📚 Year 7 Cambridge Advanced Maths: In-depth Past Paper Analysis | Year 7 剑桥进阶数学：历年真题深度解析
Welcome to an in-depth exploration of past paper questions for Year 7 Cambridge Advanced Mathematics. This article breaks down typical exam problems, reveals the reasoning behind each solution, and identifies the most common mistakes students make. Whether you are aiming for a top score in your Checkpoint test or building a strong foundation for IGCSE, these worked examples will sharpen your skills.
欢迎深度探索 Year 7 剑桥进阶数学的历年真题。本文将拆解典型考题，揭示每道题的解题思路，并指出学生最常犯的错误。无论你是想在 Checkpoint 测试中取得高分，还是为 IGCSE 打下坚实基础，这些精讲例题都将提升你的解题能力。
1. Simplifying and Expanding Algebraic Expressions | 代数表达式的化简与展开
A classic past paper question asks: Simplify and expand 3(2x – 4) + 2(x + 1).
一道经典的真题是这样问的：化简并展开 3(2x – 4) + 2(x + 1)。
Solution strategy: First apply the distributive law to remove brackets, then group like terms carefully. Many errors arise from forgetting to multiply both terms inside the bracket, especially when a negative sign is involved.
解题策略：首先运用分配律去括号，然后仔细合并同类项。许多错误源于忘记乘以括号内的每一项，尤其在涉及负号的时候。
Step 1: Multiply out 3(2x – 4) to get 6x – 12. Step 2: Multiply out 2(x + 1) to get 2x + 2. The expression becomes 6x – 12 + 2x + 2.
步骤1：将 3(2x – 4) 展开得到 6x – 12。步骤2：将 2(x + 1) 展开得到 2x + 2。表达式变为 6x – 12 + 2x + 2。
Now collect the x-terms: 6x + 2x = 8x, and the constant terms: -12 + 2 = -10. The simplified answer is:
现在合并 x 项：6x + 2x = 8x，常数项：-12 + 2 = -10。化简后的答案是：
8x – 10
Common pitfall: Writing -12 + 2 as -14 instead of -10. Always double-check signs when adding positive and negative numbers.
常见陷阱：将 -12 + 2 算成 -14 而非 -10。在加减正负数时务必反复检查符号。
2. Solving Linear Equations with Brackets | 解含括号的一元一次方程
Exam question: Solve 2(x – 3) = 10.
考题：解方程 2(x – 3) = 10。
There are two efficient approaches. Method 1: Divide both sides by 2 first, giving x – 3 = 5, then x = 8. Method 2: Expand the bracket to 2x – 6 = 10, then add 6 to get 2x = 16, so x = 8. Both lead to the same result, but Method 1 is often quicker.
有两种高效的解法。方法一：先两边除以 2，得到 x – 3 = 5，然后 x = 8。方法二：展开括号得 2x – 6 = 10，然后加 6 得到 2x = 16，所以 x = 8。两种方法结果相同，但方法一通常更快捷。
Key check: Substitute x = 8 back into the original equation: 2(8 – 3) = 2 × 5 = 10. The left-hand side equals the right-hand side, so the solution is verified.
关键验算：将 x = 8 代回原方程：2(8 – 3) = 2 × 5 = 10。左边等于右边，解得到验证。
Common mistake: When dividing only one term inside the bracket, or incorrectly handling the -3. Remember that the entire bracket is multiplied by 2, so dividing first avoids distribution errors.
常见错误：仅仅除以括号内的某一项，或者错误地处理 -3。记住整个括号都被乘以 2，因此先除以 2 可以避免分配律错误。
3. Fractions and the Order of Operations | 分数与运算顺序
Typical question from past papers: Evaluate 2/3 + 1/4 × 3/5.
真题中常见的问题：计算 2/3 + 1/4 × 3/5。
Always follow BIDMAS/BODMAS: Brackets, Indices, Division/Multiplication (left to right), Addition/Subtraction (left to right). Here, multiplication comes before addition. So first compute 1/4 × 3/5 = 3/20.
务必遵循 BIDMAS/BODMAS 规则：括号、指数、除法/乘法（从左到右）、加法/减法（从左到右）。这里乘法先于加法，所以先算 1/4 × 3/5 = 3/20。
Now the problem becomes 2/3 + 3/20. To add, find a common denominator. The LCM of 3 and 20 is 60. Convert: 2/3 = 40/60, 3/20 = 9/60. Add to get 49/60.
现在题目变为 2/3 + 3/20。要相加，先找公分母。3 和 20 的最小公倍数是 60。转换：2/3 = 40/60，3/20 = 9/60。相加得到 49/60。
Final answer is 49/60. Many students mistakenly add before multiplying, getting (2/3 + 1/4) × 3/5 = (11/12) × 3/5 = 33/60 = 11/20, which is incorrect. Always respect operation hierarchy.
最终答案是 49/60。许多学生错误地先加后乘，得到 (2/3 + 1/4) × 3/5 = (11/12) × 3/5 = 33/60 = 11/20，这是错误的。务必遵守运算层级。
4. Successive Percentage Discounts | 连续百分数折扣
A challenging exam question: A laptop costs £500. In a sale, the price is reduced by 20%, and then a further 10% off is applied at the till. What is the final price?
一道具有挑战性的考题：一台笔记本电脑售价 £500。促销期间先降价 20%，然后在收银台再享受额外 10% 的折扣。最终价格是多少？
Important: The second 10% discount applies to the reduced price, not the original. A common misconception is to add the percentages (20% + 10% = 30%) and take 30% off £500, giving £350. This is wrong because it ignores the compound effect.
重要的是：第二次 10% 折扣作用于第一次降价后的价格，而非原价。常见的误解是直接相加百分比（20% + 10% = 30%），从 £500 中减掉 30% 得到 £350。这是错误的，因为它忽略了复利效应。
Correct method: After 20% off, price becomes £500 × 0.80 = £400. Then 10% off means paying 90%, so £400 × 0.90 = £360.
正确方法：降价 20% 后，价格变为 £500 × 0.80 = £400。再打九折，即支付原价的 90%，所以 £400 × 0.90 = £360。
The final price is £360. Always apply each discount to the running total, multiply multipliers (0.80 × 0.90 = 0.72), and check that 72% of £500 is indeed £360.
最终价格为 £360。始终对当前金额依次应用折扣，将乘数相乘（0.80 × 0.90 = 0.72），并验证 £500 的 72% 确实是 £360。
5. Ratio Word Problems and Cross-Multiplication | 比率文字题与交叉相乘
From a real past paper: The ratio of boys to girls in a school is 5:7. There are 84 girls. How many boys are there?
来自真实的历年真题：某学校男生与女生的比例是 5:7。已知有 84 名女生，问男生有多少人？
Set up the proportion: boys/girls = 5/7. Let the number of boys be b. Then b/84 = 5/7. Cross-multiply to get 7b = 5 × 84, which gives 7b = 420, so b = 60.
列出比例式：男生/女生 = 5/7。设男生人数为 b，则 b/84 = 5/7。交叉相乘得 7b = 5 × 84，即 7b = 420，所以 b = 60。
Verification: The simplified ratio 60:84 divides both numbers by 12, giving 5:7, correct. Another approach: The multiplier from the given ratio is 84 ÷ 7 = 12. Multiply the boys’ part by 12: 5 × 12 = 60 boys.
验证：化简 60:84，同除以 12 得到 5:7，正确。另一种思路：给定比例的倍数因子为 84 ÷ 7 = 12。将男生的份数乘以 12：5 × 12 = 60 名男生。
Always check that you match the correct part of the ratio to the known quantity. The most frequent error is swapping the order and writing 7/5 instead of 5/7.
务必确保将比例的对应部分与已知量相匹配。最常见的错误是顺序颠倒，将比例写作 7/5 而不是 5/7。
6. Area of Composite Shapes: Rectangle and Semicircle | 复合图形的面积：矩形与半圆
Exam question: A shape is made from a rectangle of length 8 cm and width 5 cm, with a semicircle of diameter 8 cm attached to one of its longer sides. Find the total area. Use π ≈ 3.14.
考题：一个图形由一个长 8 cm、宽 5 cm 的矩形和一个附加在某条长边上的直径为 8 cm 的半圆组成。求总面积（π 取 3.14）。
First, the area of the rectangle is length × width = 8 × 5 = 40 cm². The semicircle has radius = diameter/2 = 4 cm. The area of a full circle is πr², so semicircle area = ½ × π × 4² = ½ × 3.14 × 16.
首先，矩形面积 = 长 × 宽 = 8 × 5 = 40 cm²。半圆的半径为直径的一半 = 4 cm。整圆面积为 πr²，因此半圆面积 = ½ × π × 4² = ½ × 3.14 × 16。
Compute the semicircle area: ½ × 3.14 × 16 = ½ × 50.24 = 25.12 cm². Add the rectangle: total area = 40 + 25.12 = 65.12 cm².
计算半圆面积：½ × 3.14 × 16 = ½ × 50.24 = 25.12 cm²。加上矩形：总面积 = 40 + 25.12 = 65.12 cm²。
Common mistake: Using diameter instead of radius in the area formula, or forgetting to halve the circle area. Also check that the semicircle is added, not subtracted.
常见错误：在面积公式中使用直径而非半径，或忘记将圆面积减半。还要确认半圆是增加的部分，而不是要减去的部分。
7. Arithmetic Sequences and the nth Term | 等差数列与第 n 项
Question: The nth term of a sequence is given by 3n + 2. (a) Find the 10th term. (b) Which term is equal to 50?
题目：某数列的第 n 项为 3n + 2。(a) 求第 10 项。(b) 第几项的值为 50？
For part (a), substitute n = 10: 3(10) + 2 = 30 + 2 = 32. So the 10th term is 32.
对于 (a)，代入 n = 10：3(10) + 2 = 30 + 2 = 32。所以第 10 项是 32。
For part (b), set the formula equal to the term value: 3n + 2 = 50. Solve for n: subtract 2, 3n = 48, divide by 3, n = 16. Thus the 16th term is 50.
对于 (b)，令公式等于该项的值：3n + 2 = 50。求解 n：减 2 得 3n = 48，除以 3 得 n = 16。因此第 16 项是 50。
These questions test your ability to work both forwards and backwards with an algebraic rule. Double-check your arithmetic, especially when subtracting small numbers.
这类问题测试你对代数规则正向和逆向运用的能力。仔细检查你的算术，尤其是在减去较小数字的时候。
8. Mean, Median, Mode, and Range | 平均数、中位数、众数和极差
Past paper data set: The ages of five children in a club are 8, 9, 10, 10, 13. Calculate the mean, median, mode, and range.
真题数据集：俱乐部中五个孩子的年龄分别为 8, 9, 10, 10, 13。计算平均数、中位数、众数和极差。
Mean: Sum = 8+9+10+10+13 = 50. Number of values = 5. Mean = 50 ÷ 5 = 10.
平均数：总和 = 8+9+10+10+13 = 50。数据个数 = 5。平均数 = 50 ÷ 5 = 10。
Median: List in order (already ordered). Middle value is the 3rd: 10. Mode: the most frequent value, which is 10 (appears twice). Range: maximum – minimum = 13 – 8 = 5.
中位数：数据已排序。中间值（第 3 个）是 10。众数：出现最频繁的数值，即 10（出现两次）。极差：最大值 – 最小值 = 13 – 8 = 5。
Outlier consideration: The value 13 is a bit higher, but with only five values the mean is still 10. If the set were 8, 9, 10, 10, 20, the mean would rise to 11.4, showing how an outlier pulls the mean. Always compare mean and median to sense data skew.
异常值考量：数值 13 略高，但只有五个数据时平均数仍为 10。如果数据集是 8, 9, 10, 10, 20，平均数将升至 11.4，可见异常值如何拉高平均数。始终比较平均数和中位数以感知数据偏态。
9. Negative Numbers and Order of Operations | 负数与运算顺序
Challenging question: Evaluate -4² + (-6) ÷ 2.
挑战性问题：计算 -4² + (-6) ÷ 2。
Remember that the exponent applies only to the 4, not the negative sign, unless brackets are used. Here, -4² means -(4²) = -16. If it were (-4)², it would be +16. Then division: (-6) ÷ 2 = -3. So the expression becomes -16 + (-3) = -19.
记住指数仅作用于数字 4，而非负号，除非使用了括号。在这里，-4² 表示 -(4²) = -16。如果是 (-4)²，结果则是 +16。然后是除法：(-6) ÷ 2 = -3。于是表达式变为 -16 + (-3) = -19。
Alternative check: Write as -1 × 4² + (-6)/2 = -1 × 16 – 3 = -16 – 3 = -19. Order: exponents first, then division, then addition.
另法验算：写作 -1 × 4² + (-6)/2 = -1 × 16 – 3 = -16 – 3 = -19。顺序：先指数，再除法，最后加法。
Common error: Treating -4² as (+16) and then doing 16 + (-3) = 13. This is a classic trap. Always clarify: a negative sign in front of a power without brackets means “the opposite of the square”.
常见错误：将 -4² 当作 (+16) 处理，然后计算 16 + (-3) = 13。这是一个经典陷阱。务必明确：指数前不带括号的负号表示“平方的相反数”。
10. Problem Solving: Working Backwards | 问题解决：逆向推理
Exam problem: A number is multiplied by 3, then 5 is added, and the result is 20. Find the number.
考题：一个数乘以 3，然后加 5，结果为 20。求这个数。
Let the unknown number be n. The operation describes: 3n + 5 = 20. Subtract 5 from both sides: 3n = 15. Divide by 3: n = 5. This is the forward algebraic approach.
设未知数为 n。运算描述为：3n + 5 = 20。两边减去 5：3n = 15。除以 3：n = 5。这是正向代数方法。
Working backwards is also powerful: The final result 20 was obtained after adding 5, so before adding 5 we had 20 – 5 = 15. That 15 came from multiplying by 3, so originally the number was 15 ÷ 3 = 5. This method matches the inverse operations.
逆向推理也很有效：最终结果 20 是加 5 后得到的，因此加 5 之前是 20 – 5 = 15。那个 15 是乘以 3 得到的，所以原数是 15 ÷ 3 = 5。这种方法与逆运算相符。
Always check your answer by running it through the original process: 5 × 3 = 15, + 5 = 20. Correct. This type of “think of a number” problem is very common and rewards logical step-by-step work.
务必通过原始流程验算答案：5 × 3 = 15，+ 5 = 20。正确。这类“想一个数”的问题非常常见，逻辑分步解答总能得分。
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