📚 Year 9 WJEC Maths Unit Test Mock Paper Walkthrough | 九年级 WJEC 数学单元测试模拟卷解析
This mock test walkthrough covers essential Year 9 WJEC Mathematics topics, helping you review key skills and identify areas for improvement before the real assessment.
这份模拟卷解析涵盖了九年级 WJEC 数学的核心主题，帮助你在实际评估前复习关键技能并发现薄弱环节。
1. Fractions and Decimals Calculations | 分数与小数运算
e (2/3 + 1/4) × 1.2. Give your answer as a fraction in its simplest form and as a decimal.
题目：计算 (2/3 + 1/4) × 1.2。将答案化为最简分数和小数两种形式。
Start by adding the two fractions inside the brackets. The common denominator for 3 and 4 is 12.
先计算括号内两个分数的和。3 和 4 的公分母是 12。
Convert: 2/3 = 8/12 and 1/4 = 3/12, so the sum is 8/12 + 3/12 = 11/12.
转化：2/3 = 8/12，1/4 = 3/12，因此和为 8/12 + 3/12 = 11/12。
Now multiply by 1.2. It is safer to write 1.2 as a fraction: 1.2 = 12/10 = 6/5.
接着乘以 1.2。更稳妥的做法是把 1.2 化成分数：1.2 = 12/10 = 6/5。
1 × 6)/(12 × 5) = 66/60. Simplify by dividing numerator and denominator by 6 to get 11/10.
相乘：11/12 × 6/5 = (11 × 6)/(12 × 5) = 66/60。分子分母同时除以 6 约分得 11/10。
As a decimal, 11/10 = 1.1. So the answer is 11/10 or 1.1.
化为小数为 11/10 = 1.1。因此答案为 11/10 或 1.1。
A common mistake is to multiply the decimal directly without converting, which can lead to rounding errors. Always convert to a fraction when possible.
常见错误是直接乘以小数而不化成分数，这可能导致舍入误差。只要可能，尽量先化成分数。
2. Solving Linear Equations | 解一元一次方程
Question: Solve the equation 5x – 3 = 2x + 9.
题目：解方程 5x – 3 = 2x + 9。
The goal is to isolate x. First, collect the x‑terms on one side. Subtract 2x from both sides: 5x – 2x – 3 = 9, which simplifies to 3x – 3 = 9.
目标是分离出 x。首先将含 x 的项移到同一边。两边同时减去 2x：5x – 2x – 3 = 9，化简得 3x – 3 = 9。
Now add 3 to both sides to isolate the 3x term: 3x = 12.
现在两边同时加 3，得到 3x = 12。
Finally, divide both sides by 3: x = 4.
最后两边同时除以 3：x = 4。
Always check your solution by substituting back: 5(4) – 3 = 20 – 3 = 17, and 2(4) + 9 = 8 + 9 = 17. Both sides match.
一定要将解代回原方程检验：5(4) – 3 = 20 – 3 = 17，2(4) + 9 = 8 + 9 = 17，两边相等。
Watch out for sign errors when moving terms across the equals sign. Adding or subtracting the same quantity on both sides keeps the equation balanced.
移项时要特别注意符号错误。在等式两边同时加上或减去同一个量能保持等式平衡。
3. Ratio and Proportion | 比与比例
Question: The ratio of boys to girls in a class is 3 : 5. If there are 24 students altogether, how many boys are there?
题目：某班男生与女生的比是 3 : 5。如果共有 24 名学生，男生有多少人？
A ratio of 3 : 5 means the parts can be thought of as 3 + 5 = 8 equal parts in total.
比 3 : 5 表示可将全班视为 3 + 5 = 8 个等份。
One part represents 24 ÷ 8 = 3 students.
一个等份代表 24 ÷ 8 = 3 名学生。
Boys correspond to 3 parts, so the number of boys is 3 × 3 = 9.
男生对应 3 个等份，因此男生人数是 3 × 3 = 9。
quation: let boys = 3k and girls = 5k. Then 3k + 5k = 24, so 8k = 24, k = 3, and boys = 9.
也可以列方程：设男生为 3k，女生为 5k，则 3k + 5k = 24，8k = 24，k = 3，男生为 9。
heck the total: girls would be 5 × 3 = 15, and 9 + 15 = 24, which matches the given total.
核对总数：女生为 5 × 3 = 15，9 + 15 = 24，与已知总数一致。
4. nth Term of a Linear Sequence | 线性数列的第 n 项
Question: Find the nth term of the sequence: 7, 11, 15, 19, …
题目：求数列 7, 11, 15, 19, … 的第 n 项通项公式。
First, find the common difference between consecutive terms. 11 – 7 = 4, 15 – 11 = 4, so the difference is 4.
先找出相邻两项的差。11 – 7 = 4，15 – 11 = 4，因此公差是 4。
he form an + b, where a is the common difference. Here a = 4, so the expression is 4n + b.
线性数列的第 n 项具有 an + b 的形式，其中 a 为公差。这里 a = 4，所以公式为 4n + b。
d b, use the first term (n = 1). When n = 1, the term is 7, so 4(1) + b = 7, giving b = 3.
为求 b，代入第一项（n = 1）。当 n = 1 时该项为 7，即 4(1) + b = 7，得 b = 3。
h term is 4n + 3. Check with n = 2: 4(2) + 3 = 11, correct; n = 3: 4(3) + 3 = 15, correct.
所以第 n 项是 4n + 3。检验 n = 2：4(2) + 3 = 11，正确；n = 3：4(3) + 3 = 15，正确。
A common error is forgetting to test the formula with several terms. Always verify with at least two values of n.
常见错误是忘记用多个项检验公式。一定要至少用两个 n 值来验证。
5. Angles in Parallel Lines | 平行线中的角
EF at points G and H. Angle EGB = 70°. Find the size of angle GHD, justifying your answer.
题目：两条平行线 AB 和 CD 被横截线 EF 分别交于点 G 和 H。∠EGB = 70°。求 ∠GHD 的大小并说明理由。
Angle EGB and angle GHD are in corresponding positions with respect to the parallel lines and the transversal.
∠EGB 与 ∠GHD 相对于平行线和截线处在同位角的位置。
Corresponding angles formed by parallel lines are equal. Therefore, angle GHD = 70°.
平行线形成的同位角相等。因此 ∠GHD = 70°。
site, so angle AGH = 70°, and then alternate angles between parallel lines are also equal.
如果题目要求内错角 AGH，可利用 ∠EGB 与 ∠AGH 是对顶角，得 ∠AGH = 70°，而平行线间的内错角也相等。
Always name the angle relationships clearly: corresponding, alternate, or interior. A simple diagram in your head or on scrap paper helps avoid confusion.
一定要明确标注角的关系：同位角、内错角或同旁内角。在脑中或草稿纸上简单画图有助于避免混淆。
6. Area and Perimeter of a Composite Shape | 组合图形的面积与周长
Question: A composite shape is made from a rectangle of width 6 cm and height 4 cm, with an isosceles triangle placed on top. The triangle has the same base as the width of the rectangle and a vertical height of 3 cm. Find the total area of the shape.
题目：一个组合图形由一个宽 6 cm、高 4 cm 的矩形和一个放置在其上的等腰三角形构成。三角形的底边与矩形的宽相等，垂直高度为 3 cm。求该图形的总面积。
Area of rectangle = length × width = 6 × 4 = 24 cm².
矩形面积 = 长 × 宽 = 6 × 4 = 24 cm²。
Area of triangle = ½ × base × height
三角形面积 = ½ × 底 × 高
Base of triangle = 6 cm, height = 3 cm, so area = ½ × 6 × 3 = 9 cm².
三角形底 = 6 cm，高 = 3 cm，所以面积 = ½ × 6 × 3 = 9 cm²。
Total area = 24 + 9 = 33 cm². Remember to write the correct unit (cm²) in your answer.
总面积 = 24 + 9 = 33 cm²。答案中记得写上正确的单位 cm²。
If asked for the perimeter, you would need to calculate the slant sides of the triangle using Pythagoras, but the focus here is area. Always read the question carefully.
若要求周长，需要用勾股定理计算三角形的斜边，但这里的重点是面积。务必仔细审题。
7. Probability from a Two‑Way Table | 双向表中的概率
Question: The two‑way table below shows the number of boys and girls who prefer football or tennis. Find the probability that a randomly selected student is a girl who prefers tennis.
题目：下面的双向表显示了喜欢足球或网球的男女生人数。求随机选取一名学生，她是一位喜欢网球的女生的概率。
| Football | Tennis | Total | |
|---|---|---|---|
| Boys | 15 | 10 | 25 |
| Girls | 8 | 12 | 20 |
| Total | 23 | 22 | 45 |
From the table, the number of girls who prefer tennis is 12. The total number of students is 45.
从表中可知，喜欢网球的女生有 12 人。学生总人数为 45。
Probability = (number of favourable outcomes) / (total number of outcomes) = 12/45.
概率 = （有利结果数）/（总结果数）= 12/45。
Simplify the fraction by dividing numerator and denominator by 3: 12/45 = 4/15.
分子分母同除以 3 可化简约分：12/45 = 4/15。
You can leave the probability as a fraction in simplest form, or give it as a decimal (0.266…). The question usually expects a simplified fraction.
概率可以保留为最简分数，或写成小数（0.266…）。题目通常要求最简分数形式。
8. Mean from a Frequency Table | 根据频率表求平均数
Question: The table shows the scores of 20 students in a quiz. Calculate the mean score.
题目：下表给出了 20 名学生在一次小测验中的得分。请计算平均得分。
| Score (x) | Frequency (f) |
|---|---|
| 1 | 2 |
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