GCSE WJEC Mathematics: Typical Example Questions Explained | GCSE WJEC 数学：典型例题详解

📚 GCSE WJEC Mathematics: Typical Example Questions Explained | GCSE WJEC 数学：典型例题详解

In this article, we work through a selection of typical GCSE WJEC Mathematics questions step by step, covering key topics from the specification. Each example is explained clearly in both English and Chinese to reinforce understanding.

在本文中，我们将逐步解析一系列典型的 GCSE WJEC 数学题目，涵盖考试大纲中的关键主题。每个例题都以中英双语清晰解释，以加深理解。

1. Solving Linear Equations | 解线性方程

A common type of question asks you to solve an equation like 3x + 5 = 20. Let's break it down.

一种常见题型要求解形如 3x + 5 = 20 的方程。我们来逐步分析。

First, subtract 5 from both sides to isolate the term with x:

首先，两边同时减去 5，以分离含 x 的项：

3x + 5 − 5 = 20 − 5 → 3x = 15

Then divide both sides by 3 to find x:

然后两边除以 3，求出 x：

x = 15 ÷ 3 → x = 5

Always check your answer by substituting back: 3(5) + 5 = 15 + 5 = 20 ✔.

务必通过回代检验答案：3(5) + 5 = 15 + 5 = 20 ✔。

equations with unknowns on both sides, such as 5x − 2 = 2x + 7, collect like terms first.

对于未知数在等号两边的方程，如 5x − 2 = 2x + 7，首先合并同类项。

Subtract 2x from both sides: 3x − 2 = 7. Add 2 to both sides: 3x = 9. Divide by 3: x = 3.

两边减去 2x：3x − 2 = 7。两边加 2：3x = 9。除以 3：x = 3。

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2. Percentage and Profit Problems | 百分数与利润问题

Example: A shop buys a jacket for £40 and sells it for £54. Calculate the percentage profit.

例题：商店以 40 英镑购进一件夹克，以 54 英镑卖出。计算利润百分比。

Find the actual profit: £54 − £40 = £14.

计算实际利润：54 英镑 − 40 英镑 = 14 英镑。

Percentage profit = (profit ÷ original cost) × 100%

利润百分比 = (利润 ÷ 原价) × 100%

=(14 ÷ 40) × 100% = 0.35 × 100% = 35%

Thus, the percentage profit is 35%.

因此，利润百分比为 35%。

(1 + r/100)ⁿ is often used, but WJEC may test repeated percentage change with multipliers.

对于复利，常用公式 A = P(1 + r/100)ⁿ，但 WJEC 可能会通过乘数测试重复百分比变化。

If an amount increases by 6% per year for 3 years, multiplier = 1.06, so final amount = P × (1.06)³.

如果金额每年增加 6%，持续 3 年，乘数 = 1.06，最终金额 = P × (1.06)³。

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3. Pythagoras’ Theorem | 勾股定理

In a right-angled triangle, a² + b² = c², where c is the hypotenuse.

在直角三角形中，a² + b² = c²，其中 c 为斜边。

Example: Find the length of the hypotenuse when the two shorter sides are 6 cm and 8 cm.

例题：当两条短边分别为 6 厘米和 8 厘米时，求斜边长度。

c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10 cm

To find a shorter side, rearrange: a² = c² − b².

要计算短边，重新排列：a² = c² − b²。

Example: hypotenuse = 13 cm, one side = 5 cm, find the other side.

例题：斜边长 13 厘米，一边长 5 厘米，求另一边。

x² = 13² − 5² = 169 − 25 = 144 → x = √144 = 12 cm

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4. Trigonometry: SOH CAH TOA | 三角学：SOH CAH TOA

tios: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.

使用 SOH CAH TOA 记忆比率：sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。

Example: In a right-angled triangle, the angle is 35°, the hypotenuse is 12 cm. Find the length of the opposite side.

例题：在直角三角形中，角为 35°，斜边为 12 厘米。求对边长度。

sin 35° = opposite / 12 → opposite = 12 × sin 35° ≈ 12 × 0.574 = 6.89 cm (2 d.p.)

nt are known, use tan to find the angle: tan θ = opposite/adjacent, then θ = tan⁻¹(ratio).

如果对边和邻边已知，可用正切求角：tan θ = 对边/邻边，然后 θ = tan⁻¹(比值)。

Example: opposite = 8, adjacent = 5 → tan θ = 8/5 = 1.6 → θ = tan⁻¹(1.6) ≈ 58.0°.

例题：对边=8，邻边=5 → tan θ = 8/5 = 1.6 → θ = tan⁻¹(1.6) ≈ 58.0°。

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5. Factorising Quadratic Expressions | 因式分解二次式

Factorising x² + 5x + 6 involves finding two numbers that multiply to 6 and add to 5. Those are 2 and 3.

因式分解 x² + 5x + 6，需要找到两个数，乘积为 6，和为 5。这两个数是 2 和 3。

So, x² + 5x + 6 = (x + 2)(x + 3).

因此，x² + 5x + 6 = (x + 2)(x + 3)。

For a negative constant, e.g. x² + x − 12: find numbers that multiply to −12 and add to +1: +4 and −3.

对于常数为负的情况，例如 x² + x − 12：找到两个数，乘积为 −12，和为 +1：+4 和 −3。

So, x² + x − 12 = (x + 4)(x − 3).

因此，x² + x − 12 = (x + 4)(x − 3)。

For quadratics with a coefficient of x² not equal to 1, such as 2x² + 7x + 3, use the 'ac' ), find numbers that multiply to 6 and add to 7 (6 and 1), rewrite and factor by grouping.

对于 x² 的系数不为 1 的二次式，如 2x² + 7x + 3，使用“ac”法：先将 a 和 c 相乘（2×3=6），找到乘积为 6 且和为 7 的两个数（6 和 1），重写后分组分解。

2x² + 6x + x + 3 → 2x(x+3) + 1(x+3) = (2x+1)(x+3).

2x² + 6x + x + 3 → 2x(x+3) + 1(x+3) = (2x+1)(x+3)。

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6. Probability Tree Diagrams | 概率树形图

Tree diagrams help solve problems involving successive events. Example: a bag contains 4 red and 6 blue counters. Two counters are drawn without replacement. Find the probability that both are red.

树形图有助于解决涉及连续事件的问题。例题：一个袋子装有 4 个红色和 6 个蓝色计数器。不放回地抽取两个。求两个都是红色的概率。

t draw: P(red) = 4/10 = 2/5. 2nd draw: if first was red, 3 red and 6 blue remain, so P(red|red) = 3/9 = 1/3.

第一次抽取：P(红) = 4/10 = 2/5。第二次抽取：若第一次为红，则剩余 3 红 6 蓝，因此 P(红|红) = 3/9 = 1/3。

Multiply along the branch: P(both red) = (2/5) × (1/3) = 2/15.

沿分支相乘：P(双双

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