📚 Year 7 AQA Advanced Mathematics: In-Depth Analysis of Past Paper Questions | Year 7 AQA 进阶数学：历年真题深度解析
Understanding the style and challenge of AQA Advanced Mathematics papers is essential for Year 7 students aiming for top grades. This article provides a detailed breakdown of selected past paper questions, covering key topics such as number, algebra, geometry, and problem-solving. Each question is analysed step by step, highlighting common pitfalls and effective strategies.
对于志在取得高分的 Year 7 学生来说，理解 AQA 进阶数学试卷的题型与难度至关重要。本文精选历年真题，深入解析数字、代数、几何和应用题等核心主题，逐步剖析解题思路，指出常见错误与高效策略。
1. Number and Place Value | 数字与位值
Example Question 1: Write the number 6 304 527 in words and state the place value of the digit 3.
例题 1：用英文单词写出数字 6 304 527，并指出数字 3 的位值。
Solution: The number is six million, three hundred and four thousand, five hundred and twenty-seven. The first digit 3 is in the hundred-thousands place, so its place value is hundred-thousands and its actual value is 300 000. (The second 3 in the tens place has value 30.)
解：该数字写作 six million, three hundred and four thousand, five hundred and twenty-seven。第一个数字 3 位于十万位，因此位值为十万，实际值为 300 000。（十位上的第二个 3 值为 30。）
Example Question 2: Evaluate 4² + 3 × (8 - 2³) ÷ 2.
例题 2：计算 4² + 3 × (8 - 2³) ÷ 2。
Work inside the brackets first, but inside we must do the exponent: 2³ = 8, so 8 - 8 = 0.
第一步：先算括号内，但括号内要先算乘方：2³ = 8，所以 8 - 8 = 0。
Step 2: Now the expression is 4² + 3 × 0 ÷ 2. Evaluate the exponent: 4² = 16.
第二步：此时表达式为 4² + 3 × 0 ÷ 2。计算乘方：4² = 16。
Step 3: Perform multiplication and division from left to right: 3 × 0 = 0, then 0 ÷ 2 = 0.
第三步：从左到右计算乘除：3 × 0 = 0，然后 0 ÷ 2 = 0。
Step 4: Add: 16 + 0 = 16.
第四步：加法：16 + 0 = 16。
2. Fractions, Decimals and Percentages | 分数、小数与百分比
Example Question: Calculate 2 1/3 + 1 5/6, giving your answer as a mixed number.
例题：计算 2 1/3 + 1 5/6，结果写成带分数。
Step 1: Convert mixed numbers to improper fractions: 2 1/3 = 7/3, 1 5/6 = 11/6.
第一步：将带分数化为假分数：2 1/3 = 7/3，1 5/6 = 11/6。
Step 2: Find a common denominator. The LCM of 3 and 6 is 6. Rewrite 7/3 as 14/6.
第二步：找公分母。3 和 6 的最小公倍数是 6。将 7/3 写成 14/6。
Step 3: Add: 14/6 + 11/6 = 25/6.
第三步：相加：14/6 + 11/6 = 25/6。
Step 4: Convert back to a mixed number: 25 ÷ 6 = 4 remainder 1, so 4 1/6.
第四步：化回带分数：25 ÷ 6 = 4 余 1，所以答案是 4 1/6。
Extension: Find 35% of 240. 10% = 24, 5% = 12, so 35% = 3 × 24 + 12 = 84.
拓展：求 240 的 35%。10% = 24，5% = 12，因此 35% = 3 × 24 + 12 = 84。
3. Algebraic Expressions | 代数表达式
Example Question 1: Simplify 5x + 2y - 3x + 7y - x.
例题 1：化简 5x + 2y - 3x + 7y - x。
Step 1: Group like terms for x: 5x - 3x - x = (5 - 3 - 1)x = 1x = x.
第一步：合并含 x 的项：5x - 3x - x = (5 - 3 - 1)x = 1x = x。
Step 2: Group like terms for y: 2y + 7y = 9y.
第二步：合并含 y 的项：2y + 7y = 9y。
Step 3: Write the simplified expression: x + 9y.
第三步：写出化简结果：x + 9y。
Example Question 2: Expand and simplify 3(2a - 4) + 2(5 - a).
例题 2：展开并化简 3(2a - 4) + 2(5 - a)。
Step 1: Multiply out: 3 × 2a = 6a, 3 × (-4) = -12, 2 × 5 = 10, 2 × (-a) = -2a.
第一步：展开各项：3 × 2a = 6a，3 × (-4) = -12，2 × 5 = 10，2 × (-a) = -2a。
Step 2: Write together: 6a - 12 + 10 - 2a.
第二步：合并写为 6a - 12 + 10 - 2a。
Step 3: Combine like terms: 6a - 2a = 4a; -12 + 10 = -2. Final answer: 4a - 2.
第三步：合并同类项：6a - 2a = 4a；-12 + 10 = -2。最终结果：4a - 2。
4. Solving Linear Equations | 解一元一次方程
Example Question: Solve 4(x - 3) = 2x + 10.
例题：解方程 4(x - 3) = 2x + 10。
Step 1: Expand the left side: 4x - 12 = 2x + 10.
第一步：展开左边：4x - 12 = 2x + 10。
Step 2: Bring variable terms to one side: subtract 2x from both sides → 2x - 12 = 10.
第二步：移项，将含 x 的项移到一边：两边同时减 2x → 2x - 12 = 10。
Step 3: Add 12 to both sides: 2x = 22.
第三步：两边加 12：2x = 22。
Step 4: Divide by 2: x = 11.
第四步：两边除以 2：x = 11。
Check: 4(11 - 3) = 4 × 8 = 32; right side: 2×11 + 10 = 32. Correct.
检验：4(11 - 3) = 4 × 8 = 32；右边：2×11 + 10 = 32。正确。
5. Sequences and Patterns | 数列与规律
Example Question: Find the nth term of the sequence: 3, 7, 11, 15, ... Then use it to find the 20th term.
例题：求数列 3, 7, 11, 15, ... 的第 n 项公式，并求出第 20 项。
, 11 - 7 = 4, so the common difference is 4. The sequence is linear, so nth term = 4n + c.
第一步：判断公差：7 - 3 = 4，11 - 7 = 4，公差为 4。为线性数列，因此第 n 项公式为 4n + c。
Step 2: For n = 1, term = 3, so 4(1) + c = 3 → 4 + c = 3 → c = -1.
第二步：当 n = 1 时，项为 3，因此 4(1) + c = 3 → 4 + c = 3 → c = -1。
Step 3: The nth term is 4n - 1.
第三步：第 n 项公式为 4n - 1。
Step 4: Find the 20th term: 4(20) - 1 = 80 - 1 = 79.
第四步：求第 20 项：4(20) - 1 = 80 - 1 = 79。
6. Angles and Lines | 角与线
Example Question: Two straight lines intersect. One of the angles formed is 47°. Find the sizes of the other three angles.
例题：两条直线相交，形成的一个角为 47°。求其余三个角的大小。
Step 1: Vertically opposite angles are equal. The angle opposite the 47° angle is also 47°.
第一步：对顶角相等。因此与 47° 相对的角也是 47°。
Angles on a straight line add up to 180°. The angle adjacent to 47° is 180° - 47° = 133°.
第二步：平角为 180°。与 47° 相邻的角为 180° - 47° = 133°。
Step 3: The angle opposite the 133° angle is also 133° (vertically opposite).
第三步：与 133° 相对的角也是 133°（对顶角相等）。
Final answer: 47°, 133°, 47°, 133°.
答案：47°, 133°, 47°, 133°。
7. Area and Perimeter | 面积与周长
Example Question: A rectangle has length (4y - 3) m and width 6 m. (a) Write an expression for the perimeter. (b) If the perimeter is
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