📚 Year 11 Cambridge Maths: Unit Test Mock Paper Walkthrough | 剑桥Year 11数学单元测试模拟卷解析
Welcome to our detailed walkthrough of a representative Year 11 Cambridge IGCSE Mathematics unit test mock paper. In this article we examine key question types, demonstrate efficient solution strategies and highlight common pitfalls. Each section breaks down one topic, moving from method to worked example to commentary, helping you build both accuracy and confidence ahead of your real assessment.
欢迎阅读我们对一份典型剑桥 IGCSE 数学 Year 11 单元测试模拟卷的详细解析。本文剖析常见题型，演示高效解题策略并点明易错之处。每一节聚焦一个主题，从方法到完整例题再到点评，帮助你在真正的评估到来之前提升准确度与自信心。
1. Simplifying Algebraic Expressions | 代数表达式化简
We start with a non‑calculator question asking us to simplify 3a − 2b + 5a + 7b − 4. The key is to collect like terms: terms with a, terms with b and pure numbers.
我们从一道不允许使用计算器的题目开始，要求化简 3a − 2b + 5a + 7b − 4。关键在于合并同类项：含有 a 的项、含有 b 的项以及纯数字项。
Group the a‑terms: 3a + 5a = 8a. Group the b‑terms: −2b + 7b = 5b. The constant is −4. The simplified expression is therefore 8a + 5b − 4. Always write terms in alphabetical order when no other instruction is given.
合并 a 项：3a + 5a = 8a。合并 b 项：−2b + 7b = 5b。常数项为 −4。因此化简后的表达式为 8a + 5b − 4。若无特别说明，通常按字母顺序书写各项。
2. Solving Linear Equations | 解一元一次方程
The equation is 5x + 3 = 18. Our goal is to isolate x by performing inverse operations in the correct order: undo addition/subtraction first, then multiplication/division.
方程为 5x + 3 = 18。我们的目标是通过正确运用逆运算分离出 x：先消去加减法，再处理乘除法。
5x + 3 − 3 = 18 − 3 → 5x = 15
Subtract 3 from both sides to keep the equation balanced. We obtain 5x = 15.
两边同时减去 3 以保持等式平衡，得到 5x = 15。
5x ÷ 5 = 15 ÷ 5 → x = 3
iving x = 3. Always check by substitution: 5(3) + 3 = 15 + 3 = 18, which matches the right‑hand side.
两边除以 5，得 x = 3。务必代入检验：5(3) + 3 = 15 + 3 = 18，与右边一致。
3. Quadratic Factorisation | 二次因式分解
We need to factorise x² − 7x + 10. Look for two numbers whose product is +10 and sum is −7. The pairs for 10 are (1,10) and (2,5); to achieve a negative sum both numbers must be negative.
我们需要对 x² − 7x + 10 进行因式分解。寻找两个数，乘积为 +10，和为 −7。10 的因数对有 (1,10) 和 (2,5)；要使和为负，这两个数都必须为负数。
r is −2 and −5 because (−2) × (−5) = 10 and (−2) + (−5) = −7. Hence the factorised form is (x − 2)(x − 5).
正确的数对是 −2 和 −5，因为 (−2) × (−5) = 10 且 (−2) + (−5) = −7。因此因式分解的结果为 (x − 2)(x − 5)。
) = x² − 5x − 2x + 10 = x² − 7x + 10. This step is essential to guard against sign errors.
可展开验证：(x − 2)(x − 5) = x² − 5x − 2x + 10 = x² − 7x + 10。这一步必不可少，以防符号错误。
4. Inequalities and Number Lines | 不等式与数轴表示
Consider the inequality −2 < 3x + 1 ≤ 7. We solve it by isolating x in the middle, treating it as two connected inequalities.
考虑不等式 −2 < 3x + 1 ≤ 7。我们将中间的 x 分离出来，将其视为两个相连的不等式进行处理。
Subtract 1 from all three parts: −2 − 1 < 3x ≤ 7 − 1 → −3 < 3x ≤ 6.
三部分同时减 1：−2 − 1 < 3x ≤ 7 − 1 → −3 < 3x ≤ 6。
3: −1 < x ≤ 2. The solution set is all real numbers greater than −1 and up to 2 inclusive.
除以 3：−1 < x ≤ 2。解集为所有大于 −1 且不超过 2（含 2）的实数。
On a number line we represent this with an open circle at −1 (not included) and a closed circle at 2 (included), shading the region between them. Always label the endpoints clearly.
在数轴上表示时，我们在 −1 处画空心圆（不包含），在 2 处画实心圆（包含），并涂满中间区域。端点必须清晰标注。
5. Angles in Parallel Lines | 平行线中的角
A diagram shows two parallel lines cut by a transversal. One of the interior angles is labelled 115°. We are asked to find the value of the corresponding angle and the co‑interior angle on the same side.
图中两条平行线被一条截线所截。其中一个内角标为 115°。要求找出同位角以及同旁内角的大小。
Corresponding angles are equal when lines are parallel, so the corresponding angle is also 115°. Co‑interior angles sum to 180°, so the co‑interior angle on the same side is 180° − 115° = 65°.
当直线平行时同位角相等，因此同位角也是 115°。同旁内角之和为 180°，所以同侧的同旁内角为 180° − 115° = 65°。
0°, satisfying the supplementary property. Students often confuse alternate angles with co‑interior angles; remember that alternate interior angles are equal, while co‑interior angles are supplementary.
快速检验：115° + 65° = 180°，满足互补性质。学生常混淆内错角与同旁内角；请记住内错角相等，而同旁内角互补。
6. Circle Theorems – Angle at Centre | 圆定理 – 圆心角
A circle has centre O. Points A, B and C lie on the circumference such that ∠AOB = 84°. We need to find ∠ACB, the angle subtended by the same arc AB at the circumference.
一个圆圆心为 O。点 A、B、C 在圆周上，且 ∠AOB = 84°。我们需要求同一弧 AB 所对的圆周角 ∠ACB。
at the centre is twice the angle at the circumference standing on the same arc. Therefore ∠ACB = ½ × 84° = 42°.
同一弧上的圆心角是圆周角的两倍。因此 ∠ACB = ½ × 84° = 42°。
This theorem holds regardless of where C is placed on the major arc AB. Always quote the theorem when justifying your answer: “The angle subtended by an arc at the centre is twice the angle subtended at the circumference.”
该定理与 C 在优弧 AB 上的具体位置无关。解答时务必引用定理：“同一弧所对的圆心角等于其所对圆周角的两倍。”
7. Cumulative Frequency Graphs | 累积频率图
≤ 250 (6). We first construct a cumulative frequency column and then estimate the median.
一张表格给出了 60 个苹果的质量分组：0 < m ≤ 50（8 个），50 < m ≤ 100（14 个），100 < m ≤ 150（20 个），150 < m ≤ 200（12 个），200 < m ≤ 250（6 个）。我们首先构建累积频数列，然后估算中位数。
- Cumulative frequency: 8, 22, 42, 54, 60.
- 累积频数：8，22，42，54，60。
- Plot points at upper class boundaries: (50,8), (100,22), (150,42), (200,54), (250,60) and join with a smooth curve.
- 在上组界处描点：(50,8)，(100,22)，(150,42)，(200,54)，(250,60)，并用平滑曲线连接。
0th value. From the curve, read across from 30 on the cumulative axis and down to the mass axis; the median is approximately 123 g. Using the graph, you can also find quartiles and interquartile range.
中位数对应第 ½ × 60 = 30 个值。从累积频数轴上 30 处横向读出，再下垂到质量轴，中位数约为 123 g。利用该图您还可以找出四分位数与四分位距。
8. Basic Probability | 基本概率
A bag contains 5 red, 3 blue and 2 green counters. We take one counter at random. Find the probability that it is (a) blue, (b) not red.
一个袋子里有 5 个红、3 个蓝和 2 个绿计数器。随机取出一个。求它是 (a) 蓝色，(b) 不是红色的概率。
Total counters = 5 + 3 + 2 = 10. Probability of blue = number of blue / total = 3/10 = 0.3. Probability of not red = 1 − P(red) = 1 − 5/10 = 5/10 = ½ = 0.5. This also equals the sum of blue and green probabilities: (3+2)/10 = 0.5.
总数 = 5 + 3 + 2 = 10。P(蓝) = 蓝色数量 / 总数 = 3/10 = 0.3。P(非红) = 1 − P(红) = 1 − 5/10 = 5/10 = ½ = 0.5。这也等于蓝与绿概率之和：(3+2)/10 = 0.5。
ercentages as requested. The complement rule, P(not A) = 1 − P(A), is extremely useful for “at least one” type questions.
根据题目要求，概率可用最简分数、小数或百分数表示。补集规则 P(非 A) = 1 − P(A) 对“至少一个”类问题极为有用。
9. Vectors in Geometry | 几何中的向量
Triangle ABC has vector AB = a and AC = b. M is the midpoint of BC. Express vector AM in terms of a and b.
三角形 ABC 中，向量 AB = a，AC = b。M 为 BC 中点。试用 a 和 b 表示向量 AM。
First find BC: BC = BA + AC = −a + b = b − a. Since M is the midpoint, BM = ½ BC = ½(b − a). Then AM = AB + BM = a + ½(b − a) = a + ½b − ½a = ½a + ½b = ½(a + b).
先求 BC：BC = BA + AC = −a + b = b − a。因 M 为中点，BM = ½ BC = ½(b − a)。于是 AM = AB + BM = a + ½(b − a) = a + ½b − ½a = ½a + ½b = ½(a + b)。
This shows that the median from A goes to the midpoint of BC and its vector is the average of AB and AC. Often a parallelogram or ratio method can also be used; always show clear vector addition steps.
这表明从 A 出发的中线指向 BC 的中点，其向量是 AB 与 AC 之和的一半。通常也可用平行四边形法或比例法；务必展示清晰的向量加法步骤。
10. Composite Functions | 复合函数
uate (a) fg(4) and (b) gf(x). The order of operations is crucial: fg(x) means f[g(x)], i.e. apply g first, then f.
已知 f(x) = 2x + 3，g(x) = x² − 1，求 (a) fg(4) 与 (b) gf(x)。运算顺序至关重要：fg(x) 表示 f[g(x)]，即先作用 g，再作用 f。
For fg(4): g(4) = 4² − 1 = 16 − 1 = 15. Then f(15) = 2(15) + 3 = 30 + 3 = 33.
计算 fg(4)：g(4) = 4² − 1 = 16 − 1 = 15。然后 f(15) = 2(15) + 3 = 30 + 3 = 33。
For gf(x): f(x) = 2x + 3, then g[f(x)] = (2x + 3)² − 1 = 4x² + 12x + 9 − 1 = 4x² + 12x + 8. Note how we substituted the whole of f(x) into g(x).
计算 gf(x)：先得 f(x) = 2x + 3，然后 g[f(x)] = (2x + 3)² − 1 = 4x² + 12x + 9 − 1 = 4x² + 12x + 8。请注意我们将 f(x) 整体代入 g(x)。
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