KS3 Advanced Maths: Past Paper Analysis | KS3 进阶数学：历年真题解析

📚 KS3 Advanced Maths: Past Paper Analysis | KS3 进阶数学：历年真题解析

In Key Stage 3, advanced mathematics challenges students to apply foundational skills in more complex problem-solving contexts. Past papers are invaluable for understanding question styles, common topics, and the level of reasoning expected. This article analyses typical advanced KS3 questions, providing step-by-step solutions in both English and Chinese. By working through these examples, students can build confidence and refine their techniques.

在 KS3 阶段，进阶数学要求学生将基本技能应用于更复杂的问题解决情景。历年真题是了解题型、常见主题以及所需推理层次的宝贵资源。本文分析典型的进阶 KS3 题目，提供中英双语的逐步解析。通过练习这些例题，学生可以建立信心，完善解题技巧。

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1. Solving Linear Equations with Brackets | 解含有括号的一次方程

Question: Solve 3(2x – 1) = 5x + 7.

题目：解方程 3(2x – 1) = 5x + 7。

Step 1: Expand the bracket on the left-hand side. Multiply 3 by each term inside: 3 × 2x = 6x, and 3 × (–1) = –3. The equation becomes 6x – 3 = 5x + 7.

步骤1：展开左边的括号。将3乘以括号内的每一项：3 × 2x = 6x，3 × (–1) = –3。方程变为 6x – 3 = 5x + 7。

Step 2: Collect like terms. Move the x terms to one side and the numbers to the other. Subtract 5x from both sides: 6x – 5x – 3 = 7 → x – 3 = 7.

步骤2：合并同类项。将含x的项移到一边，数字移到另一边。两边减去5x：6x – 5x – 3 = 7 → x – 3 = 7。

Step 3: Isolate x. Add 3 to both sides: x – 3 + 3 = 7 + 3 → x = 10.

步骤3：分离x。两边加3：x – 3 + 3 = 7 + 3 → x = 10。

Check: Substitute x = 10 back into the original equation: 3(2×10 – 1) = 3(20 – 1) = 3×19 = 57. Right side: 5×10 + 7 = 50 + 7 = 57. Both sides match, so x = 10 is correct.

检验：将 x = 10 代入原方程：3(2×10 – 1) = 3(20 – 1) = 3×19 = 57。右边：5×10 + 7 = 50 + 7 = 57。两边相等，所以 x = 10 正确。

This classic question tests correct expansion of brackets and careful collection of terms. Always expand before moving terms.

这类经典题目考查括号的正确展开和细心的移项。一定要先展开再移项。

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2. Area and Perimeter of Composite Shapes | 复合图形的面积与周长

Question: A composite shape is formed by cutting a 3 cm by 2 cm rectangle from one corner of an 8 cm by 5 cm rectangle. Find the total area and the perimeter.

题目：从一个8 cm × 5 cm 的矩形的一个角上切去一个3 cm × 2 cm 的小矩形，形成复合图形。求其总面积和周长。

Area calculation: Area of large rectangle = 8 × 5 = 40 cm². Area of cut-out = 3 × 2 = 6 cm². Total area = 40 – 6 = 34 cm².

面积计算：大矩形面积 = 8 × 5 = 40 cm²。切去部分的面积 = 3 × 2 = 6 cm²。总面积 = 40 – 6 = 34 cm²。

Perimeter calculation: The removal of a corner piece replaces the original two edges (3 cm and 2 cm) with two new edges (also 3 cm and 2 cm), so the perimeter remains the same as the original rectangle. Perimeter = 2 × (8 + 5) = 26 cm. Alternatively, trace all outer edges: 8 + 5 + (8 – 2) + 2 + 3 + (5 – 3) = 8 + 5 + 6 + 2 + 3 + 2 = 26 cm.

周长计算：切去角后，原来的两条边（3 cm 和 2 cm）被两条新边（同样为3 cm 和2 cm）替代，因此周长与原矩形相同。周长 = 2 × (8 + 5) = 26 cm。也可以沿外边界逐段相加：8 + 5 + (8 – 2) + 2 + 3 + (5 – 3) = 8 + 5 + 6 + 2 + 3 + 2 = 26 cm。

Many mistakes occur when students forget that cutting out a corner does not change the perimeter, or they double-count internal edges. Always trace the shape carefully.

学生常犯的错误是忘记切角不改变周长，或将内部边重复计算。一定要仔细描绘边界。

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3. Ratio and Proportion Word Problems | 比率与比例应用题

Question: The ratio of red to blue marbles in a bag is 3 : 5. If there are 72 red marbles, how many blue marbles are there?

题目：袋中红弹珠与蓝弹珠的数量之比为 3 : 5。如果有 72 颗红弹珠，那么蓝弹珠有多少颗？

The ratio 3 : 5 means that for every 3 red marbles there are 5 blue marbles. The total number of parts for red is 3. These 3 parts correspond to 72 marbles. So 1 part = 72 ÷ 3 = 24 marbles.

比率 3 : 5 表示每 3 颗红弹珠对应 5 颗蓝弹珠。红弹珠所占的份数是 3 份。这 3 份对应 72 颗。因此 1 份 = 72 ÷ 3 = 24 颗。

Blue marbles correspond to 5 parts: 5 × 24 = 120 blue marbles.

蓝弹珠占 5 份：5 × 24 = 120 颗蓝弹珠。

Quick check: The ratio 72 : 120 simplifies to 3 : 5, confirming the answer. Always identify the value of one part first.

快速检验：72 : 120 化简后为 3 : 5，验证了答案。总是先求出一份的值。

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4. Sequences and the nth Term | 数列与第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项公式。

The difference between consecutive terms is constant: 12 – 7 = 5, 17 – 12 = 5, so the common difference d = 5. This tells us the nth term has the form 5n + c, where c is a constant.

相邻项的差是常数：12 – 7 = 5，17 – 12 = 5，所以公差 d = 5。这表明第n项的形式为 5n + c，其中 c 为常数。

To find c, use the first term (n = 1): 5(1) + c = 7 → 5 + c = 7 → c = 2. So the nth term is 5n + 2. Verify for n = 2: 5×2+2 = 12, correct.

求 c，代入第一项 (n = 1)：5(1) + c = 7 → 5 + c = 7 → c = 2。因此第n项为 5n + 2。检验 n = 2：5×2+2 = 12，正确。

For linear sequences, simply multiply the common difference by n and adjust with the zero term (the term before the first). Here, term ‘0’ would be 2, so 5n + 2.

对于线性数列，只需将公差乘以 n，再用第零项（首项前的项）进行调整。此处第零项为 2，所以是 5n + 2。

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5. Pythagoras’ Theorem in 2D | 平面中的勾股定理

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

题目：一个直角三角形两条直角边长分别为 5 cm 和 12 cm。求斜边长。

Using Pythagoras’ theorem: hypotenuse² = leg₁² + leg₂² = 5² + 12² = 25 + 144 = 169. Therefore, hypotenuse = √169 = 13 cm.

使用勾股定理：斜边² = 直角边₁² + 直角边₂² = 5² + 12² = 25 + 144 = 169。所以斜边 = √169 = 13 cm。

A variation: If the hypotenuse is 15 cm and one leg is 9 cm, find the other leg. Then leg² = hypotenuse² – known leg² = 15² – 9² = 225 – 81 = 144, leg = √144 = 12 cm.

变式题：如果斜边为 15 cm，一条直角边为 9 cm，求另一条直角边。则直角边² = 斜边² – 已知直角边² = 15² – 9² = 225 – 81 = 144，直角边 = √144 = 12 cm。

Always label the sides first and decide whether you are finding the hypotenuse or a shorter side.

务必先标注各边，确定要求的是斜边还是直角边。

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6. Transformations: Reflection and Rotation | 变换：反射与旋转

Question: A triangle has vertices A(2, 3), B(4, 6) and C(6, 1). Reflect the triangle in the x-axis and write the new coordinates. Then rotate the original triangle 90° clockwise about the origin and give the new coordinates.

题目：三角形顶点为 A(2, 3), B(4, 6) 和 C(6, 1)。将该三角形关于 x 轴反射，写出新坐标。再将原三角形绕原点顺时针旋转 90°，写出新坐标。

Reflection in the x-axis maps (x, y) → (x, –y):

关于 x 轴的反射规则为 (x, y) → (x, –y)：

• A(2, 3) → A'(2, –3)
• B(4, 6) → B'(4, –6)
• C(6, 1) → C'(6, –1)

Rotation 90° clockwise about the origin uses the rule (x, y) → (y, –x):

绕原点顺时针旋转 90° 的规则为 (x, y) → (y, –x)：

• A(2, 3) → A”(3, –2)
• B(4, 6) → B”(6, –4)
• C(6, 1) → C”(1, –6)

Memorising these transformation rules helps solve such questions quickly, but also try sketching the points to visualise the movement.

熟记这些变换规则有助于快速解题，但也可画出各点以直观理解移动。

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7. Fractions, Decimals and Percentages Conversion | 分数、小数和百分比的互化

Question: Convert 7/20 to a decimal and a percentage. Then convert 0.875 to a fraction in its simplest form.

题目：将 7/20 化为小数和百分数。再将 0.875 化为最简分数。

7/20 as a decimal: Divide 7 by 20: 7 ÷ 20 = 0.35. As a percentage, multiply the decimal by 100: 0.35 × 100 = 35%.

7/20 化为小数：7 ÷ 20 = 0.35。化为百分数：0.35 × 100 = 35%。

0.875 to a fraction: 0.875 = 875/1000. Find the greatest common divisor (GCD) of 875 and 1000, which is 125. Divide numerator and denominator by 125: 875 ÷ 125 =

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