📚 WJEC Year 10 Maths: In-depth Analysis of Past Papers | WJEC 十年级数学：历年真题深度解析
Working through real WJEC past papers is one of the most effective ways for Year 10 students to build confidence, identify common traps, and become familiar with the style of questions that appear in GCSE Maths. This article provides a detailed breakdown of eight typical exam topics, each illustrated with a past paper style question and a step‑by‑step guide to the solution. Whether you are targeting a grade 4 or aiming for a grade 9, understanding how to approach these core question types will sharpen your problem‑solving skills and deepen your mathematical understanding.
钻研真实的 WJEC 历年真题是十年级学生建立信心、发现常见陷阱并熟悉 GCSE 数学考题风格的最有效途径之一。本文详细拆解了八个典型考试主题，每个主题都配有一道真题风格的例题和分步解题指南。无论你的目标是 4 分还是追求 9 分，理解如何攻克这些核心题型都能提升你的解题技巧、加深你对数学的理解。
1. Algebra: Solving Linear Equations | 代数：解线性方程
A classic WJEC Foundation tier question asks you to solve an equation that involves brackets and variables on both sides. For example: Solve 3(2x − 1) = 4x + 5.
WJEC 基础层级的一道经典题目要求你解一个含有括号且两边都有变量的方程。例如：解方程 3(2x − 1) = 4x + 5。
Step 1: Expand the left‑hand side by distributing the 3: 6x − 3 = 4x + 5.
步骤 1：将左边括号展开：6x − 3 = 4x + 5。
Step 2: Collect the x terms on one side by subtracting 4x from both sides: 2x − 3 = 5.
步骤 2：把含 x 的项移到同一边，两边同时减去 4x：2x − 3 = 5。
Step 3: Isolate the term with x by adding 3 to both sides: 2x = 8.
步骤 3：两边同时加 3，使含 x 的项单独出现：2x = 8。
Step 4: Divide both sides by 2 to obtain the solution: x = 4.
步骤 4：两边同时除以 2，得到解：x = 4。
Many students lose marks by forgetting to distribute the negative sign or by subtracting 3 instead of adding it when moving terms. Always write out every step clearly, and check your solution by substituting x = 4 back into the original equation.
许多学生因忘记分配负号或者在移项时本该加 3 却减了 3 而丢分。务必将每一步都写清楚，并将 x = 4 代回原方程进行检验。
2. Quadratic Graphs: Sketching y = x² − 4x + 3 | 二次函数图像：画出 y = x² − 4x + 3 草图
A typical past paper asks you to sketch the graph of a quadratic function by finding its key features. Consider the question: Sketch the graph of y = x² − 4x + 3, clearly indicating the x‑intercepts, y‑intercept and the coordinates of the vertex.
一道典型的真题要求你通过确定关键特征来画出二次函数的草图。考虑这道题：画出 y = x² − 4x + 3 的图像，并标出 x 轴截距、y 轴截距和顶点坐标。
First, find the y‑intercept by setting x = 0: y = 0² − 4(0) + 3 = 3, so the curve crosses the y‑axis at (0, 3).
首先，令 x = 0 求出 y 轴截距：y = 0² − 4(0) + 3 = 3，因此曲线与 y 轴交于点 (0, 3)。
Next, factorise to find the x‑intercepts: x² − 4x + 3 = (x − 1)(x − 3). Setting this equal to zero gives x − 1 = 0 or x − 3 = 0, so the roots are x = 1 and x = 3. The curve meets the x‑axis at (1, 0) and (3, 0).
接下来，通过因式分解求 x 轴截距：x² − 4x + 3 = (x − 1)(x − 3)。令其等于零得到 x − 1 = 0 或 x − 3 = 0，因此根为 x = 1 和 x = 3。曲线与 x 轴交于 (1, 0) 和 (3, 0)。
To find the vertex, use the symmetry of the parabola. The x‑coordinate of the vertex is the midpoint of the roots: x = (1 + 3)/2 = 2. Substitute x = 2 into the equation: y = 2² − 4(2) + 3 = 4 − 8 + 3 = −1. The vertex is at (2, −1).
利用抛物线的对称性求顶点。顶点的 x 坐标是两根的中点：x = (1 + 3)/2 = 2。将 x = 2 代入方程：y = 2² − 4(2) + 3 = 4 − 8 + 3 = −1。顶点坐标为 (2, −1)。
Because the coefficient of x² is positive, the parabola opens upwards. Plot these four points and draw a smooth U‑shaped curve through them. The completed sketch earns full marks.
因为 x² 的系数为正，抛物线开口向上。标出这四个点，并画出一条光滑的 U 形曲线。完成草图即可得满分。
3. Geometry: Angles in Polygons | 几何：多边形的内角
WJEC often includes a question on the interior and exterior angles of regular polygons. For instance: A regular polygon has 12 sides. Calculate the size of one interior angle.
WJEC 试卷中经常考查正多边形的内角和外角问题。例如：一个正多边形有 12 条边，求它的一个内角度数。
The formula for the sum of interior angles of an n‑sided polygon is (n − 2) × 180°. For a regular polygon, each interior angle is equal to this sum divided by n.
n 边形内角和公式为 (n − 2) × 180°。对于正多边形，每个内角等于内角和除以 n。
First calculate the sum: (12 − 2) × 180° = 10 × 180° = 1800°.
先求内角和：(12 − 2) × 180° = 10 × 180° = 1800°。
Then divide by the number of sides: 1800° ÷ 12 = 150°. Therefore one interior angle is 150°.
然后除以边数：1800° ÷ 12 = 150°。因此一个内角为 150°。
You can also use the exterior angle method: the sum of exterior angles is always 360°, so each exterior angle of a regular 12‑gon is 360° ÷ 12 = 30°. The interior angle is 180° − 30° = 150°. Both methods are equally valid and are worth practising.
你也可以用外角法：外角和恒为 360°，所以正十二边形的每个外角为 360° ÷ 12 = 30°。内角 = 180° − 30° = 150°。两种方法都有效，值得练习。
4. Probability: Tree Diagrams | 概率：树状图
Tree diagrams are a reliable way to represent two‑stage probability experiments. A common exam task: A bag contains 5 red counters and 3 blue counters. Two counters are taken at random without replacement. Use a tree diagram to find the probability that the two counters are different colours.
树状图是表示两步概率实验的可靠工具。常见考题：一个袋子装有 5 个红色计数块和 3 个蓝色计数块，随机取出两个且不放回。用树状图求出两个计数块颜色不同的概率。
Draw the first branch: probability of red first = 5/8, probability of blue first = 3/8.
画出第一层分支：先抽到红色的概率 = 5/8，先抽到蓝色的概率 = 3/8。
For the second counter, adjust the denominators because the first counter is not replaced. If the first is red, then 4 red and 3 blue remain, so P(red after red) = 4/7, P(blue after red) = 3/7. If the first is blue, then 5 red and 2 blue remain, so P(red after blue) = 5/7, P(blue after blue) = 2/7.
第二个计数块的分母需调整，因为第一个没有放回。如果第一个是红色，剩余 4 红 3 蓝，所以 P(红后红) = 4/7，P(红后蓝) = 3/7。如果第一个是蓝色，剩余 5 红 2 蓝，所以 P(蓝后红) = 5/7，P(蓝后蓝) = 2/7。
The outcomes with different colours are (red, blue) and (blue, red). Their probabilities are (5/8) × (3/7) = 15/56 and (3/8) × (5/7) = 15/56. Add these together: 15/56 + 15/56 = 30/56 = 15/28.
颜色不同的结果是（红，蓝）和（蓝，红）。它们的概率分别为 (5/8) × (3/7) = 15/56 和 (3/8) × (5/7) = 15/56。两者相加：15/56 + 15/56 = 30/56 = 15/28。
The probability that the two counters are different colours is 15/28. Always remember to multiply along the branches and add the probabilities of the relevant final outcomes.
两个计数块颜色不同的概率为 15/28。始终记住沿着分支相乘，再将相关结果的概率相加。
5. Ratio and Proportion: Direct Proportion | 比与比例：正比例
Direct proportion problems appear regularly and are often phrased as ‘y is directly proportional to x’. A typical question: y is proportional to x. When x = 3, y = 12. Find an equation linking y and x, and hence find y when x = 7.
正比例问题经常出现，题目通常表述为「y 与 x 成正比」。典型题目：已知 y 与 x 成正比，当 x = 3 时，y = 12。求 y 与 x 的关系式，并由此求出当 x = 7 时 y 的值。
Write the proportional statement as y = kx, where k is the constant of proportionality.
将比例关系写为 y = kx，其中 k 为比例常数。
Substitute the given values to find k: 12 = k × 3, so k = 12 ÷ 3 = 4. Thus the equation is y = 4x.
代入已知值求 k：12 = k × 3，故 k = 12 ÷ 3 = 4。因此方程为 y = 4x。
Now use the equation to find y when x = 7: y = 4 × 7 = 28. So y = 28.
再运用该方程求 x = 7 时的 y 值：y = 4 × 7 = 28。所以 y = 28。
Direct proportion graphs are straight lines that pass through the origin. In an exam, be careful to distinguish between direct proportion and inverse proportion, which involves equations of the form y = k/x.
正比例图像是一条过原点的直线。考试中要注意区分正比例和反比例，反比例使用形如 y = k/x 的方程。
6. Trigonometry: Finding a Side Using Sine | 三角学：用正弦求边长
WJEC Higher tier frequently tests basic right‑angled trigonometry. A question may present a right‑angled triangle and ask you to calculate a missing side using the sine ratio. For example: In a right‑angled triangle, the hypotenuse is 10 cm long, and one angle is 30°. Calculate the length of the side opposite the 30° angle.
WJEC 高阶层级频繁考查基本的直角三角形三角学。一道题可能给出一个直角三角形，要求你用正弦比计算未知边长。例如：在一个直角三角形中，斜边为 10 cm，一个锐角为 30°，求 30° 角对边的长度。
Label the sides relative to the 30° angle: the opposite is the side you need, and the hypotenuse is 10 cm. The sine ratio is sin(θ) = opposite / hypotenuse.
针对 30° 角标记各边：对边是所求边，斜边为 10 cm。正弦比为 sin(θ) = 对边 / 斜边。
You should know the exact value sin 30° = ½, which is given in the WJEC formula sheet if you need it.
你应该知道 sin 30° 的精确值为 ½，如果需要，WJEC 公式表上也会给出。
Set up the equation: ½ = opposite / 10. Multiply both sides by 10 to solve: opposite = 10 × ½ = 5 cm.
列出方程：½ = 对边 / 10。两边同时乘 10：对边 = 10 × ½ = 5 cm。
Many students mix up sine and cosine. Remember the mnemonic SOH CAH TOA: Sine = Opposite over Hypotenuse, Cosine = Adjacent over Hypotenuse, Tangent = Opposite over Adjacent. A quick sketch of the triangle and correct labelling are essential before you begin.
许多学生会混淆正弦和余弦。记住口诀 SOH CAH TOA：正弦 = 对边 / 斜边，余弦 = 邻边 / 斜边，正切 = 对边 / 邻边。解题前快速画出三角形草图并正确标记各边至关重要。
7. Statistics: Mean from a Frequency Table | 统计：从频数表求平均数
Estimating the mean from a grouped frequency table is a core statistics skill. A WJEC question might look like this: The table shows the lengths of 20 leaves, measured to the nearest cm. Calculate an estimate for the mean length.
根据分组频数表估算平均数是核心统计技能。一道 WJEC 题目可能如下：表格给出了 20 片叶子的长度，精确到 cm。计算叶长的估计平均值。
| Length, L (cm) | Frequency |
|---|---|
| 4 ≤ L < 6 | 3 |
| 6 ≤ L < 8 | 8 |
| 8 ≤ L < 10 | 7 |
| 10 ≤ L < 12 | 2 |
To estimate the mean, first find the midpoint of each class interval. The midpoints are (4+6)/2 = 5, (6+8)/2 = 7, (8+10)/2 = 9, and (10+12)/2 = 11.
要估算平均数，先找出每个区间的中点。中点分别为 (4+6)/2 = 5、(6+8)/2 = 7、(8+10)/2 = 9、(10+12)/2 = 11。
Multiply each midpoint by its frequency to get the total for that class: 5 × 3 = 15, 7 × 8 = 56, 9 × 7 = 63, 11 × 2 = 22. Add these products: 15 + 56 + 63 + 22 = 156.
将每个中点乘以其频数得到该组的总额：5 × 3 = 15，7 × 8 = 56，9 × 7 = 63，11 × 2 = 22。将这些乘积相加：15 + 56 + 63 + 22 = 156。
The total frequency is 20. So the estimated mean is 156 ÷ 20 = 7.8 cm. You must state it is an estimate because you used midpoints rather than the actual data.
总频数为 20。因此估计平均值为 156 ÷ 20 = 7.8 cm。必须指出这是估计值，因为你使用的是中点而非实际数据。
8. Number: Standard Form Calculations | 数字：标准形式计算
Standard form frequently appears in both calculator and non‑calculator papers. A typical problem: Calculate (3.2 × 10⁵) × (4 × 10⁻³). Give your answer in standard form.
标准形式经常在计算器和非计算器试卷中出现。典型题目：计算 (3.2 × 10⁵) × (4 × 10⁻³)，并将答案写成标准形式。
When multiplying numbers in standard form, multiply the front numbers and add the powers of 10: (3.2 × 4) × 10⁵⁺⁽⁻³⁾ = 12.8 × 10².
做标准形式乘法时，将前面的数字相乘，并将 10 的指数相加：(3.2 × 4) × 10⁵⁺⁽⁻³⁾ = 12.8 × 10²。
However, standard form requires the front number to be between 1 and 10 (not including 10). 12.8 is not in the correct range, so adjust it: 12.8 = 1.28 × 10¹. Now combine the powers: 1.28 × 10¹ × 10² = 1.28 × 10³.
然而，标准形式要求前面的数字在 1 到 10 之间（不包括 10）。12.8 不在正确范围内，因此进行调整：12.8 = 1.28 × 10¹。然后将指数合并：1.28 × 10¹ × 10² = 1.28 × 10³。
Thus the answer is 1.28 × 10³. When dividing, you would subtract the powers. Practise with negative powers in particular, as sign errors are a common pitfall.
因此答案为 1.28 × 10³。做除法时，指数应相减。尤其要多练习负指数题，符号错误是常见的失分点。
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