Cambridge Year 11 Maths Unit Test Mock Paper Analysis | 剑桥Year 11数学单元测试模拟卷解析

📚 Cambridge Year 11 Maths Unit Test Mock Paper Analysis | 剑桥Year 11数学单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 11 Cambridge Mathematics unit test mock paper. We cover a range of topics from algebra and geometry to probability and trigonometry, offering step-by-step solutions and common pitfalls to avoid. Use this guide to strengthen your understanding and boost your exam confidence.

本文为一份典型的Year 11剑桥数学单元测试模拟卷提供详细解析。内容涵盖代数、几何、概率、三角学等多个主题，给出逐步解题过程和常见错误提醒。通过本文复习关键概念，提升你的应试信心。

1. Solving Linear Equations | 解一元一次方程

Solve: 2x − 7 = 3x + 5

Step 1: Collect variable terms on one side. Subtract 2x from both sides to get −7 = x + 5.

步骤1：将含未知数的项移到同一边。两边同时减去 2x，得到 −7 = x + 5。

Step 2: Isolate x by subtracting 5 from both sides: −7 − 5 = x, hence x = −12.

步骤2：两边同时减去 5 以分离 x：−7 − 5 = x，因此 x = −12。

− 7 = −31; RHS = 3(−12) + 5 = −36 + 5 = −31. Both sides agree, so the solution is correct.

检验：将 x = −12 代入原方程。左边 = 2(−12) − 7 = −24 − 7 = −31；右边 = 3(−12) + 5 = −36 + 5 = −31。两边相等，解无误。

Common mistake: When moving terms across the equals sign, many students forget to change the sign. Always apply the inverse operation to both sides.

常见错误：移项时忘记变号。务必对方程两边同时施加逆运算。

---

2. Quadratic Factorisation | 二次三项式因式分解

Factorise x² − 5x + 6

−5). Here, −2 and −3 satisfy both conditions because (−2) × (−3) = 6 and (−2) + (−3) = −5.

步骤1：找出两个数，其积等于常数项（+6），其和等于 x 的系数（−5）。此处 −2 和 −3 符合要求，因为 (−2) × (−3) = 6，而 (−2) + (−3) = −5。

x − 3). You can quickly expand to verify: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6.

步骤2：写出因式分解形式 (x − 2)(x − 3)。可快速展开验证：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6。

Tip: For expressions where the coefficient of x² is not 1, use the 'ac' method or splitting the middle term. Always look for a common factor first.

技巧：若 x² 的系数不为 1，可采用“ac”法或十字相乘法。务必先检查是否有公因式可提取。

---

3. Simultaneous Equations | 联立方程组

Solve: 2x + y = 7 and x − y = 2

efficients of y are +1 and −1. Add the two equations: (2x + y) + (x − y) = 7 + 2 → 3x = 9.

步骤1：对齐方程组，考虑消去 y，因为 y 的系数分别为 +1 和 −1。将两式相加：(2x + y) + (x − y) = 7 + 2 → 3x = 9。

x: 3x = 9 → x = 3. Substitute back into the second equation: 3 − y = 2 → −y = −1 → y = 1.

步骤2：求 x：3x = 9 → x = 3。将 x = 3 代入第二个方程：3 − y = 2 → −y = −1 → y = 1。

ck in the first equation: 2(3) + 1 = 7 → 6 + 1 = 7, which is true. The solution is (3, 1).

代入第一个方程检验：2(3) + 1 = 7 → 6 + 1 = 7，成立。解为 (3, 1)。

---

4. Straight Line Graphs | 直线图像

Find the gradient and y-intercept of y = 2x + 1, and sketch the graph.

. The gradient m is 2, and the y-intercept c is 1 (the line crosses the y-axis at (0, 1)).

步骤1：方程呈斜截式 y = mx + c。斜率 m 为 2，y 截距 c 为 1（直线与 y 轴交于点 (0, 1)）。

pt, set y = 0: 0 = 2x + 1 → 2x = −1 → x = −½. So the line crosses the x-axis at (−0.5, 0).

步骤2：求 x 截距，令 y = 0：0 = 2x + 1 → 2x = −1 → x = −½。因此直线与 x 轴交于点 (−0.5, 0)。

Step 3: Plot the points (0, 1) and (−0.5, 0). Since the gradient is 2, from (0,1) move 1 unit right and 2 units up to get another point (1, 3). Draw a straight line through these points.

步骤3：标出点 (0, 1) 和 (−0.5, 0)。斜率为 2，故从 (0,1) 向右移 1 单位、向上移 2 单位，得到点 (1, 3)。过这些点画直线。

---

5. Angle Properties of Parallel Lines | 平行线角度性质

In the diagram, two parallel lines are cut by a transversal. One acute angle is 65°. Find the remaining seven angles.

Step 1: Label the given angle as 65°. The vertically opposite angle is also 65°. The corresponding angle on the other parallel line is 65° as well.

步骤1：标记已知角为 65°。对顶角同为 65°。在另一条平行线上，同位角也是 65°。

, so the adjacent angle is 180° − 65° = 115°. The co-interior (allied) angles sum to 180°, giving another 115°.

步骤2：内错角相等，平行线内侧与 65° 相对的内错角也是 65°。平角之和为 180°，故邻角为 180° − 65° = 115°。同旁内角互补，因此得到又一个 115°。

Tip: Draw a clear diagram and mark equals. Keep in mind that corresponding, alternate, and vertically opposite angles are equal; co-interior angles are supplementary.

技巧：画出清晰图示并标记等角。牢记同位角、内错角、对顶角相等；同旁内角互补。

---

6. Probability Tree Diagrams | 概率树形图

A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Find the probability that both balls are red.

(red) = 3/5, P(blue) = 2/5. If a red is taken, the bag then has 2 red and 2 blue, so P(red | first red) = 2/4 = 1/2.

步骤1：第一次抽取，P(红) = 3/5，P(蓝) = 2/5。若第一次抽到红球，袋中剩 2 红 2 蓝，故 P(红 | 第一次红) = 2/4 = ½。

obability of both red is the product along the branch: P(red, red) = (3/5) × (1/2) = 3/10.

步骤2：两次都抽到红球的概率为该分支上概率的乘积：P(红, 红) = (3/5) × (½) = 3/10。

Step 3: The probability tree also helps find other outcomes, e.g., exactly one blue. Always check that all branch probabilities sum to 1.

步骤3：概率树图还可求其他结果，例如恰有一个蓝球的概率。务必检查所有分支概率之和为 1。

---

广告 / Sponsored

7. Cumulative Frequency | 累积频率

Given a frequency table of test scores, draw the cumulative frequency curve and estimate the median and interquartile range.

Step 1: Add a cumulative frequency column. For each class interval, add the frequency to the running total. Plot upper class boundaries against cumulative frequencies, then join the points with a smooth curve.

步骤1：添加累积频率列。对每个组距，将频数累加到当前总计。以组距上限为横坐标、累积频率为纵坐标描点，并用平滑曲线连接。

) on the vertical axis, draw a horizontal line to the curve, then a vertical line down to read the score.

步骤2：求中位数，先在纵轴上找到总频数的一半（如 ½ × 总频数），画水平线交曲线，再向下画垂直线读取分数。

r quartile − lower quartile. A cumulative frequency graph also helps estimate percentiles.

步骤3：下四分位数在总频数的 ¼ 处，上四分位数在 ¾ 处。四分位距 = 上四分位数 − 下四分位数。累积频率图还可估算百分位数。

---

8. Functions | 函数

Given f(x) = 2x + 3 and g(x) = x², find fg(x) and f⁻¹(x).

Step 1: fg(x) means apply g first, then f. So fg(x) = f(x²) = 2(x²) + 3 = 2x² + 3.

步骤1：fg(x) 表示先应用 g，再应用 f。所以 fg(x) = f(x²) = 2(x²) + 3 = 2x² + 3。

wap x and y: x = 2y + 3. Solve for y: x − 3 = 2y → y = (x − 3)/2. Thus f⁻¹(x) = (x − 3)/2.

步骤2：求反函数 f⁻¹(x)，设 y = 2x + 3，交换 x 与 y：x = 2y + 3，解出 y：x − 3 = 2y → y = (x − 3)/2。故 f⁻¹(x) = (x − 3)/2。

Check: f(f⁻¹(x)) = 2((x − 3)/2) + 3 = x − 3 + 3 = x. It works.

检验：f(f⁻¹(x)) = 2((x − 3)/2) + 3 = x − 3 + 3 = x，正确。

---

9. Ratio and Proportion | 比例与比率

Divide £120 among A, B, and C in the ratio 3 : 5 : 2.

Step 1: Find the total number of parts: 3 + 5 + 2 = 10 parts. Each part is worth £120 ÷ 10 = £12.

步骤1：计算总份数：3 + 5 + 2 = 10 份。每份价值为 £120 ÷ 10 = £12。

Step 2: A gets 3 parts = 3 × £12 = £36; B gets 5 parts = £60; C gets 2 parts = £24. Check: £36 + £60 + £24 = £120.

步骤2：A 得 3 份 = 3 × £12 = £36；B 得 5 份 = £60；C 得 2 份 = £24。检验：£36 + £60 + £24 = £120。

For inverse proportion or map scales, set up equations carefully. Always express units consistently.

遇到反比例或地图比例尺时，应仔细建立方程，并始终保持单位一致。

---

10. Sequences and nth Term | 数列与第 n 项

Find the nth term of the sequence: 3, 7, 11, 15, 19, …

e is linear. The nth term has the form an + b, where a is the common difference, so a = 4.

步骤1：一阶差为常数，每次 +4。因此该数列为等差数列。第 n 项表达式为 an + b，其中 a 为公差，故 a = 4。

to find b. When n = 1, 4(1) + b = 3, so 4 + b = 3 → b = −1. Hence the nth term is 4n − 1.

步骤2：利用首项求 b。当 n = 1，4(1) + b = 3，即 4 + b = 3 → b = −1。因此第 n 项为 4n − 1。

Check for n = 2: 4(2) − 1 = 8 − 1 = 7, correct. The 10th term would be 4(10) − 1 = 39.

检验 n = 2：4(2) − 1 = 8 − 1 = 7，正确。第 10 项为 4(10) − 1 = 39。

---

11. Trigonometry – Sine and Cosine Rules | 三角学 – 正弦与余弦定理

e B = 100°, side a = 7 cm (opposite A). Find side b (opposite B) to 2 significant figures.

wn values: 7 / sin 30° = b / sin 100°. Since sin 30° = 0.5, the left side is 7 / 0.5 = 14.

步骤1：使用正弦定理：a / sin A = b / sin B。代入已知值：7 / sin 30° = b / sin 100°。因 sin 30° = 0.5，左边为 7 / 0.5 = 14。

≈ 0.9848, thus b ≈ 14 × 0.9848 = 13.7872 cm. Rounded to 2 significant figures, b ≈ 14 cm.

步骤2：于是 b = 14 × sin 100°。计算得 sin 100° ≈ 0.9848，故 b ≈ 14 × 0.9848 = 13.7872 cm。保留两位有效数字，b ≈ 14 cm。

Tip: Label sides and angles carefully. Use the cosine rule when you have two sides and the included angle, or three sides.

技巧：仔细标记边与角。当已知两边及其夹角或三边时，使用余弦定理。

---

12. Exam Tips and Common Pitfalls | 考试技巧与常见错误

Always show your working even if the question seems simple. Marks are awarded for method. Check your answers by substituting back or using a different method when possible.

无论题目多简单，都要展示解题步骤。步骤分通常存在。尽量用代入法或其他方法检验答案。

Pay close attention to units, rounding, and significant figures as specified in the question. In geometry, include the degree symbol and write reasons when asked.

留意题目要求的单位、舍入规则和有效数字。几何题中，记得标注度数符号，如要求说明理由则需写出定理依据。

When expanding or factorising, watch for sign errors. In probability, remember that 'without replacement' changes the denominators. Manage your time wisely; do not spend too long on a single question.

展开或因式分解时，提防符号错误。在概率问题中，“不放回”会改变分母。合理分配时间，不要在某一题上过度纠缠。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）