📚 Mastering Year 10 Cambridge Mathematics: A Deep Dive into Past Paper Questions | 精通剑桥Year 10数学：历年真题深度解析
This comprehensive guide dissects real past paper questions from the Year 10 Cambridge Mathematics curriculum, revealing the recurring patterns, examiner expectations, and strategic approaches needed to secure top marks. By analyzing authentic exam-style problems, we uncover the underlying concepts and common pitfalls across algebra, geometry, trigonometry, and statistics. Each section presents a typical question, a step-by-step worked solution, and key insights that bridge classroom theory with practical exam technique.
这份综合指南深入剖析Year 10剑桥数学课程中的历年真题，揭示反复出现的题型模式、考官期望以及获得高分所需的策略方法。通过分析真实的考试风格问题，我们揭示代数、几何、三角学和统计等领域的基本概念和常见陷阱。每个部分都呈现一道典型题目、逐步解题过程和关键见解，在课堂理论与实用考试技巧之间架起桥梁。
1. Algebraic Simplification and Expansion | 代数化简与展开
A common past paper question asks: ‘Simplify fully (3x – 2)(x + 5) – 2(x – 1)².’ This tests both expansion and collection of like terms under time pressure. First expand the binomial product: (3x – 2)(x + 5) = 3x² + 15x – 2x – 10 = 3x² + 13x – 10. Then square the second term: 2(x – 1)² = 2(x² – 2x + 1) = 2x² – 4x + 2. Subtracting gives 3x² + 13x – 10 – (2x² – 4x + 2) = 3x² + 13x – 10 – 2x² + 4x – 2 = x² + 17x – 12. Examiners look for correct sign handling when removing brackets preceded by a minus sign.
一道常见的历年真题要求：’完全化简 (3x – 2)(x + 5) – 2(x – 1)²。’ 这考察在时间压力下展开和合并同类项的能力。首先展开二项式乘积：(3x – 2)(x + 5) = 3x² + 15x – 2x – 10 = 3x² + 13x – 10。然后平方第二项：2(x – 1)² = 2(x² – 2x + 1) = 2x² – 4x + 2。相减得到 3x² + 13x – 10 – (2x² – 4x + 2) = 3x² + 13x – 10 – 2x² + 4x – 2 = x² + 17x – 12。考官会关注去掉前面带减号的括号时符号是否正确处理。
Marks are awarded for method, not just the final answer. A typical allocation might be M1 for expanding each part correctly, A1 for the fully simplified expression. Students lose marks by forgetting the negative sign applies to every term inside the second bracket. Practice writing out intermediate lines neatly rather than attempting mental arithmetic. In examination conditions, misplacing a single constant frequently costs a whole grade boundary.
分数不仅给最终答案，也为方法步骤打分。典型的分配可能是M1用于正确展开每个部分，A1用于完全化简的表达式。学生会因为忘记负号适用于第二个括号内的每一项而丢分。练习整洁地写出中间步骤，而不是试图心算。在考试条件下，错置一个常数就经常会导致整个等级下降。
2. Solving Quadratic Equations by Factorising | 通过因式分解解二次方程
Consider the equation: 2x² – 5x – 3 = 0. Past papers repeatedly test factorisation where the coefficient of x² exceeds 1. One effective method is the ‘ac’ approach: multiply a and c (2 × -3 = -6), then find factors of -6 that sum to b (-5). The pair -6 and 1 works because -6 + 1 = -5. Rewrite the middle term: 2x² – 6x + x – 3 = 0. Factor by grouping: 2x(x – 3) + 1(x – 3) = 0, giving (2x + 1)(x – 3) = 0. Solutions: x = -1/2 or x = 3.
考虑方程：2x² – 5x – 3 = 0。历年真题反复考察x²系数大于1时的因式分解。一种有效的方法是’ac’法：将a与c相乘（2 × -3 = -6），然后找出和为b（-5）的-6的因数。 -6和1这对因数有效，因为 -6 + 1 = -5。重写中间项：2x² – 6x + x – 3 = 0。分组分解：2x(x – 3) + 1(x – 3) = 0，得到 (2x + 1)(x – 3) = 0。解为：x = -1/2 或 x = 3。
Examiners expect the final answer clearly stated, preferably as ‘x = -1/2, x = 3’ rather than just factors. Many candidates stop after factorising and lose the final A1 mark. Some questions ask for solutions to a specific decimal place when factorisation is not possible, leading directly to the quadratic formula. Always check the discriminant (b² – 4ac) first to determine whether the quadratic factorises neatly. If the discriminant is not a perfect square, the formula is required, and marks depend on correct substitution into the formula and subsequent evaluation.
考官期望最终答案表述清晰，最好写成 ‘x = -1/2, x = 3’ 而不仅仅是因式。许多考生在完成因式分解后就停下来，从而丢失了最后的A1分。有些题目在无法因式分解时要求将解保留到指定小数位，这就直接引出了二次公式。始终先检查判别式（b² – 4ac），以确定二次式是否可整洁地分解。如果判别式不是完全平方数，就需要用公式，分数取决于正确代入公式以及随后的求值。
3. Linear Inequalities and Number Line Representation | 线性不等式与数轴表示
Solve 5 – 2x ≤ 3x + 15 and represent the solution on a number line. This type appears regularly in Paper 2. Subtract 3x from both sides: 5 – 5x ≤ 15. Subtract 5: -5x ≤ 10. Divide by -5, remembering the inequality flips: x ≥ -2. Representation on a number line uses a filled circle at -2 and an arrow extending rightwards. An open circle would be used strictly for > or < inequalities. Filled circle indicates inclusion of the boundary point.
解 5 – 2x ≤ 3x + 15 并在数轴上表示解。这类题目经常出现在Paper 2中。两边减去3x：5 – 5x ≤ 15。减去5：-5x ≤ 10。除以-5，记住不等式反向：x ≥ -2。在数轴上表示时使用在-2处的实心圆点和向右延伸的箭头。空心圆点严格用于 > 或 < 不等式。实心圆点表示包含边界点。
A common error is failing to reverse the inequality sign when multiplying or dividing by a negative quantity. Examiners penalise this heavily because it demonstrates a fundamental conceptual gap. Some questions combine inequalities into compound statements such as -3 < 2x + 1 ≤ 7, requiring solution in two parts. Solve separately: -3 < 2x + 1 gives -4 < 2x, x > -2. And 2x + 1 ≤ 7 gives 2x ≤ 6, x ≤ 3. The combined solution is -2 < x ≤ 3, shown with an open circle at -2 and a filled circle at 3 connected by a line.
一个常见错误是在乘以或除以负数时未能反转不等号。考官对此扣分很重，因为这显示了基本概念上的漏洞。有些题目会将不等式组合成复合语句，例如 -3 < 2x + 1 ≤ 7，需要分两部分求解。分别解：-3 < 2x + 1 得到 -4 < 2x，x > -2。以及 2x + 1 ≤ 7 得到 2x ≤ 6，x ≤ 3。组合解为 -2 < x ≤ 3，表示时在-2处用空心圆，在3处用实心圆，中间用线段连接。
4. Simultaneous Equations and Graphical Interpretation | 联立方程组与图像解释
Solve the system: y = 2x + 1 and 3x + 2y = 16. Substitution is the natural choice here. Replace y in the second equation: 3x + 2(2x + 1) = 16. Expand: 3x + 4x + 2 = 16, so 7x + 2 = 16, 7x = 14, x = 2. Substitute back: y = 2(2) + 1 = 5. The solution (2, 5) represents the intersection point of the two lines. Cambridge questions often ask students to verify their answers by substituting into the original equations, a habit that prevents careless errors.
解方程组：y = 2x + 1 和 3x + 2y = 16。代入法在这里是自然的选择。将y代入第二个方程：3x + 2(2x + 1) = 16。展开：3x + 4x + 2 = 16，所以 7x + 2 = 16，7x = 14，x = 2。代回：y = 2(2) + 1 = 5。解 (2, 5) 代表两条直线的交点。剑桥的题目经常要求学生通过代入原方程来验证答案，这是一个防止粗心错误的好习惯。
When the equations are both in standard form ax + by = c, the elimination method is usually faster. For example, 2x + 3y = 12 and 4x – y = 10. Multiply the second equation by 3: 12x – 3y = 30. Add to the first: 14x = 42, x = 3. Then y = 4(3) – 10 = 2. Papers include contextual problems where students must formulate the equations from a word problem before solving. Read carefully: ‘Two adult tickets and one child ticket cost £18; one adult and three child tickets cost £14.’ This translates to 2a + c = 18 and a + 3c = 14. Solving gives a = 8, c = 2.
当两个方程都已化为标准式 ax + by = c 时，消元法通常更快。例如，2x + 3y = 12 和 4x – y = 10。将第二个方程乘以3：12x – 3y = 30。与第一个相加：14x = 42，x = 3。然后 y = 4(3) – 10 = 2。试卷中还包含情境题，学生必须先根据文字题列出方程再求解。仔细阅读：’两张成人票和一张儿童票花费18英镑；一张成人票和三张儿童票花费14英镑。’ 这转化为 2a + c = 18 和 a + 3c = 14。求解得 a = 8，c = 2。
5. Functions, Domain, and Composite Functions | 函数、定义域与复合函数
Given f(x) = x² + 3 and g(x) = 2x – 1, find fg(x) and gf(x). This tests understanding of order: fg(x) means apply g first, then f. So fg(x) = f(2x – 1) = (2x – 1)² + 3 = 4x² – 4x + 1 + 3 = 4x² – 4x + 4. Conversely, gf(x) = g(x² + 3) = 2(x² + 3) – 1 = 2x² + 6 – 1 = 2x² + 5. Notice that fg(x) and gf(x) are generally different; composition is not commutative. Marks are awarded for correct substitution and simplification.
给定 f(x) = x² + 3 和 g(x) = 2x – 1，求 fg(x) 和 gf(x)。这考察对顺序的理解：fg(x) 表示先应用g，再应用f。所以 fg(x) = f(2x – 1) = (2x – 1)² + 3 = 4x² – 4x + 1 + 3 = 4x² – 4x + 4。反过来，gf(x) = g(x² + 3) = 2(x² + 3) – 1 = 2x² + 6 – 1 = 2x² + 5。注意 fg(x) 和 gf(x) 通常不同；复合运算不满足交换律。分数根据正确的代入和化简来评定。
Inverse functions also feature prominently. To find f⁻¹(x) for f(x) = (3x – 4)/2, write y = (3x – 4)/2. Swap x and y: x = (3y – 4)/2. Solve for y: 2x = 3y – 4, 3y = 2x + 4, y = (2x + 4)/3. Therefore f⁻¹(x) = (2x + 4)/3. The domain of f becomes the range of f⁻¹ and vice versa. A function only has an inverse if it is one-to-one. Past papers ask students to restrict the domain of a quadratic to make it invertible, typically by taking x ≥ 0 or x ≤ 0.
反函数也频繁出现。对于 f(x) = (3x – 4)/2 求 f⁻¹(x)，写出 y = (3x – 4)/2。交换 x 和 y：x = (3y – 4)/2。解出 y：2x = 3y – 4，3y = 2x + 4，y = (2x + 4)/3。因此 f⁻¹(x) = (2x + 4)/3。f 的定义域成为 f⁻¹ 的值域，反之亦然。一个函数只有一一对应时才有反函数。历年试卷会要求学生限制二次函数的定义域以使其可逆，通常取 x ≥ 0 或 x ≤ 0。
6. Coordinate Geometry: Gradients and Equations of Lines | 坐标几何：梯度与直线方程
Find the equation of the line passing through A(1, 2) and B(5, 10). Gradient m = (10 – 2)/(5 – 1) = 8/4 = 2. Using point-slope form with A: y – 2 = 2(x – 1), giving y = 2x. In general form: 2x – y = 0. If the question requires the perpendicular bisector of AB, first find the midpoint M: ((1+5)/2, (2+10)/2) = (3, 6). The gradient of perpendicular line = -1/2. Equation: y – 6 = -1/2(x – 3), so y = -1/2x + 15/2 or x + 2y = 15.
求通过 A(1, 2) 和 B(5, 10) 的直线方程。梯度 m = (10 – 2)/(5 – 1) = 8/4 = 2。使用点斜式以A代入：y – 2 = 2(x – 1)，得到 y = 2x。一般式为：2x – y = 0。如果题目要求AB的垂直平分线，首先求中点 M：((1+5)/2, (2+10)/2) = (3, 6)。垂直线的梯度 = -1/2。方程：y – 6 = -1/2(x – 3)，所以 y = -1/2x + 15/2 或 x + 2y = 15。
Parallel and perpendicular lines appear in many geometric contexts. Lines parallel to y = 3x + 4 have gradient 3; lines perpendicular have gradient -1/3. A typical question gives three points and asks whether they form a right-angled triangle. Calculate gradients of two sides and check if their product equals -1. For instance, P(0,0), Q(4,2), R(2,8). Gradient PQ = 1/2, gradient PR = 4. PQ and PR are not perpendicular since 1/2 × 4 = 2 ≠ -1, but QR gradient = -3. Check systematically and state your conclusion clearly.
平行线和垂直线出现在许多几何情境中。平行于 y = 3x + 4 的直线梯度为3；垂直的直线梯度为 -1/3。一道典型的题目给出三个点并询问它们是否构成直角三角形。计算两边的梯度并检查其乘积是否等于-1。例如，P(0,0), Q(4,2), R(2,8)。PQ梯度 = 1/2，PR梯度 = 4。PQ和PR不垂直，因为 1/2 × 4 = 2 ≠ -1，但QR梯度 = -3。系统地检查并清晰陈述你的结论。
7. Trigonometry: SOHCAHTOA and Special Angles | 三角学：SOHCAHTOA与特殊角
In a right-angled triangle, sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Past papers frequently test exact values for 30°, 45°, and 60°. For example, sin 30° = 1/2, cos 45° = 1/√2 = √2/2, tan 60° = √3. These derive from the standard 30-60-90 triangle (sides 1, √3, 2) and the 45-45-90 triangle (sides 1, 1, √2). A question might ask: ‘Calculate the exact length of side x in a 30-60-90 triangle with hypotenuse 10 cm opposite the right angle.’ If x is opposite the 60° angle, x = 10 × sin 60° = 10 × √3/2 = 5√3 cm. Rationalising denominators is often required for full marks.
在直角三角形中，sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。历年真题经常考察30°、45°和60°的精确值。例如，sin 30° = 1/2，cos 45° = 1/√2 = √2/2，tan 60° = √3。这些值源自标准30-60-90三角形（边长1, √3, 2）和45-45-90三角形（边长1, 1, √2）。一道题可能会问：’在一个斜边长为10 cm的30-60-90三角形中，计算边x的精确长度，该边对直角。’ 如果x对着60°角，x = 10 × sin 60° = 10 × √3/2 = 5√3 cm。要将分母有理化通常才能得全分。
Angle of elevation and depression problems combine trigonometry with real-world contexts. A ladder of length 5 metres leans against a wall, making an angle of 70° with the ground. The height reached up the wall is 5 × sin 70° ≈ 4.70 m (3 significant figures). The distance from the wall is 5 × cos 70° ≈ 1.71 m. Clearly label opposite, adjacent, and hypotenuse in your diagram before solving. Some questions require the sine rule or cosine rule for non-right-angled triangles, detailed in the next section. Always check whether your calculator is in degree mode.
仰角和俯角问题将三角学与现实情境结合。一把5米长的梯子斜靠在墙上，与地面成70°角。墙上达到的高度为 5 × sin 70° ≈ 4.70 m（三位有效数字）。离墙的距离为 5 × cos 70° ≈ 1.71 m。在解题前清晰地标注图上的对边、邻边和斜边。有些题目对于非直角三角形要求使用正弦定理或余弦定理，这将在下一节详述。始终检查你的计算器是否处于度数模式。
8. Sine and Cosine Rules for Non-Right-Angled Triangles | 非直角三角形的正弦定理与余弦定理
Sine rule: a/sin A = b/sin B = c/sin C. Cosine rule: a² = b² + c² – 2bc cos A. Use the sine rule when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA, but beware the ambiguous case). Use the cosine rule when you know three sides (SSS) or two sides and the included angle (SAS). Cambridge questions often present a triangle with sides 7 cm and 9 cm and included angle 40°. Find the third side: a² = 7² + 9² – 2(7)(9)cos 40° = 49 + 81 – 126 cos 40° ≈ 49 + 81 – 96.53 = 33.47, so a ≈ 5.78 cm.
正弦定理：a/sin A = b/sin B = c/sin C。余弦定理：a² = b² + c² – 2bc cos A。当已知两角一边（AAS或ASA）或两边及一个非夹角（SSA，但需注意歧义情况）时，使用正弦定理。当已知三边（SSS）或两边及夹角（SAS）时，使用余弦定理。剑桥的题目经常给出一个三角形，两边为7 cm和9 cm，夹角为40°。求第三边：a² = 7² + 9² – 2(7)(9)cos 40° = 49 + 81 – 126 cos 40° ≈ 49 + 81 – 96.53 = 33.47，所以 a ≈ 5.78 cm。
Area of a triangle using ½ab sin C is another key formula. If two sides are 8 cm and 11 cm with included angle 65°, area = ½ × 8 × 11 × sin 65° ≈ 39.9 cm². The ambiguous case arises when finding an angle using sine rule. For example, sin B = 0.7 gives acute angle B ≈ 44.4° or obtuse B ≈ 180° – 44.4° = 135.6°. Check whether both solutions fit the triangle context; if the sum of angles exceeds 180°, the obtuse case is impossible. Always state clearly if there are two possible triangles or just one valid solution.
使用 ½ab sin C 计算三角形面积是另一个关键公式。如果两边为8 cm和11 cm，夹角为65°，面积 = ½ × 8 × 11 × sin 65° ≈ 39.9 cm²。使用正弦定理求角时会出现歧义情况。例如，sin B = 0.7 给出锐角 B ≈ 44.4° 或钝角 B ≈ 180° – 44.4° = 135.6°。检查两个解是否都符合三角形情境；如果角度之和超过180°，则钝角情况不可能。始终清晰说明是有两个可能的三角形还是只有一个有效解。
9. Circle Theorems and Geometric Proof | 圆定理与几何证明
Past papers consistently test circle theorems in multi-step angle problems. Key theorems: angle at centre is twice angle at circumference; angle in a semicircle is 90°; angles in the same segment are equal; opposite angles of a cyclic quadrilateral sum to 180°; tangent and radius meet at 90°. A typical diagram shows a circle with centre O, points A, B, C on the circumference, and a tangent at B. Angle OBA = 20°. Find angle OAB. Since OA = OB (radii), triangle OAB is isosceles, so angle OAB = 20°. Angle AOB = 180° – 40° = 140°. Angle ACB = 70° (angle at circumference is half angle at centre).
历年真题在多步角度问题中持续考察圆定理。关键定理有：圆心角是圆周角的两倍；半圆内的角为90°；同弧上的圆周角相等；圆内接四边形对角互补；切线与半径相交成90°。一个典型的示意图显示一个圆心为O的圆，点A、B、C在圆周上，在B点有一条切线。角OBA = 20°。求角OAB。由于 OA = OB（半径），三角形OAB是等腰三角形，所以角OAB = 20°。角AOB = 180° – 40° = 140°。角ACB = 70°（圆周角是圆心角的一半）。
Examiners award marks for citing the correct theorem as justification. The phrase ‘angle at centre is twice angle at circumference’ must appear in the working to earn full marks. Another common configuration: cyclic quadrilateral ABCD with angle ABC = 80° and angle ADC = 2x. Since opposite angles sum to 180°, 80° + 2x = 180°, so x = 50°. Alternate segment theorem is also tested: angle between tangent and chord equals angle in the alternate segment. Draw clear diagrams and label all known angles. Systematically deduce unknown angles using a theorem at each step.
考官会根据引用正确定理作为理由来给分。解题过程中必须出现’圆心角是圆周角的两倍’这样的表述才能获得全分。另一种常见的构图：圆内接四边形ABCD，角ABC = 80°，角ADC = 2x。由于对角互补，80° + 2x = 180°，所以 x = 50°。弦切角定理也会考察：切线与弦之间的角等于弦另一侧的圆周角。画出清晰的图并标出所有已知角。每一步运用一个定理系统地推导未知角。
10. Probability Trees and Conditional Probability | 概率树与条件概率
A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability both are red, and the probability exactly one is blue. The tree diagram shows first draw: P(red) = 4/10, P(blue) = 6/10. Second draw branches: if first is red, P(red second) = 3/9, P(blue second) = 6/9. Both red: 4/10 × 3/9 = 12/90 = 2/15. Exactly one blue can occur via red then blue or blue then red. P(red, blue) = 4/10 × 6/9 = 24/90; P(blue, red) = 6/10 × 4/9 = 24/90. Sum = 48/90 = 8/15. Multiply along branches and add across outcomes.
一个袋子装有4个红球和6个蓝球。不放回地抽取两个球。求两个都是红球的概率，以及恰好一个是蓝球的概率。树状图显示第一次抽取：P(红) = 4/10，P(蓝) = 6/10。第二次抽取的分支：如果第一次是红球，P(第二次红) = 3/9，P(第二次蓝) = 6/9。两者皆红：4/10 × 3/9 = 12/90 = 2/15。恰好一个蓝球可通过先红后蓝或先蓝后红发生。P(红, 蓝) = 4/10 × 6/9 = 24/90；P(蓝, 红) = 6/10 × 4/9 = 24/90。总和 = 48/90 = 8/15。沿分支相乘，跨结果相加。
Conditional probability questions use notation P(A|B) meaning ‘probability of A given B has occurred.’ In formula: P(A|B) = P(A and B) / P(B). Example: the probability of rain on any day is 0.3. If it rains, the probability a bus is late is 0.8; if it does not rain, late probability is 0.1. Find P(late). Use tree: P(rain and late) = 0.3 × 0.8 = 0.24; P(no rain and late) = 0.7 × 0.1 = 0.07. Total P(late) = 0.31. If the bus is late, probability it rained is P(rain|late) = 0.24 / 0.31 ≈ 0.774. Candidates must identify the reduced sample space correctly.
条件概率题目使用符号 P(A|B) 表示’在B已发生的条件下A发生的概率’。公式：P(A|B) = P(A 和 B) / P(B)。举例：任何一天下雨的概率是0.3。如果下雨，公交车晚点的概率是0.8；如果不下雨，晚点概率是0.1。求P(晚点)。使用树状图：P(下雨且晚点) = 0.3 × 0.8 = 0.24；P(不下雨且晚点) = 0.7 × 0.1 = 0.07。总的P(晚点) = 0.31。如果公交车晚点，下雨的概率是 P(下雨|晚点) = 0.24 / 0.31 ≈ 0.774。考生必须正确识别缩减后的样本空间。
11. Statistics: Cumulative Frequency and Box Plots | 统计：累积频率与箱线图
Cumulative frequency diagrams summarise grouped data and allow estimation of median, quartiles, and percentiles. A typical question provides a table of heights of 80 students in intervals: 150 ≤ h < 160, frequency 12; 160 ≤ h < 170, frequency 28, and so on. Construct cumulative frequency by adding successively: 12, 40, etc. Plot points at the upper boundary of each interval against cumulative frequency. Draw a smooth curve. Median corresponds to the 40th value; lower quartile to 20th; upper quartile to 60th. Interquartile range = Q3 - Q1.
累积频率图总结分组数据，并可用于估算中位数、四分位数和百分位数。一道典型题目提供一张80名学生身高的分组表：150 ≤ h < 160，频数12；160 ≤ h < 170，频数28，等等。通过连续相加构建累积频率：12，40，等等。在每个区间的上限处绘制点，对应累积频率。画一条平滑曲线。中位数对应第40个值；下四分位数对应第20个；上四分位数对应第60个。四分位距 = Q3 - Q1。
Box plots (box-and-whisker diagrams) display five-number summaries: minimum, Q1, median, Q3, maximum. Examiners ask students to compare distributions using box plots by commenting on median (central tendency), interquartile range (spread), and overall range. A box plot with a longer right whisker indicates positive skew; longer left whisker indicates negative skew. When comparing two data sets, use specific numerical values from the box plot. Write comparative sentences like ‘The median score for Class A (68) is higher than for Class B (61), suggesting better overall performance.’ Avoid vague statements unsupported by data.
箱线图（盒须图）显示五数概括：最小值、Q1、中位数、Q3、最大值。考官要求学生使用箱线图通过评论中位数（集中趋势）、四分位距（离散程度）和整体范围来比较分布。右须较长的箱线图表示正偏态；左须较长表示负偏态。在比较两个数据集时，使用箱线图中的具体数值。写出比较性的句子，如’甲班的中位数得分（68）高于乙班（61），表明整体表现更好。’ 避免无数据支持的空泛表述。
12. Exam Technique and Common Pitfalls | 考试技巧与常见陷阱
Past paper analysis reveals consistent patterns in mark schemes. Always show working: a correct answer with no working earns only the accuracy mark, not the method mark. If the answer is wrong, method marks are still available for correct reasoning. Round answers only at the final step unless instructed otherwise. Intermediate rounding can cause the final answer to drift outside the acceptable tolerance. Cambridge accepts answers within a certain range, typically ±0.5 or ±0.05 depending on the context, but consistent exact working guarantees success. Read questions carefully for command words: ‘factorise fully’ differs from ‘solve’; ‘expand and simplify’ requires both steps.
历年真题分析揭示了评分方案中的一贯模式。始终展示解题步骤：仅有正确答案而无过程只能得到准确性分，而非方法分。如果答案错误，方法分仍可根据正确推理获得。除非另有指示，仅在最后一步取整答案。中间步骤取整可能导致最终答案超出可接受范围。剑桥接受在一定范围内的答案，通常根据上下文为 ±0.5 或 ±0.05，但始终精确解题可保证成功。仔细阅读题目中的指令词：’完全因式分解’ 不同于 ‘解方程’；’展开并化简
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