Year 10 CAIE Mathematics: In-Depth Past Paper Analysis | Year 10 CAIE 数学：历年真题深度解析

📚 Year 10 CAIE Mathematics: In-Depth Past Paper Analysis | Year 10 CAIE 数学：历年真题深度解析

Mastering Year 10 CAIE Mathematics requires more than just memorising formulas; it demands a thorough understanding of past paper questions. By analysing real exam questions, you can uncover patterns, common pitfalls, and examiner expectations. This guide takes you through key topics and representative past-paper-style questions, providing step-by-step solutions in both English and Chinese. Whether you are aiming for a Grade 9 or simply building a solid foundation, this deep dive will sharpen your problem-solving skills and boost your confidence.

掌握 Year 10 CAIE 数学不仅仅是记住公式，更需要对历年真题有透彻的理解。通过分析真实考题，你可以发现命题规律、常见错误和考官的要求。本指南带你逐一攻克核心主题，精选具有代表性的真题样式问题，并提供中英双语的分步解答。无论你是想冲刺 9 分，还是打好扎实基础，这份深度解析都能锻炼你的解题能力并增强信心。

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1. Number Operations and Bounds | 数字运算与界限

Example (CAIE 0580 style): The length of a rectangle is 12.6 cm correct to one decimal place. The width is 7.3 cm correct to one decimal place. Calculate the lower bound and upper bound of the perimeter.

例题：一个矩形的长是 12.6 cm（精确到一位小数）。宽是 7.3 cm（精确到一位小数）。计算周长的下界和上界。

= 2 × 19.8 = 39.6 cm. Upper bound of perimeter = 2 × (12.65 + 7.35) = 2 × 20.0 = 40.0 cm.

解：长度的范围是 12.55 ≤ l < 12.65。宽度的范围是 7.25 ≤ w < 7.35。周长的下界 = 2 × (12.55 + 7.25) = 2 × 19.8 = 39.6 cm。周长的上界 = 2 × (12.65 + 7.35) = 2 × 20.0 = 40.0 cm。

Remember: To find the lower bound, subtract half of the degree of accuracy; for the upper bound, add half of the degree of accuracy.

记住：求下界时，减去精确度的一半；求上界时，加上精确度的一半。

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2. Algebraic Simplification and Substitution | 代数化简与代入

Example: Simplify 3(2x − 5) − 2(4 − x). Then find the value of the expression when x = −2.

例题：化简 3(2x − 5) − 2(4 − x)。然后求当 x = −2 时该表达式的值。

15 − 8 + 2x. Combine like terms: 8x − 23. Substitute x = −2: 8(−2) − 23 = −16 − 23 = −39.

去括号：6x − 15 − 8 + 2x。合并同类项：8x − 23。代入 x = −2：8(−2) − 23 = −16 − 23 = −39。

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3. Solving Linear Equations | 解线性方程

Example: Solve the equation (2x + 1)/3 − (x − 2)/2 = 5.

例题：解方程 (2x + 1)/3 − (x − 2)/2 = 5。

6: 2(2x+1) − 3(x−2) = 30. Expand: 4x+2 − 3x+6 = 30. Simplify: x+8 = 30. Therefore x = 22.

每一项乘以 3 和 2 的最小公倍数 6：2(2x+1) − 3(x−2) = 30。展开：4x+2 − 3x+6 = 30。化简：x+8 = 30。因此 x = 22。

Check: LHS = (2×22+1)/3 − (22−2)/2 = 45/3 − 20/2 = 15 − 10 = 5 = RHS. ✓

验算：左边 = (2×22+1)/3 − (22−2)/2 = 45/3 − 20/2 = 15 − 10 = 5 = 右边。✓

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4. Linear Inequalities | 线性不等式

Example: Solve the inequality 3 − 2x ≤ 7 and represent the solution on a number line.

例题：解不等式 3 − 2x ≤ 7，并在数轴上表示解集。

quality sign): x ≥ −2. On a number line, a filled circle at −2 with an arrow to the right.

两边减 3：−2x ≤ 4。除以 −2（不等号方向改变）：x ≥ −2。在数轴上，在 −2 处画实心点，箭头向右。

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5. Sequences and nth Term | 数列与第 n 项

Example: The first four terms of a sequence are 7, 11, 15, 19. (a) Write down the nth term. (b) Find the 50th term.

例题：某数列的前四项为 7, 11, 15, 19。(a) 写出第 n 项。(b) 求第 50 项。

The difference is +4, so nth term = 4n + 3 (since 4×1+3=7). 50th term = 4×50+3 = 203.

公差为 +4，所以第 n 项 = 4n + 3（因为 4×1+3=7）。第 50 项 = 4×50+3 = 203。

Is 243 a term in the sequence? Set 4n+3=243 → 4n=240 → n=60, so yes, it is the 60th term.

243 是数列中的项吗？设 4n+3=243 → 4n=240 → n=60，所以是，它是第 60 项。

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6. Ratio, Proportion and Rates | 比、比例和比率

Example: Anna, Ben and Chloe share $240 in the ratio 2:3:5. (a) How much does Ben receive? (b) If Chloe spends 40% of her share, how much does she have left?

例题：Anna、Ben 和 Chloe 按 2:3:5 的比例分享 240 美元。(a) Ben 得到多少钱？(b) 如果 Chloe 花掉她份额的 40%，她还剩下多少钱？

Total parts = 2+3+5 = 10. Value of one part = $240 ÷ 10 = $24. Ben's share = 3 × $24 = $72. Chloe's share = 5 × $24 = $120. After spending 40%, she has 60% left: 0.6 × $120 = $72.

总份数 = 2+3+5 = 10。一份的金额 = 240 ÷ 10 = 24 美元。Ben 的份额 = 3 × 24 = 72 美元。Chloe 的份额 = 5 × 24 = 120 美元。花掉 40% 后，剩下 60%：0.6 × 120 = 72 美元。

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7. Percentages | 百分比

Example: A jacket originally costs $80. In a sale, the price is reduced by 15%. The following week, the sale price is reduced by a further 20%. Calculate the final price.

例题：一件夹克原价 80 美元。促销降价 15%。接下来一周，促销价再降 20%。计算最终价格。

After 15% off: $80 × 0.85 = $68. Then 20% off that: $68 × 0.8 = $54.40. Alternatively, single multiplier = 0.85 × 0.8 = 0.68, so $80 × 0.68 = $54.40.

降价 15% 后：80 × 0.85 = 68 美元。再降 20%：68 × 0.8 = 54.40 美元。或者，总乘数 = 0.85 × 0.8 = 0.68，因此 80 × 0.68 = 54.40 美元。

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8. Angles and Polygons | 角与多边形

Example: A regular polygon has 9 sides. Calculate the size of one interior angle.

例题：一个正九边形。计算一个内角的大小。

of interior angles = (9−2)×180° = 7×180° = 1260°. Each interior angle = 1260° ÷ 9 = 140°.

内角和 = (9−2)×180° = 7×180° = 1260°。每个内角 = 1260° ÷ 9 = 140°。

Alternatively, exterior angle = 360° ÷ 9 = 40°, so interior = 180° − 40° = 140°.

或者，外角 = 360° ÷ 9 = 40°，因此内角 = 180° − 40° = 140°。

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9. Pythagoras' Theorem | 勾股定理

Example: In a right-angled triangle, the two shorter sides are 5 cm and 12 cm. Calculate the length of the hypotenuse.

例题：在一个直角三角形中，两条直角边分别为 5 cm 和 12 cm。计算斜边的长度。

Using a² + b² = c²: 5² + 12² = c² → 25 + 144 = c² → 169 = c². Therefore c = √169 = 13 cm.

使用 a² + b² = c²：5² + 12² = c² → 25 + 144 = c² → 169 = c²。因此 c = √169 = 13 cm。

m and one leg is 9 m, find the other leg. 15² − 9² = 225 − 81 = 144, so leg = √144 = 12 m.

另一典型题目：如果斜边为 15 m，一直角边为 9 m，求另一直角边。15² − 9² = 225 − 81 = 144，所以直角边 = √144 = 12 m。

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10. Statistics: Mean, Median, Mode and Range | 统计：平均数、中位数、众数和极差

Example: The test scores of 10 students are: 7, 5, 8, 6, 9, 5, 10, 7, 6, 4. Find (a) the mode, (b) the median, (c) the mean, and (d) the range.

例题：10 名学生的考试成绩为：7, 5, 8, 6,

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