Year 10 WJEC Mathematics: Unit Test Mock Paper Analysis | 十年级 WJEC 数学：单元测试模拟卷解析

📚 Year 10 WJEC Mathematics: Unit Test Mock Paper Analysis | 十年级 WJEC 数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a WJEC-style Year 10 unit test mock paper. Each question is broken down step by step to help you master key topics including algebra, geometry, percentages, probability, and more. Use this analysis to identify common mistakes and build confidence for your assessments.

本文详细解析一份WJEC风格的十年级单元测试模拟卷。每个题目都逐步分解，帮助你掌握代数、几何、百分数、概率等关键主题。利用这份解析找出常见错误，为评估树立信心。

1. Simplifying Algebraic Expressions | 化简代数式

Question (a): Simplify 3p + 2q – p + 5q.

题目（a）： 化简 3p + 2q – p + 5q。

Step 1: Collect the p terms. 3p – p = 2p.

第一步：合并 p 的同类项。3p – p = 2p。

Step 2: Collect the q terms. 2q + 5q = 7q.

第二步：合并 q 的同类项。2q + 5q = 7q。

Final simplified expression: 2p + 7q.

最终化简结果：2p + 7q。

Question (b): Expand and simplify 3(2x – 1) + 2(x + 4).

题目（b）： 展开并化简 3(2x – 1) + 2(x + 4)。

Expand the first bracket: 3 × 2x = 6x, 3 × (–1) = –3. Then expand the second: 2 × x = 2x, 2 × 4 = 8. The expanded expression is 6x – 3 + 2x + 8.

展开第一个括号：3 × 2x = 6x，3 × (–1) = –3。再展开第二个：2 × x = 2x，2 × 4 = 8。展开后的表达式为 6x – 3 + 2x + 8。

Now collect x‑terms: 6x + 2x = 8x. Collect constants: –3 + 8 = 5. The simplified result is 8x + 5.

现在合并 x 项：6x + 2x = 8x。合并常数项：–3 + 8 = 5。化简结果为 8x + 5。

Simplify expressions by combining only like terms – never add x and constant terms. | 化简代数式时只合并同类项——切勿将含 x 的项与常数项相加。

2. Solving Linear Equations | 解线性方程

Question: Solve 5x – 2 = 3x + 8.

题目： 解方程 5x – 2 = 3x + 8。

Step 1: Add 2 to both sides to isolate the x‑terms on the left. 5x – 2 + 2 = 3x + 8 + 2 → 5x = 3x + 10.

第一步：两边同时加 2，将含 x 的项移至左侧。5x – 2 + 2 = 3x + 8 + 2 → 5x = 3x + 10。

Step 2: Subtract 3x from both sides. 5x – 3x = 3x + 10 – 3x → 2x = 10.

第二步：两边同时减去 3x。5x – 3x = 3x + 10 – 3x → 2x = 10。

Step 3: Divide both sides by 2 to find x. x = 5.

第三步：两边除以 2 求得 x。x = 5。

Check: substitute x = 5 into the original equation. Left: 5(5)–2=23; Right: 3(5)+8=23. Both sides are equal, so the solution is correct.

检验：将 x = 5 代回原方程。左侧：5(5)–2=23；右侧：3(5)+8=23。两边相等，因此解正确。

3. Factorising Quadratic Expressions | 二次三项式的因式分解

Question: Factorise x² + 9x + 20.

题目： 将 x² + 9x + 20 分解因式。

We need two numbers that multiply to the constant term (+20) and add to the coefficient of x (+9).

我们需要找到两个数，它们的乘积等于常数项 (20)，并且和等于 x 的系数 (9)。

The pairs of factors of 20 are: 1 and 20, 2 and 10, 4 and 5. The pair 4 and 5 have a sum of 9.

20 的因数对有：1 和 20、2 和 10、4 和 5。其中 4 和 5 的和为 9。

Therefore the factorised form is (x + 4)(x + 5). You can expand to verify: x² + 5x + 4x + 20 = x² + 9x + 20.

因此因式分解结果为 (x + 4)(x + 5)。可通过乘法展开验证：x² + 5x + 4x + 20 = x² + 9x + 20。

Tip: When the quadratic is x² + bx + c with positive c, the two numbers are both positive if b is positive.

提示：当二次式为 x² + bx + c 且 c 为正数时，如果 b 为正，则这两个数都是正数。

4. Solving Quadratic Equations by Factorising | 用因式分解法解二次方程

Question: Solve x² – 3x – 10 = 0.

题目： 解方程 x² – 3x – 10 = 0。

First factorise the quadratic. Find two numbers whose product is –10 and sum is –3. The numbers are –5 and +2 because (–5) × 2 = –10 and –5 + 2 = –3.

首先将二次式因式分解。找到两个数，乘积为 –10，和为 –3，这两个数是 –5 和 +2，因为 (–5) × 2 = –10，且 –5 + 2 = –3。

So x² – 3x – 10 = (x – 5)(x + 2) = 0.

因此 x² – 3x – 10 = (x – 5)(x + 2) = 0。

Apply the zero-product rule: either x – 5 = 0 or x + 2 = 0. Solving gives x = 5 or x = –2.

应用零乘积法则：要么 x – 5 = 0，要么 x + 2 = 0。求解得 x = 5 或 x = –2。

Solutions: x = 5, x = –2 | 解为 x = 5，x = –2

5. Linear Inequalities on a Number Line | 一元一次不等式及数轴表示

Question: Solve 4x – 3 < 9 and represent the solution on a number line.

题目： 解不等式 4x – 3 < 9，并将解表示在数轴上。

Add 3 to both sides: 4x – 3 + 3 < 9 + 3 → 4x < 12.

两边加 3：4x – 3 + 3 < 9 + 3 → 4x < 12。

Divide both sides by 4: x < 3.

两边除以 4：x < 3。

On the number line, draw an open circle at 3 (since 3 is not included) and shade the line to the left, representing all numbers less than 3.

在数轴上，在 3 处画一个空心圆圈（因为 3 本身不包含在内），并向左侧画箭头线段，表示所有小于 3 的数。

Remember: multiply or divide by a negative number reverses the inequality sign, but here we divided by a positive 4, so the direction stays the same.

牢记：乘以或除以负数要改变不等号方向，但这里除以正的 4，因此方向保持不变。

6. Compound Interest | 复利计算

Question: £1500 is invested for 4 years at an annual compound interest rate of 3.2%. Find the total amount after 4 years.

题目： 投资 1500 英镑，年复利率为 3.2%，投资期 4 年。求 4 年后的总金额。

The compound interest formula is A = P(1 + r/100)ⁿ, where P = 1500, r = 3.2, n = 4.

复利公式为 A = P(1 + r/100)ⁿ，这里 P = 1500，r = 3.2，n = 4。

Substitute the values: A = 1500 × (1 + 3.2/100)⁴ = 1500 × (1.032)⁴.

代入数值：A = 1500 × (1 + 3.2/100)⁴ = 1500 × (1.032)⁴。

Calculate the multiplier: 1.032² = 1.065024, then square again for the fourth power: 1.065024 × 1.065024 ≈ 1.13428. Now multiply by the principal: 1500 × 1.13428 ≈ 1701.42.

计算乘数：1.032² = 1.065024，再将其平方得到四次方：1.065024 × 1.065024 ≈ 1.13428。再乘以本金：1500 × 1.13428 ≈ 1701.42。

Thus the total amount is approximately £1701.42.

因此总金额约为 1701.42 英镑。

Common mistake: using simple interest instead of compound interest would incorrectly give £1500 + (1500 × 0.032 × 4) = £1692. The difference shows the effect of compounding.

常见错误：如果误用单利计算，会得到 £1500 + (1500 × 0.032 × 4) = £1692。与复利的差额体现了复利效应。

7. Area of a Compound Shape | 组合图形的面积

Question: A shape is made from a rectangle of length 10 cm and width 6 cm, with a semicircle of diameter 6 cm attached to one of the shorter sides. Find the total area (use π ≈ 3.14).

题目： 一个图形由一个长 10 cm、宽 6 cm 的矩形与一个直径 6 cm 的半圆（附着在一条短边上）组成。求总面积（π ≈ 3.14）。

Area of the rectangle: length × width = 10 × 6 = 60 cm².

矩形面积：长 × 宽 = 10 × 6 = 60 cm²。

Radius of the semicircle is half of the diameter: r = 3 cm.

半圆半径是直径的一半：r = 3 cm。

Area of a full circle = πr² = 3.14 × 9 = 28.26 cm². A semicircle is half of that: 14.13 cm².

整圆的面积 = πr² = 3.14 × 9 = 28.26 cm²，半圆面积为它的一半：14.13 cm²。

Total area = rectangle area + semicircle area = 60 + 14.13 = 74.13 cm².

总面积 = 矩形面积 + 半圆面积 = 60 + 14.13 = 74.13 cm²。

Compound area: A = 60 + ½ π (3)² ≈ 74.13 cm² | 组合图形面积：A ≈ 74.13 cm²

8. Pythagoras’ Theorem | 勾股定理

Question: A ladder of length 6.5 m leans against a vertical wall. The foot of the ladder is 2.5 m from the base of the wall. How high up the wall does the ladder reach?

题目： 一把长 6.5 m 的梯子斜靠在垂直的竖直墙上，梯脚距墙脚 2.5 m。问梯子顶端离地面的高度是多少？

By Pythagoras’ theorem: height² + 2.5² = 6.5².

根据勾股定理：高度² + 2.5² = 6.5²。

Calculate the squares: 2.5² = 6.25 and 6.5² = 42.25. So height² = 42.25 – 6.25 = 36.

计算平方：2.5² = 6.25，6.5² = 42.25。因此高度² = 42.25 – 6.25 = 36。

Taking the square root gives height = 6 m (the positive root fits the context).

开平方得到高度 = 6 m（取正值符合题意）。

The ladder reaches 6 m up the wall. Always check that the hypotenuse is the longest side, which 6.5 m is.

梯子顶端离地面 6 m。记得检查斜边是否最长，这里 6.5 m 符合。

9. Scatter Graphs and Correlation | 散点图与相关性

Question: The table shows the number of hours studied (h) and test scores (s %): (2,45), (4,55), (5,60), (7,70), (8,80). (a) Plot the scatter graph. (b) Describe the correlation. (c) Use a line of best fit to estimate the expected score for a student who studied 6 hours.

题目： 下表为学习小时数 (h) 与测验分数 (s %)： (2,45), (4,55), (5,60), (7,70), (8,80)。(a) 绘制散点图。(b) 描述相关性。(c) 借助最佳拟合线估测学习 6 小时的预期分数。

(a) From your plot, each point is positioned correctly. The x‑axis shows hours, the y‑axis shows scores.

(a) 绘制散点图时，逐点标记，横轴为小时数，纵轴为分数。

(b) The points trend upwards in a roughly straight line, indicating a strong positive correlation: as study hours increase, test scores increase.

(b) 各点整体呈向上趋势且大致在一条直线附近，表明强正相关：学习时间越长，分数越高。

(c) Draw a line of best fit passing through the general direction of points, e.g. through (2,45) and (8,80). The line has equation approximately s = 5h + 40. For h = 6, s ≈ 5(6) + 40 = 70. So the estimated score is about 70%.

Interpolation at 6 hours is reliable because it lies within the data range. Extrapolation beyond 8 hours would be less reliable.

学习 6 小时为内插，因在数据范围内，结果较可靠。外推至 8 小时以上则可靠性较低。

10. Probability from a Two-Way Table | 双向表求概率

Question: 50 students were surveyed about liking football (F) and basketball (B). 30 like football, 25 like basketball, and 15 like both. (a) Complete a two-way table. (b) Find the probability that a randomly chosen student likes neither sport

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Year 10 WJEC Mathematics: Unit Test Mock Paper Analysis | 十年级 WJEC 数学：单元测试模拟卷解析