ENGAA 2018 Section 1 Answer Key and Detailed Solutions | ENGAA 2018 S1 答案与详细解析

📚 ENGAA 2018 Section 1 Answer Key and Detailed Solutions | ENGAA 2018 S1 答案与详细解析

The Engineering Admissions Assessment (ENGAA) Section 1 is a fast-paced test of mathematical and scientific reasoning. This article provides the complete answer key for the 2018 paper, along with detailed, bilingual walkthroughs of selected problems. By studying these solutions, candidates can sharpen their problem-solving skills, avoid common traps, and approach the real exam with confidence.

工程入学评估（ENGAA）第一部分是对数学与科学推理能力的快速测试。本文提供 2018 年真题的完整答案表，并精选典型题目进行中英双语详细解析。通过学习这些解法，考生可以提升解题技巧、避开常见陷阱，满怀信心地应对真实考试。

1. Structure and Scoring of ENGAA Section 1 | 考试结构与评分

ENGAA Section 1 consists of 40 multiple-choice questions to be completed in 60 minutes. Questions are split into Part A (Mathematics, 20 questions) and Part B (Physics, 20 questions), with each question carrying one mark. There is no negative marking, so attempting every question is essential. The 2018 paper followed this format, testing a broad range of topics from algebra and geometry to mechanics and electricity.

ENGAA 第一部分包含 40 道选择题，需在 60 分钟内完成。题目分为 A 部分（数学，20 题）和 B 部分（物理，20 题），每题 1 分。答错不扣分，因此每题都值得尝试。2018 年的试卷遵循此模式，覆盖了从代数、几何到力学、电学等广泛主题。

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2. Complete Answer Key for ENGAA 2018 Section 1 | 2018 年 ENGAA S1 完整答案表

The table below lists the official correct answers for all 40 questions. Use it to self-assess your performance after attempting the past paper under timed conditions.

下表列出了全部 40 题的标准答案。在限时条件下完成真题后，可据此进行自我评估。

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

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3. Time Management and General Strategies | 时间管理与通用策略

With only 90 seconds per question on average, efficiency is vital. Skim the paper first to identify the easiest questions, then tackle them methodically. Never spend more than two minutes on a single question on the first pass; mark it and return later if time permits.

平均每题仅有 90 秒，效率至关重要。快速浏览全卷，先找出最简单的题目，再有条理地解答。第一遍做题时，任何题目都不应超过两分钟；暂时跳过并在时间允许时回头再做。

Calculator use is not allowed, so numerical answers often rely on estimation, factorisation or simplification. Learn to manipulate standard form and fractions quickly, and always double-check unit conversions. In physics questions, drawing a quick diagram can save valuable seconds.

考试不允许使用计算器，因此数值答案多依赖于估算、因式分解或化简。要熟练掌握标准形式和分数的快速运算，并始终检查单位换算。在物理题中，快速画出简图能节省宝贵时间。

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4. Worked Example: Unit Conversion and Rates (Q5 style) | 例题精讲：单位换算与比率（Q5 题型）

A typical question asks: ‘A fuel consumption rate is 8.0 litres per 100 km. Express this in miles per gallon. Use 1 mile = 1.6 km, 1 gallon = 4.5 litres.’ First, convert km to miles: 100 km = 100/1.6 = 62.5 miles. The consumption is 8.0 litres per 62.5 miles. To find miles per gallon, we need the reciprocal: miles per litre = 62.5/8.0 = 7.8125 miles/litre. Multiply by litres per gallon: 7.8125 × 4.5 = 35.15625 ≈ 35 miles per gallon. The correct choice is the one closest to 35.

一道典型题目为：“某车的油耗为 8.0 升/100 公里。将其表示为英里/加仑。已知 1 英里 = 1.6 公里，1 加仑 = 4.5 升。”首先将公里转换为英里：100 km = 100/1.6 = 62.5 miles。油耗为 8.0 升对应 62.5 英里。为求英里/加仑，需要倒数：每升英里数 = 62.5/8.0 = 7.8125 miles/litre。再乘以加仑对应的升数：7.8125 × 4.5 = 35.15625，约等于 35 英里/加仑。正确答案应为最接近 35 的选项。

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5. Worked Example: Algebraic Simplification (Q7 style) | 例题精讲：代数化简（Q7 题型）

Consider simplifying the expression (x² – 9)/(x – 3). Recognise the numerator as a difference of two squares: x² – 9 = (x – 3)(x + 3). Then the fraction becomes (x – 3)(x + 3)/(x – 3). Cancel the common factor (x – 3), leaving x + 3, provided x ≠ 3. The trap is forgetting the restriction or misapplying the difference of squares; always factorise fully before cancelling.

考虑化简表达式 (x² – 9)/(x – 3)。分子是平方差：x² – 9 = (x – 3)(x + 3)。于是分式变为 (x – 3)(x + 3)/(x – 3)。约去公因式 (x – 3)，得到 x + 3，前提是 x ≠ 3。常见的错误是忘记定义域限制或误用平方差公式；务必先完全分解因式再约分。

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6. Worked Example: Graph Interpretation (Q12 style) | 例题精讲：图像解读（Q12 题型）

A velocity–time graph shows a straight line from the origin to (t, v), followed by a horizontal line. The area under the graph gives displacement. For the first part, displacement = ½ × t × v. For the second horizontal part, displacement = v × (total time – t). Add these areas to find total distance. Many students confuse gradient with area; clarify that gradient gives acceleration, area gives displacement.

速度–时间图像显示一条从原点出发到 (t, v) 的直线，随后是一段水平线。图像下的面积表示位移。第一部分位移 = ½ × t × v。水平段位移 = v ×（总时间 – t）。将这两块面积相加即得总距离。许多学生混淆斜率与面积；要明确斜率代表加速度，而面积代表位移。

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7. Worked Example: Forces and Equilibrium (Q19 style) | 例题精讲：力与平衡（Q19 题型）

A block rests on a rough inclined plane. Resolve forces parallel to the slope: the component of weight down the slope is mg sin θ, and friction acts up the slope. At the point of slipping, friction = μR, where R = mg cos θ. Setting mg sin θ = μ mg cos θ gives μ = tan θ. If θ = 30°, then μ = tan 30° = 1/√3 ≈ 0.58. Answer look-up shows that the coefficient of friction is required, not the force.

一物块静置于粗糙斜面上。沿斜面分解力：重力沿斜面的分量为 mg sin θ，摩擦力沿斜面向上。即将滑动时，摩擦力 = μR，而支持力 R = mg cos θ。令 mg sin θ = μ mg cos θ 可得 μ = tan θ。若 θ = 30°，则 μ = tan 30° = 1/√3 ≈ 0.58。查表可知需要的是摩擦系数，而非力的大小。

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8. Worked Example: Ratio and Proportion (Q25 style) | 例题精讲：比与比例（Q25 题型）

The masses of three objects are in the ratio 2:3:4. The total mass is 180 g. To find the largest mass, set the parts as 2k, 3k, 4k. Then 2k + 3k + 4k = 9k = 180, so k = 20. The largest mass is 4k = 80 g. Alternatively, use fractions: largest fraction = 4/9 of total = (4/9)×180 = 80 g. Always verify the sum of the parts equals the given total.

三个物体的质量比为 2:3:4，总质量为 180 g。为求最大质量，设各份为 2k、3k、4k。于是 2k + 3k + 4k = 9k = 180，得 k = 20。最大质量为 4k = 80 g。也可用分数法：最大份额为总质量的 4/9，(4/9)×180 = 80 g。务必检验各份之和等于给定总量。

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9. Worked Example: Trigonometry and Pythagoras (Q30 style) | 例题精讲：三角与勾股定理（Q30 题型）

A right-angled triangle has legs of length 5 cm and 12 cm. Find the smallest angle. The hypotenuse = √(5² + 12²) = √(25+144) = √169 = 13 cm. The smallest angle is opposite the shortest side (5 cm), so sin θ = 5/13. Using known approximations or exact values, θ ≈ 22.6°. In a multiple-choice format, you may need to identify the angle whose sine or tangent matches the ratio.

一直角三角形两直角边长为 5 cm 和 12 cm，求最小角。斜边 = √(5² + 12²) = √(25+144) = √169 = 13 cm。最小角对最短边（5 cm），故 sin θ = 5/13。根据已知近似值或精确值，θ ≈ 22.6°。在选择题中，你需要识别出正弦或正切值与给定比值匹配的角。

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10. Worked Example: Circuit Analysis (Q35 style) | 例题精讲：电路分析（Q35 题型）

Two resistors, 4 Ω and 6 Ω, are connected in parallel, and this combination is connected in series with a 2 Ω resistor. Find the equivalent resistance. Calculate parallel part: 1/Rp = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so Rp = 12/5 = 2.4 Ω. Then add the series resistor: Rtotal = 2.4 + 2 = 4.4 Ω. Remember parallel resistance is always less than the smallest individual resistor.

两个电阻 4 Ω 和 6 Ω 并联，再与一个 2 Ω 电阻串联。求等效电阻。先计算并联部分：1/Rp = 1/4 + 1/6 = 3/12 + 2/12 = 5/12，所以 Rp = 12/5 = 2.4 Ω。然后加上串联电阻：总电阻 = 2.4 + 2 = 4.4 Ω。记住并联电阻总是小于最小的单个电阻。

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11. Common Pitfalls and How to Avoid Them | 常见易错点与对策

Many errors arise from misreading units (e.g., cm instead of m), forgetting to square conversion factors in area and volume conversions, or incorrectly applying sign conventions in equations of motion. Double-check the final answer by plugging it back into the original conditions or estimating the expected magnitude. In algebra, watch for hidden division by zero possibilities.

许多错误源于单位看错（如 cm 与 m 混淆）、在面积与体积换算中遗漏平方或立方系数，或在运动方程中误用符号约定。可通过将答案代回原条件或估算预期数量级来复核最终结果。代数题中，注意隐藏的除以零的可能性。

On the physics side, mixing up mass and weight, confusing heat capacity with specific heat capacity, or reversing ray diagram constructions are frequent mistakes. Always write down the relevant formula before substituting numbers, and keep all working neat to allow quick checking.

在物理部分，混淆质量与重量、分不清热容与比热容、或画反光路图是常见错误。代入数字前先写出相关公式，并保持解题过程整洁以便快速检查。

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12. Further Practice and Resource Recommendations | 进一步练习与资源推荐

After reviewing the 2018 paper, attempt additional ENGAA past papers under timed conditions. Focus on weak areas identified from the answer key. Complement your preparation with A-Level Mathematics and Physics problem sets that emphasise speed and accuracy without a calculator.

复习完 2018 年试卷后，请在计时条件下尝试更多 ENGAA 历年真题。根据答案表找出薄弱环节并重点突破。用强调无计算器下速度与准确性的 A-Level 数学和物理习题集进行补充训练。

Create a formula sheet with essential relationships: kinematics equations, Ohm’s law, trigonometric identities, and common geometry facts. Review it daily so that recall becomes automatic during the exam.

制作一张包含关键关系的公式表：运动学方程、欧姆定律、三角恒等式和常用几何结论。每日复习，使考试时能够条件反射式地回忆。

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