📚 GCSE AQA Maths: Typical Exam Questions Explained | GCSE AQA 数学：典型例题详解
This article walks you through some of the most common question types in the AQA GCSE Mathematics exams. Each section covers a different topic area, breaking down how to approach the problem and what examiners look for. Whether you need a quick revision or a deeper understanding of key techniques, these worked examples will sharpen your skills.
本文带你逐一攻克 AQA GCSE 数学考试中最常见的题型。每个部分覆盖一个不同的知识领域，拆解解题思路，分析考官评分要点。无论是快速复习还是深入理解核心技巧，这些典型例题的详解都能帮你提升得分能力。
1. Solving Linear Equations with Brackets | 含括号的一元一次方程
A common exam question asks you to solve equations like 3(2x – 1) + 4 = 19. Always expand brackets first, then collect like terms and isolate the variable.
常见考题是求解 3(2x – 1) + 4 = 19 这类方程。一定要先展开括号，然后合并同类项，再移项解出变量。
Step 1: Expand 3(2x – 1) to get 6x – 3, so the equation becomes 6x – 3 + 4 = 19.
步骤 1：展开 3(2x – 1) 得到 6x – 3，方程变为 6x – 3 + 4 = 19。
Step 2: Simplify the left side: 6x + 1 = 19.
步骤 2：化简左边：6x + 1 = 19。
Step 3: Subtract 1 from both sides: 6x = 18.
步骤 3：两边减去 1：6x = 18。
Step 4: Divide by 6: x = 3. Always check by substituting back into the original equation.
步骤 4：除以 6：x = 3。务必代回原方程验证。
AQA examiners award method marks for correct expansion and simplification, even if the final answer is wrong. Show every step clearly.
AQA 考官对正确的展开和化简步骤会给出方法分，即使最终答案有误。确保每一步书写清晰。
2. Factorising Quadratics & Finding Roots | 二次函数因式分解与求根
Factorising x² + 5x + 6 into (x + 2)(x + 3) is routine, but watch for negative signs and coefficients greater than 1, e.g. 2x² + 7x – 15.
将 x² + 5x + 6 分解为 (x + 2)(x + 3) 很基础，但要警惕负号以及首项系数大于 1 的式子，如 2x² + 7x – 15。
For 2x² + 7x – 15, multiply a and c: 2 × (–15) = –30. Find two numbers that multiply to –30 and add to 7: +10 and –3.
对于 2x² + 7x – 15，先将 a 和 c 相乘：2 × (–15) = –30。找到两个数，积为 –30，和为 7：+10 和 –3。
Split the middle term: 2x² + 10x – 3x – 15, then factor by grouping: 2x(x + 5) – 3(x + 5) = (2x – 3)(x + 5).
拆分中间项：2x² + 10x – 3x – 15，然后分组分解：2x(x + 5) – 3(x + 5) = (2x – 3)(x + 5)。
If the question then asks for roots, set each bracket to zero: x = 3/2 and x = –5. This double-step often appears in a 5-mark question.
如果题目接着要求求根，令每个括号等于零：x = 3/2 和 x = –5。这种两步设问通常出现在 5 分题中。
3. Ratio and Proportion – Sharing & Three-Way Ratios | 比例与比率——分配与三项比
A typical problem: ‘Divide £240 between Alice, Ben and Chloe in the ratio 3 : 5 : 2.’ Find the total number of parts: 3 + 5 + 2 = 10 parts. One part = £240 ÷ 10 = £24.
典型问题：‘将 £240 按照 3 : 5 : 2 的比例分给 Alice、Ben 和 Chloe。’ 先求总份数：3 + 5 + 2 = 10 份。一份 = £240 ÷ 10 = £24。
Then Alice gets 3 × £24 = £72, Ben gets 5 × £24 = £120 and Chloe gets 2 × £24 = £48. Show the check: £72 + £120 + £48 = £240.
因此 Alice 得 3 × £24 = £72，Ben 得 5 × £24 = £120，Chloe 得 2 × £24 = £48。展示验证：£72 + £120 + £48 = £240。
When the ratio includes fractions or decimals, scale it up to whole numbers first. For instance, 0.5 : 2 : 1.5 can be multiplied by 2 to become 1 : 4 : 3.
如果比例中含有分数或小数，先将其放大为整数比。例如 0.5 : 2 : 1.5 可以乘以 2，变为 1 : 4 : 3。
Ratio also appears in map scales and recipes. Always convert units to the same type before simplifying, and give the final ratio in its simplest form.
比例还出现在地图比例尺和食谱类问题中。化简前务必将单位统一，最终结果以最简比形式呈现。
4. Pythagoras’ Theorem & Trigonometry in Right-Angled Triangles | 勾股定理与直角三角形中的三角比
Given a right-angled triangle with legs 5 cm and 12 cm, Pythagoras gives the hypotenuse c: 5² + 12² = c² → 25 + 144 = c² → c = √169 = 13 cm.
已知直角三角形两条直角边为 5 cm 和 12 cm，由勾股定理求斜边 c：5² + 12² = c² → 25 + 144 = c² → c = √169 = 13 cm。
In another question, you may need to find an angle. For a triangle with opposite 8 and hypotenuse 10, sin θ = 8/10, so θ = sin⁻¹(0.8) ≈ 53.1°.
另一类题需求角度。若对边为 8，斜边为 10，sin θ = 8/10，则 θ = sin⁻¹(0.8) ≈ 53.1°。
A common trap is labelling the sides incorrectly. Label the hypotenuse first, then identify opposite (across from given angle) and adjacent (next to angle, not hypotenuse).
常见错误是标注边长位置出错。先标出斜边，然后区分对边（已知角所对的边）和邻边（紧邻已知角但不是斜边的边）。
Many exam questions embed Pythagoras in a real-life context, like a ladder against a wall. Draw a diagram and label lengths clearly before starting calculations.
很多考试题将勾股定理嵌入实际情境，如梯子靠墙问题。开始计算前要画出草图并清晰标注长度。
5. Simultaneous Equations – Elimination Method | 联立方程——消元法
Solve 3x + 2y = 12 and 5x – 2y = 4. Adding the equations eliminates y: 8x = 16 → x = 2. Substitute back: 3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3.
解方程组 3x + 2y = 12 和 5x – 2y = 4。两式相加消去 y：8x = 16 → x = 2。代回：3(2) + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3。
If coefficients are not already matching, multiply one or both equations. Example: 4x + 3y = 23 and 2x + 5y = 22. Multiply the second equation by 2 to get 4x + 10y = 44.
如果系数不相匹配，则对方程组进行倍乘。例：4x + 3y = 23 和 2x + 5y = 22。将第二个方程乘以 2，得到 4x + 10y = 44。
Subtracting gives 7y = 21 → y = 3, then x = 3.5. Always give the solution as a pair x = … , y = … . Check both original equations.
相减得 7y = 21 → y = 3，进而 x = 3.5。答案务必写为 x = … , y = … 的形式，并代入原方程组检验。
6. Probability – Expected Outcomes & Combined Events | 概率——期望次数与复合事件
A bag contains 3 red, 2 blue and 5 green counters. In 60 trials, the expected number of blue counters pulled is (2/10) × 60 = 12.
袋中有 3 个红球、2 个蓝球和 5 个绿球。若进行 60 次试验，预期抽出蓝球的次数为 (2/10) × 60 = 12。
For combined events, such as rolling a die and flipping a coin, list all outcomes systematically. P(even and heads) = (3/6) × (1/2) = 3/12 = 1/4.
对于复合事件，如掷骰子并抛硬币，要系统列出所有结果。P(偶数及正面) = (3/6) × (1/2) = 3/12 = 1/4。
Sometimes you must complete a probability tree. Multiply along branches and add across branches for ’or’ probabilities. Label each branch with clear probabilities, even if some are fractions and some decimals.
有时需要完成概率树。沿树枝相乘，对于“或”的概率则将不同路径相加。每条树枝上要清晰标出概率，即便有些是分数有些是小数。
7. Statistical Graphs – Cumulative Frequency & Box Plots | 统计图——累积频率与箱线图
A cumulative frequency table shows running totals. Plot upper class boundaries on the x‑axis and cumulative frequencies on the y‑axis; then read the median and quartiles from the curve.
累积频率表展示累计和。以区间上限为 x 轴，累积频率为 y 轴描点，然后从曲线上读取中位数和四分位数。
For a box plot, you need minimum, Q1, median, Q3, and maximum. The interquartile range (IQR) = Q3 – Q1, used to measure spread.
绘制箱线图需要最小值、Q1、中位数、Q3 和最大值。四分位距 (IQR) = Q3 – Q1，用来度量离散程度。
A typical question: ‘Draw a box plot from this cumulative frequency graph.’ Remember to draw the box with Q1 and Q3, the median line inside, and whiskers extending to the min and max. Label the axis properly.
典型考题：‘根据累积频率图画出箱线图。’ 记得用 Q1 和 Q3 绘制箱子，中位线置于箱内，触须延伸到最小值和最大值。坐标轴要规范标注。
8. Algebraic Fractions & Simplifying Expressions | 代数分式与化简表达式
Simplify (x² – 9)/(x² – x – 6). Factor numerator: (x – 3)(x + 3); denominator: (x – 3)(x + 2). Cancel (x – 3) to get (x + 3)/(x + 2), with restriction x ≠ 3, –2.
化简 (x² – 9)/(x² – x – 6)。分子因式分解：(x – 3)(x + 3)；分母因式分解：(x – 3)(x + 2)。约去 (x – 3) 得到 (x + 3)/(x + 2)，并标注 x ≠ 3，–2。
When adding or subtracting like (2/x) + (3/(x+1)), find a common denominator: 2(x+1)/(x(x+1)) + 3x/(x(x+1)) = (2x+2+3x)/(x(x+1)) = (5x+2)/(x(x+1)).
进行加减法如 (2/x) + (3/(x+1))，先求公分母：2(x+1)/(x(x+1)) + 3x/(x(x+1)) = (2x+2+3x)/(x(x+1)) = (5x+2)/(x(x+1))。
AQA mark schemes reward showing the factorised forms and clearly stating restrictions. Never skip the restriction step when cancelling variables.
AQA 评分标准鼓励展示因式分解形式并清晰写出限制条件。约掉变量时切忌省略限制条件这一步。
9. Area & Volume of Compound Shapes | 组合图形的面积与体积
Find the area of a shape formed by a rectangle and a semicircle. Split the shape, calculate each area separately, then add or subtract as needed.
求由一个矩形和一个半圆组成的图形面积。先分割图形，分别计算各部分的面积，然后根据需要加减。
Always write the formula first. For a semicircle of diameter 14 cm, radius = 7 cm, full circle area = π × 7² = 49π, semicircle area = 24.5π cm². Give the answer in terms of π unless a decimal is asked.
写上公式。一个直径为 14 cm 的半圆，半径 = 7 cm，整圆面积 = π × 7² = 49π，半圆面积 = 24.5π cm²。除非要求用小数，否则答案保留 π。
Volume of a prism = area of cross‑section × length. In problems with a composite cross‑section (e.g. L‑shape), work out the area step‑by‑step and remember to convert all units to the same system.
棱柱体积 = 横截面积 × 长度。当横截面由复合图形组成（如 L 形），逐步计算面积，并记住将所有单位统一成同一系统。
10. Vectors – Adding, Subtracting & Scalar Multiplication | 向量——加减法与数乘
If a = (2, 3) and b = (–1, 4), then 3a + 2b = 3(2, 3) + 2(–1, 4) = (6, 9) + (–2, 8) = (4, 17).
若 a = (2, 3) 且 b = (–1, 4)，则 3a + 2b = 3(2, 3) + 2(–1, 4) = (6, 9) + (–2, 8) = (4, 17)。
Questions often ask you to prove that three points lie on a straight line. Show that vector AB is a scalar multiple of vector BC. Include the scalar factor and a conclusion.
考题常要求证明三点共线。证明向量 AB 是向量 BC 的标量倍数。写出倍数关系并给出结论。
Column vectors can be combined with geometry. For a parallelogram, opposite sides are equal as vectors, so you can use them to find unknown coordinates.
列向量可与几何综合考查。在平行四边形中，对边作为向量相等，因此可以利用此性质求未知坐标。
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