📚 IGCSE WJEC Maths: Common Mistakes and Tricky Questions | IGCSE WJEC 数学：易错题精讲
Many students preparing for the IGCSE WJEC Mathematics exam lose marks not because they do not understand the topics, but because they fall into predictable traps. This article highlights some of the most common mistakes and tricky questions across the WJEC specification, with worked solutions and explanations. By learning to recognise these pitfalls, you can avoid careless errors and boost your final grade. Each section focuses on one typical error, explains why it occurs, and shows the correct approach.
许多备考 IGCSE WJEC 数学考试的学生丢分并不全是因为不懂知识点，而是掉入了可预见的陷阱。本文梳理了 WJEC 考纲中最常见的一些错误和易错题，并配有详细解答和解释。学会识别这些坑，你就能避免粗心失分，提升最终成绩。每个小节聚焦一种典型错误，解释其产生原因，并展示正确解法。
1. Inverse Proportion Confusion | 反比关系的混淆
ty constant. In WJEC questions, you are often told that y is inversely proportional to x, and given a pair of values. The common error is to write y = x/k instead of y = k/x. Remember, inverse proportion means the product xy is constant: xy = k. Always find k first using the given values, then use this constant to find the missing value.
一个经典错误是学生认为如果一量加倍，另一量就减半，却没有正确建立比例常数。在 WJEC 题目中，常告知 y 与 x 成反比，并给出一组对应值。常见错误是将关系写成 y = x/k 而非 y = k/x。记住，反比意味着乘积 xy 为常数：xy = k。一定要先用给定值求出 k，然后再用这个常数求未知量。
Example: portional to x. When x = 4, y = 6. Find y when x = 3. Common wrong answer: y = (4/3)×6 = 8? k is mistakenly taken as y/x, then it would be wrong. A better example: If y is inversely proportional to x², errors multiply. Always write the full expression: y = k / x².
例子：y 与 x 成反比，当 x = 4 时 y = 6。求 x = 3 时的 y 值。常见错误答案：会算对但方法错。如果换成 y 与 x² 成反比，错误就明显了。正确步骤：由 y = k/x² 求出 k，再代入新值。一定写出完整表达式。
2. Successive Percentage Changes | 连续百分比增减
ot add percentages when a quantity is increased and then decreased. If a price rises by 20% and then falls by 20%, the final price is not the original. The common mistake is to think the net change is 0%. The correct approach uses multipliers: a 20% increase means multiply by 1.20; a 20% ns multiply by 0.80. So the combined multiplier is 1.20 × 0.80 = 0.96, a net decrease of 4%.
当一量先增加再减少相同的百分比时，千万不要直接加减百分数。如果价格先涨 20% 再降 20%，最终价格并不等于原价。常见错误是认为净变化为 0%。正确做法使用乘数：涨 20% 是乘以 1.20；降 20% 是乘以 0.80。综合乘数为 1.20 × 0.80 = 0.96，相当于净减少 4%。
Tricky WJEC question: A salary of £24 000 is increased by 5% and then by a further 3%. What is the final salary? Inexperienced students add 5% + 3% = 8% and apply a single 8% increase. That gives £25 920, but the correct amount is £24 000 × 1.05 × 1.03 = £25 956. The difference becomes significant when many successive changes are applied.
WJEC易错题：一份 24 000 英镑的薪水先涨 5% 再涨 3%。最终薪水是多少？经验不足的学生会将 5% 与 3% 相加得到 8%，然后只乘一次 1.08，得到 25 920 英镑。正确答案为 24 000 × 1.05 × 1.03 = 25 956 英镑。当涉及多次连续变化时，这种差异会变得很大。
3. Expanding Brackets with Negative Signs | 去括号时的符号错误
When expanding brackets with a negative multiplier, such as –2(3x – 5), many students forget to multiply the –5 by –2, leaving it as – ake. The correct expansion multiplies every term inside the bracket by the factor outside: –2 × 3x = –6x, and –2 × (–5) = +10. So the result is –6x + 10. Similarly, be careful with expressions like (x – 3)(2 – Expanding systematically using FOIL or grid method is safer than trying to do it mentally.
当括号前有负号乘数时，例如 –2(3x – 5)，许多学生忘记将 –5 与 –2 相乘，错误地保留成 –5 或 +5。正确展开应将括号内每一项与外部的因数相乘：–2 × 3x = –6x，–2 × (–5) = +10。结果为 –6x + 10。同样，对于 (x – 3)(2 – x) 这类式子，有条不紊地使用首外内尾（FOIL）或格点法展开，比心算更稳妥。
Also watch out for squaring brackets: (x + 4)² is not x² + 16. You must write it as (x + 4)(x + 4) and expand to x² + 8x + 16. Missing the middle term is a pervasive error in WJEC algebra questions.
还要当心完全平方：(x + 4)² 不等于 x² + 16。必须将其写成 (x + 4)(x + 4) 展开，得到 x² + 8x + 16。漏掉中间项是 WJEC 代数题中普遍的错误。
4. Solving Linear Equations: Moving Terms | 解线性方程：移项错误
A frequent slip occurs when students try to isolate the variable. For the equation 5x – to one side, you must do to the other. A safer method is to write the step explicitly: 5x – 3 – 2x = 2x + 9 – 2x, giving 3x – 3 = 9. Then add 3: 3x = 12, so x = 4.
学生在移项时常犯的错误：对于方程 5x – 3 = 2x + 9，有人从右边减 2x 却忘记左边也要减，或者只在一侧加 3。黄金法则：对等式一侧做的任何运算，另一侧必须同样操作。更稳妥的方法是明确写出步骤：5x – 3 – 2x = 2x + 9 – 2x，得到 3x – 3 = 9。然后加 3：3x = 12，x = 4。
nators, here 6: 6(x/3) + 6(1) = 6(x/2) → 2x + 6 = 3x. Then solve: x = 6. Avoid the mistake of multiplying only the fractions and forgetting the constant term.
另一个易错情况：含有分数的方程，如 x/3 + 1 = x/2。学生经常乘错。应将每一项乘以分母的最小公倍数（此处为 6）：6(x/3) + 6(1) = 6(x/2) → 2x + 6 = 3x。解得 x = 6。切勿只乘含 x 的项而漏掉常数项。
5. Pythagoras' Theorem: Identifying the Hypotenuse | 勾股定理：找错斜边
te a² + b² = c² but assign c to one of the shorter legs. When the hypotenuse is given as one of the known sides, students sometimes incorrectly set up the equation. If the hypotenuse is given and one leg is missing, use h² – l² = l²? No: the correct form is hypotenuse² = leg² + leg², so missing leg = √hypotenuse² – known leg².
在任何直角三角形中，斜边总是直角的对边，也是最长的边。WJEC 考试中一个常见错误是将 a² + b² = c² 写错，把 c 当成较短的一条直角边。当斜边已知而一条直角边缺失时，学生有时会列错方程。正确形式是：斜边² = 直角边₁² + 直角边₂²，因此缺失直角边 = √斜边² – 已知直角边²。
Example: leg 5 cm. Find the other leg. Incorrect: 13² + 5² = x². Correct: 13² = 5² + x² → x² = 169 – 25 = 144 → x = 12 cm. Always label the hypotenuse first before applying the theorem.
例子：一直角三角形斜边长为 13 cm，一直角边为 5 cm，求另一直角边。错误列式：13² + 5² = x²。正确列式：13² = 5² + x² → x² = 169 – 25 = 144 → x = 12 cm。先标出斜边再用定理。
6. Probability: With and Without Replacement | 概率：放回与不放回
he mistake is to treat sampling without replacement as independent events, using the same denominator for the second draw. If a bag contains 5 red and 3 blue discs, the probability of drawing two red discs without replacement is (5/8) × (4/7) = 20/56 = 5/14. Many students write (5/8) × (5/8) = 25/64. Always check the wording: 'without replacement' or 'not replaced' means the total and the count of that colour both decrease.
关于从袋中取两个或多个物品的概率题，常考的关键是放回还是不放回。典型错误是把不放回抽样当成独立事件，第二次抽取时分母还用原来的总数。如果一个袋中有 5 红 3 蓝，不放回地抽两个红球的概率是 (5/8) × (4/7) = 20/56 = 5/14。许多学生却写成 (5/8) × (5/8) = 25/64。务必留意题干措辞：“不放回”或“未放回”意味着总数和对应颜色的个数都减少了。
Also, when questions ask for 'at least one' event, remember to use the complement method: P(at least one) = 1 – P(none). This is far less error-prone than adding individual probabilities.
另外，当题目问到“至少一个”事件时，记得用补集法：P(至少一个) = 1 – P(一个都没有)。这比逐个概率相加要稳妥得多。
7. Mean, Median, and Misleading Averages | 平均数、中位数与误导性均值
liers. If a data set is highly skewed, the median is more representative. A common task: given a small set like 3, 4, 5, 6, 42, the mean is (3+4+5+6+42)/5 = 12, but the median is 5. The mean is pulled up by the outlier 42. In such cases, the median gives a better picture of the typical value.
WJEC 统计题有时会问哪种平均数最能代表数据。学生常常不加思考就选择平均数（均值），而不考虑异常值。如果数据高度偏斜，中位数更具代表性。常见题型：给定一组数据如 3, 4, 5, 6, 42，均值为 (3+4+5+6+42)/5 = 12，但中位数为 5。均值被异常值 42 拉高了。这种情况下，中位数更能体现典型水平。
Also be precise with calculations: the median of an even number of data values is the mean of the two middle numbers. Many candidates lose a mark by not dividing by 2 correctly.
计算也要精确：偶数个数据的中位数是中间两个数的平均值。许多考生因为忘记除以 2 而丢分。
8. Unit Conversions for Area and Volume | 面积与体积单位换算
0 cm². Similarly, 1 m³ = 1 000 000 cm³. When converting units of area or volume, always square or cube the linear conversion factor. This mistake often appears in contexts like converting densities or filling containers.
WJEC 度量题中最顽固的错误之一是把 1 平方米当成 100 平方厘米。事实上，1 米 = 100 厘米，所以 1 平方米 = (100 厘米)² = 10 000 平方厘米，而不是 100 平方厘米。同样，1 立方米 = 1 000 000 立方厘米。在换算面积或体积单位时，一定要对线性换算系数进行平方或立方。此类错误常出现在密度换算或容器填充题中。
Example: so area = 200 × 50 = 10 000 cm². Alternatively, work in m: 0.5 m × 2 m = 1 m² = 10 000 cm².
例子：一个矩形长 2 米，宽 50 厘米，求面积（平方厘米）。常见错误：面积 = 2 × 50 = 100 平方厘米。但必须先将单位统一：2 米 = 200 厘米，所以面积 = 200 × 50 = 10 000 平方厘米。或者用米计算：0.5 米 × 2 米 = 1 平方米 = 10 000 平方厘米。
9. Index Laws: Multiplying Powers and Roots | 指数律：幂的乘方与开方
e is multiply the indices: (a²)³ = a²×³ = a⁶. For a² × a³, you do add: a²⁺³ = a⁵. Mixing these two rules is a typical slip. Also, negative indices represent reciprocals: a⁻² = 1/a², not –a². Fractional indices link to roots: a1/2 = √a, and am/n = (ⁿ√a)ᵐ.
指数记号错误非常常见。对于 (a²)³，很多学生将其错写成 a⁵，把指数相加而不是相乘。正确规则是：幂的乘方，指数相乘：(a²)³ = a²×³ = a⁶。而对于 a² × a³，指数确实是相加：a²⁺³ = a⁵。混淆这两种规则是典型错误。另外，负指数表示倒数：a⁻² = 1/a²，而不是 –a²。分数指数与根式相联系：a1/2 = √a，am/n = (ⁿ√a)ᵐ。
WJEC style question: Simplify (8x⁶)2/3. Step 1: Apply index to each factor: 82/3 × (x⁶)2/3. 82/3 = (∛8)² = 2² = 4. (x⁶)2/3 = x6 × 2/3 = x⁴. Answer: 4x⁴. Many candidates forget to take the cube root of 8 correctly.
WJEC 风格题目：化简 (8x⁶)2/3。步骤：将指数应用到每个因子：82/3 × (x⁶)2/3。82/3 = (∛8)² = 2² = 4。(x⁶)2/3 = x6 × 2/3 = x⁴。答案：4x⁴。很多考生忘记正确计算 8 的立方根。
10. Straight Line Graphs: Gradient and Intercept | 直线图：斜率与截距
-intercept is not clearly shown on the grid. The gradient m = change in y / change in x. If the line slopes downwards, the gradient is negative. The intercept c is the y-value where the line crosses the y-axis. If the graph does not clearly show the y-axis crossing, use the point-slope form: y – y₁ = m(x – x₁) with a known point.
要求学生从图形求直线方程的题目，遇到斜率为负或图像未清晰显示 y 轴交点时，学生常犯错。斜率 m = y 的变化量 / x 的变化量。如果直线向下倾斜，斜率为负。截距 c 是直线与 y 轴交点的 y 值。如果图上没有明显显示 y 轴交点，可以使用点斜式：y – y₁ = m(x – x₁)，代入一个已知点。
Common error: n of : mixing up the sign of the gradient when calculating from two points, say (1, 5) and (3, 1). Correct: m = (1 – 5)/(3 – 1) = –4/2 = –2. Using the points in the wrong order gives (5 – 1)/(1 – 3) = 4/–2 = – but a sign slip often happens when subtracting coordinates carelessly. Consistency is key.
常见错误：由两点计算斜率时搞错符号，例如点 (1, 5) 和 (3, 1)。正确：m = (1 – 5)/(3 – 1) = –4/2 = –2。如果用不同顺序 (5 – 1)/(1 – 3) = 4/–2 = –2，结果仍正确，但粗心相减时常会弄错符号。保持一致性是关键。
11. Interpreting Histograms and Frequency Polygons | 直方图与频数多边形的解读
WJEC often tests the understanding that in a histogram, the area of the bar represents the frequency, not the height, when the class widths are unequal. If the class intervals are of different widths, you must work out the frequency density (frequency ÷ class width) for each bar. Students who assume frequency equals height will get the wrong answers when interpreting or drawing histograms with unequal intervals.
WJEC 常考查以下理解：在直方图中，当组距不等时，是用每个直方的面积（而非高度）表示频数。若组宽度不相同，必须计算每个直方的频数密度（频数 ÷ 组距）。学生在解读或绘制区间不等宽的直方图时，如果误以为频数就是高度，就会得出错误答案。
Similarly, a frequency polygon is created by plotting the midpoints of class intervals against frequency and joining with straight lines. Do not plot against the class boundaries. Always use midpoints.
同样，频数多边形是将各组的组中值对频数描点，并用直线连接。切勿用组边界值描点。一定要用组中值。
12. Algebraic Fractions and Cancelling | 代数分式与约分
When simplifying algebraic fractions like (x² – 9)/(x² – x – 12), students often cancel terms incorrectly across the numerator and denominator without factorising. The wrong move: seeing x² in both and cancelling, leaving something meaningless. Correct method: factorise completely. Numerator: (x – 3)(x + 3); Denominator: (x – 4)(x + 3). Cancel the common factor (x + 3) to get (x – 3)/(x – 4), provided x ≠ –3. Always remember: you cannot cancel through addition or subtraction, only through multiplication factors.
化简代数分式如 (x² – 9)/(x² – x – 12) 时，学生常不经因式分解就错误地进行约分。错误操作：看到分子分母都有 x² 就约去，留下无意义的式子。正确方法：完全因式分解。分子：(x – 3)(x + 3)；分母：(x – 4)(x + 3)。约去公因式 (x + 3) 得到 (x – 3)/(x – 4)，且 x ≠ –3。始终牢记：只能是相乘的因子才能约分，不能在加减法之间直接约分。
iding both sides by x, losing the solution x = 0. Instead, bring all terms to one side: x² – 9x = 0, then factorise: x(x – . Never divide by a variable unless you know it is non-zero and you retain that condition.
这也涉及解方程如 x² = 9x 时的常见错误。学生两边同除以 x，从而丢失了 x = 0 这个解。正确的做法是移项：x² – 9x = 0，然后因式分解：x(x – 9) = 0，得到 x = 0 或 x = 9。除非已知变量非零并保留该条件，否则绝不要除以变量。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图