📚 Practical Problems in IGCSE Maths | IGCSE数学中的实际应用问题
Practical problems lie at the heart of IGCSE Mathematics. They take the abstract skills of algebra, geometry and statistics and place them into real‑world contexts — from calculating a restaurant tip and comparing mobile phone tariffs to estimating the paint needed for a bedroom wall. Mastering these questions means you can read a scenario, decide which mathematical tool to use, carry out the calculations and then interpret your answer in the original situation. This article walks you through the main types of practical problems you will meet in Edexcel IGCSE Maths, with paired explanations in English and Chinese to strengthen your understanding.
实际问题处于 IGCSE 数学的核心位置。它们将代数、几何和统计的抽象技能置于真实情境中——从计算餐厅小费、比较手机套餐到估算卧室墙面所需的油漆量。掌握这类题目意味着你能读懂情境、决定使用哪种数学工具、完成计算并在原情境中解释答案。本文将带你梳理 Edexcel IGCSE 数学中主要的实际问题类型，并用英中对照的方式帮助加深理解。
1. Reading and Understanding the Problem | 阅读与理解问题
Before doing any calculation, read the question at least twice. Underline the quantities you are given and circle what you are asked to find. Pay attention to units, the degree of accuracy required and any hints like “give your answer to 2 decimal places.” Many marks are lost simply because a student solves the wrong problem.
在做任何计算之前，至少把题目读两遍。在给出的数量下方画线，圈出需要求的量。注意单位、要求的精确度以及类似“答案保留两位小数”的提示。许多失分仅仅是因为学生解错了问题。
Convert the wordy description into a simple mathematical statement. For example, “Sophie buys 3 packs of stickers for £2.49 each and pays with a £10 note. How much change does she get?” becomes: total cost = 3 × 2.49, then change = 10.00 – total cost.
把冗长的文字描述转化成简单的数学表达式。例如，“Sophie 购买了 3 包贴纸，每包 2.49 英镑，她用一张 10 英镑的纸币付款。她找回多少钱？”转化为：总花费 = 3 × 2.49，然后找零 = 10.00 – 总花费。
2. Ratios and Proportions | 比率与比例
ratio a : b means for every a parts of one quantity there are b parts of another. To share £120 in the ratio 3 : 5, add the parts (3 + 5 = 8), find one part (120 ÷ 8 = £15), then multiply: 3 parts = £45, 5 parts = £75.
比率用于比较数量，出现在食谱、地图比例尺和分钱问题中。比例 a : b 意味着每一个量 a 份，另一个量就有 b 份。把 120 英镑按 3:5 分配：先把份数相加 (3+5=8)，求出一份 (120÷8=15 英镑)，再相乘：3 份 = 45 英镑，5 份 = 75 英镑。
or inverse relationships. For direct proportion, y = kx. Example: 5 litres of petrol cost £7.80. Find the cost of 13 litres. First find k = price per litre = 7.80 ÷ 5 = £1.56, then multiply by 13 to get £20.28.
比例问题常涉及正比或反比关系。正比例关系为 y = kx。例如：5 升汽油售价 7.80 英镑，求 13 升的价钱。先求出每升价钱 k = 7.80 ÷ 5 = 1.56 英镑，然后乘以 13 得到 20.28 英镑。
3. Percentages in Context | 情境中的百分数
Percentage problems go far beyond simple “find 15% of 80”. You will need to calculate percentage increase and decrease, profit and loss, discounts, tax and repeated changes. To increase 200 by 30%, multiply by 1.30. To decrease it by 30%, multiply by 0.70. For a discount followed by a tax, apply the multiplier chain.
百分数问题远远超出简单的“求 80 的 15%”。你需要计算百分比增减、盈亏、折扣、税款以及重复变化。将 200 增加 30%，乘以 1.30；减少 30%，则乘以 0.70。对于折扣后再加税的情况，使用连续乘数。
A classic practical problem: a sofa priced at £480 excluding VAT. VAT is 20%. Find the price including VAT. You multiply by 1.20: 480 × 1.20 = £576. Or a sale gives 15% off, so a £60 jacket costs 60 × 0.85 = £51.
一个经典的实际问题：沙发不含增值税标价为 480 英镑，增值税为 20%，求含税价。乘以 1.20：480 × 1.20 = 576 英镑。或者打八五折，一件 60 英镑的夹克卖 60 × 0.85 = 51 英镑。
4. Simple and Compound Interest | 单利与复利
l, r is annual interest rate as a decimal, t is time in years. The total amount A = P + I.
金融数学是实际问题的重要组成部分。单利仅按初始本金计算：I = P × r × t，其中 P 为本金，r 为年利率（小数），t 为时间（年）。总金额 A = P + I。
1 + r/100)ⁿ for percentage rate, or A = P(1 + i)ⁿ where i is the decimal equivalent. E.g., £700 invested at 4% compound interest for 3 years: A = 700 × (1.04)³ = £787.40 (to nearest penny).
复利不仅对本金计算利息，还对已产生的利息计算利息。公式为 A = P(1 + r/100)ⁿ（百分率形式），或 A = P(1 + i)ⁿ，其中 i 为小数形式。例如，700 英镑以 4% 复利投资 3 年：A = 700 × (1.04)³ = 787.40 英镑（精确到分）。
eciation (loss of value) uses a similar formula with a multiplier less than 1. A car worth £15,000 depreciates by 18% each year. After n years its value is 15000 × (0.82)ⁿ.
贬值（价值损失）使用类似的公式，但乘数小于 1。一辆价值 15000 英镑的汽车每年贬值 18%，n 年后其价值为 15000 × (0.82)ⁿ。
5. Speed, Distance and Time | 速度、距离与时间
stance ÷ Speed. The hardest part is often handling mixed units: if speed is in km/h and time is in minutes, convert minutes to hours first (divide by 60).
这一关系可以用公式三角形表示：距离 = 速度 × 时间，速度 = 距离 ÷ 时间，时间 = 距离 ÷ 速度。最棘手的部分往往是单位不统一：如果速度是 km/h，时间是分钟，先把分钟转换为小时（除以 60）。
Example: A cyclist covers 45 km in 2 hours 30 minutes. Average speed? 8 km/h. Another common question involves two parts of a journey with different speeds; you must find the total distance and total time before computing average speed.
例如：一名自行车手在 2 小时 30 分钟内骑行 45 km。平均速度？时间以小时计为 2.5 h，速度 = 45 ÷ 2.5 = 18 km/h。另一种常见题目是分段旅程，速度不同；你必须先算总距离和总时间，再求平均速度。
6. Area, Perimeter and Volume | 面积、周长与体积
Practical measurement problems ask you to find how much turf to buy for a garden, how many tiles to cover a floor, or how much water a tank can hold. Remember the formulas for rectangles, triangles, circles, prisms and cylinders. Often you need to combine shapes or subtract one area from another.
实际测量问题会让你计算花园需要多少草皮、铺设地面需要多少瓷砖、或者水箱能容纳多少水。记住矩形、三角形、圆形、棱柱和圆柱的公式。通常你需要组合图形或从某块面积中减去另一块。
rea = 4 × 2.5 = 10 m², so 1 litre is enough. If two coats are required, double the amount.
一个典型题目：一面长 4 m、高 2.5 m 的墙要粉刷。油漆每升可涂 10 m²。求需要多少升油漆。面积 = 4 × 2.5 = 10 m²，所以 1 升足够。如果需要刷两层，用量翻倍。
× 40 = 96,000 cm³. Since 1 litre = 1000 cm³, the tank holds 96 litres of water when full.
体积方面，一个鱼缸尺寸为 80 cm × 30 cm × 40 cm。体积（cm³）= 80 × 30 × 40 = 96,000 cm³。因为 1 升 = 1000 cm³，鱼缸装满可容纳 96 升水。
7. Unit Conversions | 单位换算
54 cm, 1 kg = 2.2 pounds, 1 gallon ≈ 4.5 litres, 1 litre = 1000 cm³, 1 m³ = 1,000,000 cm³.
实际问题经常故意混用公制和英制单位，或需要进行面积和体积单位的换算。请记住这些关键换算关系：1 英里 = 1.6 公里，1 英寸 = 2.54 厘米，1 千克 = 2.2 磅，1 加仑 ≈ 4.5 升，1 升 = 1000 cm³，1 m³ = 1,000,000 cm³。
mon error is to think 1 m² = 100 cm², which leads to answers that are 100 times too small.
换算平方或立方单位时，必须将线性换算系数平方或立方。例如，1 m = 100 cm，因此 1 m² = (100)² = 10,000 cm²。常见错误是认为 1 m² = 100 cm²，这会导致答案小 100 倍。
8 m wide. Express its area in cm². Area = 12 × 8 = 96 m², then 96 × 10,000 = 960,000 cm².
例子：一个花园长 12 m、宽 8 m，用 cm² 表示其面积。面积 = 12 × 8 = 96 m²，那么 96 × 10,000 = 960,000 cm²。
8. Budgeting and Financial Planning | 预算与财务规划
sizes, calculate the unit price (price per gram, per litre, etc.). For exchange rates, use proportions: if £1 = $1.27, then £400 = 400 × 1.27 = $508.
这类问题要求你比较成本、处理汇率并根据最佳性价比做出决定。比较不同规格的产品时，计算单位价格（每克、每升的价格等）。对于汇率，使用比例：如果 1 英镑 = 1.27 美元，那么 400 英镑 = 400 × 1.27 = 508 美元。
Best‑buy questions: a 750 g box of cereal costs £3.45, and a 1.2 kg box costs £5.28. Which is better value? Find the price per 100 g for each. For the first: (3.45 ÷ 7.5) = £0.46 per 100 g. For the second: (5.28 ÷ 12) = £0.44 per 100 g, so the larger box is cheaper per 100 g.
最佳购买问题：一盒 750 g 麦片售价 3.45 英镑，另一盒 1.2 kg 售价 5.28 英镑。哪种更划算？分别计算每 100 g 的价钱。第一盒：(3.45 ÷ 7.5) = 0.46 英镑/100 g。第二盒：(5.28 ÷ 12) = 0.44 英镑/100 g，所以大盒每 100 g 更便宜。
Budgeting tasks might involve monthly bills, income and savings. A student earns £320 a month from a part‑time job and spends 65% on rent and utilities, 20% on food and saves the rest. Amount saved = 320 × 0.15 = £48.
预算任务可能涉及月账单、收入和储蓄。一名学生每月兼职赚 320 英镑，65% 用于房租和水电，20% 用于食物，其余存起来。储蓄金额 = 320 × 0.15 = 48 英镑。
9. Interpreting Graphs and Charts | 解释图图表
Many practical problems present data in a visual form: distance‑time graphs, conversion graphs, pie charts and frequency tables. You need to read axes carefully, extract information and draw conclusions. On a distance‑time graph, a horizontal line means the object is stationary; the gradient represents speed.
许多实际问题以可视化的形式呈现数据：距离–时间图、转换图、饼图和频率表。你需要仔细读取坐标轴，提取信息并得出结论。在距离–时间图中，水平线段表示物体静止；斜率代表速度。
A conversion graph can be used to convert between currencies or units. If the graph of pounds against dollars is a straight line through the origin, the gradient gives the exchange rate. Always check which variable is on which axis.
转换图可用于货币或单位之间的转换。如果英镑对美元的图像是一条通过原点的直线，其斜率就是汇率。务必检查哪个变量在哪条坐标轴上。
a pie chart, calculate the fraction: (sector angle ÷ 360) × total. If the total budget is £36,000 and the “transport” sector has an angle of 55°, the amount spent on transport is (55 ÷ 360) × 36,000 = £5,500.
饼图展示比例。要从饼图中求出某个量，计算分数：(扇区角度 ÷ 360) × 总量。如果总预算为 36,000 英镑，“交通”扇区角度为 55°，那么交通支出为 (55 ÷ 360) × 36,000 = 5,500 英镑。
10. Multi‑step Problem Solving | 多步骤问题求解
Real‑world problems rarely involve a single calculation. You might need to convert units, find a percentage, apply a ratio and then check for rounding — all within one question. Break the problem into small, logical steps and write them down in order. Keep your working neat so you can spot mistakes.
现实世界的问题很少只涉及单一计算。你可能需要换算单位、求一个百分比、应用比例，然后检查舍入——所有步骤都在同一道题里。把问题分解为小而合逻辑的步骤，并按顺序写下。保持解题过程整洁，以便发现错误。
ght 1.5 m. Water flows in at 8 litres per minute. Estimate the time to fill the tank to 80% × 0.36 × 1.5 = 1.6956 m³. Step 2: volume in litres = 1.6956 × 1000 = 1695.6 L. Step 3: 80% 56.48 L. Step 4: time = volume ÷ rate = 1356.48 ÷ 8 = 169.56 minutes ≈ 2 hours 50 minutes.
例子：一个圆柱形水箱，直径 1.2 m，高 1.5 m。水以每分钟 8 升的速度流入。估算注满 80% 容量所需的时间。步骤 1：半径 = 0.6 m，圆柱体积 = π × r² × h = 3.14 × 0.36 × 1.5 = 1.6956 m³。步骤 2：体积换算为升 = 1.6956 × 1000 = 1695.6 L。步骤 3：80% 体积 = 0.8 × 1695.6 = 1356.48 L。步骤 4：时间 = 体积 ÷ 流量 = 1356.48 ÷ 8 = 169.56 分钟 ≈ 2 小时 50 分钟。
Always remember to present your final answer in the context of the question, with correct units and a suitable degree of accuracy. A complete sentence like “It will take about 2 hours and 50 minutes to fill the tank to 80% capacity” often scores the final mark.
永远记得在题目情境下给出最终答案，并带上正确的单位和适当的精确度。像“注满水箱 80% 大约需要 2 小时 50 分钟”这样完整的句子往往能拿到最后一分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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