📚 KS3 Edexcel Further Mathematics: Mock Unit Test Analysis | KS3 Edexcel 进阶数学：单元测试模拟卷解析
This article provides a step-by-step walkthrough of a mock unit test designed for KS3 Edexcel Further Mathematics. It covers essential topics including algebra, geometry, statistics, and transformations, helping students consolidate key skills and avoid common mistakes.
本文为 KS3 Edexcel 进阶数学的单元测试模拟卷提供逐步解析。试题涵盖代数、几何、统计与图形变换等核心主题，旨在帮助同学们巩固重点技能并规避常见错误。
1. Question 1: Solving Linear Equations | 第1题：解线性方程
Question: Solve 2(3x − 1) + 5 = 7x − 2, showing all steps clearly.
题目：解方程 2(3x − 1) + 5 = 7x − 2，并写出完整步骤。
Step 1: Expand the bracket on the left-hand side. Multiply 2 by each term inside: 2(3x − 1) = 6x − 2.
第1步：展开左边括号。将 2 乘以括号内每一项：2(3x − 1) = 6x − 2。
Step 2: Rewrite the equation with the expanded form. Now we have 6x − 2 + 5 = 7x − 2.
第2步：用展开后的形式重写方程。现在得到 6x − 2 + 5 = 7x − 2。
Step 3: Combine the constant terms on the left side: −2 + 5 = 3. The equation simplifies to 6x + 3 = 7x − 2.
第3步：合并左边的常数项：−2 + 5 = 3。方程简化为 6x + 3 = 7x − 2。
Step 4: Collect the x terms on one side and the constants on the other. Subtract 6x from both sides: 3 = x − 2.
第4步：将含 x 项移到一边，常数项移到另一边。两边同时减去 6x，得 3 = x − 2。
Step 5: Add 2 to both sides to isolate x: 3 + 2 = x, so x = 5.
第5步：两边加 2 解出 x：3 + 2 = x，所以 x = 5。
Step 6: Verify the solution by substituting x = 5 into the original equation. Left side: 2(3×5 − 1) + 5 = 2(14) + 5 = 33. Right side: 7×5 − 2 = 33. Both sides match.
第6步：将 x = 5 代入原方程验证。左边：2(3×5 − 1) + 5 = 2(14) + 5 = 33。右边：7×5 − 2 = 33。两边相等，解正确。
Answer: x = 5
答案：x = 5
2. Question 2: Solving Inequalities | 第2题：解不等式
Question: Solve the inequality 4x + 7 ≥ 3x − 2 and represent the solution set on a number line.
题目：解不等式 4x + 7 ≥ 3x − 2，并在数轴上表示解集。
Step 1: Gather the terms containing x on the left side. Subtract 3x from both sides: 4x − 3x + 7 ≥ −2.
第1步：将含 x 项移到左边。两边同时减去 3x：4x − 3x + 7 ≥ −2。
Step 2: Simplify to obtain x + 7 ≥ −2.
第2步：化简得 x + 7 ≥ −2。
Step 3: Isolate x by subtracting 7 from both sides: x ≥ −2 − 7, which gives x ≥ −9.
第3步：两边减 7，解出 x：x ≥ −2 − 7，即 x ≥ −9。
Step 4: Draw a number line. Mark a closed circle at −9 (because the inequality includes ‘equal to’) and shade the line to the right to show all values greater than or equal to −9.
第4步：画数轴。在 −9 处画一个实心圆（因为包含等号），并向右画阴影线表示所有大于等于 −9 的数。
Solution: x ≥ −9
解集：x ≥ −9
3. Question 3: Equation of a Straight Line | 第3题：直线方程
Question: A straight line passes through the point (2, 5) and has a gradient of 3. Find the equation of the line in the form y = mx + c.
题目：一条直线经过点 (2, 5) 且斜率为 3。求该直线的方程，写成 y = mx + c 的形式。
Step 1: Write down the slope-intercept form y = mx + c. Here the gradient m = 3.
第1步：写出斜截式 y = mx + c。本题斜率 m = 3。
Step 2: Substitute the given coordinates into the equation. For x = 2, y = 5: 5 = 3(2) + c.
第2步：将已知点坐标代入方程。x = 2，y = 5：5 = 3(2) + c。
Step 3: Solve for c. 3 × 2 = 6, so 5 = 6 + c, hence c = 5 − 6 = −1.
第3步：解出 c。3 × 2 = 6，因此 5 = 6 + c，所以 c = 5 − 6 = −1。
Step 4: Write the full equation: y = 3x − 1.
第4步：写出完整的直线方程：y = 3x − 1。
Answer: y = 3x − 1
答案：y = 3x − 1
4. Question 4: Pythagoras’ Theorem | 第4题：毕达哥拉斯定理
Question: In a right-angled triangle, the two shorter sides are 5 cm and 12 cm. Calculate the length of the hypotenuse, giving your answer with an appropriate unit.
题目：一个直角三角形中，两条直角边分别为 5 cm 和 12 cm。计算斜边的长度，并带上合适的单位。
Step 1: Recall Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse.
第1步：回忆毕达哥拉斯定理：a² + b² = c²，其中 c 为斜边。
Step 2: Substitute the given side lengths. Let a = 5 and b = 12. Then c² = 5² + 12².
第2步：代入已知边长。令 a = 5，b = 12。则 c² = 5² + 12²。
Step 3: Calculate the squares: 5² = 25, 12² = 144. So c² = 25 + 144 = 169.
第3步：计算平方：5² = 25，12² = 144。于是 c² = 25 + 144 = 169。
Step 4: Take the positive square root of 169 to find c. √169 = 13. Therefore, the hypotenuse is 13 cm.
第4步：取 169 的正平方根求 c。√169 = 13。因此，斜边长为 13 cm。
Answer: 13 cm
答案：13 cm
5. Question 5: Percentage Increase | 第5题：百分数增加
Question: A bicycle originally costs £200. Its price is increased by 15%. Find the new price of the bicycle.
题目：一辆自行车原价 200 英镑，价格上调 15%。求自行车的新价格。
Method 1: Find the increase first. Step 1: Calculate 15% of £200: (15/100) × 200 = 30.
方法一：先求增加量。第1步：计算 200 英镑的 15%：(15/100) × 200 = 30。
Step 2: Add the increase to the original price: 200 + 30 = 230. So the new price is £230.
第2步：将增加量加到原价上：200 + 30 = 230。新价格为 230 英镑。
Method 2: Use a multiplier. An increase of 15% means the new price is 115% of the original, i.e., multiplier = 1.15.
方法二：使用乘数。增加 15% 意味着新价格是原价的 115%，即乘数 = 1.15。
Step: New price = 200 × 1.15 = 230. Both methods yield the same result.
步骤：新价格 = 200 × 1.15 = 230。两种方法结果一致。
Answer: £230
答案：230 英镑
6. Question 6: Arithmetic Sequences | 第6题：等差数列
Question: The first four terms of a sequence are 5, 9, 13, 17. Find an expression for the nth term of this sequence.
题目：一个数列的前四项为 5, 9, 13, 17。求该数列第 n 项的表达式。
Step 1: Identify the type of sequence. The difference between consecutive terms is constant: 9 − 5 = 4, 13 − 9 = 4, so it is arithmetic with common difference d = 4.
第1步：识别数列类型。相邻两项的差为常数：9 − 5 = 4，13 − 9 = 4，因此是等差数列，公差 d = 4。
Step 2: Write the general form of an arithmetic sequence: nth term = a + (n − 1)d, where a is the first term.
第2步：写出等差数列通项公式的一般形式：第 n 项 = a + (n − 1)d，其中 a 为首项。
Step 3: Substitute a = 5 and d = 4: nth term = 5 + (n − 1) × 4.
第3步：代入 a = 5，d = 4：第 n 项 = 5 + (n − 1) × 4。
Step 4: Simplify: 5 + 4n − 4 = 4n + 1.
第4步：化简：5 + 4n − 4 = 4n + 1。
Step 5: Check with known terms. For n = 1: 4×1 + 1 = 5. n = 2: 9. Correct.
第5步：用已知项检验。n = 1：4×1 + 1 = 5；n = 2：9。正确。
Answer: nth term = 4n + 1
答案：第 n 项 = 4n + 1
7. Question 7: Probability with Tree Diagrams | 第7题：概率与树状图
Question: A bag contains 3 red counters and 2 blue counters. Two counters are drawn at random without replacement. Draw a tree diagram and calculate the probability that both counters are the same colour.
题目：一个袋子里装有 3 个红色计数器和 2 个蓝色计数器。随机抽出两个，不放回。画出树状图，并计算两个计数器颜色相同的概率。
Step 1: Set up the first draw. P(Red) = 3/5, P(Blue) = 2/5.
第1步：设定第一次抽取。P(红) = 3/5，P(蓝) = 2/5。
Step 2: After drawing a red first, there remain 2 red and 2 blue (total 4). So P(Red | Red) = 2/4 = 1/2, P(Blue | Red) = 2/4 = 1/2.
第2步：若第一次抽到红色，剩余 2 红 2 蓝（共 4 个）。因此，P(第二次红 | 第一次红) = 2/4 = 1/2，P(第二次蓝 | 第一次红) = 2/4 = 1/2。
Step 3: After drawing a blue first, there remain 3 red and 1 blue (total 4). So P(Red | Blue) = 3/4, P(Blue | Blue) = 1/4.
第3步：若第一次抽到蓝色，剩余 3 红 1 蓝（共 4 个）。因此，P(第二次红 | 第一次蓝) = 3/4，P(第二次蓝 | 第一次蓝) = 1/4。
Step 4: The probability of two reds is (3/5) × (1/2) = 3/10. The probability of two blues is (2/5) × (1/4) = 2/20 = 1/10.
第4步：两次均为红的概率为 (3/5) × (1/2) = 3/10。两次均为蓝的概率为 (2/5) × (1/4) = 2/20 = 1/10。
Step 5: Add these probabilities to find the probability of both being the same colour: 3/10 + 1/10 = 4/10 = 2/5.
第5步：将两概率相加得到颜色相同的概率：3/10 + 1/10 = 4/10 = 2/5。
Answer: 2/5
答案：2/5
8. Question 8: Transformation – Reflection | 第8题：图形变换——反射
Question: Describe fully the single transformation that maps shape A onto shape B when shape B is the image of shape A after a reflection in the line y = x.
题目：图形 A 经过关于直线 y = x 的反射后得到图形 B。请完整描述这个单一变换。
Step 1: A reflection in the line y = x swaps the x- and y-coordinates of every point.
第1步：关于直线 y = x 的反射会交换每个点的 x 坐标和 y 坐标。
Step 2: For any point (x, y) on shape A, its image on shape B will be (y, x). For example, (3, 1) → (1, 3).
第2步：对于图形 A 上的任意点 (x, y)，其在图形 B 上的像为 (y, x)。例如 (3, 1) → (1, 3)。
Step 3: The line y = x acts as a mirror with a slope of 1 passing through the origin at 45° to both axes. The perpendicular distance from a point to the line is preserved, but the side is reversed.
第3步：直线 y = x 作为镜面，斜率为 1，通过原点，与两坐标轴均构成 45° 角。点到该直线的垂直距离保持不变，但方向翻转。
Step 4: To fully describe the transformation, state: ‘Reflection in the line y = x’.
第4步：完整描述该变换时说明：“关于直线 y = x 的反射”。
Transformation: Reflection in the line y = x
变换：关于直线 y = x 的反射
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