📚 Year 11 AQA Maths: Unit Test Mock Paper Analysis | Year 11 AQA 数学：单元测试模拟卷解析
This article provides detailed worked solutions for a mock unit test covering key topics from the Year 11 AQA GCSE Mathematics syllabus. Each question is broken down step by step to reinforce understanding and exam technique, helping you gain confidence for your assessments.
本文详细解析了一份覆盖 Year 11 AQA GCSE 数学核心单元的模拟测试卷。每道题都分步骤讲解，旨在强化理解与应试技巧，助力你自信迎接考试。
1. Solving Quadratic Equations | 解二次方程
Question: Solve x² – 7x + 12 = 0.
题目：解方程 x² – 7x + 12 = 0。
Method: Factorise the quadratic. We need two numbers that multiply to 12 and add to –7. These are –3 and –4.
解法：对二次式进行因式分解。我们需要两个数，乘积为 12，和为 –7，这两个数是 –3 和 –4。
Hence, (x – 3)(x – 4) = 0. Setting each factor equal to zero gives x – 3 = 0 or x – 4 = 0.
因此，(x – 3)(x – 4) = 0。令每个因式为零，得 x – 3 = 0 或 x – 4 = 0。
Solutions: x = 3 or x = 4. Always check by substitution: 3² – 7×3 + 12 = 9 – 21 + 12 = 0, and similarly for 4.
解为：x = 3 或 x = 4。务必代入原方程检验：3² – 7×3 + 12 = 9 – 21 + 12 = 0，4 同理。
2. Graph Transformations | 图形变换
Question: The graph of y = f(x) is transformed to y = 3f(x – 2). Describe fully the two transformations applied.
题目：y = f(x) 的图像经过变换得到 y = 3f(x – 2)。完整描述所施行的两个变换。
The transformation from f(x) to f(x – 2) is a horizontal translation 2 units to the right. The ‘3’ outside the function represents a vertical stretch by a factor of 3.
从 f(x) 到 f(x – 2) 的变换是将图像沿 x 轴向右平移 2 个单位。函数外的 ‘3’ 表示沿 y 轴方向伸缩，放大为原来的 3 倍。
Order matters: first translate right by 2, then apply the vertical stretch. Writing it as y = 3f(x – 2) is clear.
变换顺序重要：首先向右平移 2 个单位，再进行垂直放大。表达式 y = 3f(x – 2) 已经清晰地指明了顺序。
3. Trigonometric Ratios | 三角比
Question: In a right-angled triangle, the side opposite an acute angle θ is 3 cm and the adjacent side is 4 cm. Calculate θ, correct to 1 decimal place.
题目：在一个直角三角形中，锐角 θ 的对边长为 3 cm，邻边长为 4 cm。求 θ 的大小，结果精确到 1 位小数。
Using the tangent ratio: tan θ = opposite / adjacent = 3/4 = 0.75.
使用正切比：tan θ = 对边 / 邻边 = 3/4 = 0.75。
Therefore, θ = tan⁻¹(0.75). Using a calculator, tan⁻¹(0.75) ≈ 36.8698976…°
因此，θ = tan⁻¹(0.75)。用计算器求值得 tan⁻¹(0.75) ≈ 36.8698976…°。
Rounded to 1 decimal place: θ = 36.9°.
四舍五入保留 1 位小数：θ = 36.9°。
4. Circle Theorems | 圆定理
Question: Points A, B and C lie on the circumference of a circle with centre O. Angle AOB = 100°. Find the size of angle ACB.
题目：点 A、B、C 在以 O 为圆心的圆周上。已知角 AOB = 100°，求角 ACB 的大小。
The angle at the centre is twice any angle at the circumference subtended by the same arc. Here, arc AB subtends angle AOB at the centre and angle ACB at the circumference.
同一段弧所对的圆心角是圆周角的 2 倍。此处，弧 AB 所对的圆心角为 AOB，圆周角为 ACB。
So, angle ACB = 1/2 × angle AOB = 1/2 × 100° = 50°.
因此，角 ACB = 1/2 × 角 AOB = 1/2 × 100° = 50°。
5. Probability Tree Diagrams | 概率树形图
Question: A bag contains 5 red sweets and 3 blue sweets. Two sweets are picked at random without replacement. Find the probability that the sweets are of different colours.
题目：一个袋子里有 5 颗红色糖果和 3 颗蓝色糖果。随机依次取出两颗且不放回。求两颗糖果颜色不同的概率。
Draw a tree diagram: First pick: P(Red) = 5/8, P(Blue) = 3/8. Second pick probabilities change depending on the first outcome.
画出树形图：第一次抽取：P(Red) = 5/8, P(Blue) = 3/8。第二次抽取的概率根据第一次结果变化。
For different colours, we need (Red then Blue) or (Blue then Red).
不同颜色的情况为（先红后蓝）或（先蓝后红）。
P(Red then Blue) = (5/8) × (3/7) = 15/56. P(Blue then Red) = (3/8) × (5/7) = 15/56.
P(红然后蓝) = (5/8) × (3/7) = 15/56。P(蓝然后红) = (3/8) × (5/7) = 15/56。
Total probability = 15/56 + 15/56 = 30/56 = 15/28.
总概率 = 15/56 + 15/56 = 30/56 = 15/28。
6. Equation of a Straight Line | 直线方程
Question: Find the equation of the straight line passing through the points (2, 5) and (6, 13).
题目：求过点 (2, 5) 和 (6, 13) 的直线方程。
Step 1: Calculate the gradient m = (13 – 5)/(6 – 2) = 8/4 = 2.
步骤 1：计算斜率 m = (13 – 5)/(6 – 2) = 8/4 = 2。
Using point-slope form with (2, 5): y – 5 = 2(x – 2).
利用点斜式，代入点 (2, 5)：y – 5 = 2(x – 2)。
Simplify: y – 5 = 2x – 4 → y = 2x + 1.
化简：y – 5 = 2x – 4 → y = 2x + 1。
The equation of the line is y = 2x + 1. Check with the other point: when x = 6, y = 2×6 + 1 = 13, correct.
直线方程为 y = 2x + 1。用另一个点检验：当 x = 6 时，y = 2×6 + 1 = 13，符合。
7. Factorising Harder Quadratics | 因式分解复杂二次式
Question: Factorise fully 3x² + 8x + 4.
题目：将 3x² + 8x + 4 彻底因式分解。
Look for two numbers that multiply to ac = 3×4 = 12 and add to b = 8. The numbers are 6 and 2.
寻找两个数，使它们的乘积为 ac = 3×4 = 12，和为 b = 8。这两个数是 6 和 2。
Rewrite the middle term: 3x² + 6x + 2x + 4.
将一次项拆分：3x² + 6x + 2x + 4。
Factor by grouping: 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2).
分组分解：3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2)。
Thus, 3x² + 8x + 4 = (3x + 2)(x + 2).
因此，3x² + 8x + 4 = (3x + 2)(x + 2)。
8. Vectors | 向量
Question: Given a = 3i – j and b = –2i + 4j, find (a) a – b, (b) |a + b|.
题目：已知 a = 3i – j，b = –2i + 4j，求 (a) a – b, (b) |a + b|。
(a) a – b = (3i – j) – (–2i + 4j) = 3i – j + 2i – 4j = 5i – 5j.
(a) a – b = (3i – j) – (–2i + 4j) = 3i – j + 2i – 4j = 5i – 5j。
(b) First, a + b = (3i – j) + (–2i + 4j) = i + 3j.
(b) 先计算 a + b = (3i – j) + (–2i + 4j) = i + 3j。
Magnitude: |a + b| = √(1² + 3²) = √(1 + 9) = √10.
模长：|a + b| = √(1² + 3²) = √(1 + 9) = √10。
9. Averages from Grouped Frequency Tables | 分组频数表求平均数
Question: The table shows the heights of 25 plants. Estimate the mean height.
题目：下表显示了 25 株植物的高度。估计平均高度。
| Height, h (cm) | Frequency |
|---|---|
| 140 ≤ h < 150 | 5 |
| 150 ≤ h < 160 | 7 |
| 160 ≤ h < 170 | 10 |
| 170 ≤ h < 180 | 3 |
Add a midpoint column and an f × midpoint column.
添设中点列和频数 × 中点列。
Midpoints: 145, 155, 165, 175. f×midpoint: 5×145=725, 7×155=1085, 10×165=1650, 3×175=525.
中点：145, 155, 165, 175。频数 × 中点：5×145=725, 7×155=1085, 10×165=1650, 3×175=525。
Sum of f×midpoint = 725 + 1085 + 1650 + 525 = 3985. Total frequency = 25.
频数 × 中点之和 = 725 + 1085 + 1650 + 525 = 3985。总频数 = 25。
Estimated mean = 3985 ÷ 25 = 159.4 cm.
估计平均高度 = 3985 ÷ 25 = 159.4 cm。
10. Solving Inequalities | 解不等式
Question: Solve the inequality –3 < 2x + 1 ≤ 7.
题目：解不等式 –3 < 2x + 1 ≤ 7。
Subtract 1 from all three parts: –3 – 1 < 2x ≤ 7 – 1 → –4 < 2x ≤ 6.
将所有部分同时减去 1：–3 – 1 < 2x ≤ 7 – 1 → –4 < 2x ≤ 6。
Divide by 2: –2 < x ≤ 3.
除以 2：–2 < x ≤ 3。
The solution set is all x such that x is greater than –2 and less than or equal to 3.
解集为所有满足 x > –2 且 x ≤ 3 的 x 值。
11. Simultaneous Equations | 联立方程组
Question: Solve the simultaneous equations: y = x + 2 and y = x² – 2x – 3.
题目：解联立方程组：y = x + 2, y = x² – 2x – 3。
Since both expressions equal y, set them equal: x + 2 = x² – 2x – 3.
因为两式都等于 y，可令它们相等：x + 2 = x² – 2x – 3。
Rearrange to form a quadratic: 0 = x² – 3x – 5, or x² – 3x – 5 = 0.
移项得到二次方程：0 = x² – 3x – 5，即 x² – 3x – 5 = 0。
Use the quadratic formula:
x = ( –(–3) ± √((–3)² – 4×1×(–5)) ) / (2×1) = (3 ± √(9 + 20)) / 2 = (3 ± √29) / 2.
使用求根公式：
x = ( 3 ± √(9 + 20) ) / 2 = (3 ± √29) / 2。
Thus, x = (3 + √29)/2 or x = (3 – √29)/2. Substitute back into y = x + 2 to find y values.
因此，x = (3 + √29)/2 或 x = (3 – √29)/2。代回 y = x + 2 求相应的 y 值。
Solutions are ((3+√29)/2, (7+√29)/2) and ((3–√29)/2, (7–√29)/2).
解为 ((3+√29)/2, (7+√29)/2) 和 ((3–√29)/2, (7–√29)/2)。
12. Direct and Inverse Proportion | 正比例与反比例
Question: y is inversely proportional to x. When x = 4, y = 6. Find y when x = 6.
题目：y 与 x 成反比例。当 x = 4 时，y = 6。求当 x = 6 时的 y 值。
Inverse proportion means y = k / x for some constant k. Substitute known values: 6 = k / 4 → k = 24.
反比例意味着 y = k / x，其中 k 为常数。代入已知值：6 = k / 4 → k = 24。
Now, when x = 6, y = 24 / 6 = 4.
则当 x = 6 时，y = 24 / 6 = 4。
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