📚 Year 8 OCR Maths: Unit Test Mock Paper Walkthrough | 8年级OCR数学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 8 OCR Maths unit test mock paper focused on algebra and equations. This paper covers linear equations, expanding brackets, factorising, sequences, coordinates, direct proportion, substitution, inequalities, and problem-solving with equations. Each question is followed by a step-by-step solution with key tips to help you master the methods and avoid common mistakes. Use this resource to check your understanding and build confidence for your actual unit test.
欢迎收看这份针对8年级OCR数学单元测试模拟卷的详细解析，重点涵盖代数与方程。本试卷包括线性方程、去括号、因式分解、数列、坐标、正比例、代入求值、不等式以及用方程解应用题。每个题目后都配有分步解析和关键提示，帮助你掌握方法并避免常见错误。使用这份资源检查你的理解，为真正的单元测试建立信心。
The mock paper contains the following 10 questions, which we will work through one by one:
模拟卷包含以下10道题，我们将逐一解析：
- 1. Solve 3x + 5 = 20
- 2. Solve 4(2y – 3) = 28
- 3. Expand and simplify 2(3a + 4) – (a – 5)
- 4. Factorise 12b – 8
- 5. Find the next term and nth term of the sequence: 5, 11, 17, 23, …
- 6. Write the equation of the straight line passing through (1,4) with gradient 2.
- 7. If 3 apples cost 90p, how much do 7 apples cost? Give your answer in pounds.
- 8. Evaluate x² – 3x + 7 when x = –2.
- 9. Solve 2n – 3 ≤ 7 and represent the solution on a number line.
- 10. The length of a rectangle is three times its width. The perimeter is 56 cm. Find the length and width.
Note: Exchange rate is 100p = £1.
1. Question 1: Solving a Simple Linear Equation | 第1题：解简单线性方程
Solve 3x + 5 = 20.
. Always check your answer by substituting back: 3(5) + 5 = 15 + 5 = 20, which is correct.
解方程 3x + 5 = 20。
要孤立 x，首先两边同时减去5：3x = 15。然后两边同时除以3，得 x = 5。记得代回检验：3×5 + 5 = 15 + 5 = 20，正确。
2. Question 2: Solving an Equation with Brackets | 第2题：含括号的方程
Solve 4(2y – 3) = 28.
Method 1: Expand the bracket first: 8y – 12 to both sides: 8y = 40. Divide by 8: y = 5. Method 2: Divide both sides by 4 to give 2y – 3 = 7, then add 3: 2y = 10, so y = 5. Both methods are valid and yield the same answer.
解方程 4(2y – 3) = 28。
方法1：先展开括号：8y – 12 = 28。两边加12：8y = 40。除以8得 y = 5。方法2：两边同时除以4，得到2y – 3 = 7，再加3：2y = 10，y = 5。两种方法都可行，结果相同。
3. Question 3: Expanding and Simplifying | 第3题：展开并化简
Expand and simplify 2(3a + 4) – (a – 5).
xpand the first bracket: 2 × 3a = 6a, 2 × 4 = 8, giving 6a + 8. Expand the second bracket: –(a – 5) = –a + 5 (careful with the minus sign). Combine like terms: 6a – a = 5a, and 8 + 5 = 13. The simplified expression is 5a + 13.
展开并化简 2(3a + 4) – (a – 5)。
展开第一个括号：2×3a = 6a，2×4 = 8，得 6a + 8。展开第二个括号：–(a – 5) = –a + 5（注意符号变化）。合并同类项：6a – a = 5a，8 + 5 = 13。化简结果为 5a + 13。
4. Question 4: Factorising a Linear Expression | 第4题：因式分解一次式
Factorise 12b – 8 fully.
d the greatest common factor (GCF) of 12 and 8, which is 4. Write the expression as 4 × 3b – 4 × 2. Factor out the 4: 4(3b – 2). Check by expanding: 4(3b – 2) = 12b – 8.
因式分解 12b – 8。
找出12和8的最大公因数（GCF）为4。将表达式写成 4×3b – 4×2。提取公因数4：4(3b – 2)。展开检验：4(3b – 2) = 12b – 8。
5. Question 5: Sequences and nth Term | 第5题：数列与第n项
The sequence begins: 5, 11, 17, 23, …
a) Find the next term. The difference between terms is +6. Next term: 23 + 6 = 29.
term. Since the common difference is 6, the nth term is of the form 6n ± c. For n=1, 6(1) – 1 = 5. So nth term = 6n – 1.
已知数列：5, 11, 17, 23, …
a) 求下一项。相邻项的差为+6。下一项：23 + 6 = 29。
b) 求第n项。因为公差为6，第n项形式为 6n ± 常数。当 n=1 时，6×1 – 1 = 5。所以第n项为 6n – 1。
6. Question 6: Equation of a Straight Line | 第6题：直线方程
Find the equation of the line with gradient 2 passing through the point (1, 4).
= 2 + c → c = 2. So the equation is y = 2x + 2. Alternatively, use point-gradient form: y – 4 = 2(x – 1) → y = 2x + 2.
求过点(1, 4)且斜率为2的直线方程。
使用斜截式 y = mx + c，其中 m=2。代入已知点：4 = 2×1 + c → 4 = 2 + c → c = 2。因此直线方程为 y = 2x + 2。也可用点斜式：y – 4 = 2(x – 1) → y = 2x + 2。
7. Question 7: Direct Proportion | 第7题：正比例问题
3 apples cost 90p. Work out the cost of 7 apples. Give your answer in pounds (£).
pple. Then cost for 7 apples: 7 × 30p = 210p. Convert to pounds by dividing by 100: 210p = £2.10. So 7 apples cost £ Always pay attention to units; you may also set up a proportion: (cost/quantity) constant.
3个苹果售价90p。求7个苹果的价格，结果以英镑(£)表示。
先求一个苹果价格：90p ÷ 3 = 30p。7个苹果价格：7 × 30p = 210p。转换为英镑：210p ÷ 100 = £2.10。因此7个苹果要£2.10。注意单位换算；也可以用比例关系：价格/数量为常数。
8. Question 8: Substituting Negative Numbers | 第8题：代入负数求值
Evaluate x² – 3x + 7 when x = –2.
Replace x with –2 carefully, using brackets where helpful: (–2)² – 3(–2) + 7. Calculate: (–2)² = 4, –3(–2) = +6, plus 7 gives 4 + 6 + 7 = 17. Common mistake: writing –2² as –4 instead of 4; always square the entire negative number as (–2)².
当x = –2时，求 x² – 3x + 7 的值。
将x替换为–2，建议加括号：(–2)² – 3(–2) + 7。计算：(–2)² = 4，–3(–2) = +6，再加7得 4 + 6 + 7 = 17。常见错误：把 –2² 当作 –4 而不是 4；一定要将整个负数平方，即 (–2)²。
9. Question 9: Solving an Inequality and Number Line | 第9题：解不等式并在数轴上表示
Solve 2n – 3 ≤ 7 and show the solution on a number line.
cludes 5) and shade the line to the left to represent all numbers less than or equal to 5.
解不等式 2n – 3 ≤ 7 并在数轴上表示解集。
两边加3：2n ≤ 10。除以2：n ≤ 5。在数轴上，在n=5处画一个实心圆（因为≤包含5），并向左画阴影线，表示所有小于或等于5的数。
10. Question 10: Forming and Solving an Equation | 第10题：列方程解应用题
The length of a rectangle is three times its width. The perimeter is 56 cm. Find the length and width.
= 2(3w + w) = 2(4w) = 8w. Set 8w = 56, so w = 56 ÷ 8 = 7 cm. Then length = 3 × 7 = 21 cm.
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