📚 Year 11 Eduqas Maths: Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析
A unit test mock paper is one of the most effective tools for consolidating your knowledge and identifying exactly which topics need more attention before the real exam. This walkthrough breaks down a typical Year 11 Eduqas Mathematics mock paper question by question, explaining the required methods, common pitfalls, and how marks are awarded. By working through each problem with us, you will build confidence in your problem-solving routine and learn how to present your working clearly to maximise your score.
单元测试模拟卷是巩固知识、精准定位薄弱点的最佳工具之一。本文详细解析一份典型的 Year 11 Eduqas 数学模拟卷，逐题讲解解题方法、常见错误及评分要点。跟随我们的步伐逐一攻克，你将建立起稳定的解题思路，并掌握清晰展示步骤以争取高分的技巧。
1. Paper Overview and Time Management | 试卷概览与时间管理
The mock paper consists of two sections: a non-calculator section worth 50 marks and a calculator section also worth 50 marks. You will have 60 minutes for each section. Before you start, quickly scan the entire paper to identify the question types and their mark allocations. Aim to spend roughly 1 minute per mark, leaving some time at the end for checking.
模拟卷分为不能使用计算器的部分（50分）和可使用计算器的部分（50分），每部分时间均为60分钟。作答前，快速浏览整份试卷，判断题型与分值分布。大致按“1分钟1分”的节奏推进，并预留末尾检查时间。
Always read the question prompt at least twice before writing. Underline command words like ‘simplify’, ‘factorise’, ‘solve’, ‘show that’, or ‘estimate’. Annotate diagrams immediately and write down any formula you think you might need at the top of the page to avoid forgetting under pressure.
动笔前至少读题两遍。圈出“化简”“因式分解”“求解”“证明”“估算”等指令词。立刻在图上标注信息，并在草稿区记下可能需要的公式，以防紧张时遗忘。
2. Problem 1: Solving Linear Equations | 问题1：解线性方程
Question: Solve 2x + 5 = 3x – 7. (3 marks)
题目：解方程 2x + 5 = 3x – 7。（3分）
Step 1: Bring the variable terms to one side. Subtract 2x from both sides: 5 = x – 7.
第一步：将含变量的项移到一边。两边同时减去2x，得 5 = x – 7。
Step 2: Isolate x. Add 7 to both sides: 12 = x.
第二步：分离x。两边同时加7，得 12 = x。
Answer: x = 12. Always write your final answer clearly and consider substituting back into the original equation to check: 2(12) + 5 = 24 + 5 = 29; 3(12) – 7 = 36 – 7 = 29. Both sides are equal, so the solution is correct.
答案：x = 12。务必清晰写出最终答案，并考虑代回原方程检验：2(12) + 5 = 24 + 5 = 29；3(12) – 7 = 36 – 7 = 29，左右相等，解法正确。
3. Problem 2: Factorising Quadratics | 问题2：因式分解二次三项式
Question: Factorise x² – 5x – 14. (2 marks)
题目：因式分解 x² – 5x – 14。（2分）
We need two numbers that multiply to give the constant term -14 and add to give the coefficient of x, which is -5. List the factor pairs of -14: (-1, 14), (1, -14), (-2, 7), (2, -7). The pair with a sum of -5 is 2 and -7.
我们需要找出两个数，它们的乘积等于常数项 -14，且和等于 x 的系数 -5。列出 -14 的因数对：(-1, 14), (1, -14), (-2, 7), (2, -7)。其中和为 -5 的是 2 和 -7。
Therefore, the factorised form is (x + 2)(x – 7). Expand to verify: x² – 7x + 2x – 14 = x² – 5x – 14. The +2 and -7 directly form the brackets.
因此，因式分解结果为 (x + 2)(x – 7)。展开检验：x² – 7x + 2x – 14 = x² – 5x – 14。直接使用 +2 和 -7 构建括号即可。
4. Problem 3: Operations with Fractions | 问题3：分数运算
Question: Evaluate 3/4 + 2/5, giving your answer as a mixed number in its simplest form. (2 marks)
题目：计算 3/4 + 2/5，结果用最简带分数表示。（2分）
Find a common denominator. The lowest common multiple of 4 and 5 is 20. Convert: 3/4 = 15/20 and 2/5 = 8/20. Add: 15/20 + 8/20 = 23/20.
先确定公分母。4和5的最小公倍数是20。转换：3/4 = 15/20，2/5 = 8/20。相加得 15/20 + 8/20 = 23/20。
23/20 is an improper fraction. Convert to a mixed number: 20 goes into 23 once with a remainder of 3. So, 23/20 = 1 3/20. This fraction is already in its simplest form because 3 and 20 share no common factors other than 1.
23/20 是假分数。化为带分数：20除23得1余3，因此 23/20 = 1 3/20。3与20互质，已是最简形式。
5. Problem 4: Percentages and Proportion | 问题4：百分比与比例
Question: A jacket originally costing £80 is reduced by 15% in a sale. The sale price is then increased by 15% after the sale. Is the final price higher, lower, or the same as the original? Justify your answer. (3 marks)
题目：一件夹克原价£80，促销降价15%。促销结束后价格又上涨15%。最终价格与原价相比是升高、降低还是不变？请证明。（3分）
Sale price: 85% of £80 = 0.85 × 80 = £68. After the increase: 115% of £68 = 1.15 × 68 = £78.20.
促销价：£80的85% = 0.85 × 80 = £68。涨价后：£68的115% = 1.15 × 68 = £78.20。
The final price is £78.20, which is lower than £80. The reason is that the 15% increase is applied to a smaller amount (£68), so the gain is smaller than the original discount. Percentages on different bases are not symmetric.
最终价格为£78.20，低于原价£80。因为15%的加幅作用于较小的基数（£68），其增加额小于最初的折扣额。在不同基数上施加相同百分比并不对称。
6. Problem 5: Equation of a Straight Line | 问题5：直线方程
Question: Find the equation of the line that passes through the points (2, 5) and (4, 11). Give your answer in the form y = mx + c. (4 marks)
题目：求经过点(2, 5)和(4, 11)的直线方程，结果以 y = mx + c 形式表示。（4分）
Step 1: Calculate the gradient m. m = (y₂ – y₁) / (x₂ – x₁) = (11 – 5) / (4 – 2) = 6 / 2 = 3.
第一步：计算斜率 m。m = (y₂ – y₁) ÷ (x₂ – x₁) = (11 – 5) ÷ (4 – 2) = 6 ÷ 2 = 3。
Step 2: Substitute one point into y = 3x + c. Using (2, 5): 5 = 3(2) + c → 5 = 6 + c → c = -1.
第二步：将一点代入 y = 3x + c。用(2,5)：5 = 3(2) + c → 5 = 6 + c → c = -1。
Equation: y = 3x – 1. Check with the other point: when x = 4, y = 3(4) – 1 = 12 – 1 = 11. Both points satisfy the equation.
方程为 y = 3x – 1。用另一点检验：x=4时，y=3(4)-1=12-1=11，符合。
7. Problem 6: Area and Perimeter of Compound Shapes | 问题6：复合图形的面积与周长
Question: The shape is formed by a rectangle and a semicircle on one end. The rectangle measures 10 cm by 6 cm, and the semicircle has a diameter of 6 cm. Calculate the total area and the perimeter of the shape, giving both answers to 3 significant figures. (5 marks)
题目：图形由一个矩形及一端的一个半圆组成。矩形尺寸为10 cm × 6 cm，半圆直径为6 cm。计算该图形的总面积和周长，答案均保留三位有效数字。（5分）
Area of rectangle: 10 × 6 = 60 cm². Radius of semicircle = 3 cm. Area of semicircle = (1/2) × π × r² = 0.5 × π × 9 ≈ 14.137… cm². Total area ≈ 60 + 14.137 = 74.137… ≈ 74.1 cm² (3 s.f.).
矩形面积：10×6=60 cm²。半圆半径=3 cm。半圆面积=(1/2) × π × 3² = 0.5 × π × 9 ≈ 14.137… cm²。总面积≈60+14.137=74.137…≈74.1 cm²（三位有效数字）。
Perimeter: The perimeter includes three sides of the rectangle (two lengths of 10 cm and one side of 6 cm, but the side with the semicircle is not included) plus the curved length of the semicircle. The curved part = (1/2) × circumference = 0.5 × π × 6 = 3π ≈ 9.4248 cm. Straight parts: 10 + 10 + 6 = 26 cm. Total perimeter ≈ 26 + 9.4248 = 35.4248 ≈ 35.4 cm (3 s.f.).
周长：周长包含矩形的三个边（两个10 cm长边和一个6 cm边，附带半圆的边不计入直线部分）加上半圆弧长。弧长=(1/2) × 圆周长 = 0.5 × π × 6 = 3π ≈ 9.4248 cm。直线部分：10+10+6=26 cm。总周长≈26+9.4248=35.4248≈35.4 cm（三位有效数字）。
8. Problem 7: Probability and Tree Diagrams | 问题7：概率与树状图
Question: A bag contains 3 red sweets and 5 blue sweets. A sweet is taken at random, its colour noted, and then it is not replaced. A second sweet is then taken. Draw a tree diagram and find the probability that both sweets taken are the same colour. (4 marks)
题目：袋中有3颗红色糖果和5颗蓝色糖果。随机取出一颗，记录颜色后不放回，再取第二颗。画出树状图，并求两颗糖果颜色相同的概率。（4分）
First pick: P(Red) = 3/8, P(Blue) = 5/8. After one red is taken, 2 red and 5 blue remain: P(Red | Red) = 2/7, P(Blue | Red) = 5/7. After one blue is taken, 3 red and 4 blue remain: P(Red | Blue) = 3/7, P(Blue | Blue) = 4/7.
第一次抽取：P(红)=3/8，P(蓝)=5/8。取出一颗红后，余2红5蓝：P(红|红)=2/7，P(蓝|红)=5/7。取出一颗蓝后，余3红4蓝：P(红|蓝)=3/7，P(蓝|蓝)=4/7。
Same colour means both red or both blue. P(both red) = (3/8) × (2/7) = 6/56 = 3/28. P(both blue) = (5/8) × (4/7) = 20/56 = 5/14. Total probability = 3/28 + 5/14 = 3/28 + 10/28 = 13/28.
相同颜色意为双双红或双双蓝。P(双双红)= (3/8)×(2/7)=6/56=3/28。P(双双蓝)= (5/8)×(4/7)=20/56=5/14。总概率= 3/28 + 5/14 = 3/28 + 10/28 = 13/28。
Simplify or leave as fraction; 13/28 is already in simplest form. A common mistake is to treat the second pick as independent; tree diagrams help avoid that.
可保留分数形式，13/28已最简。常见错误是把第二次抽取当作独立事件处理，树状图有助于避免这一错误。
9. Problem 8: Interpreting Statistical Diagrams | 问题8：解读统计图表
Question: A cumulative frequency graph shows the masses of 80 apples. Use the graph to estimate the median and the interquartile range (IQR). (3 marks)
题目：一份累积频数图显示了80个苹果的质量。利用该图估算中位数和四分位距（IQR）。（3分）
To find the median, go to half of the total frequency (80/2 = 40) on the cumulative frequency axis, draw a horizontal line to the curve, then drop a vertical line to the mass axis. Suppose this gives 142 g. For the lower quartile (Q1), use 1/4 of total = 20, read off ≈ 128 g. For the upper quartile (Q3), use 3/4 of total = 60, read off ≈ 158 g.
求中位数时，在累积频数轴上找到总频数一半（80/2=40）处，作水平线与曲线相交，再作垂直线至质量轴。设读得142 g。下四分位数(Q1)取总频数1/4=20，读出≈128 g。上四分位数(Q3)取总频数3/4=60，读出≈158 g。
Interquartile range IQR = Q3 – Q1 ≈ 158 – 128 = 30 g. Always state the units. The cumulative frequency diagram makes it easy to see the spread of the middle 50% of data.
四分位距 IQR = Q3 – Q1 ≈ 158 – 128 = 30 g。务必注明单位。累积频数图可以直观展示中间50%数据的分散程度。
10. Common Mistakes and Pitfalls | 常见错误与陷阱
Many marks are lost due to simple presentation errors. For example, forgetting to include units in area or perimeter questions, or writing the final answer with incorrect significant figures. Always double-check how many marks are allocated and ensure your working is sufficient to justify each step.
很多失分源于简单的表达错误。例如，面积或周长题目漏写单位，或最终答案的有效数字位数错误。务必再次检查每小题分值，并确保过程足够详细，能支撑每一步推导。
Avoid mixing up ‘factorise’ and ‘solve’. Factorising x² – 5x – 14 gives (x+2)(x-7), while solving x² – 5x – 14 = 0 would then give x = -2 or x = 7. Misreading the command word leads straight to zero marks.
切勿混淆“因式分解”与“求解”。因式分解 x²-5x-14 得到 (x+2)(x-7)，而求解方程 x²-5x-14=0 则得到 x=-2 或 x=7。误读指令词将直接导致零分。
In probability, always consider whether events are independent or dependent. A ‘not replaced’ means the probabilities change for the second trial, so a tree diagram with updated denominators is essential.
概率题中，务必判断事件是独立还是相关。“不放回”意味着第二次试验的概率发生改变，因此需使用更新分母的树状图，这一点至关重要。
When drawing graphs, label axes clearly, use a sensible scale, and plot points with small crosses. A quick sketch without scales can lose marks. In cumulative frequency questions, you must use the graph, not raw data, to estimate quartiles, so annotate the graph to show your construction lines.
绘制图表时，坐标轴要清晰标注，使用合理刻度，并用小叉号描点。没有刻度的快速草图会失分。在累积频数题中，必须依据图形而非原始数据估算四分位数，所以图上要保留辅助线痕迹。
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