📚 Number 7: Ratio and Proportion | 第七单元：比率与比例
Ratio and proportion form the backbone of many real‑world applications in IGCSE Mathematics. From mixing paints and sharing prize money to scaling recipes and reading maps, these concepts help you compare quantities and understand how changes in one quantity affect another. In this unit, aligned with the Edexcel IGCSE specification, we will explore how to write and simplify ratios, divide amounts in a given ratio, work with direct and inverse proportion, and solve problems involving scale and best buys.
比率和比例是 IGCSE 数学中许多实际应用的基石。从调配颜料、分配奖金，到缩放食谱和阅读地图，这些概念帮助你比较数量，并理解一个量的变化如何影响另一个量。在本单元中，我们将依据 Edexcel IGCSE 的考纲，探讨如何书写和化简比、按给定比例分配数量、处理正比例和反比例，以及解决涉及比例尺和最优化购买的问题。
1. Understanding Ratios | 理解比
A ratio is a way of comparing two or more quantities of the same kind. It shows how many times one quantity contains another. For example, if a fruit bowl contains 3 apples and 5 oranges, the ratio of apples to oranges is written as 3 : 5. The order of the numbers is crucial – reversing the order changes the meaning completely.
比是一种比较两个或多个同类数量的方式。它表示一个数量包含另一个数量的多少倍。例如，如果水果碗中有 3 个苹果和 5 个橙子，苹果与橙子的比写作 3 : 5。数字的顺序至关重要——颠倒顺序会完全改变含义。
A ratio has no units because it compares quantities measured in the same units. You can also express a ratio as a fraction: the ratio a : b can be written as a/b, but this representation usually implies a part‑to‑whole relationship, whereas the colon notation emphasises a part‑to‑part comparison.
比没有单位，因为它比较的是同类测量单位的数量。你也可以将比写成分数形式：比 a : b 可以写成 a/b，但这种表示法通常暗示部分与整体的关系，而冒号记法则更强调部分与部分的比较。
2. Simplifying Ratios | 化简比
To simplify a ratio, divide every term by their highest common factor (HCF). For example, to simplify 8 : 12, divide both numbers by 4 to obtain 2 : 3. The simplified ratio 2 : 3 has the same proportional relationship as the original – for every 2 units of the first quantity, there are 3 of the second.
化简比时，将每一项除以它们的最大公因数（HCF）。例如，化简 8 : 12，将两数都除以 4 得到 2 : 3。化简后的比 2 : 3 与原始比具有相同的比例关系——第一个量每有 2 个单位，第二个量就有 3 个单位。
If a ratio involves fractions or decimals, multiply every term by a common multiple to clear them. For instance, 0.4 : 1.2 can be multiplied by 10 to give 4 : 12, which simplifies to 1 : 3. You can also simplify three‑term ratios, such as 6 : 9 : 12 dividing by 3 gives 2 : 3 : 4.
如果比含有分数或小数，可将每一项乘以一个公倍数以消除它们。例如，0.4 : 1.2 乘以 10 得到 4 : 12，化简为 1 : 3。你也可以化简三项的比，例如 6 : 9 : 12 除以 3 得到 2 : 3 : 4。
3. Dividing Quantities in a Given Ratio | 按给定比例分配数量
To share an amount using a ratio, first add the parts of the ratio together to find the total number of parts. Then divide the total quantity by this sum to obtain the value of one part. Multiply this single‑part value by each respective ratio number to find the individual shares. For example, divide £400 between two people in the ratio 3 : 5.
要按比例分配一笔金额，首先将比的各部分相加，求出总份数。然后用总数量除以这个和，得到一份所代表的值。将一份的值分别乘以比的每一项，即可得到各自应得的份额。例如，将 400 英镑按 3 : 5 分配给两个人。
Total parts = 3 + 5 = 8. One part = £400 ÷ 8 = £50. The first person receives 3 × £50 = £150, and the second receives 5 × £50 = £250. Always check that the individual shares add back to the original total: £150 + £250 = £400. This method works for mixtures, ingredients, and prize money.
总份数为 3 + 5 = 8。一份是 400 英镑 ÷ 8 = 50 英镑。第一个人分得 3 × 50 英镑 = 150 英镑，第二个人分得 5 × 50 英镑 = 250 英镑。始终检查各份额之和是否等于原始总数：150 英镑 + 250 英镑 = 400 英镑。此方法同样适用于混合物、配料和奖金。
4. Ratios and Fractions | 比与分数
A ratio can be expressed as a fraction to highlight part‑to‑ s 4/7. This conversion is especially useful when moving from ratios to probability or percentages.
比可以表示为分数，以突出部分与整体的关系。如果班级中男生与女生的比是 3 : 4，那么男生占全班的比例为 3/(3+4) = 3/7，女生的比例为 4/7。当需要将比转化为概率或百分比时，这种转换尤其有用。
the ratio. If we know that 2/5 of a solution is alcohol and the rest is water, the alcohol‑to‑water ratio is 2 : (5−2) = 2 : 3. Be careful with wording: “there are twice as many apples as oranges” implies the ratio apples : oranges = 2 : 1.
反之，给定分数，通常也可以还原出比。如果已知某种溶液中 2/5 是酒精，其余为水，则酒精与水的比为 2 : (5−2) = 2 : 3。注意措辞：“苹果的数量是橙子的两倍”意味着苹果与橙子的比为 2 : 1。
5. Direct Proportion | 正比例
a constant k. The graph of direct proportion is a straight line passing through the origin. Real‑ t of apples (the more you buy, the more you pay) and distance travelled at constant speed.
当两个量以相同的速率增加或减少时，它们成正比例。数学上，如果 y = kx（k 为常数），则 y 与 x 成正比例。正比例的图像是一条通过原点的直线。生活中的实例包括苹果的价格（买得越多，付得越多）以及以恒定速度行驶的距离。
t of proportionality k using a given pair of values. For example, if 5 identical toys cost £12.50, then cost C is directly proportional to the number of toys n: C = kn. Since £12.50 = k × 5, k = £2.50 per toy. Then the cost of 8 toys is C = 2.50 × 8 = £20.00.
解正比例问题时，先用给定的一对数值求出比例常数 k。例如，若 5 个相同的玩具售价 12.50 英镑，则费用 C 与玩具数量 n 成正比例：C = kn。由 12.50 英镑 = k × 5，得 k = 每个玩具 2.50 英镑。那么 8 个玩具的费用为 C = 2.50 × 8 = 20.00 英镑。
6. Inverse Proportion | 反比例
amples include the number of workers and the time to complete a job (more workers complete the work in fewer days), and speed with time for a fixed distance.
当两个量一个增加，另一个相应减少，且它们的乘积保持恒定时，它们成反比例。其方程为 y = k/x，或等价地 xy = k。实例包括工人数量与完成一项工作所需的时间（工人越多，所需天数越少），以及固定路程中的速度与时间。
that increasing one quantity causes the other to decrease to confirm the relationship is inverse.
解反比例问题时，使用乘积常数。例如，若 6 名工人粉刷栅栏需要 10 小时，则工人数与所需小时数的乘积恒定：6 × 10 = 60。如果有 8 名工人，所需时间 t 满足 8t = 60，因此 t = 7.5 小时。始终检查增加一个量是否会导致另一个量减少，以确认该关系是反比例关系。
7. Proportion and Best Buys | 比例与最优化购买
mparing prices, unit cost is your best friend. To decide whether a 350 g box of cereal for £2.80 or a 500 g box for £3.80 is the better buy, find the price per 100 g or per gram. For the first box: £2.80 ÷ 3.5 = £0.80 per 100 g. For the second: £3.80 ÷ 5 = £0.76 per 100 g, so the larger box offers better value.
比较价格时，单位成本是最好的工具。要判断一盒 350 克售价 2.80 英镑的麦片和一盒 500 克售价 3.80 英镑的麦片哪个更划算，可以求出每 100 克或每克的价格。第一盒：2.80 英镑 ÷ 3.5 = 每 100 克 0.80 英镑。第二盒：3.80 英镑 ÷ 5 = 每 100 克 0.76 英镑，因此大盒更划算。
You can also express the cost‑to‑weight ratio: for the first box the ratio cost : weight is 280 : 350, which simplifies to 4 : 5, meaning you pay 4 pence for every 5 g. For the second box, 380 : 500 simplifies to 19 : 25, which is about 0.76 pence per gram – lower. Always use the same units when forming ratios in best‑buy problems.
你也可以用费用与重量的比来表示：第一盒的费用 : 重量比为 280 : 350，化简为 4 : 5，即每 5 克支付 4 便士。第二盒比率为 380 : 500，化简为 19 : 25，约每克 0.76 便士——更低。在最优化购买问题中形成比时，务必使用相同单位。
8. Scale Drawings and Maps | 比例尺与地图
a model car is 8 cm long to a scale of 1 : 25, the real car is 8 × 25 = 200 cm = 2 m long.
比例尺图按比例缩小或放大来显示真实物体。比例尺以比的形式表示，如 1 : 50，意味着图上 1 厘米代表实际 50 厘米。要计算实际长度，将图上长度乘以比例因子。若一辆模型车按 1 : 25 的比例制作，长度为 8 厘米，则真车长 8 × 25 = 200 厘米 = 2 米。
on the plan is 5 cm², the real area is 5 × 200² = 5 × 40000 = 200000 cm², which is 20 m².
地图使用的比例尺如 1 : 25000，图上 1 厘米代表地面 25000 厘米（250 米）。由比例尺计算面积时，必须将线性比例因子平方。对于一个比例尺为 1 : 200 的花园平面图，若图上面积为 5 平方厘米，则实际面积为 5 × 200² = 5 × 40000 = 200000 平方厘米，即 20 平方米。
9. Equations and Algebraic Ratios | 用方程处理比
io or use the sum condition: (x+10) + (2x−5) = 90 → 3x + 5 = 90 → x = 85/3. Then substitute back to find the angles.
有时比会以代数形式给出。例如，若三角形中两个角的比为 (x + 10) : (2x − 5)，且它们的和为 90 度，你可以列方程。将两角表示为 k(x+10) 和 k(2x−5)，但既然它们处于给定的比中，你也可以令分数 (x+10)/(2x−5) 等于一个已知的比值，或使用和的条件：(x+10) + (2x−5) = 90 → 3x + 5 = 90 → x = 85/3。然后代回求出角度。
4m, 5m. Then 3m + 4m + 5m = 36 → 12m = 36 → m = 3. The sides are 9 cm, 12 cm, 15 cm. This ‘multiplier method’ simplifies many ratio problems involving perimeters or volumes.
另一种常见情形：三角形三边之比为 3 : 4 : 5，周长为 36 厘米。引入乘数 m，则三边长分别为 3m、4m、5m。于是 3m + 4m + 5m = 36 → 12m = 36 → m = 3。边长分别为 9 厘米、12 厘米、15 厘米。这种“乘数法”可简化许多涉及周长或体积的比的问题。
10. Solving Ratio Problems with Tables and Unitary Method | 用表格和归一法解比例题
The unitary method involves finding the value of one unit first and then scaling to the required number of units. This is especially powerful when working with rates, such as fuel consumption or speed. Arrange data in a table to organise your thinking. For example, a car uses 8 litres of petrol to travel 100 km. How far can it go on 14 litres?
归一法先求出一个单位对应的值，然后再缩放到所需的单位数。在处理速率问题（如油耗或速度）时，这一方法尤其有效。可以用表格来整理思路。例如，一辆汽车行驶 100 公里消耗 8 升汽油，那么 14 升汽油可以行驶多远？
| Litres (L) | Distance (km) |
| 8 | 100 |
| 1 | 100 ÷ 8 = 12.5 |
| 14 | 12.5 × 14 = 175 km |
Thus, the car can travel 175 km on 14 litres. This method is a direct application of direct proportion. Whenever you see a problem that says “how many …”, “how much …”, the unitary method often provides a clear path to the answer.
因此，这辆车用 14 升汽油可以行驶 175 公里。这一方法是正比例的直接应用。每当遇到“多少……（how many …）”之类的问题时，归一法往往能提供一条清晰的解题途径。
11. Map Scales and Distance Conversions | 地图比例尺与距离换算
0 cm. Since 100 cm = 1 m, and 1000 m = 1 km, we divide by 100 000: 600000 ÷ 100000 = 6 km.
地图比例尺的题目在 IGCSE 考试中经常出现。一个典型问题：在比例尺为 1 : 50000 的地图上，两镇之间的距离为 12 厘米。求实际距离（以千米为单位）。实际距离 = 12 × 50000 = 600000 厘米。因为 100 厘米 = 1 米，1000 米 = 1 千米，所以除以 100000：600000 ÷ 100000 = 6 千米。
: 8 km = 800000 cm. Then map distance = 800000 ÷ 200000 = 4 cm. Check your units carefully – mixing cm, m, and km is a very common mistake.
如果已知实际距离而要求图上距离，只需将实际距离除以比例因子。要在 1 : 200000 的地图上放置 8 千米，先将千米转换为厘米：8 千米 = 800000 厘米。然后图上距离 = 800000 ÷ 200000 = 4 厘米。仔细检查单位——混淆厘米、米和千米是一个非常常见的错误。
12. Proportion in Recipes and Mixtures | 食谱与混合物中的比例
÷ 8 = 2.5. Multiply each ingredient by 2.5: flour = 500 g, sugar = 250 g, butter = 200 g.
食谱问题考查你利用给定比例来缩放数量的能力。假设一个给 8 人份的饼干食谱需要 200 克面粉、100 克糖和 80 克黄油。要为 20 人制作食谱，先求缩放因子：20 ÷ 8 = 2.5。将每种原料乘以 2.5：面粉 = 500 克，糖 = 250 克，黄油 = 200 克。
l in the ratio 1 : 2 : 4. If you have 50 kg of cement, how much sand and gravel are needed? , one part is 50 kg. Sand = 2 × 50 = 100 kg, gravel = 4 × 50 = 200 kg. Total mix = 350 kg.
有时，缩放因子来自一种原料与整个混合物的比。一种混凝土由水泥、沙子和砾石按 1 : 2 : 4 的比例配制。如果你有 50 千克水泥，需要多少沙子和砾石？水泥 : 沙子 : 砾石 = 1 : 2 : 4，因此 1 份水泥对应 50 千克，一份即为 50 千克。沙子 = 2 × 50 = 100 千克，砾石 = 4 × 50 = 200 千克。总混合料 = 350 千克。
Always express all quantities in the same unit when working with ratios in recipes. If the ratio mentions millilitres and litres, convert first. The unitary method and multiplier approach work hand in hand here, and using a clear step‑by‑step layout reduces errors significantly.
在食谱中处理比时，始终用相同单位表示所有数量。如果比同时涉及毫升和升，要先进行单位换算。归一法和乘数法在这里可以结合使用，清晰的分步推导能显著减少错误。
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