📚 Year 9 Edexcel Maths: Unit Test Mock Paper Analysis | Edexcel 九年级数学单元测试模拟卷解析
This article takes you through a full mock unit test designed for Year 9 students following the Edexcel Maths curriculum. Every question is broken down with clear explanations, covering essential topics such as number, algebra, geometry, statistics and probability. Use this walkthrough to identify common mistakes, strengthen your problem‑solving skills and build confidence for real assessments.
本文为Edexcel九年级数学学生解析一份完整的单元测试模拟卷。每一道题目都配有清晰的中英文讲解，涵盖数、代数、几何、统计与概率等核心主题。通过这份解析，你可以发现常见错误、强化解题思路，为真正的考试积累信心。
1. Number: Operations with Fractions and Decimals | 数字：分数与小数的混合运算
Question: Work out 2 ⅓ × (1.5 + ¾). Give your answer as a mixed number in its simplest form.
题目：计算 2 ⅓ × (1.5 + ¾)，并以最简带分数形式给出答案。
First, write 2 ⅓ as an improper fraction: 2 ⅓ = 7/3. Write 1.5 as a fraction: 1.5 = 3/2. The bracket becomes 3/2 + 3/4. The common denominator is 4: 3/2 = 6/4, so 6/4 + 3/4 = 9/4. Multiply: 7/3 × 9/4 = 63/12. Simplify by dividing numerator and denominator by 3: 63/12 = 21/4. Convert to a mixed number: 21 ÷ 4 = 5 remainder 1, giving 5 ¼.
首先将 2 ⅓ 化为假分数：2 ⅓ = 7/3。将 1.5 写成分数：1.5 = 3/2。括号内变为 3/2 + 3/4。公分母为 4：3/2 = 6/4，所以 6/4 + 3/4 = 9/4。相乘：7/3 × 9/4 = 63/12。分子分母同时除以 3 约分：63/12 = 21/4。转换为带分数：21 ÷ 4 = 5 余 1，结果是 5 ¼。
2. Percentages and Percentage Change | 百分数与百分数变化
Question: A laptop originally costs £540. In a sale, its price is reduced by 15%. After the sale, the price is increased by 20%. What is the final price of the laptop?
题目：一台笔记本电脑原价 540 英镑。促销期间降价 15%。促销结束后又提价 20%。笔记本电脑的最终价格是多少？
Reduction of 15% means the multiplier is 1 − 0.15 = 0.85. Sale price = 540 × 0.85 = £459. Then a 20% increase uses the multiplier 1 + 0.20 = 1.20. Final price = 459 × 1.20 = £550.80. Be careful not to simply add the percentages: the second change applies to the reduced price, not the original.
降价 15% 意味着乘数为 1 − 0.15 = 0.85。促销价 = 540 × 0.85 = 459 英镑。随后提价 20% 使用的乘数是 1 + 0.20 = 1.20。最终价格 = 459 × 1.20 = 550.80 英镑。注意不能简单地将百分比相加：第二次变化是针对降价后的价格，而非原价。
3. Ratio and Proportion | 比与比例
Question: The ratio of boys to girls in a school is 5 : 7. When 18 new boys join, the ratio becomes 7 : 8. How many students were originally in the school?
题目：一所学校男生与女生的比为 5 : 7。当 18 名新男生加入后，比率变为 7 : 8。学校原有多少名学生？
Let the original number of boys be 5x and girls be 7x. After 18 boys join, boys become 5x + 18, girls remain 7x. New ratio: (5x + 18) / 7x = 7/8. Cross-multiply: 8(5x + 18) = 7 × 7x → 40x + 144 = 49x. Solve: 144 = 9x → x = 16. Originally, total students = 5x + 7x = 12x = 12 × 16 = 192.
设原有男生人数为 5x，女生为 7x。加入 18 名男生后，男生变为 5x + 18，女生仍为 7x。新比例：(5x + 18) / 7x = 7/8。交叉相乘：8(5x + 18) = 7 × 7x → 40x + 144 = 49x。解方程：144 = 9x → x = 16。原有学生总数 = 5x + 7x = 12x = 12 × 16 = 192。
4. Algebraic Simplification and Expansion | 代数化简与展开
Question: Expand and simplify: 3(2a − 5) − 2(a + 4) + 7a.
题目：展开并化简：3(2a − 5) − 2(a + 4) + 7a。
Expand each bracket: 3(2a − 5) = 6a − 15, and −2(a + 4) = −2a − 8. Write the expression: 6a − 15 − 2a − 8 + 7a. Group like terms: (6a − 2a + 7a) = 11a; constants: −15 − 8 = −23. Final simplified expression: 11a − 23.
分别展开各项：3(2a − 5) = 6a − 15，−2(a + 4) = −2a − 8。表达式变为：6a − 15 − 2a − 8 + 7a。合并同类项：含 a 的项为 (6a − 2a + 7a) = 11a；常数项为 −15 − 8 = −23。化简结果：11a − 23。
5. Solving Linear Equations | 解一元线性方程
Question: Solve the equation: (5y − 3)/2 = 4y + 1.
题目：解方程：(5y − 3)/2 = 4y + 1。
Multiply both sides by 2 to eliminate the denominator: 5y − 3 = 2(4y + 1) → 5y − 3 = 8y + 2. Bring y terms to one side: 5y − 8y = 2 + 3 → −3y = 5. Divide by −3: y = −5/3. Check by substitution: left side (5(−5/3) − 3)/2 = (−25/3 − 9/3)/2 = (−34/3)/2 = −17/3; right side 4(−5/3) + 1 = −20/3 + 3/3 = −17/3. Both sides match.
等式两边同时乘以 2 消去分母：5y − 3 = 2(4y + 1) → 5y − 3 = 8y + 2。将含 y 的项移到一边：5y − 8y = 2 + 3 → −3y = 5。两边除以 −3，得 y = −5/3。代入检验：左边 (5(−5/3) − 3)/2 = (−25/3 − 9/3)/2 = (−34/3)/2 = −17/3；右边 4(−5/3) + 1 = −20/3 + 3/3 = −17/3。左右相等。
6. Sequences: Finding the nth Term | 数列：求第 n 项公式
Question: The first three terms of a sequence are 5, 11, 17, 23. Write an expression for the nth term. Is 203 a term in this sequence? Show your reasoning.
题目：某数列的前四项为 5, 11, 17, 23。写出第 n 项的表达式。判断 203 是否是该数列的一项，并说明理由。
The sequence goes up by 6 each time, so it is linear with common difference d = 6. The nth term formula has the form an + b. When n = 1, a(1) + b = 6 + b = 5, so b = −1. Thus the nth term is 6n − 1. Check: n = 2 gives 12 − 1 = 11, correct. To test if 203 is a term, set 6n − 1 = 203 → 6n = 204 → n = 34. Since 34 is a positive integer, 203 is the 34th term.
数列每次增加 6，为公差 d = 6 的线性数列。第 n 项公式形式为 an + b。当 n = 1，a(1) + b = 6 + b = 5，得 b = −1。因此第 n 项为 6n − 1。验证：n = 2 时得 12 − 1 = 11，正确。判断 203：令 6n − 1 = 203 → 6n = 204 → n = 34。由于 34 是正整数，故 203 是第 34 项。
7. Geometry: Angles in Polygons | 几何：多边形的内角
Question: A regular polygon has interior angles of 156°. How many sides does it have?
题目：一个正多边形的每个内角为 156°。求它的边数。
The sum of interior angles of an n‑sided polygon is (n − 2) × 180°. For a regular polygon each interior angle = (n − 2) × 180° / n. Set this equal to 156°: (n − 2) × 180 / n = 156. Multiply both sides by n: 180(n − 2) = 156n → 180n − 360 = 156n. Subtract 156n: 24n = 360 → n = 15. The polygon has 15 sides. Alternatively, use the exterior angle: 180° − 156° = 24°. Sum of exterior angles is 360°, so number of sides = 360 / 24 = 15.
n 边形内角和为 (n − 2) × 180°。正多边形每个内角等于 (n − 2) × 180° / n。令其等于 156°：(n − 2) × 180 / n = 156。两边同乘 n：180(n − 2) = 156n → 180n − 360 = 156n。移项：24n = 360 → n = 15。该多边形有 15 条边。还可以用外角计算：外角 = 180° − 156° = 24°，外角和为 360°，因此边数 = 360 / 24 = 15。
8. Area and Perimeter of Compound Shapes | 组合图形的面积与周长
Question: The diagram shows an L‑shape formed by two rectangles. The larger rectangle measures 10 cm by 6 cm and has a smaller rectangle of 3 cm by 4 cm cut from one corner. Calculate the total perimeter and the shaded area of the L‑shape.
题目：下图为两个长方形组成的 L 形。大长方形长 10 cm、宽 6 cm，从一角切掉一个 3 cm × 4 cm 的小长方形。求该 L 形的总周长和阴影部分面积。
Area: total large rectangle area = 10 × 6 = 60 cm². Cut‑out area = 3 × 4 = 12 cm². Remaining area = 60 − 12 = 48 cm². Perimeter: draw the shape and trace all outer edges. The large rectangle perimeter would be 2 × (10 + 6) = 32 cm. Removing the corner creates new edges: one horizontal 4 cm and one vertical 3 cm, but we lose 4 cm and 3 cm of the original outline. Effectively the perimeter becomes 32 − (4 + 3) + (4 + 3) = 32 cm. Actually the overall boundary length remains unchanged because the cut‑out replaces removed edges of the same total length. So perimeter is still 32 cm.
面积：大长方形面积 = 10 × 6 = 60 cm²。切掉的面积为 3 × 4 = 12 cm²。剩余面积 = 60 − 12 = 48 cm²。周长：画出图形，沿外边界测量。大长方形周长为 2 × (10 + 6) = 32 cm。切掉一角后增加了水平边 4 cm 和竖直边 3 cm，但减少了原外轮廓中的 4 cm 和 3 cm。因此周长变为 32 − (4 + 3) + (4 + 3) = 32 cm。实际上边界总长度不变，因为切掉的部分被相同长度的新边替代。故周长仍为 32 cm。
9. Statistics: Mean from a Frequency Table | 统计：根据频数表求平均数
Question: The frequency table shows the number of books read by 25 students last month. Estimate the mean number of books read.
| Books read | Frequency |
|---|---|
| 0–4 | 6 |
| 5–9 | 10 |
| 10–14 | 7 |
| 15–19 | 2 |
题目：频数表显示了上月 25 名学生的阅读书本数量。估算阅读书本数量的平均数。
| 书本数量 | 频数 |
|---|---|
| 0–4 | 6 |
| 5–9 | 10 |
| 10–14 | 7 |
| 15–19 | 2 |
Use the midpoint of each interval: (0+4)/2 = 2, (5+9)/2 = 7, (10+14)/2 = 12, (15+19)/2 = 17. Multiply each midpoint by its frequency: 2 × 6 = 12; 7 × 10 = 70; 12 × 7 = 84; 17 × 2 = 34. Sum of these products = 12 + 70 + 84 + 34 = 200. Total frequency = 25. Estimated mean = 200 ÷ 25 = 8 books.
取每个区间的中点值：0–4 的中点 = 2，5–9 的中点 = 7，10–14 的中点 = 12，15–19 的中点 = 17。各中点值乘以对应频数：2 × 6 = 12；7 × 10 = 70；12 × 7 = 84；17 × 2 = 34。乘积总和 = 200。总频数 = 25。估算平均数 = 200 ÷ 25 = 8 本。
10. Probability: Combined Events and Expected Outcomes | 概率：组合事件与期望次数
Question: A fair six‑sided dice is rolled and a fair coin is tossed. Calculate the probability of getting a number greater than 4 on the dice and heads on the coin. If the experiment is repeated 120 times, how many times would you expect both conditions to occur?
题目：投掷一枚均匀的六面骰子和一枚均匀硬币。求骰子点数大于 4 且硬币正面朝上的概率。如果该试验重复 120 次，预期两者同时发生的次数是多少？
P(greater than 4 on dice) = P(5 or 6) = 2/6 = 1/3. P(heads) = 1/2. Since the dice and coin are independent, multiply: P(both) = (1/3) × (1/2) = 1/6. Expected number in 120 trials = 120 × (1/6) = 20. Thus you would expect this outcome about 20 times.
掷骰子点数大于 4 的概率 = P(5 或 6) = 2/6 = 1/3。硬币正面概率 = 1/2。由于骰子和硬币相互独立，相乘得：P(同时发生) = (1/3) × (1/2) = 1/6。120 次试验中期望次数 = 120 × (1/6) = 20。因此可预期大约出现 20 次。
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