KS3 OCR Advanced Mathematics: Unit Test Mock Paper Analysis | KS3 OCR 进阶数学：单元测试模拟卷解析

📚 KS3 OCR Advanced Mathematics: Unit Test Mock Paper Analysis | KS3 OCR 进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a sample Unit Test for the OCR KS3 Advanced Mathematics course. Each question is broken down with step-by-step solutions and key learning points, designed to help students consolidate their understanding and improve exam technique.

本文详细解析了一份针对OCR KS3进阶数学课程的单元测试模拟卷。每道题目都配有逐步解答和关键知识点，旨在帮助学生巩固理解并提升考试技巧。

1. Overview of the Mock Paper | 模拟卷概览

The mock paper consists of 10 questions covering core KS3 topics: algebra, number, geometry, data handling, and probability. It is designed to be completed in 45 minutes, with marks allocated according to difficulty. The questions range from straightforward calculations to multi-step problem solving, reflecting the OCR assessment style.

本模拟卷包含10道题，覆盖核心KS3知识点：代数、数字、几何、数据处理和概率。试卷设计为45分钟完成，分数按难度分配。题目从直接计算到多步问题解决，反映了OCR的评估风格。

2. Question 1: Solving Linear Equations | 第1题：解线性方程

Question: Solve the equation 3x – 7 = 2x + 5.

题目：解方程 3x – 7 = 2x + 5。

Solution: Start by collecting like terms. Subtract 2x from both sides: 3x – 7 – 2x = 2x + 5 – 2x, giving x – 7 = 5. Then add 7 to both sides: x – 7 + 7 = 5 + 7, so x = 12.

解析：首先合并同类项。两边同时减去2x：3x – 7 – 2x = 2x + 5 – 2x，得到 x – 7 = 5。然后两边加7：x – 7 + 7 = 5 + 7，因此 x = 12。

Always check your answer by substituting back: 3(12) – 7 = 36 – 7 = 29, and 2(12) + 5 = 24 + 5 = 29. The equation balances.

务必通过代入检验答案：3(12) – 7 = 36 – 7 = 29，2(12) + 5 = 24 + 5 = 29，等式成立。

3. Question 2: Operations with Fractions | 第2题：分数运算

Question: Evaluate 2½ ÷ 1⅓, giving your answer as a mixed number in its simplest form.

题目：计算 2½ ÷ 1⅓，答案以最简带分数形式表示。

Solution: Convert mixed numbers to improper fractions: 2½ = (2×2+1)/2 = 5/2; 1⅓ = (1×3+1)/3 = 4/3. Division by a fraction is equivalent to multiplying by its reciprocal: 5/2 ÷ 4/3 = 5/2 × 3/4 = (5×3)/(2×4) = 15/8. Convert back to a mixed number: 15 ÷ 8 = 1 remainder 7, so 1⁷/₈.

解析：将带分数转化为假分数：2½ = (2×2+1)/2 = 5/2；1⅓ = (1×3+1)/3 = 4/3。除以一个分数等于乘以它的倒数：5/2 ÷ 4/3 = 5/2 × 3/4 = (5×3)/(2×4) = 15/8。再转化为带分数：15 ÷ 8 = 1 余 7，所以答案为 1⁷/₈。

Remember to always simplify your final answer. Here 15 and 8 share no common factors other than 1, so 1⁷/₈ is already in simplest form.

记住最终答案一定要化简。这里15和8除了1以外没有公因数，因此1⁷/₈已经是最简形式。

4. Question 3: Percentage Increase and Decrease | 第3题：百分比增减

Question: A coat originally costs £80. In a sale, its price is reduced by 15%. After the sale, the price is increased by 10%. What is the final price?

题目：一件外套原价80英镑。促销时降价15%，促销结束后又提价10%。最终的售价是多少？

Solution: First, find the sale price: 15% of £80 = 0.15 × 80 = £12; sale price = £80 – £12 = £68. Then increase this by 10%: 10% of £68 = 0.10 × 68 = £6.80; final price = £68 + £6.80 = £74.80.

解析：首先计算促销价：80英镑的15%为 0.15 × 80 = 12英镑；促销价 = 80 – 12 = 68英镑。然后在此基础上涨价10%：68英镑的10%为 0.10 × 68 = 6.80英镑；最终价格 = 68 + 6.80 = 74.80英镑。

It is important to apply percentage changes sequentially, not simply combine them into a net change. The base changes after the first operation.

重要的是按顺序应用百分比变化，而不是简单合并为净变化。第一次运算后基数已经改变。

5. Question 4: Area and Perimeter of Composite Shapes | 第4题：组合图形的面积与周长

Question: The diagram shows an L-shaped figure formed by two rectangles. One rectangle measures 6 cm by 4 cm, and the other attached rectangle measures 3 cm by 5 cm. The longer side of the small rectangle aligns with part of the larger one. Calculate the total area and the perimeter of the shape. (Assume all angles are right angles.)

题目：下图是由两个矩形组成的L形图形。一个矩形长6 cm、宽4 cm，另一个相连的矩形长3 cm、宽5 cm。小矩形的长边与较大矩形的部分边对齐。计算该图形的总面积和周长。（所有角均为直角。）

Solution: Place the larger rectangle with vertices (0,0), (6,0), (6,4), (0,4). The smaller rectangle attaches along the right side, sharing the segment from (6,0) to (6,4), and extends to the right. Its vertices are (6,0), (9,0), (9,5), (6,5). The combined shape is an L-shape polygon. Total area = area of large rectangle + area of small rectangle = 6×4 + 3×5 = 24 + 15 = 39 cm². To find the perimeter, trace the outer boundary: start at (0,0) → (9,0) (length 9) → (9,5) (length 5) → (6,5) (length 3) → (6,4) (length 1) → (0,4) (length 6) → back to (0,0) (length 4). Perimeter = 9 + 5 + 3 + 1 + 6 + 4 = 28 cm.

解析：将大矩形顶点设为(0,0), (6,0), (6,4), (0,4)。小矩形沿右侧连接，共用从(6,0)到(6,4)的线段，并向右延伸，顶点为(6,0), (9,0), (9,5), (6,5)。组合形状为L形多边形。总面积 = 大矩形面积 + 小矩形面积 = 6×4 + 3×5 = 24 + 15 = 39 cm²。计算周长

Published by TutorHao | KS3 进阶数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）