配时间段避免冲突。3. 输出时间表。从具体名称中抽象出来。
12. Exam-Style Integrated Questions | 考试风格整合题
Finally, here is a multi-part problem that combines several topics: binary, spreadsheet, and flowcharts. This mirrors typical CIE Year 7 assessments.
最后，这里有一道整合多个主题的多部分问题，结合了二进制、电子表格和流程图。这反映了典型的 CIE 七年级考核方式。
Problem: Part (a): Convert the denary number 29 into binary. (b) In a spreadsheet, cell C2 contains = (A2 + B2) * 2. If A2=5 and B2=the binary value from (a) converted back to denary, what is C2? (c) Draw a flowchart to decide if the result in C2 is even or odd, outputting ‘Even’ or ‘Odd’.
题目：(a) 将十进制数 29 转换为二进制。(b) 在电子表格中，单元格 C2 包含公式 = (A2 + B2) * 2。如果 A2=5，B2 等于 (a) 中的二进制数再转回十进制的结果，C2 是多少？(c) 绘制一个流程图，判断 C2 的结果是偶数还是奇数，并输出 ‘Even’ 或 ‘Odd’。
Solution: (a) 29₁₀ = 11101₂ (16+8+4+0+1). (b) B2 = 29, so C2 = (5+29)*2 = 68. (c) Flowchart: Start → Set num = 68 → Decision ‘Is num MOD 2 = 0?’ → Yes: Output ‘Even’, No: Output ‘Odd’ → Stop. Since 68 MOD 2 = 0, output ‘Even’.
解答：(a) 29₁₀ = 11101₂ (16+8+4+0+1)。(b) B2 = 29, 因此 C2 = (5+29)*2 = 68。(c) 流程图：开始 → 设置 num = 68 → 判断 ‘num MOD 2 = 0?’ → 是: 输出 ‘Even’，否: 输出 ‘Odd’ → 停止。因为 68 MOD 2 = 0，输出 ‘Even’。
These integrated questions train you to think across boundaries, exactly as required in the CIE Computer Science curriculum.
这些综合题训练你跨越边界思考，这正是 CIE 计算机科学课程所要求的。
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