📚 Common Question Types and Problem-Solving Strategies for Math Word Problems | 数学文字题常见考点与解题策略
Mathematics word problems are not simply arithmetic exercises wrapped in sentences. They test a student’s ability to translate real-world situations into mathematical structures, apply relevant formulas, and interpret results in context. In examinations, word problems often appear in algebra, number, ratio, and geometry sections, and they reward candidates who read carefully and plan before calculating.
数学文字题并不是披着句子的算术练习。它们考查学生将现实情境转化为数学结构、运用相关公式并在具体情境中解读结果的能力。在考试中，文字题常出现在代数、数、比例和几何板块，那些仔细阅读并在计算前先作规划的学生通常能取得更好的成绩。
1. Translating Language into Equations | 将语言转化为方程
The first core skill is converting English phrases into algebraic expressions. Keywords such as “sum,” “difference,” “product,” “more than,” “times,” and “is” indicate specific operations. For example, “three more than twice a number” becomes 2x + 3, while “the product of a number and five decreased by two” becomes 5x − 2.
第一项核心技能是将英文短语转化为代数表达式。诸如 “sum”、“difference”、“product”、“more than”、“times” 和 “is” 等关键词对应着特定的运算。例如，“three more than twice a number” 应写为 2x + 3，而 “the product of a number and five decreased by two” 应写为 5x − 2。
A useful method is to define the unknown first. Let x represent the quantity the question asks about, then build the equation from the remaining information. Always read the final sentence before starting: it usually tells you what variable or quantity is actually required.
一个有效的方法是先设未知数。让 x 代表题目所问的量，再利用其余信息构建方程。动笔前务必先读最后一句，因为它通常说明真正需要求的变量或量是什么。
Additionally, pay attention to comparison phrases like “as old as,” “half of,” and “increased by.” These expressions create relationships that must be written precisely. Misplacing one word can change the entire equation.
此外，要留意诸如 “as old as”、“half of” 和 “increased by” 等比较性短语。这些表达方式建立了需要精确写出的关系。一个词的位置错误就可能改变整个方程。
2. Solving Age Problems | 年龄问题
Age problems are among the most common word problems in algebra. They usually involve two people whose ages are related now and at some time in the past or future. A table that separates “now” and “then” is especially helpful.
年龄问题是代数中最常见的文字题之一。它们通常涉及两个人的年龄，在现在以及过去或未来的某个时刻存在某种关系。将 “现在” 和 “那时” 分开的表格尤其有用。
For example, suppose a father is three times as old as his son. In 6 years, he will be twice as old as his son. Let the son’s current age be x. Then the father’s current age is 3x. In 6 years, the son is x + 6 and the father is 3x + 6. The equation is:
例如，假设父亲年龄是儿子的三倍。6 年后，父亲年龄将是儿子的两倍。设儿子现在年龄为 x，则父亲现在年龄为 3x。6 年后，儿子为 x + 6，父亲为 3x + 6。方程为：
3x + 6 = 2(x + 6)
Solving gives x = 6, so the son is 6 and the father is 18. The key is consistency: every expression must be measured from the same point in time.
解得 x = 6，因此儿子 6 岁，父亲 18 岁。关键在于一致性：每个表达式必须以同一时间点作为参照。
3. Consecutive Integer Problems | 连续整数问题
Consecutive integer problems ask for numbers that follow one another, such as 5, 6, 7, or even numbers like 2, 4, 6. The most difficult part is choosing the correct variable representation.
连续整数问题要求找出依次排列的数，例如 5、6、7，或偶数如 2、4、6。最困难的部分在于选择正确的变量表示方式。
For consecutive integers, use x, x + 1, x + 2. For consecutive even or odd integers, use x, x + 2, x + 4, because the difference between consecutive even numbers is 2 and the difference between consecutive odd numbers is also 2.
对于连续整数，使用 x、x + 1、x + 2。对于连续偶数或连续奇数，使用 x、x + 2、x + 4，因为相邻偶数之差为 2，相邻奇数之差也为 2。
Example: The sum of three consecutive odd integers is 57. Let the integers be x, x + 2, x + 4. Then:
示例：三个连续奇数的和为 57。设这三个数为 x、x + 2、x + 4，则：
x + (x + 2) + (x + 4) = 57
This simplifies to 3x + 6 = 57, so x = 17. The numbers are 17, 19, and 21. Always check whether “inclusive” or “exclusive” wording affects the sequence.
化简得 3x + 6 = 57，所以 x = 17。这三个数是 17、19、21。始终检查题目中 “含端点” 或 “不含端点” 的表述是否会影响到数列。
4. Ratio and Proportion Problems | 比例与正比问题
Ratio problems often describe a mixture of quantities, such as nuts in a blend or students in a school. The standard method is to multiply each part of the ratio by a common factor k.
比例问题常描述若干数量的混合，例如混合坚果或学校中的学生人数。标准方法是把比值的每一部分乘以公共因子 k。
If a recipe requires flour and sugar in the ratio 3 : 2, let the actual amounts be 3k and 2k. If the total amount is 500 g, then 3k + 2k = 500, so k = 100. Thus flour is 300 g and sugar is 200 g.
如果某配方要求面粉与糖按 3 : 2 的比例混合，令实际用量分别为 3k 和 2k。若总质量为 500 g，则 3k + 2k = 500，所以 k = 100。因此面粉为 300 g，糖为 200 g。
Proportion problems also include direct and inverse proportion. If y is directly proportional to x, then y = kx. If y is inversely proportional to x, then y = k ÷ x. Recognising which type is described is essential: “increases with” suggests direct proportion, while “as one doubles the other halves” suggests inverse proportion.
正比问题还包括正比例和反比例。若 y 与 x 成正比，则 y = kx。若 y 与 x 成反比，则 y = k ÷ x。识别题目描述的是哪种关系至关重要：“随……增加” 暗示正比例，而 “一个翻倍另一个减半” 则暗示反比例。
5. Percentage Increase and Decrease | 百分数增减问题
Percentage problems frequently appear in financial contexts such as discounts, salaries, and population changes. The central relation is that an amount after an increase or decrease equals the original amount multiplied by a decimal factor.
百分数问题常出现在折扣、工资、人口变化等与金融相关的语境中。核心关系是：增减后的量等于原来的量乘一个小数因子。
A 15% increase means the final amount is 115% of the original, so multiply by 1.15. A 15% decrease means the final amount is 85% of the original, so multiply by 0.85. For example, a price of £80 increased by 15% becomes:
增加 15% 意味着最终金额是原来的 115%，因此乘以 1.15。减少 15% 意味着最终金额是原来的 85%，因此乘以 0.85。例如，一件价格 £80 的商品增加 15% 后变为：
80 × 1.15 = 92
When a question asks for the original value after a change, divide by the factor instead. If a new salary is £920 after a 15% increase, the original salary is 920 ÷ 1.15 = £800. Many students incorrectly multiply by 0.15; remember that the percentage change factor is always based on the original, not the new value.
当题目要求变化之后原来的值时，应除以该因子。如果加薪 15% 后新工资为 £920，则原工资为 920 ÷ 1.15 = £800。许多学生误乘 0.15；请记住，百分比变化因子总是基于原来的量，而不是新值。
6. Speed, Distance, and Time Problems | 速度、距离与时间问题
Speed, distance, and time problems follow the formula distance = speed × time. These questions often involve two objects moving toward or away from each other, or a single object travelling in two parts.
速度、距离与时间问题遵循公式：距离 = 速度 × 时间。这类题常涉及两个物体相向或相背而行，或者一个物体分两段行程。
For two objects moving toward each other, the combined distance they cover equals the total separation. If a car and a bus start 240 km apart and travel toward each other at 60 km/h and 40 km/h, then after t hours, the equation is:
对于两个相向而行的物体，它们共同行驶的距离等于总间隔。如果一辆汽车和一辆公交车相距 240 km，分别以 60 km/h 和 40 km/h 相向而行，则 t 小时后方程为：
60t + 40t = 240
Thus 100t = 240, so t = 2.4 hours. Always state whether the time is in hours or minutes, and convert units consistently: if speed is in km/h, time must be in hours and distance in km.
因此 100t = 240，所以 t = 2.4 小时。务必说明时间单位是小时还是分钟，并一致地进行单位换算：速度单位为 km/h 时，时间必须用小时，距离必须用 km。
7. Work and Rate Problems | 工作量与速率问题
Work problems involve the rate at which a person or machine completes a task. The basic principle is that rate = work ÷ time. If one pipe fills a tank in 4 hours, its rate is ¼ tank per hour. If another fills it in 6 hours, its rate is ⅙ tank per hour.
工程问题涉及人或机器完成一项任务的速率。基本原则是：速率 = 工作量 ÷ 时间。如果一根水管 4 小时注满水池，其速率为每小时 ¼ 池；如果另一根 6 小时注满，则其速率为每小时 ⅙ 池。
When working together, rates add. Let t be the time for both pipes together:
共同工作时，速率相加。设两根水管共同工作所需时间为 t：
(¼ + ⅙) × t = 1
The common denominator 12 gives 3/12 + 2/12 = 5/12. Therefore t = 12/5 = 2.4 hours. This “add the rates” strategy works for any situation where the independent rates are known.
公分母 12 可得 3/12 + 2/12 = 5/12。因此 t = 12/5 = 2.4 小时。这种 “速率相加” 的策略适用于任何已知独立速率的情境。
8. Geometry Word Problems | 几何文字题
Geometry word problems require translating descriptions of shapes into equations using perimeter, area, or volume formulas. A classic type involves a rectangle where the length is expressed in terms of the width.
几何文字题要求利用周长、面积或体积公式，将形状的描述转化为方程。一个典型类型是长用宽来表示的矩形问题。
Example: A rectangle is 3 cm longer than it is wide. Its perimeter is 26 cm. Let the width be w and the length be w + 3. Then:
示例：一个矩形的长比宽多 3 cm，其周长为 26 cm。设宽为 w，长为 w + 3，则：
2(w + 3) + 2w = 26
This gives 4w + 6 = 26, so w = 5 and the length is 8. For compound shapes, it is often wise to separate the shape into simpler figures and label all unknown sides before setting up equations.
这得到 4w + 6 = 26，所以 w = 5，长为 8。对于复合图形，较为明智的做法是先将图形拆分为更简单的图形，并在建立方程前标注所有未知边。
9. Money and Coin Problems | 货币与硬币问题
Coin and money problems build equations from the value of each item. The number of coins and the total value form two equations. For example, a collection of 20 coins consists only of 10p and 50p pieces with a total value of £5.20, which is 520 pence.
硬币与货币问题根据每件物品的面值建立方程。硬币数量和总价值构成两个方程。例如，某收藏有 20 枚硬币，只包含 10p 和 50p 硬币，总面值为 £5.20，即 520 便士。
Let x be the number of 10p coins and y be the number of 50p coins. Then:
设 x 为 10p 硬币数量，y 为 50p 硬币数量，则：
x + y = 20 and 10x + 50y = 520
From the first equation, y = 20 − x. Substituting gives 10x + 50(20 − x) = 520. Simplifying, 10x + 1000 − 50x = 520, so −40x = −480 and x = 12. Therefore there are 12 ten-pence coins and 8 fifty-pence coins.
由第一个方程得 y = 20 − x。代入得 10x + 50(20 − x) = 520。化简：10x + 1000 − 50x = 520，因此 −40x = −480，x = 12。所以有 12 枚 10 便士硬币和 8 枚 50 便士硬币。
10. Mixture Problems | 混合问题
Mixture problems involve combining substances with different concentrations or prices. A standard method is to set up an equation for the total amount of a key component, such as salt, acid, or cost.
混合问题涉及合并具有不同浓度或单价的物质。标准方法是为关键成分建立总量方程，例如盐、酸或总成本。
If 200 ml of a 10% salt solution is mixed with 300 ml of a 20% salt solution, the total salt is 0.10 × 200 + 0.20 × 300 = 20 + 60 = 80 ml. The total mixture is 500 ml, so the final concentration is:
如果将 200 ml 浓度为 10% 的盐溶液与 300 ml 浓度为 20% 的盐溶液混合，则盐的总量为 0.10 × 200 + 0.20 × 300 = 20 + 60 = 80 ml。混合液总量为 500 ml，因此最终浓度为：
80 ÷ 500 × 100% = 16%
Always write the concentration as a decimal in calculations, and verify that the units for volume or mass are consistent across all terms.
计算时始终把浓度写成小数形式，并检查所有项的体积或质量单位是否一致。
11. Data Interpretation and Statistics | 数据解读与统计
Word problems in statistics often describe a data set and ask for the mean, median, mode, or range. The wording matters: “average” usually means mean, but a question may explicitly mention the median or mode.
统计文字题通常描述一个数据集并求平均数、中位数、众数或极差。措辞至关重要：“average” 通常指平均数，但也可能明确提及中位数或众数。
A common algebraic trick involves the mean. If the mean of five numbers is 12, their sum is 5 × 12 = 60. If four of the numbers are known, the fifth can be found by subtraction. For grouped or summarised data, write out the sum formula before calculating:
一个常见的代数技巧涉及平均数。如果五个数的平均数为 12，则它们的总和为 5 × 12 = 60。若已知其中四个数，第五个数可通过减法求得。对于分组或汇总数据，先写出求和公式再计算：
Mean = Σx ÷ n
When a question asks how a new data point changes the mean, use the current sum and count, then add the new value and divide by the new count.
当题目询问一个新数据点如何改变平均数时，使用当前的总和与个数，加上新值后再除以新的个数。
12. General Problem-Solving Strategies | 通用解题策略
All mathematical word problems benefit from a structured approach. First, read the whole problem slowly and underline the question. Second, identify what data is given and what is unknown. Third, define variables clearly, including their units. Fourth, build one or more equations from the relationships described. Fifth, solve and then check the answer in the original wording.
所有数学文字题都能从结构化的方法中受益。首先，慢速阅读整道题并在问题处下划线。其次，识别已知数据和未知量。第三，清晰定义变量并注明单位。第四，根据所描述的关系建立一个或多个方程。第五，求解并在原题表述中检验答案。
A useful habit is the “unit sanity check”: if a question asks for a time, the answer should include units such as hours or minutes, not a length. Another habit is to look for hidden assumptions, such as whether a percentage is applied once or twice in a row. Finally, practise translating sentences into expressions regularly; this reduces the risk of errors during high-pressure exams.
一个有用的习惯是 “单位合理性检查”：如果题目求时间，答案应包含小时或分钟等单位，而不是长度。另一个习惯是寻找隐含假设，例如百分比是应用一次还是连续应用两次。最后，定期练习将句子转化为表达式；这能降低高压考试中出错的风险。
By mastering these common question types, students can transform word problems from a source of anxiety into a predictable set of tools. The same logical structure underlies nearly every word problem: define, relate, solve, and verify.
通过掌握这些常见题型，学生可以将文字题从焦虑来源转化为一套可预测的工具。几乎每一道文字题背后都有着相同的逻辑结构：设未知、找关系、求解答、再验证。
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