📚 Interdisciplinary Integrated Exercises for Year 7 CIE Computing | 跨学科综合题型训练：Year 7 CIE 计算机
Computing does not live in isolation. In Year 7 CIE Computing, you will often encounter problems that blend digital skills with mathematics, science, geography, art and even language. This article provides a set of integrated exercises to help you practise computational thinking across subjects. Every exercise is designed to mirror the style of CIE assessment, combining clear English explanations with Chinese translations to support bilingual learners. Work through these challenges, and you will not only sharpen your coding and algorithmic skills but also see how computing enriches every area of study.
计算机并不是一门孤立的学科。在 Year 7 CIE 计算机课程中，你经常会遇到将数字技能与数学、科学、地理、艺术甚至语言融合在一起的题目。本文提供了一套跨学科综合练习题，帮助你训练计算思维。每道题都模拟了 CIE 评估的风格，同时提供清晰的英文解释和中文翻译，为双语学习者提供支持。逐一攻克这些挑战，你不仅能提升编程和算法能力，还能体会到计算如何让每一个学习领域更加丰富。
1. Binary Counting and Mathematics | 二进制计数与数学
Computers use binary (base‑2) to represent numbers. In this exercise you will convert a decimal number into binary and then use that binary value to solve a simple arithmetic puzzle. Imagine you are helping a science lab record the number of bacteria colonies. The count is 29. Express 29 as an 8‑bit binary number. Now imagine the number doubles every hour. If you left‑shift the binary number by one place, what decimal value do you obtain? A left shift on a binary number multiplies it by 2.
计算机使用二进制（基数为 2）表示数字。在这道练习题中，你需要把一个十进制数转换为二进制，然后利用这个二进制值解决一个简单的算术谜题。假设你正在帮助一个科学实验室记录细菌菌落的数量，计数值为 29。请将 29 表示为 8 位二进制数。现在，假设细菌数量每小时翻倍。如果你将二进制数左移一位，会得到什么十进制数值？二进制数左移相当于原数乘以 2。
Solution step: 29 in binary is 00011101. Left shift gives 00111010, which is 58 in decimal. The exercise links place value in mathematics with binary shifts, reinforcing the concept of multiplication by powers of two.
解题步骤：29 的二进制形式是 00011101。左移一位得到 00111010，即十进制 58。这道题将数学中的位值与二进制移位联系起来，强化了乘以 2 的幂次的概念。
2. Algorithm Design for a Science Experiment | 科学实验中的算法设计
In a chemistry lesson, you need to record the temperature of a solution every minute for five minutes. Design an algorithm, using a simple flowchart or pseudocode, that reads five temperature inputs and then displays the average temperature. Your algorithm must use a loop and a variable to accumulate the sum.
在一节化学课上，你需要每隔一分钟记录一次溶液的温度，共记录五次。请设计一个算法（用简单流程图或伪代码），能够读取五个温度输入值，然后显示平均温度。你的算法必须使用循环和一个变量来累加总和。
For example, pseudocode might look like:
SET total TO 0
FOR count FROM 1 TO 5
INPUT temp
SET total TO total + temp
END FOR
SET average TO total / 5
OUTPUT average
例如，伪代码可以写成：
将 total 设为 0
对于 count 从 1 到 5
输入 temp
将 total 设为 total + temp
结束循环
将 average 设为 total / 5
输出 average
This kind of task appears frequently in CIE questions and encourages you to think procedurally about collecting data, just as a scientist does.
这类任务在 CIE 考题中经常出现，它鼓励你像科学家一样，用过程化的方式思考数据收集。
3. Geography Mapping with Coordinates | 地理坐标与映射
A geographer uses a grid to locate landmarks. In computing, we can store coordinates (x, y) to represent points on a map. Imagine a treasure map where the starting point is at (0,0). The directions are given as a sequence of moves: ‘N’ increases y by 1, ‘S’ decreases y by 1, ‘E’ increases x by 1, ‘W’ decreases x by 1. Write a simple program or trace a set of instructions to find the final coordinates after the sequence N, E, E, S, W, N, N. Show the final position and calculate the straight‑line distance from the starting point using the Pythagorean theorem.
地理学家使用网格来确定地标位置。在计算机科学中，我们可以用坐标 (x, y) 来表示地图上的点。假设有一张藏宝图，起点为 (0,0)。给出的指令是一系列移动：’N’ 使 y 增加 1，’S’ 使 y 减少 1，’E’ 使 x 增加 1，’W’ 使 x 减少 1。请编写一段简短的程序，或者逐步追踪指令，找出序列 N, E, E, S, W, N, N 之后的最终坐标。然后计算最终位置到起点的直线距离（使用勾股定理）。
Trace: start (0,0). N → (0,1). E → (1,1). E → (2,1). S → (2,0). W → (1,0). N → (1,1). N → (1,2). Final position (1,2). Distance = √((1-0)² + (2-0)²) = √(1 + 4) = √5 ≈ 2.236 units. This exercise bridges coordinate geometry with sequential processing, a core computing idea.
追踪：起点 (0,0)。N → (0,1)。E → (1,1)。E → (2,1)。S → (2,0)。W → (1,0)。N → (1,1)。N → (1,2)。最终位置 (1,2)。距离 = √((1-0)² + (2-0)²) = √(1 + 4) = √5 ≈ 2.236 单位长度。这个练习将坐标几何与顺序处理联系在一起，这是计算机科学的核心思想。
4. Art and Turtle Graphics | 艺术与海龟绘图
Turtle graphics is an excellent way to combine geometry and programming. Using commands like FORWARD 100, RIGHT 90, you can draw regular polygons. Write a sequence of LOGO‑style turtle commands to draw a square of side length 80 units. Then modify your sequence to draw an equilateral triangle. Each interior angle of an equilateral triangle is 60°, so the turtle must turn by 120° at each vertex. Remember that the turtle turns by the external angle.
海龟绘图是将几何与编程结合起来的绝佳方式。通过 FORWARD 100、RIGHT 90 等命令，你可以绘制出正多边形。请写出一组 LOGO 风格的海龟命令，绘制一个边长为 80 单位的正方形。然后修改你的命令序列，绘制一个等边三角形。等边三角形的每个内角为 60°，所以海龟在每个顶点处需要转向外部角 120°。记住，海龟转动的是外角。
For a square: REPEAT 4 [FORWARD 80 RIGHT 90]. For a triangle: REPEAT 3 [FORWARD 100 RIGHT 120]. This task blends artistic design with logical sequencing and reinforces your understanding of angles.
正方形的命令：REPEAT 4 [FORWARD 80 RIGHT 90]。三角形的命令：REPEAT 3 [FORWARD 100 RIGHT 120]。这项任务将艺术设计与逻辑序列融合在一起，并加深了你对角度的理解。
5. Language, Pseudocode and Sequencing | 语言、伪代码与排序
Understanding a set of instructions written in a natural language and then translating them into pseudocode is a key skill. Read this English description: ‘Ask the user for their name. If the name is “Alice”, greet her with “Hello, Alice! Welcome back.” Otherwise, greet the user with “Nice to meet you, [name].”‘ Convert this into pseudocode using INPUT, IF, THEN, ELSE and OUTPUT.
理解用自然语言编写的一组指令，然后将其转换为伪代码，是一项关键技能。请阅读下面的英文描述：’Ask the user for their name. If the name is “Alice”, greet her with “Hello, Alice! Welcome back.” Otherwise, greet the user with “Nice to meet you, [name].”‘ 使用 INPUT、IF、THEN、ELSE 和 OUTPUT 将其转换为伪代码。
Sample solution:
INPUT name
IF name = “Alice” THEN
OUTPUT “Hello, Alice! Welcome back.”
ELSE
OUTPUT “Nice to meet you, ” + name
ENDIF
示例解决方案：
输入 name
如果 name = “Alice” 那么
输出 “Hello, Alice! Welcome back.”
否则
输出 “Nice to meet you, ” + name
结束如果
This type of cross‑disciplinary activity sharpens both your comprehension and your ability to formalise logic, a skill needed in languages and in computer programming.
这类跨学科活动既能提高你的理解能力，又能锻炼你将逻辑形式化的能力，这既是语言学习所需，也是计算机编程的基础。
6. Data Handling with History Timelines | 历史时间线中的数据处理
In history class, you may create timelines. In computing, you can sort dates and analyse patterns. Suppose you have the following historical events with their years: Battle of Hastings (1066), Magna Carta (1215), and the Great Fire of London (1666). Write an algorithm that places these events in chronological order using a simple sorting technique, like bubble sort or insert sort. Then explain how a computer could check whether there is an event exactly in the middle of the timeline (median).
在历史课上，你可能会制作时间线。在计算机科学中，你可以对日期进行排序并分析模式。假设有以下历史事件及其年份：黑斯廷斯战役（1066 年）、大宪章（1215 年）、伦敦大火（1666 年）。请写出一段算法，使用简单的排序技术（如冒泡排序或插入排序）将这些事件按时间顺序排列。然后解释计算机如何检查时间线上是否存在恰好位于正中间的事件（中位数）。
For three items, after sorting (1066, 1215, 1666) the median is 1215. The algorithm could simply compare dates step by step. This integrates computational sorting with chronological reasoning.
对于三个项目，排序后为 (1066, 1215, 1666)，中位数为 1215。算法可以逐步比较日期。这将计算排序与时间顺序推理结合在一起。
7. Network Communication and Social Studies | 网络通信与社会研究
Social studies topics often include how people share information. In computing, we model this with networks. Draw a simple network diagram with four nodes labelled A, B, C, D. Assume information can flow from A to B, B to C, C to D, and also directly from A to C. A piece of news starts at node A. List all the different routes the news can take to reach node D. Explain which route you think is the fastest and why, in terms of computer network speed (number of hops).
社会研究课题常常涉及人们如何共享信息。在计算机领域，我们用网络来对此建模。请绘制一个简单的网络图，包含四个节点，分别标记为 A、B、C、D。假设信息可以从 A 流向 B，从 B 流向 C，从 C 流向 D，同时还可以直接从 A 流向 C。一条消息从节点 A 出发。列出消息到达节点 D 的所有可能路径。根据计算机网络速度（跳数），解释你认为哪条路径最快，并说明理由。
Routes: A→B→C→D (3 hops), A→C→D (2 hops), maybe A→B→C→D only. The direct A→C→D route has only 2 hops and should be faster, assuming no congestion. This introduces the concept of routing and network topology through a social‑studies lens.
路径有：A→B→C→D（3 跳）、A→C→D（2 跳）。直接的 A→C→D 路径只有两跳，在没有拥塞的情况下应该更快。这通过社会研究的视角引入了路由和网络拓扑的概念。
8. Logic Puzzles and Boolean Algebra | 逻辑谜题与布尔代数
Boolean logic underpins all computer decision making. Combine logic with a puzzle: In a game, a character can enter a treasure room only if they have a golden key AND a silver crystal, OR if they speak the secret password. Let A = has golden key, B = has silver crystal, C = speaks password. Write a Boolean expression for entering the room: (A AND B) OR C. Now create a truth table showing all possible combinations of A, B, C (true/false) and whether entry is allowed. This teaches logical operators, which are also used in mathematical reasoning.
布尔逻辑是计算机进行所有决策的基础。将逻辑与谜题结合起来：在一个游戏中，角色只有在拥有金钥匙并且拥有银水晶，或者会说秘密口令的情况下，才能进入宝藏室。设 A = 拥有金钥匙，B = 拥有银水晶，C = 会说口令。请写出允许进入房间的布尔表达式：(A AND B) OR C。现在，创建一个真值表，显示 A、B、C 所有可能的组合（真/假），以及是否允许进入。这可以教授逻辑运算符，它们在数学推理中同样会用到。
For example, when A=true, B=true, C=false, expression evaluates to true. When A=false, B=true, C=false, expression false. This activity connects computing logic with systematic truth evaluation used in philosophy and maths.
例如，当 A=真，B=真，C=假时，表达式的结果为真。当 A=假，B=真，C=假时，结果为假。这项活动将计算逻辑与哲学和数学中常见的系统性真值评估联系了起来。
9. Flowcharts for Decision Making in Design & Technology | 设计与技术中的决策流程图
In design and technology, you often follow a process to build a product. Similarly, computing uses flowcharts to map decisions. Design a flowchart for choosing materials: if the product must be waterproof, use plastic; otherwise, if it needs to be strong, use metal; otherwise, use wood. Use standard flowchart symbols (oval for start/end, diamond for decision, rectangle for process). This helps you visualise conditional logic, a concept shared between design and programming.
在设计与技术课上，你常常要按照一定的流程来制作产品。同样，计算机则使用流程图来规划决策。请为材料选择设计一个流程图：如果产品必须防水，就使用塑料；否则，如果需要坚固，就使用金属；否则，使用木材。请使用标准流程图符号（椭圆形表示开始/结束，菱形表示判断，矩形表示处理）。这有助于你可视化条件逻辑，这是设计与编程共享的一个概念。
Your flowchart should start with ‘Start’, then a decision box ‘Waterproof?’, with ‘Yes’ leading to ‘Use plastic’, ‘No’ to another decision ‘Strong?’, with ‘Yes’ to ‘Use metal’, ‘No’ to ‘Use wood’, all ending at ‘End’. This explicit mapping of conditions is fundamental in writing IF‑THEN‑ELSE structures.
你的流程图应该以“开始”为起点，然后是一个判断框“是否防水？”，“是”指向“使用塑料”，“否”指向另一个判断框“是否需要坚固？”，“是”指向“使用金属”，“否”指向“使用木材”，所有分支最终汇聚到“结束”。这种对条件的显式映射是编写 IF‑THEN‑ELSE 结构的基础。
10. Spreadsheet Modelling a School Fair Budget | 用电子表格为学校义卖会预算建模
Spreadsheets are powerful tools for mathematical modelling. For a school fair, you plan to sell cupcakes. Each cupcake costs £0.40 to make and sells for £1.00. You also have a fixed cost of £15 for stall rental. Create a spreadsheet model that calculates profit for a given number of cupcakes sold. In column A, list possible sales numbers (10, 20, 30, …). In column B, write a formula using cell references showing total revenue. In column C, total cost (variable + fixed). In column D, profit. Explain what formula you would place in cell D2. This links computing with business maths and budgeting.
电子表格是进行数学建模的强大工具。学校义卖会上，你计划售卖纸杯蛋糕。每个蛋糕的制作成本为 0.40 英镑，售价为 1.00 英镑。此外，你还需要支付 15 英镑的固定摊位租赁费。请创建一个电子表格模型，计算出售给定数量纸杯蛋糕后的利润。在 A 列列出可能的销售数量（10、20、30……）。在 B 列书写使用单元格引用的公式，用于显示总收入。在 C 列显示总成本（可变成本 + 固定成本）。在 D 列显示利润。请解释你会在 D2 单元格中输入什么公式。这将计算与商业数学和预算编制联系起来。
Assuming A2 holds the number sold, B2 formula: =A2*1. C2: =A2*0.4 + 15. D2: =B2 – C2. This exercise mirrors a typical CIE spreadsheet question and reinforces algebraic thinking.
假设 A2 存放销售数量，B2 公式为：=A2*1，C2 为：=A2*0.4 + 15，D2 为：=B2 – C2。这道题模拟了典型的 CIE 电子表格考题，并强化了代数思维。
11. Pseudocode and Music Rhythms | 伪代码与音乐节奏
Musical patterns can be represented using sequences of symbols. Computing algorithms can generate or analyse such patterns. Suppose a rhythm is encoded as a string of characters: ‘B’ for beat, ‘R’ for rest. A simple melody repeats: B B R B. Write a pseudocode algorithm that takes such a string and counts the number of beats. Then, using a loop, output the rhythm, but replacing every ‘B’ with a clap sound (represented by the word ‘clap’) and every ‘R’ with ‘silence’. This encourages pattern recognition and string processing, both important in computing and in music theory.
音乐模式可以用符号序列表示，计算算法可以生成或分析这些模式。假设一个节奏被编码为字符串：’B’ 代表节拍，’R’ 代表休止。一段简单的旋律重复着：B B R B。请编写一段伪代码算法，输入这样一个字符串，统计节拍的数量。然后，使用循环输出该节奏，但将每个 ‘B’ 替换为拍手声（用单词 ‘clap’ 表示），将每个 ‘R’ 替换为 ‘silence’。这可以鼓励模式识别和字符串处理，这两者在计算和音乐理论中都很重要。
Pseudocode: SET count TO 0. FOR each character in string: IF character = ‘B’ THEN count = count + 1; OUTPUT ‘clap’ ELSE OUTPUT ‘silence’. END FOR. This blend of arts and computing showcases how algorithms can manipulate creative content.
伪代码：将 count 设为 0。对于字符串中的每个字符：如果字符 = ‘B’，那么 count = count + 1；输出 ‘clap’，否则输出 ‘silence’。结束循环。这种艺术与计算的融合展示了算法如何操控创造性内容。
12. Debugging a Story Sequence | 给故事排序纠错
In English, you learn to sequence events logically. In computing, a program with incorrect order fails. Read this set of mixed‑up instructions meant to describe making a cup of tea: ‘Pour tea into cup. Boil water. Put tea bag in cup. Fill kettle with water.’ Identify the logical error. Rewrite the instructions in the correct order and explain why sequence matters in both storytelling and coding. Debugging is a crucial skill across all subjects.
在英语课上，你会学习如何逻辑地排列事件顺序。在计算机领域，顺序错误的程序会运行失败。请阅读下面这套描述泡茶步骤的混乱指令：’将茶水倒入杯中。烧开水。将茶包放入杯中。往水壶里加水。’ 找出其中的逻辑错误。按正确顺序重写指令，并解释为什么顺序在讲故事和编程中都至关重要。纠错（调试）是一门跨所有学科的重要技能。
Correct order: Fill kettle with water → Boil water → Put tea bag in cup → Pour water into cup. In programming, if you pour before boiling, the program logic fails. Similarly, a narrative with jumbled events confuses the reader. This task highlights the universal importance of correct sequencing.
正确顺序：往水壶里加水 → 烧开水 → 将茶包放入杯中 → 将热水倒入杯中。在编程中，如果你先倒水再烧水，程序逻辑就会出错。同样，事件混乱的叙述会让读者感到困惑。这项任务突显了正确排序的普遍重要性。
Published by TutorHao | Computing Revision Series | aleveler.com
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