Year 11 CAIE Mathematics: Unit Test Mock Paper Analysis | Year 11 CAIE 数学：单元测试模拟卷解析

📚 Year 11 CAIE Mathematics: Unit Test Mock Paper Analysis | Year 11 CAIE 数学：单元测试模拟卷解析

This article walks you through a full mock unit test designed for Year 11 CAIE IGCSE Mathematics. Each question is broken down with clear steps, common pitfalls, and essential revision notes to help you secure top marks in your actual exam.

本文带你逐题解析一套为 Year 11 CAIE IGCSE 数学设计的单元测试模拟卷。每道题都配以清晰的解题步骤、常见错误提示和核心考点总结，帮助你在真正的考试中稳稳拿下高分。

1. Percentages and Compound Interest | 百分数与复利

Question: A jacket is priced at $120. During a sale, it is reduced by 15%. A customer must then pay a sales tax of 8% on the reduced price. Find the final amount the customer pays. Additionally, $2500 is invested at a compound interest rate of 3.6% per annum, compounded annually. Calculate the value of the investment after 4 years, giving your answer to the nearest cent.

题目：一件夹克标价 120 美元。促销期间降价 15%，之后顾客还需按折扣价支付 8% 的消费税。求顾客最终支付的金额。另将 2500 美元以年利率 3.6% 每年复利一次投资，计算 4 年后的投资价值，答案精确到美分。

Solution – discount and tax: First, calculate the sale price: 15% off means multiplier = 1 − 0.15 = 0.85. Sale price = 120 × 0.85 = $102. Then apply tax: multiplier = 1 + 0.08 = 1.08. Final price = 102 × 1.08 = $110.16.

解答 – 折扣与税：首先计算折扣价：8.5 折即乘数 = 1 − 0.15 = 0.85，折扣价为 120 × 0.85 = 102 美元。再加税：乘数 = 1 + 0.08 = 1.08，最终价格 = 102 × 1.08 = 110.16 美元。

Compound interest: mpute stepwise: (1.036)² = 1.073296, then (1.073296)² ≈ 1.15196. Thus A ≈ 2500 × 1.15196 = $2879.90. (More precisely $2879.92 when using unrounded intermediate values.)

复利计算：使用公式 A = P(1 + r/100)ⁿ。此处 P = 2500, r = 3.6, n = 4。A = 2500 × (1.036)⁴。逐步计算：(1.036)² = 1.073296，(1.073296)² ≈ 1.15196，因此 A ≈ 2500 × 1.15196 = 2879.90 美元（若不中途舍入则为 2879.92 美元）。

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2. Algebraic Fractions | 代数分式

Question: Simplify fully: (x² − 9) / (x² − 2x − 3) ÷ (x + 3) / (x + 1).

题目：化简： (x² − 9) / (x² − 2x − 3) ÷ (x + 3) / (x + 1)。

Step 1 – Factorise all polynomials: − 3)(x + 1). The expression becomes [(x − 3)(x + 3)] / [(x − 3)(x + 1)] ÷ (x + 3)/(x + 1).

第一步 – 因式分解：x² − 9 = (x − 3)(x + 3)；x² − 2x − 3 = (x − 3)(x + 1)。原式变为 [(x − 3)(x + 3)] / [(x − 3)(x + 1)] ÷ (x + 3)/(x + 1)。

Step 2 – Change division to multiplication: rs: (x − 3) cancels, (x + 3) cancels, (x + 1) cancels. Result = 1, provided x ≠ 3, −3, −1.

第二步 – 除变乘：÷ (x + 3)/(x + 1) 变为 × (x + 1)/(x + 3)。得到 [(x − 3)(x + 3)] / [(x − 3)(x + 1)] × (x + 1)/(x + 3)。约去公因式：(x − 3)、(x + 3) 和 (x + 1) 全部抵消，结果为 1，但需注明 x ≠ 3, −3, −1。

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3. Solving Quadratic Equations | 解二次方程

Question: nd the x-coordinates of the points where the graph of y = x² − 7x + 10 crosses the x-axis.

题目：用因式分解法解方程 x² − 7x + 10 = 0。由此或另法，求出抛物线 y = x² − 7x + 10 与 x 轴交点的 x 坐标。

Solution – Factorising: and add to −7: −5 and −2. So x² − 7x + 10 = (x − 5)(x − 2) = 0. Therefore x = 5 or x = 2.

解答 – 因式分解：寻找积为 +10、和为 −7 的两个数：−5 和 −2。故 x² − 7x + 10 = (x − 5)(x − 2) = 0，解得 x = 5 或 x = 2。

Graph interpretation: The solutions to the quadratic equation are exactly the x-intercepts. Hence the curve crosses the x-axis at (2, 0) and (5, 0).

图像解释：二次方程的解正是图像与 x 轴交点的横坐标。因此抛物线与 x 轴交于 (2, 0) 和 (5, 0)。

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4. Simultaneous Equations | 联立方程组

Question: d x − y = 1. Give both the algebraic solution and a rough sketch showing the intersection.

题目：解方程组：2x + 3y = 12 与 x − y = 1。要求给出代数解并绘制简图表示交点。

Algebraic method – substitution: + 1) + 3y = 12 → 2y + 2 + 3y = 12 → 5y = 10 → y = 2. Then x = 2 + 1 = 3. Solution: (3, 2).

代数法 – 代入：由 x − y = 1 得 x = y + 1。代入第一个方程：2(y + 1) + 3y = 12 → 2y + 2 + 3y = 12 → 5y = 10 → y = 2。于是 x = 3。解为 (3, 2)。

Graphical check: and y = x − 1. Plotting shows lines crossing at (3,2). Always check by substituting back.

图形检验：将两方程改写为斜截式：y = (12 − 2x)/3 和 y = x − 1。绘图可见两直线交于 (3, 2)。务必代回原方程检验。

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5. Trigonometry in Right-Angled Triangles | 直角三角形中的三角学

Question: e C = 90°, AB = 13 cm (hypotenuse), and BC = 5 cm. Find angle A and the length of side AC.

题目：在直角三角形 ABC 中，∠C = 90°，斜边 AB = 13 cm，BC = 5 cm。求 ∠A 和边 AC 的长度。

Finding AC using Pythagoras: AC² = AB² − BC² = 13² − 5² = 169 − 25 = 144. So AC = √144 = 12 cm.

用勾股定理求 AC：AC² = AB² − BC² = 13² − 5² = 169 − 25 = 144，故 AC = √144 = 12 cm。

Finding angle A: 22.6°. You could also use cos A = adjacent/hyp = AC/AB = 12/13 → A = cos⁻¹(12/13) ≈ 22.6°.

求 ∠A：sin A = 对边/斜边 = BC/AB = 5/13，因此 A = sin⁻¹(5/13) ≈ 22.6°。也可用 cos A = 邻边/斜边 = 12/13 求得相同角度。

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6. Vectors and Magnitude | 向量与模长

Question: Given vector a = (3, −4) and vector b = (−2, 5), find: (a) 2a − b; (b) |b|, the magnitude of b; (c) the unit vector in the direction of a.

题目：已知向量 a = (3, −4)，b = (−2, 5)，求：(a) 2a − b；(b) |b|，即 b 的模长；(c) a 方向上的单位向量。

2a − b: 2a = (6, −8). Subtract b: (6 − (−2), −8 − 5) = (8, −13).

2a − b：2a = (6, −8)。减去 b： (6 − (−2), −8 − 5) = (8, −13)。

Magnitude of b: |b| = √[(-2)² + 5²] = √4 + 25 = √29. That is the exact form; approximate 5.385.

b 的模长：|b| = √[(-2)² + 5²] = √4 + 25 = √29（精确值，约等于 5.385）。

Unit vector of a: First, |a| = √(3² + (−4)²) = √9 + 16 = √25 = 5. Unit vector = (1/5)·a = (3/5, −4/5) or (0.6, −0.8).

a 的单位向量：先求 |a| = √(3² + (−4)²) = √25 = 5。单位向量 = (1/5)·a = (3/5, −4/5) 或写作 (0.6, −0.8)。

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7. Probability Tree Diagrams | 概率树形图

Question: A bag contains 3 red and 5 blue marbles. Two marbles are drawn without replacement. Draw a tree diagram and find the probability that both marbles are the same colour.

题目：一个袋子里有 3 个红球和 5 个蓝球，不放回地连续摸出两个球。画出树形图，并求两球颜色相同的概率。

Tree diagram and probabilities: e = 5/7. If first is Blue, remaining: 3 Red, 4 Blue → second Red = 3/7, second Blue = 4/7.

树形图与概率：第一次：P(红) = 3/8，P(蓝) = 5/8。若第一次为红，剩下 2 红 5 蓝 → 第二次红 = 2/7，蓝 = 5/7。若第一次为蓝，剩下 3 红 4 蓝 → 第二次红 = 3/7，蓝 = 4/7。

Same colour = both red OR both blue: /8) × (2/7) = 6/56. P(BB) = (5/8) × (4/7) = 20/56. Sum = 26/56, which simplifies to 13/28.

同色 = 两红或两蓝：P(RR) = (3/8) × (2/7) = 6/56，P(BB) = (5/8) × (4/7) = 20/56。相加得 26/56，化简为 13/28。

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8. Statistics: Mean, Median and Cumulative Frequency | 统计：平均数、中位数与累积频数

Question: The table below shows the test scores of 40 students. Estimate the mean, the median class, and draw a cumulative frequency curve to find the median and interquartile range.

题目：下表为 40 名学生的考试成绩。估计平均分、中位数所在组，并绘制累积频数曲线以求中位数和四分位距。

Scores: 0–10 (freq 4), 10–20 (f=8), 20–30 (f=12), 30–40 (f=10), 40–50 (f=6).

分数段：0–10（频数 4），10–20（8），20–30（12），30–40（10），40–50（6）。

Estimated mean: ×8 + 25×12 + 35×10 + 45×6 = 20 + 120 + 300 + 350 + 270 = 1060. Divide by 40 → mean = 26.5.

估计平均分：取组中值：5, 15, 25, 35, 45。∑(组中值×频数)= 5×4 + 15×8 + 25×12 + 35×10 + 45×6 = 1060。除以 40，得平均分 26.5。

Cumulative frequency and median: CFs: 4, 12, 24, 34, 40. Median position = 20.5th value → lies in 20– ead median ≈ 25.5. Lower quartile (10.25th) ≈ 17; upper quartile (30.75th) ≈ 35. IQR = 18.

累积频数与中位数：累积频数：4, 12, 24, 34, 40。中位数位置 = 20.5，位于 20–30 组。由累积频数曲线读得中位数 ≈ 25.5，下四分位数 ≈ 17，上四分位数 ≈ 35，四分位距 = 18。

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9. Circle Theorems | 圆定理

Question: h centre O. Angle AOB = 110°. Find angle ACB. Also, if AB is a diameter, what is angle ACB?

题目：如图，A、B、C 为以 O 为圆心的圆上的点，∠AOB = 110°。求 ∠ACB。若 AB 为直径，∠ACB 又是多少？

Angle at the centre theorem: The angle subtended by arc AB at the centre is twice the angle at the circumference. So ∠AOB = 2 × ∠ACB. Thus 110° = 2 × ∠ACB → ∠ACB = 55°.

圆心角定理：弧 AB 所对的圆心角是圆周角的两倍。故 ∠AOB = 2 × ∠ACB，110° = 2 × ∠ACB → ∠ACB = 55°。

Angle in a semicircle: ameter, then arc AB is a semicircle. The angle in a semicircle is always a right angle, so ∠ACB = 90°. This is a special case of the theorem above (2 × 90° = 180°).

半圆上的圆周角：若 AB 为直径，则弧 AB 是半圆，半圆所对的圆周角为直角，即 ∠ACB = 90°。这是上述定理的特例（2 × 90° = 180°）。

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10. Functions and Inverse Functions | 函数与反函数

Question: and verify using the inverse. If g(x) = x² + 1 for x ≥ 0, find fg(2) and the range of fg.

题目：已知 f(x) = (2x − 1)/3，求 f⁻¹(x)。解方程 f(x) = 7 并用反函数验证。设 g(x) = x² + 1，定义域 x ≥ 0，求 fg(2) 以及 fg 的值域。

Finding inverse: y − 1)/3. Solve for y: 3x = 2y − 1 → 2y = 3x + 1 → y = (3x + 1)/2. So f⁻¹(x) = (3x + 1)/2.

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

Solve f(x) = 7: 2x − 1)/3 = 7 → 2x − 1 = 21 → x = 11. Check: f⁻¹(7) = (3×7 + 1)/2 = 22/2 = 11. Consistent.

解方程：(2x − 1)/3 = 7 → 2x − 1 = 21 → x = 11。验证：f⁻¹(7) = (21+1)/2 = 11，一致。

Composite fg(2): First find g(2) = 2² + 1 = 5. Then f(5) = (2×5 − 1)/3 = 9/3 = 3.

fg(2)：先算 g(2) = 4 + 1 = 5，再算 f(5) = (10−1)/3 = 3。

Range of fg: f(g(x)) will produce all values ≥ f(1). f(1) = (2−1)/3 = 1/3. Hence range of fg is y ≥ ⅓.

fg 的值域：x ≥ 0 时 g(x) ≥ 1，f 为斜率为正的线性函数，因此 f(g(x)) 的最小值为 f(1) = (2−1)/3 = ⅓，值域为 y ≥ ⅓。

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11. Straight Line Graphs and Gradient | 直线图像与斜率

Question: the equation in the form y = mx + c, and the coordinate where the line crosses the y-axis.

题目：直线 L 经过点 (2, 5) 和 (6, 17)。求斜率、以 y = mx + c 形式写出直线方程，并求出直线与 y 轴交点的坐标。

Gradient: m = (y₂ − y₁) / (x₂ − x₁) = (17 − 5) / (6 − 2) = 12/4 = 3.

斜率：m = (17 − 5) / (6 − 2) = 12/4 = 3。

Equation using point-slope: 3(x − 2) → y = 3x − 6 + 5 → y = 3x − 1. So c = −1. The line crosses the y-axis at (0, −1).

方程：用点斜式：y − 5 = 3(x − 2) → y = 3x − 1。因此 c = −1，与 y 轴交于 (0, −1)。

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12. Area and Perimeter of Composite Shapes | 组合图形的面积与周长

Question: he shorter sides. Calculate the total area and the perimeter of the shape (take π = 3.14).

题目：一个图形由一个矩形和一个半圆组成，矩形长 10 cm、宽 4 cm，半圆在矩形的一条短边上。求该组合图形的总面积和周长（π 取 3.14）。

Area: radius = 4/2 = 2 cm → area = ½ × π × 2² = ½ × 3.14 × 4 = 6.28 cm². Total area = 46.28 cm².

面积：矩形面积 = 10 × 4 = 40 cm²。半圆半径 2 cm，面积 = ½ × 3.14 × 4 = 6.28 cm²。总面积 = 46.28 cm²。

Perimeter: : The straight edges include three sides of the rectangle: 10 + 4 + 10 = 24 cm. The curved part is half circumference: ½ × 2πr = πr = 3.14

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