📚 Year 10 CCEA Mathematics: In-Depth Analysis of Past Paper Questions | Year 10 CCEA 数学：历年真题深度解析
Welcome to this comprehensive revision guide focusing on past paper questions from CCEA Year 10 Mathematics. Working through past papers is one of the most effective ways to prepare for examinations, as it helps you become familiar with the style of questions, common pitfalls, and time management. This article will analyse key topics frequently tested in CCEA Year 10 exams, providing step-by-step solutions and expert tips to boost your confidence and performance.
欢迎阅读这篇专注于CCEA Year 10数学历年真题的全面复习指南。通过真题练习是备考的最有效方法之一，它能帮助你熟悉题型、常见错误和时间管理。本文将深入分析CCEA Year 10考试中经常出现的重点题目，提供逐步解析和专家提示，助你提升自信与成绩。
1. Solving Linear Equations | 解线性方程
The question is: Solve 3(x – ) = 2x + 1. This type of linear equation appears frequently in CCEA non-calculator papers.
题目为：解方程 3(x – 4) = 2x + 1。这类线性方程在CCEA非计算器试卷中经常出现。
Step 1: Expand the bracket. Multiply 3 by each term inside: 3x – 12 = 2x + 1.
第一步：展开括号。将3乘以括号内每一项：3x – 12 = 2x + 1。
Step 2: Collect x terms on one side. Subtract 2x from both sides: 3x – 2x – 12 = 1, which simplifies to x – 12 = 1.
第二步：将含x的项移到同一边。两边减去2x：3x – 2x – 12 = 1，简化为 x – 12 = 1。
Step 3: Isolate x. Add 12 to both sides: x = 13.
第三步：求出x。两边加12：x = 13。
Always check by substituting back: 3(13 – 4) = 3×9 = 27, and 2×13 + 1 = 27. Correct!
一定要代入检验：3(13 – 4) = 3×9 = 27，而 2×13 + 1 = 27。正确！
Common mistake: forgetting to distribute the negative sign or mishandling the constant term. Write each step carefully.
常见错误：忘记分配负号或处理常数项不当。务必仔细写出每一步。
x = 13
2. Percentage Increase and Decrease | 百分比增减
A jacket is priced at £85. In a sale, it is reduced by 15%. Find the sale price. Percentages are a staple of CCEA foundation and higher papers.
一件夹克售价85英镑。促销期间降价15%。求促销价。百分比是CCEA基础与进阶试卷都必考的内容。
Method 1: Calculate 15% of £85: 0.15 × 85 = 12.75. Subtract from original: 85 – 12.75 = £72.25.
方法一：计算85英镑的15%：0.15 × 85 = 12.75。从原价中减去：85 – 12.75 = 72.25英镑。
Method 2: A 15% reduction means you pay 100% – 15% = 85% of the original price. So sale price = 0.85 × 85 = £72.25.
方法二：降价15%意味着你支付原价的100% – 15% = 85%。因此促销价 = 0.85 × 85 = 72.25英镑。
Using the multiplier method is faster and reduces errors. Always show your working to gain method marks even if the final answer is slightly off.
使用乘数法更快且不易出错。即使最终答案略有偏差，也要展示解题过程以获得步骤分。
Sale price = £72.25
3. Pythagoras' Theorem | 勾股定理
classic tests your ability to identify the hypotenuse and apply the formula c² = a² + b².
一个直角三角形两条直角边分别为6 cm和8 cm。计算斜边长度。这道CCEA经典考题检验你识别斜边并应用公式 c² = a² + b² 的能力。
Step 1: Write the theorem: In a right triangle, the square of the hypotenuse equals the sum of squares of the other two sides.
第一步：写出定理：在直角三角形中，斜边的平方等于两条直角边的平方和。
Step 2: Substitute: c² = 6² + 8² = 36 + 64 = 100.
第二步：代入：c² = 6² + 8² = 36 + 64 = 100。
Step 3: Take the square root: c = √100 = 10 cm.
第三步：开平方：c = √100 = 10 cm。
Always check that the hypotenuse is the longest side. 10 cm is greater than 6 and 8, so the answer is reasonable.
总是检查斜边是否最长边。10 cm 大于 6 和 8，所以答案合理。
Hypotenuse = 10 cm
4. Trigonometry in Right-Angled Triangles | 直角三角形中的三角函数
rrect to one decimal place. This is a typical CCEA question requiring use of the tangent ratio.
在三角形PQR中，角Q = 90°，边PQ = 12米，角P = 35°。求边QR的长度，精确至一位小数。这是一道典型的CCEA题，需要使用正切比。
angle P (35°): PQ is adjacent, QR is opposite. So tan 35° = opposite / adjacent = QR / 12.
确定相对角P（35°）的边：PQ是邻边，QR是对边。因此 tan 35° = 对边 / 邻边 = QR / 12。
tan 35°. Use a calculator to find tan 35° ≈ 0.7002075. Multiply: 12 × 0.7002075 ≈ 8.40249.
移项：QR = 12 × tan 35°。用计算器求 tan 35° ≈ 0.7002075。相乘：12 × 0.7002075 ≈ 8.40249。
Round to one decimal place: 8.4 m. Always check your calculator is in degree mode, not radians.
四舍五入至一位小数：8.4 m。务必检查计算器是否处于度数模式，而非弧度模式。
QR ≈ 8.4 m
5. Simultaneous Equations | 联立方程
Solve the simultaneous equations: 2x + y = 7, x – 2. CCEA exams often reward candidates who can efficiently use elimination or substitution.
解联立方程：2x + y = 7, x – y = 2。CCEA考试通常会奖励那些能有效运用消元法或代入法的考生。
Method (Elimination): Add the two equations to eliminate y: (2x + y) + (x – y) = 7 + 2 → 3x = 9 → x = 3.
方法（消元法）：将两式相加消去y：(2x + y) + (x – y) = 7 + 2 → 3x = 9 → x = 3。
Substitute x = 3 into the second equation: 3 – y = 2 → –y = 2 – 3 = –1 → y = 1.
将 x = 3 代入第二个方程：3 – y = 2 → –y = 2 – 3 = –1 → y = 1。
) + 1 = 6 + 1 = 7, which is correct. Always verify your solution satisfies both equations.
代入第一个方程检验：2(3) + 1 = 6 + 1 = 7，正确。始终验证解是否满足两个方程。
x = 3, y = 1
6. Probability from Two-Way Tables | 双向表概率
A two-way table shows 50 students choosing between Art and Music. 12 study both, 10 study Art only, 18 study Music only, and 10 study neither. If a student is randomly selected, find the probability that they study Art. This type of data handling question is very common in CCEA past papers.
一张双向表显示了50名学生在美术与音乐之间的选择。12人两科都学，10人只学美术，18人只学音乐，10人都不学。如果随机选一名学生，求他学习美术的概率。这类数据处理题在CCEA历年真题中非常常见。
+ 10 + 18 + 10 = 50. Then find the number who study Art (both + Art only) = 12 + 10 = 22.
首先，计算学生总数：12 + 10 + 18 + 10 = 50。然后计算学美术的人数（两科都学 + 只学美术）= 12 + 10 = 22。
omes / total outcomes = 22/50. Simplify by dividing numerator and denominator by 2: 11/25.
概率 = 有利结果数 / 总结果数 = 22/50。分子分母同除以2化简：11/25。
Remember to simplify fractions where possible; CCEA mark schemes often award the final mark only for the simplest form.
记得尽可能化简分数；CCEA评分方案通常只给最简形式的答案满分。
P(studies Art) = 11/25
7. Area and Circumference of a Circle | 圆的面积与周长
A circle has a diameter of 14 cm. Find its area and circumference, leaving both answers in terms of π. CCEA frequently tests working with exact answers involving π.
一个圆的直径为14厘米。求其面积与周长，答案保留π。CCEA经常考查与π相关的精确值计算。
Radius r = diameter / 2 = 14 / 2 = 7 cm. Circumference = π × diameter = π × 14 = 14π cm.
半径 r = 直径 / 2 = 14 / 2 = 7 cm。周长 = π × 直径 = π × 14 = 14π cm。
π × r² = π × 7² = π × 49 = 49π cm². Remember the units: circumference in cm, area in cm².
面积 = π × r² = π × 7² = π × 49 = 49π cm²。注意单位：周长以cm为单位，面积以cm²为单位。
If the question asks to leave answers in terms of π, do not use 3.14 or the calculator value; otherwise you may lose accuracy marks.
如果题目要求保留π，就不要使用3.14或计算器近似值，否则可能丢失精确度分数。
Circumference = 14π cm, Area = 49π cm²
8. Standard Form Calculations | 标准形式计算
our answer in standard form. This topic blends indices and decimals, a favourite in CCEA calculator and non-calculator papers.
计算 (3.6 × 10⁴) × (5 × 10⁻²)。将答案用标准形式表示。该知识点融合了指数与小数，是CCEA计算器与非计算器试卷中的热门考点。
tiply the number parts: 3.6 × 5 = 18.0. Add the powers of 10: 10⁴ × 10⁻² = 10⁴⁺⁽⁻²⁾ = 10².
数字部分相乘：3.6 × 5 = 18.0。10的幂相加：10⁴ × 10⁻² = 10⁴⁺⁽⁻²⁾ = 10²。
e the first part must be between 1 and 10. Adjust: 18 × 10² = 1.8 × 10¹ × 10² = 1.8 × 10³.
结果：18 × 10²。这还不是标准形式，因为第一部分必须在1到10之间。调整：18 × 10² = 1.8 × 10¹ × 10² = 1.8 × 10³。
Standard form is a × 10ⁿ, where 1 ≤ a < 10. Always check this condition.
标准形式为 a × 10ⁿ，其中 1 ≤ a < 10。务必检查此条件。
(3.6 × 10⁴) × (5 × 10⁻²) = 1.8 × 10³
9. nth Term of a Linear Sequence | 线性序列的通项
Find the nth term of the sequence: 7, 12, 17, 22, … Hence find the 60th term. CCEA expects you to recognise arithmetic sequences and use the formula a + (n – 1)d or the difference method.
求序列 7, 12, 17, 22, … 的通项，并据此求出第60项。CCEA要求你识别等差数列，并使用 a + (n – 1)d 或差分法。
Find the common difference: 12 – 7 = 5, 17 – 12 = 5. So the coefficient of n is 5. The sequence can be written as 5n + c.
求公差：12 – 7 = 5，17 – 12 = 5。所以n的系数为5。序列可表示为 5n + c。
When n = 1, 5(1) + c = 7 → 5 + c = 7 → c = 2. Thus nth term = 5n + 2.
当 n = 1 时，5(1) + c = 7 → 5 + c = 7 → c = 2。因此通项 = 5n + 2。
0: 5 × 60 + 2 = 300 + 2 = 302. Always test with a known term, e.g. n=3: 5×3+2=17, correct.
第60项：代入n=60：5 × 60 + 2 = 300 + 2 = 302。总是用已知项验证，例如n=3：5×3+2=17，正确。
nth term = 5n + 2, 60th term = 302
10. Angles in Parallel Lines |
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