📚 Year 9 Edexcel Mathematics: In-Depth Analysis of Past Papers | Edexcel 九年级数学：历年真题深度解析
Past papers reveal exactly how Edexcel structures its Year 9 assessments, blending straightforward recall with multi-step problem solving. This article dissects ten common question types, shows full worked solutions, and highlights the techniques and pitfalls that make the difference between a solid pass and a top score.
历年真题清晰地展示了 Edexcel 九年级数学的命题思路——既考查基础知识点的直接应用，也要求在多步骤问题中展现推理能力。本文拆解十类高频真题，给出完整的解答步骤，并标出那些决定普通分数与高分之间差距的技巧和易错点。
1. Reverse Percentage Problems | 逆向百分比问题
Question: A laptop is reduced by 15% in a sale. The sale price is '510. Calculate the original price before the reduction.
题目：一台笔记本电脑在促销中降价 15%，促销价为 '510。计算降价前的原价。
If a price is reduced by 15%, the sale price represents 85% of the original. Let the original price be x. Then 0.85x = 510.
如果某物降价 15%，那么促销价就是原价的 85%。设原价为 x，那么 0.85x = 510。
an rewrite 0.85 as 85/100, so x = 510 × 100/85 = 510 × 20/17 = 600. The original price was '600.
两边同时除以 0.85：x = 510 ÷ 0.85。可以把 0.85 写成 85/100，这样 x = 510 × 100/85 = 510 × 20/17 = 600。原价是 '600。
A common mistake is to find 15% e percentage reduction applies to the original, not the sale price. Always identify which quantity matches 100%.
常见错误是求出 510 的 15% 后直接加上去，得到 586.50，但这是错误的，因为降价的百分比是基于原价而非促销价。要始终明确哪个量对应 100%。
2. Solving Linear Equations with Unknowns on Both Sides | 解两边含未知数的线性方程
Question: Solve 4(x - 3) = 2x + 8.
题目：解方程 4(x - 3) = 2x + 8。
First expand the bracket: 4x - 12 = 2x + 8.
首先展开括号：4x - 12 = 2x + 8。
Collect like terms by subtracting 2x from both sides: 2x - 12 = 8.
移项，两边同时减去 2x：2x - 12 = 8。
Add 12 to both sides: 2x = 20. Divide by 2: x = 10.
两边同时加上 12：2x = 20。除以 2 得 x = 10。
check by substituting back: 4(10 - 3) = 4 × 7 = 28, and 2 × 10 + 8 = 28. Both sides match.
务必回代检验：4(10 - 3) = 4 × 7 = 28，2 × 10 + 8 = 28。两边相等。
3. Equation of a Straight Line from Two Points | 由两点求直线方程
Question: : Find the equation of the line that passes through (1, 4) and (3, 10). Give your answer in the form y = mx + c.
题目：求经过点 (1, 4) 与 (3, 10) 的直线方程，结果写成 y = mx + c 的形式。
dient m is the change in y divided by the change in x: m = (10 - 4) / (3 - 1) = 6 / 2 = 3.
斜率 m 等于 y 的变化量除以 x 的变化量：m = (10 - 4) / (3 - 1) = 6 / 2 = 3。
Now substitute one point into y = 3x + c. Using (1, 4): 4 = 3 × 1 + c, so c = 1.
然后将一个点代入 y = 3x + c。用 (1, 4)：4 = 3 × 1 + c，得 c = 1。
The equation is y = 3x + 1.
直线方程为 y = 3x + 1。
Checking with the second point: when x = 3, y = 3 × 3 + 1 = 10, correct.
用第二个点检验：当 x = 3 时，y = 3 × 3 + 1 = 10，正确。
4. Ratio Sharing with Three Parts | 三部分比例分配
Question: Share '800 between three people in the ratio 3 : 5 : 2. How much does each person receive?
题目：将 '800 按 3 : 5 : 2 的比例分给三人，每人分得多少钱？
Add the parts of the ratio: 3 + 5 + 2 = 10 parts in total.
将比例的项数相加：3 + 5 + 2 = 10 份。
One part is worth 800 ÷ 10 = '80.
一份的价值是 800 ÷ 10 = '80。
Person A gets 3 parts: 3 × 80 = '240. Person B gets 5 parts: 5 × 80 = '400. Person C gets 2 parts: 2 × 80 = '160.
A 得 3 份：3 × 80 = '240；B 得 5 份：5 × 80 = '400；C 得 2 份：2 × 80 = '160。
Quick check: 240 + 400 + 160 = 800 ' exactly.
快速验证：240 + 400 + 160 = 800 '，完全吻合。
5. Interior Angles of Regular Polygons | 正多边形的内角
Question: Each interior angle of a regular polygon is 144°. How many sides does the polygon have?
题目：一个正多边形的每个内角都是 144°，这个多边形有多少条边？
terior angle formula: Interior angle = (n - 2) × 180° / n, where n is the number of sides.
使用内角公式：内角 = (n - 2) × 180° / n，其中 n 为边数。
Set up the equation: (n - 2) × 180 / n = 144.
列出方程：(n - 2) × 180 / n = 144。
Multiply both sides by n: 180(n - 2) = 144n.
两边同乘 n：180(n - 2) = 144n。
Expand: 180n - 360 = 144n.
展开：180n - 360 = 144n。
: 180n - 144n = 360, so 36n = 360. Divide by 36: n = 10. The polygon is a regular decagon.
移项得 180n - 144n = 360，即 36n = 360。除以 36：n = 10，该多边形是正十边形。
ular polygon, sum of exterior angles is 360°, so n = 360 / 36 = 10. This often saves time.
更巧妙的解法是用外角：外角 = 180° - 144° = 36°。正多边形外角和为 360°，所以 n = 360 / 36 = 10。这种方法通常更省时。
6. Area of a Compound Shape | 组合图形的面积
Question: The diagram shows a rectangle of length 12 cm and width 5 cm joined to a right-angled triangle. The triangle shares the 5 cm side, and its other leg is 6 cm. Calculate the total area. (Diagram not shown)
题目：示意图由一个长 12 cm、宽 5 cm 的长方形和一个直角三角形组合而成。三角形与长方形共用 5 cm 边，另一条直角边为 6 cm。求总面积。（图略）
Area of rectangle = length × width = 12 × 5 = 60 cm².
长方形面积 = 长 × 宽 = 12 × 5 = 60 cm²。
iangle = ½ × base × height. Here base = 6 cm, height = 5 cm, so area = ½ × 6 × 5 = 15 cm².
三角形面积 = ½ × 底 × 高。这里底 = 6 cm，高 = 5 cm，所以面积 = ½ × 6 × 5 = 15 cm²。
Total area = 60 + 15 = 75 cm².
总面积 = 60 + 15 = 75 cm²。
Be careful: if the shared side is the height of the triangle, the base must be the other given length perpendicular to it. Identifying the two perpendicular lengths is key.
注意：若共有边是三角形的高，那么底必须是另一条与之垂直的已知长度。准确找出两条垂直边的长度是关键。
7. Mean from a Frequency Table | 由频数表求平均数
Question: The table shows the number of books read by 20 students: 0 books (freq 3), 1 book (freq 7), 2 books (freq 6), 3 books (freq 4). Calculate the mean number of books read.
题目：下表统计了 20 名学生的阅读本数：0 本（频数 3），1 本（7），2 本（6），3 本（4）。计算平均阅读本数。
for (number of books) × (frequency). Multiply each: 0×3 = 0, 1×7 = 7, 2×6 = 12, 3×4 = 12.
首先添加一列“本数 × 频数”。依次相乘：0×3 = 0，1×7 = 7，2×6 = 12，3×4 = 12。
Sum of these products = 0 + 7 + 12 + 12 = 31.
这些乘积之和 = 0 + 7 + 12 + 12 = 31。
Total frequency = 3 + 7 + 6 + 4 = 20.
总频数 = 3 + 7 + 6 + 4 = 20。
Mean = sum of products / total frequency = 31 / 20 = 1.55 books.
平均数 = 乘积总和 / 总频数 = 31 / 20 = 1.55 本。
Even though a student cannot read 1.55 books, the mean is a useful average. A common error is to divide by the number of categories (4) instead of total frequency. Always divide by total frequency.
虽然学生实际不可能读 1.55 本书，但平均数是有用的代表值。常见错误是除以类别数（4）而非总频数。务必除以总频数。
8. Probability Tree Diagram without Replacement | 无放回情况下的概率树图
Question: A bag contains 4 red and 5 blue counters. Two counters are taken at random without replacement. Draw a tree diagram and find the probability that both counters are red.
题目：一个袋子里有 4 个红色和 5 个蓝色筹码。随机取出两个筹码，每次不放回。画出树图，并求两个都是红色的概率。
First draw: P(red) = 4/9, P(blue) = 5/9.
第一次抽取：P(红) = 4/9，P(蓝) = 5/9。
After drawing one red, 3 red and 5 blue remain, so P(red|first red) = 3/8.
第一次抽到红后，还剩 3 红 5 蓝，所以 P(第二次红 | 第一次红) = 3/8。
Probability both red = (4/9) × (3/8) = 12/72 = 1/6 after simplifying.
两次均为红色的概率 = (4/9) × (3/8) = 12/72 = 1/6（化简后）。
If the question asked for 'at least one blue' (both red) = 1 - 1/6 = 5/6. Use complementary probability wherever it simplifies the work.
如果问题要求“至少一个蓝色”，通常更快捷的方法是用互补概率：1 - P(两个红) = 1 - 1/6 = 5/6。只要能让计算简化，就应使用互补概率。
9. nth Term of a Linear Sequence | 线性数列的第 n 项
Question: Find the nth term of the sequence: 7, 12, 17, 22, 27, ...
题目：求数列 7, 12, 17, 22, 27, ... 的第 n 项。
on difference d between consecutive terms is 12 - 7 = 5. The nth term has the form 5n + k.
相邻两项的差 d 为 12 - 7 = 5。因此第 n 项形式为 5n + k。
For n = 1, the term should be 7: 5 × 1 + k = 7, so k = 2.
当 n = 1 时，项应为 7：5 × 1 + k = 7，所以 k = 2。
Hence the nth term is 5n + 2.
所以第 n 项为 5n + 2。
Verify with n = 3: 5 × 3 + 2 = 17, which matches. The method 'difference × n + zero term' can also be used, but the n=1 substitution is reliable.
用 n = 3 验证：5 × 3 + 2 = 17，吻合。也可以用“公差 × n + 第零项”的方法，但代入 n=1 的方法更可靠。
10. Describing a Single Transformation | 描述单一变换
Question: Triangle A is mapped onto triangle B by a single transformation. Coordinates: A(2,1), (4,1), (2,5) and B(6,1), (8,1), (6,5). Describe the transformation fully.
题目：三角形 A 通过一次变换得到三角形 B。坐标：A(2,1), (4,1), (2,5)；B(6,1), (8,1), (6,5)。完整描述该变换。
Compare corresponding points: (2,1) maps to (6,1); (4,1) maps to (8,1); (2,5) maps to (6,5). The y-coordinates stay the same, while the x-coordinates increase by 4.
比较对应点：(2,1) → (6,1)；(4,1) → (8,1)；(2,5) → (6,5)。纵坐标不变，横坐标增加 4。
This is a translation by the vector (4, 0). Write: Translation by the vector (4, 0) (or 4 units to the right).
这是一个平移，平移向量为 (4, 0)。可以写作：沿向量 (4, 0) 平移（或向右平移 4 个单位）。
In the exam, always state the type of transformation and give all necessary detail. For a translation, that means stating the vector or the direction and distance in words.
考试中，务必说明变换类型并给出所有必要细节。对于平移，需要写明向量或用文字表述方向与距离。
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