📚 In-depth Analysis of Past Papers: Year 7 WJEC Further Maths | Year 7 WJEC 进阶数学：历年真题深度解析
Working through past exam papers is one of the most effective ways to build confidence and accuracy in WJEC Further Mathematics at Year 7. These papers test not just recall but the ability to apply concepts in unfamiliar contexts. By examining real questions, students can identify common question types, avoid typical mistakes, and develop a structured approach to problem-solving that is essential for further study.
刷历年真题是提升 Year 7 WJEC 进阶数学信心与准确率最有效的方法之一。这些试卷不仅考察知识记忆，更考查在陌生情境中运用概念的能力。通过分析真实考题，学生能识别常见题型、避开典型错误，并培养有条理的解题思路，为后续学习打下坚实基础。
1. Solving Linear Equations | 解一元一次方程
ways check your solution by substituting back into the original equation: 4(7) - 7 = 28 - 7 = 21, so it is correct.
常见真题如：解方程 4x - 7 = 21。先分离含 x 的项，两边同时加 7 得 4x = 28，再两边除以 4 得到 x = 7。一定要代回原方程检验：4(7) - 7 = 28 - 7 = 21，结果正确。
mber the order of operations, and do not forget to multiply every term inside the bracket.
还有题型包含括号，如 3(2x + 1) = 27。先去括号：6x + 3 = 27，然后两边减 3：6x = 24，得 x = 4。注意运算顺序，括号内每一项都要乘到。
2. Sequences: Finding the nth Term | 数列：求第 n 项
n difference and a is the first term. Here d = 4, a = 5, so the nth term = 4n + (5 - 4) = 4n + 1. Test with n=1: 4×1+1=5, n=2: 9, correct.
考虑一个线性数列：5, 9, 13, 17, ... 相邻两项的差是 4。第 n 项公式为 an = dn + (a - d)，其中 d 是公差，a 是首项。此处 d = 4，a = 5，所以第 n 项 = 4n + (5 - 4) = 4n + 1。检验 n=1 得 5，n=2 得 9，正确。
A harder past-paper problem gives: "The nth term of a sequence is 7n - 3. Which term has the value 39?" Set 7n - 3 = 39, add 3: 7n = 42, n = 6. So the 6th term is 39.
较难的真题：“某数列第 n 项为 7n - 3，哪一项的值为 39？” 设 7n - 3 = 39，加 3 得 7n = 42，n = 6，即第 6 项为 39。
3. Angles in Triangles and Parallel Lines | 三角形内角与平行线
A typical question: "In triangle ABC, angle A = 32°, angle B = 2x°, angle C = 3x°. Find x." 32 + 2x + 3x = 180 → 32 + 5x = 180 → 5x = 148 → x = 29.6. Then check angles are positive.
典型题：“三角形 ABC 中，∠A=32°，∠B=2x°，∠C=3x°，求 x。” 三角形内角和为 180°，故 32+2x+3x=180 → 32+5x=180 → 5x=148 → x=29.6。确认角度为正。
Questions on parallel lines often use alternate and corresponding angles. For instance, if two parallel lines are cut by a transversal and one angle is 75°, find the alternate angle. It is also 75°. Interior angles sum to 180° when lines are parallel.
涉及平行线的题常考错角和同位角。例如，两平行线被横线所截，一角为 75°，求其错角，也是 75°。平行时同旁内角之和为 180°。
4. Ratio and Proportion: Dividing Quantities | 比和比例：分配数量
Past paper: "Divide £96 in the ratio 3:5." The total parts = 3 + 5 = 8. One part = £96 ÷ 8 = £12. The first share = 3 × £12 = £36, second share = 5 × £12 = £60. Always check the sum: £36 + £60 = £96.
真题：“按 3:5 的比例分配 £96。” 总份数 = 3+5=8，每份为 £96÷8=£12。第一部分 3×12=£36，第二部分 5×12=£60。务必验证总和为 £96。
Proportion problems also appear: "A recipe for 4 people requires 300 g of flour. How much flour is needed for 10 people?" people need 75×10 = 750 g. Alternatively, use a scale factor 10/4 = 2.5; 300×2.5 = 750 g.
比例问题也常见：“一份 4 人食谱需面粉 300 g，10 人需要多少？” 每人需 300÷4=75 g，10 人需 75×10=750 g。或用倍数 10/4=2.5，300×2.5=750 g。
5. Probability from Two-Way Tables | 双向表求概率
Example: " ootball and tennis. Find the probability that a randomly chosen student likes both sports." fill in missing values, then probability = number liking both / total number of students.
示例：“表格显示喜欢足球和网球的学生人数。求随机选一名学生既喜欢足球又喜欢网球的概率。” 先补全表格总数，概率 = 两项都喜欢的人数 ÷ 总人数。
| Like Tennis | Dislike Tennis | Total | |
| Like Football | 12 | 8 | 20 |
| Dislike Football | 6 | 4 | 10 |
| Total | 18 | 12 | 30 |
ikes both football and tennis is 12/30 = 2/5. Express the answer as a simplified fraction.
从上表，学生既喜欢足球又喜欢网球的概率为 12/30 = 2/5。答案要写成最简分数。
6. Expanding Brackets and Factorising | 去括号与因式分解
double bracket: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12, using the FOIL method.
展开 5(3x - 2) = 15x - 10，用 5 乘以括号内每一项。处理两个括号：(x+3)(x+4) = x²+4x+3x+12 = x²+7x+12，用首外内尾法则。
bers that multiply to 12 and add to 7. Past papers often ask to factorise expressions like x² - 9, recognising it as the difference of two squares: (x+3)(x-3).
逆过程是因式分解：x²+7x+12 = (x+3)(x+4)，寻找相乘得 12、相加得 7 的两个数。真题常考 x²-9 这种形式，识别为平方差：(x+3)(x-3)。
7. Rules of Indices | 指数运算法则
rst multiply 2³ × 2⁴ = 2⁷, then divide by 2² gives 2⁵ = 32. The general rules: aᵐ × aⁿ = am+n and aᵐ ÷ aⁿ = am-n.
化简：2³ × 2⁴ ÷ 2²。用指数法则：先乘 2³×2⁴=2⁷，再除以 2² 得 2⁵=32。一般法则：aᵐ × aⁿ = am+n，aᵐ ÷ aⁿ = am-n。
Another common question: "Write 3⁵ ÷ 3² as a single power." anything to the power 0 equals 1, so 5⁰ = 1. Negative indices mean reciprocal: 2⁻¹ = 1/2.
另一常见题：“将 3⁵ ÷ 3² 写成单一幂形式。” 答案：3³。切记任何数的 0 次方均为 1，如 5⁰=1。负指数表示倒数：2⁻¹ = 1/2。
8. Introduction to Pythagoras' Theorem | 勾股定理基础
6 + 64 = 100, so h = √100 = 10 cm. The formula is a² + b² = c², where c is the hypotenuse.
一直角三角形两直角边为 6 cm 和 8 cm，求斜边 h。用 h² = 6² + 8² = 36+64=100，故 h=√100=10 cm。公式为 a² + b² = c²，c 是斜边。
A WJEC past paper might ask: " st a wall. The foot of the ladder is 5 m from the wall. How high up the wall does it reach?" This forms a right triangle: height² = 13² - 5² = 169 - 25 = 144, height = √144 = 12 m.
WJEC 真题可能问：“长 13 m 的梯子斜靠在墙上，梯脚距墙 5 m。梯子顶端距地面多高？” 构成直角三角形：高度² = 13² - 5² = 169-25=144，高度 = √144 = 12 m。
c² = a² + b²
9. Area and Perimeter of Composite Shapes | 复合图形的面积与周长
ther method is to split the shape into two non-overlapping rectangles and sum their areas.
求一个 L 形图形面积，该图形由 5 m×7 m 的矩形挖去一个 2 m×3 m 的矩形得到。总面积=5×7 - 2×3=35-6=29 m²。另一种方法是分割成两个不重叠矩形再求和。
7-3) = 5+7+3+3+2+4 = 24 m. Always label sides and use subtraction to find missing lengths.
求周长需仔细把所有外边相加。同一图形，沿外边相加：5+7+(5-2)+3+2+(7-3)=5+7+3+3+2+4=24 m。务必标注各边，用减法求出未知边长。
10. Function Machines and Inverse Operations | 函数机器与逆运算
Question: "I think of a number, multiply by 3, subtract 7, and get 20. What was the original number?" h 20, add 7 → 27, divide by 3 → 9. The original number is 9. This uses inverse operations.
题目：“我想了一个数，乘 3，减 7，得到 20。原来是多少？” 逆向运算：从 20 开始，加 7 → 27，除以 3 → 9。原数是 9。这里运用了逆运算。
s 20, solve 3x - 7 = 20. The backward method is often faster for simple mental arithmetic.
函数机器可用图示表示。输入 x，输出为 3x - 7。若输出 20，解 3x - 7 = 20。对简单的心算，逆推法通常更快。
11. Data Handling: Mean, Median, Mode | 数据处理：平均数、中位数、众数
ith 6 values, median = (8+8)/2 = 8. Mean = (3+5+8+8+9+11) ÷ 6 = 44 ÷ 6 = 7.33 (to 2 d.p.).
给定数据集：5, 8, 3, 8, 11, 9。众数为 8（出现最频繁）。中位数：排序为 3, 5, 8, 8, 9, 11，共 6 个数，中位数=(8+8)/2=8。平均数=(3+5+8+8+9+11)÷6=44÷6≈7.33（保留两位小数）。
Past papers may ask to choose the most appropriate average. When there is an outlier, the median is often better. If the data is 1, 1, 2, 3, 50, the mean is 11.4 which is affected by 50; median is 2, more representative.
真题会要求选择最合适的平均数。存在异常值时中位数往往更佳。若数据为 1,1,2,3,50，平均数为 11.4 受到 50 影响；中位数为 2，更具代表性。
12. Problem Solving with Money and Decimals | 货币与小数应用题
A common multistep problem: "Lucy buys 3 notebooks at £2.55 each and a pack of pens for £4.30. She pays with a £20 note. How much change does she receive?" Cost = 3 × 2.55 + 4.30 = 7.65 + 4.30 = 11.95. Change = 20.00 - 11.95 = £8.05.
常见的多步题：“露西买 3 本笔记本，每本 £2.55，再加一包笔 £4.30。她付 £20 纸币，找回多少？” 花费=3×2.55+4.30=7.65+4.30=11.95。找回=20.00-11.95=£8.05。
Work systematically and show all steps. Use column addition or a number line for subtraction to avoid careless errors. Always include the unit (£) in the final answer.
有条理地计算并展示所有步骤。用竖式加法或数轴减法以避免粗心错误。最终答案务必包含单位（£）。
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