📚 Cross-Curricular Integrated Question Practice for Year 11 Eduqas Computer Science | 跨学科综合题型训练：Eduqas 计算机科学
In Year 11 Eduqas Computer Science, exam questions increasingly weave together concepts from multiple subjects—mathematics, physics, logic and electronic principles—to test your ability to apply knowledge in unfamiliar contexts. This article presents twelve cross-disciplinary problem scenarios that mirror the style of integrated questions found in Components 1 and 2. Working through these will sharpen your computational thinking and help you make connections between theory and practical problem-solving.
在 Year 11 Eduqas 计算机科学中，考题越来越多地将数学、物理、逻辑和电子学等多个学科的概念交织在一起，考查你在陌生情境中应用知识的能力。本文提供了十二个跨学科问题情境，还原了试卷一和试卷二中常见综合题型的特点。通过练习这些题目，你可以磨炼计算思维，并在理论与实践之间建立起牢固的联系。
1. Data Representation: Binary, Denary and Hexadecimal Conversions with Arithmetic | 数据表示：二进制、十进制与十六进制转换及算术运算
Question: Add the hexadecimal value B4 to the binary value 11001010. Express the final result in both denary and hexadecimal, treating both numbers as unsigned 8-bit integers. Identify whether an overflow occurs.
问题： 将十六进制值 B4 与二进制值 11001010 相加。假设两者均为无符号 8 位整数，用十进制和十六进制表示最终结果，并判断是否发生溢出。
Step 1 – Convert all values to binary: B4₁₆ = 1011 0100₂ (B = 1011, 4 = 0100). The given binary number is 11001010₂.
步骤1 – 将所有数值转换为二进制： B4₁₆ = 1011 0100₂（B 是 1011，4 是 0100）。给定的二进制数是 11001010₂。
Step 2 – Perform binary addition: Use column addition. The sum is 1 01111110₂, which requires 9 bits.
步骤2 – 执行二进制加法： 列竖式计算，结果为 1 01111110₂，需要 9 位来表示。
| Carry | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 10110100 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| + 11001010 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| Sum | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 |
Step 3 – Interpret the result: al is 17E₁₆ (group as 0001 0111 1110). The denary equivalent is 1×256 + 7×16 + 14 = 382₁₀.
步骤3 – 解释结果： 9 位结果 1 01111110₂ 表明在 8 位寄存器中发生了溢出。若按 9 位值处理，十六进制为 17E₁₆（分组成 0001 0111 1110）。十进制值为 1×256 + 7×16 + 14 = 382₁₀。
Cross-disciplinary link: This task blends number base conversions (mathematics) with the hardware limitation of fixed-width registers (computer architecture). Recognising overflow is essential when programming with finite integer types.
跨学科链接： 本题将数制转换（数学）与固定宽度寄存器的硬件限制（计算机体系结构）融合。理解溢出对于使用有限位宽整数的编程至关重要。
2. Binary Addition and Overflow Detection in Two's Complement | 二进制补码加法与溢出检测
Question: Using 8-bit two's complement representation, add +109 (01101101) and +92 (01011100). State whether overflow occurs and justify your answer by examining the sign bits.
问题： 在 8 位补码表示下，将 +109（01101101）和 +92（01011100）相加。判断是否发生溢出，并通过检查符号位说明理由。
Step 1 – Perform the addition: 101 + 01011100 = 11001001. The result has a leading 1, indicating a negative number in two's complement.
步骤1 – 执行加法： 01101101 + 01011100 = 11001001。结果的首位是 1，表明在补码中这是一个负数。
Step 2 – Analyse overflow: h : Both operands are positive (MSB = 0), but the sum is negative (MSB = 1). According to the two' w because two positives cannot yield a negative without exceeding the representable range.
步骤2 – 分析溢出： 两个操作数都是正数（MSB = 0），但和却是负数（MSB = 1）。根据补码溢出规则，这表明发生了溢出——两个正数相加不可能得到一个负数，除非超出了可表示的范围。
Step 3 – Cross-disciplinary reasoning: y falls outside this interval. The hardware signals overflow by comparing carry into and out of the sign bit. Mathematically, we are testing inequalities and modular arithmetic boundaries.
步骤3 – 跨学科推理： 8 位有符号整数的范围是 −128 到 +127。+109 + +92 = +201 显然超出了这一区间。硬件通过比较符号位的进位输入与输出来发出溢出信号。从数学角度看，这正是在检验不等式边界和模运算的界限。
3. Logic Circuits, Truth Tables and Boolean Algebra | 逻辑电路、真值表与布尔代数
Question: A AND B) OR (NOT A AND B), then draw a logic circuit and confirm with a truth table. State the law you used.
问题： 化简布尔表达式 X = (A AND B) OR (NOT A AND B)，画出逻辑电路并用真值表验证。说明你使用的定律。
Step 1 – Algebraic manipulation: is always TRUE (law of the excluded middle, a mathematical tautology), X = B AND TRUE = B.
步骤1 – 代数化简： 提取公因子 B：X = B AND (A OR NOT A)。因为 (A OR NOT A) 恒为真（排中律，数学上的重言式），X = B AND TRUE = B。
Step 2 – Truth table verification: List all combinations of A and B. Observe that X always equals B, confirming the simplification.
步骤2 – 真值表验证
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