Year 10 CAIE Mathematics: Key Concepts Overview | Year 10 CAIE 数学：核心知识点梳理

📚 Year 10 CAIE Mathematics: Key Concepts Overview | Year 10 CAIE 数学：核心知识点梳理

Welcome to this review of the key concepts in Year 10 CAIE Mathematics. This guide covers the essential topics that form the foundation for your IGCSE studies, including number operations, algebra, geometry, and statistics. We will walk through each major area with clear explanations and worked examples to help you build confidence.

欢迎来到本文 Year 10 CAIE 数学核心知识点梳理。本指南涵盖构成 IGCSE 学习基础的关键主题，包括数字运算、代数、几何与统计。我们将通过清晰的解释和实际示例逐一梳理每个重点领域，帮助你建立信心。

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1. Numbers and Operations | 数字与运算

Number skills are the bedrock of all mathematics. You must be confident working with integers, fractions, decimals, and percentages, switching fluently between them. Everyday calculations, such as finding 15% of a restaurant bill or increasing a quantity by a ratio, rely on these skills.

数字技巧是所有数学的基石。你必须熟练处理整数、分数、小数和百分数，并在这些形式之间自如转换。日常计算，例如计算餐厅账单的 15% 或按比例增加数量，都依赖于这些技能。

Standard form (scientific notation) is used to express very large or very small numbers efficiently. The number 0.000034 becomes 3.4 × 10⁻⁵, while 67 000 000 is written as 6.7 × 10⁷. Remember to use upper and lower bounds when measurements are given to a certain degree of accuracy.

标准形式（科学记数法）用于高效表示极大或极小的数字。0.000034 写成 3.4 × 10⁻⁵，而 67 000 000 则写成 6.7 × 10⁷。当测量值给出一定精度时，请记得使用上界和下界。

Ratios and proportions appear in recipes, maps, and scale drawings. If a mixture needs cement, sand, and gravel in the ratio 1 : 2 : 4, doubling the total requires multiplying each part by the same scale factor. Direct and inverse proportion are also examined, so be prepared to solve problems like ‘y is directly proportional to x²’.

比率和比例出现在食谱、地图和比例图中。如果一种混合物需要水泥、沙子和石子按 1 : 2 : 4 的比例，那么总量加倍就需要将每部分乘以相同的倍数。正比和反比也会考察，因此要准备好解决类似“y 与 x² 成正比”的问题。

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2. Algebraic Expressions | 代数表达式

You will expand and simplify expressions involving single and double brackets. For instance, 3(2x – 5) gives 6x – 15, and (x + 4)(x – 3) expands to x² + x – 12. Always re‑multiply to check your sign work.

你将展开并化简含有单项和两项括号的表达式。例如，3(2x – 5) 得出 6x – 15，而 (x + 4)(x – 3) 展开得到 x² + x – 12。每次都要重新相乘来检查符号是否正确。

Factorising is the reverse process: turning a sum into a product. For quadratics like x² + 7x + 10, find two numbers whose product is 10 and sum is 7 — they are 2 and 5, giving (x + 2)(x + 5). When the coefficient of x² is not 1, such as 2x² + 7x + 3, you need to split the middle term before factorising by grouping.

因式分解是逆过程：将求和转化为求积。对于像 x² + 7x + 10 这样的二次式，找出两个数，其乘积为 10，和为 7 —— 它们是 2 和 5，得到 (x + 2)(x + 5)。当 x² 的系数不是 1 时，例如 2x² + 7x + 3，需要先拆分中间项再进行分组因式分解。

Algebraic fractions are treated like numeric fractions. To simplify 1/(x – 2) + 3/(x + 2), write a common denominator (x – 2)(x + 2). After expanding the numerator and simplifying, you obtain a single fraction. Remember to state that x ≠ 2, –2.

代数分式可按数值分式的方式处理。要化简 1/(x – 2) + 3/(x + 2)，写出公分母 (x – 2)(x + 2)。分子展开并化简后，可得到单个分式。切记注明 x ≠ 2、–2。

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3. Linear Equations and Inequalities | 线性方程与不等式

Solving an equation such as 4x – 9 = 2x + 3 involves collecting like terms: 4x – 2x = 3 + 9 → 2x = 12 → x = 6. Always verify by substitution. When fractions appear, multiply every term by the lowest common denominator.

解方程 4x – 9 = 2x + 3 需要合并同类项：4x – 2x = 3 + 9 → 2x = 12 → x = 6。务必代入验证。当出现分数时，每一项都要乘以最小公分母。

Linear inequalities are solved similarly, but remember to reverse the sign if you multiply or divide by a negative number. The solution set of –2x ≤ 10 is x ≥ –5. Inequalities are often displayed on a number line with an open or closed circle.

线性不等式的解法类似，但切记：若乘以或除以负数，不等号方向需反转。–2x ≤ 10 的解集为 x ≥ –5。不等式常用数轴表示，并配以空心或实心圆点。

Simultaneous equations arise when two conditions link the same variables. With y = 3x – 2 and x + y = 10, substituting gives x + (3x – 2) = 10, leading to x = 3 and y = 7. The elimination method is useful when coefficients align neatly.

当两个条件涉及相同变量时，便会引出联立方程。已知 y = 3x – 2 和 x + y = 10，代入后得 x + (3x – 2) = 10，从而求出 x = 3、y = 7。若系数对齐整齐，消元法十分好用。

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4. Sequences | 数列

A linear sequence changes by a constant difference. The nth term of 5, 8, 11, 14, … is 3n + 2. Once you have the nth term, you can find any term, such as the 100th term: 3(100) + 2 = 302.

线性数列按恒定公差变化。数列 5, 8, 11, 14, … 的第 n 项为 3n + 2。一旦有了第 n 项，你就可以求出任意项，例如第 100 项：3(100) + 2 = 302。

Quadratic sequences have a constant second difference. For 2, 7, 14, 23, 34, …, the first differences are 5, 7, 9, 11, and the second difference is 2, so the nth term involves n². Setting up the form an² + bn + c and solving with known terms gives a = 1, b = 2, c = –1, so the nth term is n² + 2n – 1.

二次数列的二阶差恒定。对于数列 2, 7, 14, 23, 34, …，一阶差为 5, 7, 9, 11，二阶差为 2，因此第 n 项含 n²。设第 n 项为 an² + bn + c，代入已知项求解，得 a = 1、b = 2、c = –1，故第 n 项为 n² + 2n – 1。

Recognise special sequences such as square numbers (1, 4, 9, 16, …), triangular numbers (1, 3, 6, 10, …), and Fibonacci‑type patterns where each term is the sum of the two preceding it.

识别特殊数列，如平方数（1, 4, 9, 16, …）、三角形数（1, 3, 6, 10, …）以及每项均为前两项之和的斐波那契型模式。

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5. Angles and Polygons | 角度与多边形

Knowing angle facts saves time: angles on a straight line add up to 180°, angles around a point total 360°, and vertically opposite angles are equal. With parallel lines, alternate angles are equal, corresponding angles are equal, and co‑interior angles sum to 180°.

熟记角度事实可节省时间：直线上的角度和为 180°，一点周角为 360°，对顶角相等。平行线中，内错角相等，同位角相等，同旁内角互补（和为 180°）。

For any polygon, the sum of interior angles = (n – 2) × 180°. A regular polygon has equal interior angles, each measuring (n – 2) × 180° ÷ n. Always remember that the exterior angles, taken one at each vertex, sum to 360°, regardless of the number of sides.

对于任意多边形，内角和 = (n – 2) × 180°。正多边形的每个内角相等，度数为 (n – 2) × 180° ÷ n。始终牢记，每个顶点取一个外角时，外角和总是 360°，与边数无关。

These properties are often combined in diagrams with triangles and quadrilaterals. For example, to find a missing angle in a pentagon, use the interior angle sum and subtract the known angles.

这些性质常与三角形和四边形的图示结合。例如，要求五边形中的未知角，可借助内角和再减去已知角度。

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6. Pythagoras’ Theorem and Trigonometry | 勾股定理与三角学

Pythagoras’ theorem states that in any right‑angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.

a² + b² = c²

If a = 6 cm and b = 8 cm, then c = √(6² + 8²) = 10 cm. Use this to check if a triangle is right‑angled or to find distances on a coordinate grid.

若 a = 6 cm，b = 8 cm，则 c = √(6² + 8²) = 10 cm。用此定理检验三角形是否为直角三角形，或求坐标网格上的距离。

The three trigonometric ratios relate an acute angle to two sides:

sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent

Given one side and one angle, you can find any other side. For example, if a supporting cable makes a 40° angle with the ground and the horizontal distance is 15 m, the cable length L satisfies cos 40° = 15/L, so L = 15 / cos 40°.

已知一边一角，可求任意其他边。例如，一根支撑缆绳与地面成 40° 角，且水平距离为 15 m，缆绳长度 L 满足 cos 40° = 15/L，因此 L = 15 / cos 40°。

Bearings are measured clockwise from north, written as three‑digit angles (e.g.

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