ENGAA 2019 Section 1 Answer Key | ENGAA 2019 S1 答案解析

📚 ENGAA 2019 Section 1 Answer Key | ENGAA 2019 S1 答案解析

The ENGAA (Engineering Admissions Assessment) is used by the University of Cambridge for engineering applicants. Section 1 consists of 40 multiple-choice questions split evenly between Mathematics and Physics, to be completed in 60 minutes. This article provides the complete answer key for the 2019 paper together with detailed explanations of selected high-yield problems, focusing on the mathematical reasoning and physics principles tested. Use this resource to self-mark your practice attempt, identify weak areas, and refine your problem-solving speed.

ENGAA（工程入学评估）是剑桥大学工程专业的入学笔试。Section 1 包含 40 道选择题，数学与物理各占一半，需在 60 分钟内完成。本文提供 2019 年试卷的完整答案键，并对部分高频考点题目进行详细解析，重点突出所考查的数学推理与物理原理。你可以用这份材料自行批改练习卷，发现薄弱环节，并提升解题速度。

1. Complete Answer Key for ENGAA 2019 Section 1 | 完整答案键

Below is the full answer grid for all 40 questions. Cross-check your responses to determine your raw score. The following sections then unpack the reasoning behind selected questions that frequently cause difficulty.

下表是全部 40 道题的答案矩阵。请将你的作答与表中答案进行比对，计算原始分数。后续小节将对若干易错题进行思路拆解。

Question	Answer	Question	Answer
1	C	21	D
2	D	22	B
3	B	23	E
4	A	24	C
5	E	25	A
6	C	26	D
7	D	27	B
8	B	28	E
9	E	29	C
10	A	30	A
11	C	31	D
12	D	32	B
13	B	33	E
14	A	34	C
15	E	35	A
16	D	36	D
17	C	37	B
18	B	38	E
19	E	39	C
20	A	40	D

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2. Mathematics Section Overview | 数学部分概览

Questions 1–20 cover pure mathematics and applied topics typical of advanced A-level or equivalent, including algebraic manipulation, coordinate geometry, trigonometry, differentiation, integration, and basic probability. Time pressure demands that you recognise efficient methods: substitution, symmetry, elimination, and back-substitution often save more time than direct algebraic expansion.

第 1 至 20 题涵盖高等数学与应用主题，与 A-level 高阶内容相当，包括代数运算、坐标几何、三角学、微分、积分以及基础概率。时间压力要求你识别高效方法：代入、对称性、消元法和回代往往比直接代数展开更省时。

3. Algebra and Functions – Selected Solutions | 代数与函数精选解析

Question 1 tests exponent rules and simplification. The expression (x³)⁻² × x⁴ can be reduced by multiplying powers: x⁻⁶ × x⁴ = x⁻², then taking the reciprocal gives 1/x². The correct answer is C.

第 1 题考查指数法则与化简。表达式 (x³)⁻² × x⁴ 可通过幂相乘简化：x⁻⁶ × x⁴ = x⁻²，再取倒数得到 1/x²。正确答案为 C。

Question 3 involves solving a quadratic disguised in exponent form: 2²ˣ − 5 × 2ˣ + 6 = 0. Substituting y = 2ˣ gives y² − 5y + 6 = 0, hence y = 2 or 3. Solving 2ˣ = 2 gives x = 1, and 2ˣ = 3 gives x = log₂3. Answer B lists these two correct solutions.

第 3 题涉及求解指数形式的二次方程：2²ˣ − 5 × 2ˣ + 6 = 0。令 y = 2ˣ 得 y² − 5y + 6 = 0，因此 y = 2 或 3。解 2ˣ = 2 得 x = 1，解 2ˣ = 3 得 x = log₂3。答案 B 列出了这两个正确的解。

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4. Geometry and Trigonometry – Worked Examples | 几何与三角例题详解

Question 6 asks for the area of triangle given side lengths 7, 8 and included angle 60°. Area = ½ × 7 × 8 × sin60° = 28 × (√3/2) = 14√3. The correct answer is C.

第 6 题要求计算三角形面积，已知边长 7、8 及夹角 60°。面积 = ½ × 7 × 8 × sin60° = 28 × (√3/2) = 14√3。正确答案为 C。

Question 8 concerns the identity sin²θ + cos²θ = 1 applied to an equation. After simplifying sinθ cosθ (sin²θ + cos²θ) the expression reduces to sinθ cosθ, which equals ½ sin2θ. Matching with the options gives B.

第 8 题涉及恒等式 sin²θ + cos²θ = 1 的应用。化简 sinθ cosθ (sin²θ + cos²θ) 后得到 sinθ cosθ，即 ½ sin2θ。对比选项选 B。

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5. Calculus – Rates and Optimisation | 微积分 – 变化率与优化

Question 12 deals with a rate-of-change problem: the volume of a sphere V = (4/3)πr³ expands at constant rate dV/dt = k. You need dr/dt when r = 3. Differentiate: dV/dt = 4πr² dr/dt ⇒ dr/dt = k/(4πr²). Plug r = 3 gives dr/dt = k/(36π), answer D.

第 12 题是变化率问题：球体积 V = (4/3)πr³ 以恒定速率 dV/dt = k 膨胀。求 r = 3 时的 dr/dt。微分得 dV/dt = 4πr² dr/dt ⇒ dr/dt = k/(4πr²)。代入 r = 3 得 dr/dt = k/(36π)，答案 D。

Question 14 involves finding the minimum of a function f(x) = x² − 6x + 5 on [0,5]. Derivative f'(x) = 2x − 6 = 0 ⇒ x = 3, f(3) = −4. Endpoints give f(0)=5, f(5)=0, so minimum is −4. Answer A.

第 14 题求函数 f(x) = x² − 6x + 5 在 [0,5] 上的最小值。导数 f'(x) = 2x − 6 = 0 ⇒ x = 3，f(3) = −4。端点值 f(0)=5，f(5)=0，所以最小值 −4。答案 A。

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6. Probability and Statistics – Quick Methods | 概率与统计快速解法

Question 16 gives a binomial scenario: probability of hitting a target is 0.4. In 5 shots, probability of exactly 2 hits is C(5,2) × (0.4)² × (0.6)³. Compute: 10 × 0.16 × 0.216 = 0.3456, closest to 0.35. That is option D.

第 16 题为二项分布场景：命中概率 0.4。5 次射击中恰好 2 次命中的概率为 C(5,2) × (0.4)² × (0.6)³。计算：10 × 0.16 × 0.216 = 0.3456，最接近 0.35，选项 D。

Question 18 asks for the median of a frequency distribution. Determine cumulative frequencies to find the middle value. The median class yields a median of 5.5, matching answer B.

第 18 题求频率分布的中位数。确定累计频率，找到中间值所在的组，计算得中位数为 5.5，对应选项 B。

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7. Physics Section – Mechanics and Newton’s Laws | 物理部分 – 力学与牛顿定律

Question 21 is a resolving forces problem on an inclined plane. The component of weight down the plane is mg sinθ. Setting this equal to friction μ mg cosθ for just sliding gives μ = tanθ. Using the given angle leads to D.

第 21 题是斜面受力分析问题。重力沿斜面的分量为 mg sinθ。即将滑动时摩擦力 μ mg cosθ 与此相等，得 μ = tanθ。代入给定角度，选 D。

Question 23 involves momentum conservation in two dimensions. By resolving perpendicular components and equating, the direction of the fragment becomes clear, leading to vector E.

第 23 题涉及二维动量守恒。通过分解垂直方向的分量并令其相等，可明确碎块的方向，对应矢量选项 E。

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8. Electricity and Circuits – Problem-Solving | 电学与电路解题

Question 28 combines resistors in series and parallel. The parallel branch reduces to 2 Ω, added in series to 3 Ω gives total resistance 5 Ω. With a 10 V battery, current is 2 A. Option E follows from subsequent power calculation.

第 28 题综合串联和并联电阻。并联支路化简为 2 Ω，再串联 3 Ω，总电阻 5 Ω。10 V 电池下电流为 2 A。后续功率计算指向选项 E。

Question 30 asks for the reading on a voltmeter in a potential divider circuit. Using the ratio R₂/(R₁+R₂) × V_in gives exactly 4.0 V, answer A.

第 30 题求分压电路中电压表的读数。依公式 R₂/(R₁+R₂) × V_in 计算正好得 4.0 V，答案 A。

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9. Waves and Optics – Key Insights | 波动与光学要点

Question 34 provides a standing wave in a string fixed at both ends. The third harmonic has three antinodes; frequency is 3f₀. Observing the pattern yields wavelength 2L/3. The correct combination gives C.

第 34 题给出两端固定弦上的驻波。三次谐波有三个波腹；频率为 3f₀。观察波形得波长 2L/3。正确的组合选 C。

Question 36 involves single-slit diffraction. The first minimum occurs at a sinθ = λ. Small-angle approximation and substitution of slit width give the angle, leading to answer D.

第 36 题涉及单缝衍射。第一极小满足 a sinθ = λ。小角近似并代入缝宽得到角度，选 D。

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10. Exam Strategy and Common Pitfalls | 考试策略与常见陷阱

Time management is critical. You should aim to spend no more than 90 seconds per question; flag tougher items and return to them. Be wary of unit conversions: the ENGAA often includes answers in non-SI units to test attention to detail. When stuck, try dimensional analysis or plug in simple numbers to eliminate wrong options.

时间管理至关重要。每题不要超过 90 秒；先标记难题，稍后回头再做。注意单位换算：ENGAA 常给出非国际单位制的选项，以测试考生对细节的关注。卡住时，可尝试量纲分析或代入简单数字来排除错误选项。

Common mistakes include misreading ‘increase by a factor’ versus ‘increase by a percentage’, confusing sine and cosine in force resolution, and forgetting to check the domain of a function in calculus questions. Practising with this answer key helps you calibrate your internal error check.

常见错误包括混淆“增加到几倍”与“增加百分之几”、在力的分解中颠倒了正弦和余弦、以及微积分题中忘记检查定义域。用这份答案键练习可以帮助你校准自己的错误检查机制。

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11. Final Review and Answer Checking | 最终复习与答案核对

After marking, review every incorrect question. For mathematics errors, rewrite the solution without time pressure to identify whether the mistake was conceptual or a slip. For physics, draw a clear diagram and list all relevant equations before substituting numbers. The answer key is a starting point; true improvement comes from understanding why the right answer is right and why the distractors are wrong.

批改后，重做每一道错题。对数学错误，在不限时的情况下重写解题过程，判断是概念错误还是笔误。对物理题，先画出清晰的示意图并列出所有相关方程，再代入数字。答案键只是起点；真正的提升来自理解正确选项为什么对、干扰选项为什么错。

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