📚 Year 11 AQA Maths: Unit Test Mock Paper Solutions | 十一年级AQA数学：单元测试模拟卷解析
This article walks you through a complete AQA-style unit test mock paper designed for Year 11 students. Each section focuses on a typical exam question, providing step-by-step solutions, common pitfalls, and revision tips to strengthen your understanding and confidence before the real assessment.
本文为你详细解析一份为十一年级设计的AQA风格单元测试模拟卷。每个部分围绕一道典型考题，提供逐步解题过程、常见错误以及复习技巧，帮助你在正式测评前加深理解、建立信心。
1. Foundation Skills: Number and Arithmetic | 基础技巧：数与算术
Question: Evaluate (3/5) × (2/7) + 1/2. Give your answer as a fraction in its simplest form.
题目：计算 (3/5) × (2/7) + 1/2，并以最简分数形式给出答案。
Multiply the fractions first: (3/5) × (2/7) = (3×2)/(5×7) = 6/35.
先进行分数乘法：(3/5) × (2/7) = (3×2)/(5×7) = 6/35。
Now add 1/2 to 6/35. Find a common denominator: the LCM of 35 and 2 is 70.
接下来将 1/2 与 6/35 相加。找出公分母：35 和 2 的最小公倍数是 70。
Rewrite both fractions: 6/35 = 12/70, 1/2 = 35/70. Then 12/70 + 35/70 = 47/70.
将两个分数改写：6/35 = 12/70，1/2 = 35/70。然后 12/70 + 35/70 = 47/70。
The fraction 47/70 is already in simplest form because 47 is a prime number and does not divide 70.
分数 47/70 已经是最简形式，因为 47 是质数且不能整除 70。
Answer: 47/70. A common mistake is forgetting to find a common denominator before adding, or adding numerators and denominators directly.
答案：47/70。一个常见错误是相加前忘记通分，或是直接将分子与分母各自相加。
2. Algebraic Equations: Linear Mastery | 代数方程：一次方程精通
Question: Solve the equation: 3(x − 2) = 2x + 5.
题目：解方程：3(x − 2) = 2x + 5。
First expand the bracket on the left-hand side: 3 × (x − 2) becomes 3x − 6.
首先展开左边括号：3 × (x − 2) 得到 3x − 6。
The equation is now 3x − 6 = 2x + 5. Collect x-terms on one side by subtracting 2x from both sides: x − 6 = 5.
方程现在变为 3x − 6 = 2x + 5。通过两边同时减去 2x 将含 x 的项移到一边：x − 6 = 5。
Add 6 to both sides to isolate x: x = 11.
两边加 6 解出 x：x = 11。
Always check your answer by substituting back: 3(11 − 2) = 3 × 9 = 27, and 2(11) + 5 = 22 + 5 = 27. Both sides match.
永远记得代入原方程验证：3(11 − 2) = 3 × 9 = 27，2(11) + 5 = 22 + 5 = 27，两边相等。
Final answer: x = 11. Many errors occur when expanding brackets with a negative sign; here it was straightforward, but double-check signs.
最终答案：x = 11。许多错误发生在括号外有负号的情况下；此处虽然简单，也请再次检查符号。
3. Ratio and Proportion: Recipe-Style Problems | 比和比例：配方类问题
Question: A cake recipe uses sugar and flour in the ratio 2 : 5. You have 300 g of flour. What mass of sugar is needed?
题目：一份蛋糕配方中糖和面粉的比例为 2 : 5。现有 300 克面粉，需要多少克糖？
The ratio 2 : 5 means for every 2 parts of sugar, there are 5 parts of flour. The flour part corresponds to 5 shares.
比例 2 : 5 表示每 2 份糖对应 5 份面粉。面粉对应 5 份。
Find the value of one share: 300 g ÷ 5 = 60 g per share.
求出一份的重量：300 克 ÷ 5 = 60 克/份。
Sugar has 2 shares, so sugar needed = 2 × 60 g = 120 g.
糖占 2 份，所以需要糖量 = 2 × 60 克 = 120 克。
Another approach is setting up a proportion: sugar/flour = 2/5, so sugar = (2/5) × 300 = 120 g.
另一种方法是列出比例方程：糖/面粉 = 2/5，因此糖 = (2/5) × 300 = 120 克。
Answer: 120 g. Watch out for reversing the ratio order; always identify which quantity matches which term.
答案：120 克。注意不要颠倒比例顺序，始终确认哪个量对应比例中的哪一项。
4. Geometry: Angle Facts in Triangles and Polygons | 几何：三角形与多边形的角度性质
Question: In triangle ABC, angle A = 45° and angle B = 60°. Find angle C. Hence state whether the triangle is acute, right-angled or obtuse.
题目：在三角形 ABC 中，∠A = 45°，∠B = 60°。求 ∠C，并判断该三角形是锐角、直角还是钝角三角形。
Angles in a triangle sum to 180°. So angle C = 180° − (45° + 60°) = 180° − 105° = 75°.
三角形内角和为 180°。因此 ∠C = 180° − (45° + 60°) = 180° − 105° = 75°。
All three angles (45°, 60°, 75°) are less than 90°, so the triangle is acute.
三个角（45°、60°、75°）都小于 90°，因此这是一个锐角三角形。
Now consider a regular pentagon. Calculate the interior angle. The sum of interior angles = (5 − 2) × 180° = 540°. Each interior angle = 540° ÷ 5 = 108°.
再考虑一个正五边形。计算其内角。内角和 = (5 − 2) × 180° = 540°。每个内角 = 540° ÷ 5 = 108°。
Remember that exterior angles of any convex polygon always sum to 360°, which is a useful shortcut for regular polygons.
记住，任何凸多边形的外角和始终为 360°，这对计算正多边形是很有用的捷径。
Answers: angle C = 75°, triangle is acute; regular pentagon interior angle = 108°.
答案：∠C = 75°，三角形为锐角三角形；正五边形内角 = 108°。
5. Pythagoras’ Theorem: Right-Angled Triangles | 勾股定理：直角三角形
Question: The two shorter sides of a right-angled triangle are 3 cm and 4 cm. Calculate the length of the hypotenuse. Give an exact answer.
题目：一个直角三角形的两条直角边分别为 3 cm 和 4 cm。计算斜边的长度，给出精确值。
According to Pythagoras’ theorem: a² + b² = c², where c is the hypotenuse.
根据勾股定理：a² + b² = c²，其中 c 为斜边。
Substitute the values: 3² + 4² = c² → 9 + 16 = c² → 25 = c².
代入数值：3² + 4² = c² → 9 + 16 = c² → 25 = c²。
Taking the square root gives c = √25 = 5 cm. (The negative root is ignored as length is positive.)
开平方得 c = √25 = 5 cm。（负根忽略，因为长度为正。）
The triangle is a classic 3-4-5 Pythagorean triple. In exams, recognising common triples like (5, 12, 13) and (8, 15, 17) saves time.
这是一个经典的 3-4-5 勾股数。考试中识别常见勾股数（如 5、12、13 和 8、15、17）能节省时间。
Answer: 5 cm. When the hypotenuse is given and a shorter side is unknown, rearrange to a² = c² − b².
答案：5 cm。如果已知斜边求直角边，需将公式变形为 a² = c² − b²。
6. Statistics: Mean, Median and Mode | 统计：平均数、中位数与众数
Question: The data set shows the number of books read by students in a month: 5, 7, 8, 5, 9, 10, 5. Find the mean, median and mode.
题目：某数据集显示学生一个月内阅读书籍数量：5、7、8、5、9、10、5。求平均数、中位数和众数。
First list the values in ascending order: 5, 5, 5, 7, 8, 9, 10. There are 7 data points.
先将数值按升序排列：5、5、5、7、8、9、10。共有 7 个数据点。
Mode is the most frequent value: 5 appears three times, so mode = 5.
众数是出现频率最高的值：5 出现三次，因此众数 = 5。
Median is the middle value: the 4th number in the ordered list is 7, so median = 7.
中位数为中间值：排序后的第 4 个数是 7，因此中位数 = 7。
Mean = sum of all values ÷ number of values. Sum = 5+7+8+5+9+10+5 = 49. Mean = 49 ÷ 7 = 7.
平均数 = 所有值之和 ÷ 数据个数。总和 = 5+7+8+5+9+10+5 = 49。平均数 = 49 ÷ 7 = 7。
Answers: mean = 7, median = 7, mode = 5. Note that the mean and median can be equal even when the distribution is not symmetric; the mode reveals the most typical value.
答案：平均数 = 7，中位数 = 7，众数 = 5。注意即使分布不对称，平均数和中位数也可能相等；众数则揭示了最普遍的数值。
7. Probability: Tree Diagrams Without Replacement | 概率：无放回树状图
Question: A bag contains 3 red balls and 5 blue balls. Two balls are drawn at random without replacement. Draw a tree diagram and calculate the probability that both balls are red.
题目：袋子里有 3 个红球和 5 个蓝球。随机连续抽取两个球且不放回。画出树状图，并求两个球都是红色的概率。
Total balls = 8. For the first draw, P(red) = 3/8, P(blue) = 5/8.
总球数 = 8。第一次抽取，P(红) = 3/8，P(蓝) = 5/8。
If the first ball is red, the bag now has 2 red and 5 blue (7 balls left). The second-draw probabilities after a red are: P(red|red) = 2/7, P(blue|red) = 5/7.
如果第一次抽到红球，袋中剩 2 红 5 蓝（共 7 球）。在第一次是红球的条件下第二次抽取概率为：P(红|红) = 2/7，P(蓝|红) = 5/7。
If the first ball is blue, the bag contains 3 red and 4 blue. Then P(red|blue) = 3/7, P(blue|blue) = 4/7.
如果第一次抽到蓝球，袋中剩 3 红 4 蓝。那么 P(红|蓝) = 3/7，P(蓝|蓝) = 4/7。
Probability both are red: multiply along the red-red branch: (3/8) × (2/7) = 6/56 = 3/28.
两个都是红球的概率：沿红-红分支相乘：(3/8) × (2/7) = 6/56 = 3/28。
It is essential to reduce fractions to simplest form unless asked otherwise. A tree diagram helps visualise all outcomes and is often required for full marks.
除非另有要求，分数必须化为最简形式。树状图有助于可视化所有结果，考试中通常要求画出以得满分。
Answer: P(both red) = 3/28.
答案：P(两个红球) = 3/28。
8. Rates of Change: Distance-Time Graphs | 变化率：距离-时间图
Question: A car travels 150 km in 2 hours 30 minutes at constant speed. Calculate the speed in km/h. Then sketch a distance-time graph and describe its key features.
题目：一辆汽车以恒定速度行驶 150 km，用时 2 小时 30 分钟。计算速度（单位 km/h）。然后画出距离-时间图的草图并描述其主要特征。
Convert time to hours: 2 hours 30 minutes = 2.5 hours. Speed = distance ÷ time = 150 ÷ 2.5 = 60 km/h.
将时间转换为小时：2 小时 30 分钟 = 2.5 小时。速度 = 距离 ÷ 时间 = 150 ÷ 2.5 = 60 km/h。
A distance-time graph for constant speed is a straight line passing through the origin, with gradient equal to speed. The line goes from (0,0) to (2.5, 150).
恒定速度下的距离-时间图是一条过原点的直线，斜率即为速度。该直线从 (0,0) 到 (2.5, 150)。
Gradient = rise/run = 150/2.5 = 60, confirming the speed. If the line were horizontal, it would indicate the object is stationary.
斜率 = 垂直增量/水平增量 = 150/2.5 = 60，与速度相符。如果直线水平，则表示物体静止。
Acceleration would be shown by a curve with increasing gradient. In unit tests, you may be asked to interpret such graphs or calculate speeds from tangents.
加速运动则表现为一条斜率逐渐增大的曲线。单元测试中可能要求解读此类图像，或通过作切线计算速度。
Answer: speed = 60 km/h. Graph: straight line from origin to (2.5, 150).
答案：速度 = 60 km/h。图像：一条从原点到 (2.5, 150) 的直线。
9. Standard Form: Multiplication and Division | 标准形式：乘除运算
Question: Calculate (2 × 10³) × (3 × 10⁴). Give your answer in standard form.
题目：计算 (2 × 10³) × (3 × 10⁴)，并用标准形式表示答案。
Multiply the number parts: 2 × 3 = 6. Multiply the powers of 10: 10³ × 10⁴ = 10³⁺⁴ = 10⁷.
将数字部分相乘：2 × 3 = 6。将 10 的指数相加：10³ × 10⁴ = 10³⁺⁴ = 10⁷。
The result is 6 × 10⁷, which is already in standard form because 1 ≤ 6 < 10.
结果为 6 × 10⁷，它已经是标准形式，因为系数 6 满足 1 ≤ 6 < 10。
Now consider division: (8 × 10⁵) ÷ (4 × 10²). Divide number parts: 8 ÷ 4 = 2. Divide powers: 10⁵ ÷ 10² = 10⁵⁻² = 10³. Answer: 2 × 10³.
再考虑除法：(8 × 10⁵) ÷ (4 × 10²)。数字部分相除：8 ÷ 4 = 2。指数相减：10⁵ ÷ 10² = 10⁵⁻² = 10³。答案：2 × 10³。
If the number part is outside the range 1–10, adjust it: e.g. 15 × 10³ becomes 1.5 × 10⁴. Make sure you can enter and read standard form on your calculator.
如果数字部分超出 1–10 的范围，需要调整：如 15 × 10³ 化为 1.5 × 10⁴。确保你能够熟练使用计算器输入和显示标准形式。
Answer: 6 × 10⁷.
答案：6 × 10⁷。
10. Algebra: Expanding Products and Simplifying | 代数：乘积展开与化简
Question: Expand and simplify (x + 3)(x − 2) + 4x.
题目：展开并化简 (x + 3)(x − 2) + 4x。
Expand the double brackets using the FOIL method. First: x × x = x². Outer: x × (−2) = −2x. Inner: 3 × x = 3x. Last: 3 × (−2) = −6.
使用 FOIL 方法展开双括号。首项：x × x = x²。外项：x × (−2) = −2x。内项：3 × x = 3x。尾项：3 × (−2) = −6。
Combine the terms from expansion: x² − 2x + 3x − 6 = x² + x − 6.
合并展开后的各项：x² − 2x + 3x − 6 = x² + x − 6。
Now add the +4x from the original expression: x² + x − 6 + 4x = x² + 5x − 6.
再将原式中的 +4x 加上：x² + x − 6 + 4x = x² + 5x − 6。
The expression cannot be factorised further with integer coefficients, so this is the final simplified form.
此表达式无法进一步用整数系数分解，因此这就是最终化简结果。
Answer: x² + 5x − 6. Always check by substituting a small value, e.g. x = 1: original gives (4)(−1) + 4 = 0; simplified gives 1+5−6=0.
答案：x² + 5x − 6。建议代入一个小数值检验，例如 x = 1：原式得 (4)(−1) + 4 = 0；化简后得 1+5−6=0。
11. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Many marks are lost through avoidable errors. One frequent issue is sign errors when expanding brackets, especially with negative terms. Always rewrite the expression carefully, putting brackets around negative numbers if needed.
许多失分源于可避免的错误。一个常见问题是展开括号时的符号错误，尤其是带有负项的情况。始终仔细重写表达式，必要时给负数加上括号。
Another common slip is forgetting to convert units—for example, grams to kilograms or minutes to hours—before substituting into formulas. Underline the units given in the question and use conversion factors systematically.
另一个常见疏忽是忘记单位换算——例如在代入公式前没有将克转换为千克，或分钟转换为小时。用下划线标出题目中的单位，并系统使用换算因子。
In probability without replacement, students often forget to update the denominator and numerator for the second event. Always imagine the real-life situation: ‘If I have taken one red ball out, what is left in the bag?’
在无放回概率问题中，学生常会忘记为第二个事件更新分母和分子。时刻在脑海中模拟真实情景：“如果我拿走了一个红球，袋中还剩下什么？”
When drawing graphs or diagrams, use a pencil and ruler, label axes, and plot points accurately. A quick sketch without tools can lose method marks even if the reasoning is correct.
画图或示意图时，使用铅笔和直尺，标注坐标轴，并准确地描点。不用工具随手画草图，即使推理正确，也可能丢失方法分。
12. Final Tips and Exam Technique | 最终提示与考试技巧
Read each question twice: once to grasp the topic, and once to underline key information and command words such as ‘expand’, ‘simplify’, ‘give your answer in standard form’.
每道题读两遍：第一遍把握题目主题，第二遍划出关键信息和指令词，如“展开”、“化简”、“用标准形式给出答案”。
Show all working clearly. In a unit test, the majority of marks are awarded for the correct method, not just the final answer. Even if your final answer is wrong, a logical method can earn most of the marks.
清晰展示所有解题步骤。在单元测试中，大部分分数是给予正确方法，而不仅仅是最终答案。即使最终答案有误，逻辑清晰的方法也能获得绝大部分分数。
Manage your time wisely. If you get stuck on a question, put a star next to it, move on, and return at the end. A unit test usually allocates about one minute per mark, so pace yourself accordingly.
合理管理时间。如果卡在某道题上，在旁边做个星号标记，先往下做，最后再回来。单元测试通常按 1 分对应 1 分钟来命题，请据此把握节奏。
Finally, use your calculator effectively only when allowed. For non-calculator papers, practise mental arithmetic and written methods to build speed and accuracy.
最后，只有在允许时才有效使用计算器。对于非计算器试卷，多练习心算和笔算方法，以提高速度与准确性。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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