📚 Year 8 OCR Further Mathematics: Unit Test Mock Paper Walkthrough | Year 8 OCR 进阶数学：单元测试模拟卷解析
This walkthrough takes you through a full mock paper designed to mirror the style of a Year 8 OCR Further Mathematics unit test. Each section presents a typical exam-style question, followed by a clear, step-by-step solution. The topics covered include algebra, geometry, probability, ratio and statistics — all key areas in the OCR Further Mathematics programme for Year 8.
本文详细解析一套模拟单元测试卷，题目风格与 Year 8 OCR 进阶数学考试一致。每个小节都会展示一道典型考题，并给出清晰的逐步解答。覆盖的主题包括代数、几何、概率、比例和统计，均为 Year 8 OCR 进阶数学课程的核心内容。
1. Simplifying Algebraic Expressions | 简化代数表达式
Question: Simplify 5a + 3b − 2a + 7b − 4a.
题目：化简 5a + 3b − 2a + 7b − 4a。
Start by identifying like terms. Terms containing ‘a’ are 5a, −2a and −4a. Terms containing ‘b’ are 3b and 7b.
首先识别同类项。含有 a 的项为 5a、−2a 和 −4a。含有 b 的项为 3b 和 7b。
Now group and combine the coefficients: (5 − 2 − 4)a gives −1a, which is written as −a. For the b terms, (3 + 7)b simplifies to 10b.
接着分组并合并系数：(5 − 2 − 4)a 得到 −1a，写作 −a。对于 b 项，(3 + 7)b 化简为 10b。
The final simplified expression is −a + 10b. It is conventional to write terms with a positive coefficient first if possible, but both forms are acceptable.
最终化简结果为 −a + 10b。习惯上尽可能将正系数项写在前面，但两种形式均可接受。
2. Solving Linear Equations | 解一元一次方程
Question: Solve 4(x − 3) + 2x = 5x + 8.
题目：解方程 4(x − 3) + 2x = 5x + 8。
First expand the bracket: 4 × x = 4x and 4 × (−3) = −12, so the left-hand side becomes 4x − 12 + 2x.
首先展开括号：4 × x = 4x，4 × (−3) = −12，因此左边变为 4x − 12 + 2x。
Simplify the left by combining like terms: 4x + 2x gives 6x. The equation is now 6x − 12 = 5x + 8.
合并左边的同类项：4x + 2x 得到 6x。方程现在为 6x − 12 = 5x + 8。
Subtract 5x from both sides to collect x terms together: 6x − 5x − 12 = 8, so x − 12 = 8.
两边同时减去 5x 将含 x 的项移到一边：6x − 5x − 12 = 8，即 x − 12 = 8。
Finally add 12 to both sides: x = 20. Substitute back to check: 4(20−3)+40 = 4×17+40 = 68+40=108, and 5×20+8=100+8=108.
最后两边同时加 12：x = 20。代入检验：4(20−3)+40=4×17+40=68+40=108，5×20+8=100+8=108，结果正确。
3. Inequalities on a Number Line | 数轴上的不等式
Question: Represent the inequality −2 < x ≤ 3 on a number line and list the integer values that satisfy it.
题目：在数轴上表示不等式 −2 < x ≤ 3，并列出所有满足条件的整数值。
An open circle is placed at −2 to show that x is strictly greater than −2 (not equal). A closed circle is placed at 3 to show that x can equal 3. Shade the region between them.
在 −2 处画空心圆，表示 x 严格大于 −2（不能等于）。在 3 处画实心圆，表示 x 可以等于 3。将两点之间的区域涂上阴影。
The integers that satisfy −2 < x ≤ 3 are those whole numbers strictly greater than −2 and up to 3 inclusive. So x can be −1, 0, 1, 2, 3.
满足 −2 < x ≤ 3 的整数是那些严格大于 −2 且不超过 3 的整数。因此 x 可以是 −1、0、1、2、3。
Note that −2 itself is not included because of the strict inequality sign. Students often forget to exclude the boundary when drawing open circles.
注意 −2 本身并不包含在内，因为使用了严格不等号。学生在画空心圆时经常忘记排除边界。
4. nth Term of a Linear Sequence | 线性序列的第 n 项
Question: The first four terms of a sequence are 7, 11, 15, 19. Find the nth term and use it to calculate the 50th term.
题目：某序列的前四项为 7、11、15、19。求第 n 项公式，并计算第 50 项。
Determine the common difference by subtracting consecutive terms: 11 − 7 = 4, 15 − 11 = 4. This tells us the coefficient of n is 4, so the nth term is of the form 4n + c.
通过相邻项相减求公差：11 − 7 = 4，15 − 11 = 4。这说明 n 的系数为 4，因此第 n 项的形式为 4n + c。
To find c, use the first term (n = 1). If 4(1) + c = 7, then 4 + c = 7, so c = 3. Hence the nth term is 4n + 3.
为求 c，代入第一项（n = 1）。若 4(1) + c = 7，则 4 + c = 7，得 c = 3。因此第 n 项公式为 4n + 3。
For the 50th term, substitute n = 50: 4(50) + 3 = 200 + 3 = 203. Always verify with an early term: n=2 gives 4×2+3=11, which matches.
第 50 项需代入 n = 50：4(50) + 3 = 200 + 3 = 203。务必用早期项验证：n=2 得 4×2+3=11，与序列一致。
5. Angles in Polygons | 多边形内角
Question: A regular polygon has interior angle 156°. How many sides does it have?
题目：一个正多边形的每个内角为 156°。它有多少条边？
Use the relationship between interior and exterior angles. The sum of an interior and its corresponding exterior angle is 180°, so exterior angle = 180° − 156° = 24°.
利用内角和外角的关系。任一个内角与其相邻外角之和为 180°，因此外角 = 180° − 156° = 24°。
For any regular polygon, the sum of all exterior angles is 360°. If each exterior angle is 24°, the number of sides n satisfies 24n = 360. Divide both sides by 24: n = 360 ÷ 24 = 15.
任意正多边形的外角和恒为 360°。若每个外角为 24°，则边数 n 满足 24n = 360。两边除以 24：n = 360 ÷ 24 = 15。
Thus the polygon has 15 sides — a pentadecagon. This method avoids memorising the interior angle formula and reduces arithmetic errors.
因此该多边形有 15 条边，即十五边形。这种方法无需死记内角和公式，并能减少计算错误。
6. Probability of Combined Events | 组合事件概率
Question: A fair six-sided die is rolled and a fair coin is flipped. What is the probability of rolling a number greater than 4 and getting a head?
题目：同时掷一枚公平的六面骰子和抛一枚公平的硬币。求掷出大于 4 的点数且硬币正面朝上的概率。
The two events are independent. The probability of rolling a number greater than 4 (i.e., 5 or 6) is 2 out of 6, which simplifies to 1/3.
两个事件相互独立。掷出大于 4 的点数（即 5 或 6）的概率为 2/6，化简为 1/3。
The probability of getting a head is 1/2. Multiply the separate probabilities: P(greater than 4 AND head) = (1/3) × (1/2) = 1/6.
得到正面的概率为 1/2。将两个概率相乘：P(大于 4 且正面) = (1/3) × (1/2) = 1/6。
We can verify by listing outcomes: possible favourable pairs are (5,H) and (6,H) out of 12 equally likely outcomes (6 × 2). 2/12 = 1/6, confirming the result.
我们可以通过列举结果来验证：有利组合为 (5,H) 和 (6,H)，总共 12 种等可能结果（6×2）。2/12 = 1/6，与结果一致。
7. Ratio and Proportion | 比例与比率
Question: The ratio of boys to girls in a school is 5:4. There are 270 boys. How many students are there in total?
题目：某校男生与女生的数量比为 5:4。已知男生有 270 人，求全校学生总数。
The ratio 5:4 means for every 5 boys there are 4 girls. The number of ‘parts’ for boys is 5, which corresponds to 270. So 1 part = 270 ÷ 5 = 54.
比例 5:4 表示每 5 名男生对应 4 名女生。男生的“份数”为 5，对应 270 人。因此 1 份 = 270 ÷ 5 = 54。
Girls correspond to 4 parts: 4 × 54 = 216. Total students = boys + girls = 270 + 216 = 486.
女生对应 4 份：4 × 54 = 216。总学生数 = 男生 + 女生 = 270 + 216 = 486。
An alternative method is to set up the proportion 5/(5+4) = 270/total, but the unitary method keeps the working clear for Year 8 students.
另一种方法是列出比例式 5/(5+4) = 270/总数，但“每份计算法”对 Year 8 学生而言更加清晰。
8. Simple Interest and Compound Interest | 单利与复利
Question: £800 is invested at 3% per annum simple interest for 4 years. Calculate the total amount after 4 years.
题目：将 800 英镑以年利率 3% 的单利投资 4 年。计算 4 年后的总金额。
Simple interest means the interest earned each year is based only on the original principal. Interest per year = 3% of £800 = 0.03 × 800 = £24.
单利意味着每年的利息仅基于原始本金计算。每年利息 = 800 英镑的 3% = 0.03 × 800 = 24 英镑。
Over 4 years, total interest = 4 × £24 = £96. The total amount after 4 years = principal + interest = £800 + £96 = £896.
4 年内总利息 = 4 × 24 英镑 = 96 英镑。4 年后的总金额 = 本金 + 利息 = 800 英镑 + 96 英镑 = 896 英镑。
If the same principal were invested at compound interest, the calculation would involve multiplying by (1+0.03) each year, yielding a slightly higher amount due to interest on interest.
若以复利计算，则需要每年乘以 (1+0.03)，由于“利滚利”效应，最终金额会略高一些。
9. Pythagoras’ Theorem | 勾股定理
Question: A right-angled triangle has legs of length 6 cm and 8 cm. Find the length of the hypotenuse and the area of the triangle.
题目：一个直角三角形的两条直角边分别为 6 cm 和 8 cm。求斜边长和三角形面积。
By Pythagoras, hypotenuse² = 6² + 8². Calculate: 6² = 36, 8² = 64. Sum = 100.
根据勾股定理，斜边² = 6² + 8²。计算：6² = 36，8² = 64。和为 100。
Take the square root: hypotenuse = √100 = 10 cm. The hypotenuse is the longest side, always opposite the right angle.
取平方根：斜边 = √100 = 10 cm。斜边是最长边，总是直角所对的边。
Area of a triangle = ½ × base × height. Here, the legs serve as base and height: area = ½ × 6 × 8 = 24 cm².
三角形面积 = ½ × 底 × 高。直角边可互为底和高：面积 = ½ × 6 × 8 = 24 cm²。
Some students confuse area with perimeter. Keep them separate — perimeter would be 6+8+10 = 24 cm, which coincidentally equals the area number but with different units.
部分学生会将面积与周长混淆。应加以区分——周长为 6+8+10 = 24 cm，数值巧合地等于面积值，但单位不同。
10. Scatter Graphs and Correlation | 散点图与相关性
Question: The table shows test scores in Maths and Science for five students: (12,14), (15,18), (18,22), (21,24), (24,28). Plot a scatter graph and describe the correlation.
题目：下表列出五名学生的数学与科学成绩：(12,14), (15,18), (18,22), (21,24), (24,28)。绘制散点图并描述相关性。
Plot Maths scores on the horizontal axis (x-axis) and Science scores on the vertical axis (y-axis). Each pair becomes a point.
将数学成绩标在横轴（x 轴），科学成绩标在纵轴（y 轴）。每一对对应一个点。
Looking at the trend, as Maths scores increase, Science scores also increase. The points lie roughly along a straight line rising to the right.
观察趋势，随着数学成绩提高，科学成绩也相应提高。这些点大致沿一条向右上升的直线分布。
This indicates a strong positive correlation. ‘Positive’ means the variables move in the same direction; ‘strong’ means the points are closely clustered around an imagined line.
这表明强正相关。“正”意味着两个变量同方向变化；“强”意味着点紧密聚集在一条想象中的直线附近。
We can also draw an approximate line of best fit to make predictions, but be cautious when extrapolating beyond the data range.
我们还可以画一条近似的拟合直线用于预测，但在数据范围外进行外推时需格外谨慎。
11. Solving Linear Inequalities | 解线性不等式
Question: Solve 3(2x − 1) ≤ 5x + 7 and represent the solution on a number line.
题目：解不等式 3(2x − 1) ≤ 5x + 7，并在数轴上表示解集。
Expand the left side: 3×2x = 6x, 3×(−1)=−3, giving 6x − 3 ≤ 5x + 7.
展开左边：3×2x = 6x，3×(−1)=−3，得到 6x − 3 ≤ 5x + 7。
Subtract 5x from both sides: 6x − 5x − 3 ≤ 7, so x − 3 ≤ 7.
两边减去 5x：6x − 5x − 3 ≤ 7，即 x − 3 ≤ 7。
Add 3 to both sides: x ≤ 10. The solution set includes all numbers less than or equal to 10.
两边加 3：x ≤ 10。解集包含所有小于或等于 10 的数。
On a number line, place a closed circle at 10 and shade to the left. A closed circle is used because of the ‘or equal to’ symbol.
在数轴上，在 10 处画实心圆并向左涂阴影。由于含有“或等于”符号，需使用实心圆。
12. Reflections and Key Takeaways | 总结与关键要点
Final tips: Always show clear working — even for simple calculations. Check that negative signs are carried correctly when simplifying expressions. When dealing with inequalities, remember to reverse the inequality sign if multiplying or dividing by a negative number, though that did not arise in this paper.
最后提示：计算步骤务必书写清晰——即使是简单计算也不例外。化简表达式时要确保负号正确无误。处理不等式时，若乘以或除以负数需反转不等号方向，虽然本套试卷未涉及此类情况。
Practise drawing graphs and number lines accurately, as visual representations often earn method marks even if the final answer is slightly off. Use the markscheme-style checking habit: substitute your solution back into the original equation or inequality whenever possible.
平时要练习准确绘制图形和数轴，因为即便最终答案稍有偏差，视觉呈现通常也能获得方法分。养成对照评分标准的检查习惯：尽可能将解代入原方程或原不等式进行验证。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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