Year 11 WJEC Maths: Unit Test Mock Paper Walkthrough | WJEC 11年级数学单元测试模拟卷解析

📚 Year 11 WJEC Maths: Unit Test Mock Paper Walkthrough | WJEC 11年级数学单元测试模拟卷解析

This article provides a detailed walkthrough of a mock paper for the WJEC Year 11 Mathematics Unit Test. We break down key question types, step-by-step solutions, and essential exam techniques to help you master algebra, geometry, statistics, and more. Each section models how to present your working clearly and avoid common pitfalls.

本文详细讲解了一套 WJEC 11 年级数学单元测试模拟卷。我们分解了典型题型、逐步解法以及关键的考试技巧，帮助你掌握代数、几何、统计等内容。每一节都示范了如何清晰地展示解题过程并避开常见错误。

1. Simplifying Algebraic Expressions | 代数式的化简

Question 1: Expand and simplify (3x - 2)(x + 5).

问题 1：展开并化简 (3x - 2)(x + 5)。

We use the FOIL method: multiply the First terms, Outer terms, Inner terms, and Last terms.

我们使用 FOIL 方法：首项相乘，外项相乘，内项相乘，末项相乘。

First: 3x × x = 3x²

首项：3x × x = 3x²

Outer: 3x × 5 = 15x

外项：3x × 5 = 15x

Inner: -2 × x = -2x

内项：-2 × x = -2x

Last: -2 × 5 = -10

末项：-2 × 5 = -10

Add the results: 3x² + 15x - 2x - 10 = 3x² + 13x - 10.

把各项相加：3x² + 15x - 2x - 10 = 3x² + 13x - 10。

Key point: Always combine like terms to simplify fully. A common mistake is forgetting to multiply the negative signs correctly.

关键：始终合并同类项以完全化简。常见错误是忘记正确处理负号的乘法。

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2. Factorising and Solving Quadratics | 因式分解与解二次方程

Question 2: (a) Factorise x² - x - 6. (b) Solve x² - 5x + 6 = 0.

问题 2：(a) 因式分解 x² - x - 6。(b) 解方程 x² - 5x + 6 = 0。

For (a): We need two numbers that multiply to -6 and add to -1. The numbers are -3 and 2.

对于 (a)：我们需要两个数，乘积为 -6，且和为 -1。这两个数是 -3 和 2。

So, x² - x - 6 = (x - 3)(x + 2).

因此，x² - x - 6 = (x - 3)(x + 2)。

For (b): Factorising x² - 5x + 6, we look for numbers that multiply to 6 and add to -5: -2 and -3.

对于 (b)：对 x² - 5x + 6 进行因式分解，我们寻找乘积为 6 且和为 -5 的数：-2 和 -3。

Thus, (x - 2)(x - 3) = 0. Setting each factor to zero gives x = 2 or x = 3.

因此，(x - 2)(x - 3) = 0。令每个因式为零，得到 x = 2 或 x = 3。

Remember: Always set the quadratic equal to zero before factorising. Writing the solutions as coordinates is a common error – just state the x-values.

记住：在因式分解之前，一定要先将二次方程设为零。把解写成坐标是一个常见错误——只需写出 x 的值。

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3. Solving Linear Equations and Inequalities | 解线性方程与不等式

x + 10. Also, solve the inequality 5 - 2x > 1 and represent the solution on a number line.

问题 3：解方程 2(3x - 1) = 4x + 10。同时，解不等式 5 - 2x > 1 并把解表示在数轴上。

Equation: Expand the left side: 6x - 2 = 4x + 10.

方程：展开左边：6x - 2 = 4x + 10。

Subtract 4x from both sides: 2x - 2 = 10.

两边同时减去 4x：2x - 2 = 10。

Add 2 to both sides: 2x = 12, so x = 6.

两边加 2：2x = 12，因此 x = 6。

Inequality: Subtract 5 from both sides: -2x > -4.

不等式：两边同时减去 5：-2x > -4。

Divide by -2 and reverse the inequality sign: x < 2.

除以 -2 并反转不等号：x < 2。

On a number line, draw an open circle at 2 and shade to the left.

在数轴上，在 2 处画一个空心圆，并向左涂阴影。

Common pitfall: Forgetting to flip the inequality sign when multiplying or dividing by a negative number costs marks.

常见陷阱：当乘以或除以一个负数时忘记反转不等号会导致失分。

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4. Applying Pythagoras' Theorem and Trigonometry | 勾股定理与三角学的应用

Question 4: A right-angled triangle has legs of 5 cm and 12 cm. Find the length of the hypotenuse and the size of the smallest angle.

问题 4：一个直角三角形的两条直角边分别为 5 cm 和 12 cm。求斜边的长度以及最小角的大小。

Using Pythagoras: hypotenuse² = 5² + 12² = 25 + 144 = 169, so hypotenuse = √169 = 13 cm.

利用勾股定理：斜边² = 5² + 12² = 25 + 144 = 169，因此斜边 = √169 = 13 cm。

est angle is opposite the shortest side (5 cm). We use sin θ = opposite/hypotenuse = 5/13.

最小的角对着最短边（5 cm）。我们使用 sin θ = 对边/斜边 = 5/13。

So, θ = sin⁻¹(5/13). Using a calculator, θ ≈ 22.6° (to 1 d.p.).

因此，θ = sin⁻¹(5/13)。使用计算器得出 θ ≈ 22.6°（精确到1位小数）。

Always check that your calculator is in degree mode and label the angle correctly with the degree symbol.

始终检查计算器是否处于角度模式，并用度数符号正确标记角度。

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5. Area of Sectors and Arcs | 扇形面积与弧长

Question 5: A circle has radius 8 cm. A sector is cut with a central angle of 60°. Calculate the area of the sector and the length of its arc.

问题 5：一个圆的半径为 8 cm。切出一个圆心角为 60°的扇形。计算扇形的面积及其弧长。

360) × π × r² = (60/360) × π × 8² = (1/6) × 64π = (32/3)π cm² (or approximately 33.5 cm²).

扇形面积 = (θ / 360) × π × r² = (60/360) × π × 8² = (1/6) × 64π = (32/3)π cm²（或约 33.5 cm²）。

c length = (θ / 360) × 2πr = (1/6) × 2π × 8 = (16/6)π = (8/3)π cm (approximately 8.38 cm).

弧长 = (θ / 360) × 2πr = (1/6) × 2π × 8 = (16/6)π = (8/3)π cm（约 8.38 cm）。

It is acceptable to leave answers in terms of π unless the question specifies a decimal. Always write the units.

除非题目要求化为小数，否则可以将答案用 π 表示。一定要写明单位。

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6. Ratio and Proportion Problems | 比例问题

Question 6: The ratio of flour to sugar in a cake recipe is 5 : 3. If 400 g of flour is used, how much sugar is needed? Mix 60 kg of concrete using sand, cement, and gravel in the ratio 2 : 1 : 3. How much cement is required?

问题 6：蛋糕配方中面粉与糖的比例为 5 : 3。如果用了 400 g 面粉，需要多少糖？用沙子、水泥和石子按 2 : 1 : 3 的比例配制 60 kg 混凝土，需要多少水泥？

arts correspond to 400 g, so 1 part = 400 ÷ 5 = 80 g. Sugar is 3 parts, so 3 × 80 = 240 g.

第一部分：5 份对应 400 g，所以 1 份 = 400 ÷ 5 = 80 g。糖占 3 份，因此 3 × 80 = 240 g。

: Total parts = 2 + 1 + 3 = 6. Cement is 1 part in 6, so its mass = (1/6) × 60 kg = 10 kg.

第二部分：总份数 = 2 + 1 + 3 = 6。水泥占 6 份中的 1 份，所以它的质量 = (1/6) × 60 kg = 10 kg。

Both questions rely on finding the value of one share first. Always check that your shares add up to the given total.

这两道题都依赖于先求出一份的数量。务必检查各份之和与给定的总数相符。

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7. Averages from Frequency Tables | 从频数表求平均数

Question 7: The table below shows the scores of 30 students. Estimate the mean.

问题 7：下表显示了 30 名学生的分数。请估算平均数。

Score interval	Frequency (f)
0–10	4
10–20	6
20–30	9
30–40	8
40–50	3

We add a midpoint column (x) and an f × x column.

我们添加中点列 (x) 和 f × x 列。

s: 5, 15, 25, 35, 45. Then f × x: 4×5 = 20; 6×15 = 90; 9×25 = 225; 8×35 = 280; 3×45 = 135.

中点值：5, 15, 25, 35, 45。然后 f × x：4×5 = 20；6×15 = 90；9×25 = 225；8×35 = 280；3×45 = 135。

Sum of f × x = 20 + 90 + 225 + 280 + 135 = 750. Total frequency = 4+6+9+8+3 = 30.

f × x 的总和 = 20 + 90 + 225 + 280 + 135 = 750。总频数 = 4+6+9+8+3 = 30。

Published by TutorHao | Year 11 Mathematics Revision Series | aleveler.com

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