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IGCSE 数学 · 知识点 1

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数论

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英语旁白 · 英中内嵌字幕

English

This handout covers Topic 1, Number. Cambridge Maths has two levels: Core and Extended. Parts marked (Extended) are only tested on the Extended papers; everything else is for both levels.

1.1

数的类型

教学大纲
学科内容 备注与示例
识别并使用:• 自然数 • 整数(正整数、零和负整数) • 质数 • 平方数 • 立方数 • 公因数 • 公倍数 • 有理数和无理数 • 倒数。 示例任务包括: • 将数字与单词相互转换,例如:六十亿是 6000000000;10007 是一万零七 • 将 72 表示为其质因数的乘积 • 求两个数的最大公因数 (HCF) • 求两个数的最小公倍数 (LCM)。
学科内容 注释与示例
识别并运用: • 自然数 • 整数(正整数、零和负整数) • 质数 • 完全平方数 • 完全立方数 • 公因数 • 公倍数 • 有理数和无理数 • 倒数。 示例任务包括: • 在数字与单词之间进行转换,例如:六十亿是 6000000000;10007 是一万零七 • 将 72 表示为其质因数的乘积 • 求两个数的最大公因数 (HCF) • 求两个数的最小公倍数 (LCM)。

来源:Cambridge International 教学大纲

English

You must know the words for the different kinds of number. Examiners give marks for using them correctly.

Nested boxes showing natural numbers inside integers inside rational numbers inside real numbers, with irrational numbers in a separate box
Each set sits inside the next: every natural number is an integer, every integer is rational, every rational is real. Irrational numbers are real but not rational.

Counting numbers and integers

  • natural numbers 自然数 — the counting numbers $1, 2, 3, 4, \dots$
  • integers 整数 — whole numbers, positive, negative or zero: $\dots, -2, -1, 0, 1, 2, \dots$

Factors and multiples

  • A factor 因数 of a number divides into it exactly, leaving no remainder 余数. The factors of $18$ are $1, 2, 3, 6, 9, 18$.
  • A multiple 倍数 of a number is that number times an integer. Multiples of $6$ are $6, 12, 18, 24, \dots$
  • A common factor 公因数 of two numbers is a factor of both.
  • A common multiple 公倍数 of two numbers is a multiple of both.

Prime, square and cube numbers

  • A prime number 质数 has exactly two factors: $1$ and itself. The first primes are $2, 3, 5, 7, 11, 13, \dots$ Note that $1$ is not prime.
  • A square number 平方数 is a whole number times itself: $1, 4, 9, 16, 25, \dots$
  • A cube number 立方数 uses a whole number three times: $1, 8, 27, 64, \dots$

Rational, irrational and reciprocal

  • A rational number 有理数 can be written as a fraction 分数 $\frac{a}{b}$ of two integers, with $b \ne 0$. Examples: $\frac{3}{4}$, $5$, $0.7$.
  • An irrational number 无理数 cannot be written this way. Examples: $\pi$ and $\sqrt{2}$.
  • The reciprocal 倒数 of a non-zero number is $1$ divided by that number. The reciprocal of $4$ is $\frac{1}{4}$; the reciprocal of $0.25$ is $4$; the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Prime factors, HCF and LCM

Every integer above $1$ is prime, or can be written as a product 乘积 of prime numbers. To write a number as a product of its prime factors 质因数, keep dividing by the smallest prime that fits.

Worked example. Write $72$ as a product of its prime factors.

$$72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3 = 2^{3} \times 3^{2}.$$
A factor tree for 72: it splits into 8 and 9, then down to the circled primes 2, 2, 2 and 3, 3
Keep splitting until every branch ends on a prime (circled); collecting them gives $72 = 2^{3} \times 3^{2}$.

The highest common factor (HCF) 最大公因数 of two numbers is the largest factor they share. The lowest common multiple (LCM) 最小公倍数 is the smallest multiple they share. Prime factors give a quick method.

Worked example. Find the HCF and LCM of $72$ and $120$.

First write each as a product of primes:

$$72 = 2^{3} \times 3^{2}, \qquad 120 = 2^{3} \times 3 \times 5.$$
  • HCF: take the lowest power of each prime that appears in both: $2^{3} \times 3 = 24$.
  • LCM: take the highest power of every prime that appears: $2^{3} \times 3^{2} \times 5 = 360$.
探索

数的集合

每个计数数也是整数,每个整数也是有理数——看数集如何层层嵌套,以及并集和交集如何把它们组合起来。

词汇 训练
English 中文 拼音
natural number/ˈnætʃərəl ˈnʌmbə/ 自然数 zì rán shù
integer/ˈɪntɪdʒə/ 整数 zhěng shù
factor/ˈfæktə/ 因数 yīn shù
remainder/rɪˈmeɪndə/ 余数 yú shù
multiple/ˈmʌltɪpl/ 倍数 bèi shù
common factor/ˈkɒmən ˈfæktə/ 公因数 gōng yīn shù
common multiple/ˈkɒmən ˈmʌltɪpl/ 公倍数 gōng bèi shù
prime number/praɪm ˈnʌmbə/ 质数 zhì shù
square number/skweə ˈnʌmbə/ 平方数 píng fāng shù
cube number/kjuːb ˈnʌmbə/ 立方数 lì fāng shù
rational number/ˈræʃənl ˈnʌmbə/ 有理数 yǒu lǐ shù
fraction/ˈfrækʃn/ 分数 fēn shù
irrational number/ɪˈræʃənl ˈnʌmbə/ 无理数 wú lǐ shù
reciprocal/rɪˈsɪprəkl/ 倒数 dào shù
product/ˈprɒdʌkt/ 乘积 chéng jī
prime factor/praɪm ˈfæktə/ 质因数 zhì yīn shù
highest common factor/ˈhaɪɪst ˈkɒmən ˈfæktə/ 最大公因数 zuì dà gōng yīn shù
lowest common multiple/ˈləʊɪst ˈkɒmən ˈmʌltɪpl/ 最小公倍数 zuì xiǎo gōng bèi shù
1.2

集合

教学大纲
学科内容 说明与示例
理解并使用集合语言、符号和韦恩图来描述集合。 韦恩图仅适用于两个集合。将使用以下集合符号: • $n(A)$ 集合 $A$ 中的元素数量 • $A'$ 集合 $A$ 的补集 • $\mathscr{E}$ 全集 • $A \cup B$ $A$ 与 $B$ 的并集 • $A \cap B$ $A$ 与 $B$ 的交集。集合定义的示例: $A = \{x : x \text{ is a natural number}\}$ $B = \{a, b, c, \dots\}$ $C = \{x : a \leqslant x \leqslant b\}$。
学科内容 注释与示例
理解并使用集合语言、符号和韦恩图来描述集合以及表示集合之间的关系。 韦恩图仅限于两个或三个集合。将使用以下集合符号:• $n(A)$ 集合 $A$ 中的元素个数 • $\in$ “...是...的一个元素” • $\notin$ “...不是...的一个元素” • $A'$ 集合 $A$ 的补集 • $\varnothing$ 空集 • $\mathscr{E}$ 全集 • $A \subseteq B$ $A$ 是 $B$ 的子集 • $A \nsubseteq B$ $A$ 不是 $B$ 的子集 • $A \cup B$ $A$ 与 $B$ 的并集 • $A \cap B$ $A$ 与 $B$ 的交集。集合的定义示例:$A = \{x : x \text{ is a natural number}\}$ $B = \{(x, y) : y = mx + c\}$ $C = \{x : a \leqslant x \leqslant b\}$ $D = \{a, b, c, \dots\}$。

来源:Cambridge International 教学大纲

English

A set 集合 is a collection of objects. Each object in the set is an element 元素 of the set. You should know this notation:

Symbol Meaning
$n(A)$ the number of elements in set $A$
$x \in A$ $x$ is an element of $A$
$x \notin A$ $x$ is not an element of $A$
$\mathscr{E}$ the universal set 全集 — everything being talked about
$A'$ the complement 补集 of $A$ — everything not in $A$
$\varnothing$ the empty set 空集 — a set with no elements
$A \subseteq B$ $A$ is a subset 子集 of $B$ — every element of $A$ is also in $B$
$A \cup B$ the union 并集 — elements in $A$ or $B$ or both
$A \cap B$ the intersection 交集 — elements in both $A$ and $B$

A Venn diagram 维恩图 draws each set as a circle inside a rectangle (the universal set). Core uses two sets; Extended may use three.

Worked example. $\mathscr{E} = \{1,2,3,4,5,6,7,8,9,10\}$, $A = \{\text{even numbers}\}$, $B = \{\text{multiples of } 3\}$.

  • $A = \{2,4,6,8,10\}$ and $B = \{3,6,9\}$.
  • $A \cap B = \{6\}$ — the only number in both.
  • $A \cup B = \{2,3,4,6,8,9,10\}$ — the numbers in either set.
  • $n(A \cup B) = 7$.
Two overlapping circles in a rectangle: circle A holds the even numbers, circle B the multiples of 3, with 6 in the overlap and 1, 5, 7 outside both
$A \cap B = \{6\}$ is the only number in both circles; $A \cup B$ is everything inside either circle.

You may also see a set written as a rule, e.g. $C = \{x : 1 \leqslant x \leqslant 5\}$ means "all values $x$ such that $1 \leqslant x \leqslant 5$".

探索

韦恩图

点击区域以查看并集、交集和补集——集合论的语言。

词汇 训练
English 中文 拼音
set/set/ 集合 jí hé
element/ˈelɪmənt/ 元素 yuán sù
universal set/ˌjuːnɪˈvɜːsl set/ 全集 quán jí
complement/ˈkɒmplɪmənt/ 补集 bǔ jí
empty set/ˈempti set/ 空集 kōng jí
subset/ˈsʌbset/ 子集 zǐ jí
union/ˈjuːnɪən/ 并集 bìng jí
intersection/ˌɪntəˈsekʃn/ 交集 jiāo jí
Venn diagram/ven ˈdaɪəɡræm/ 维恩图 wéi ēn tú
1.3

幂与根

教学大纲
学科内容 注释与示例
进行以下运算: • 平方 • 平方根 • 立方 • 立方根 • 数的其他幂和根。 包括回忆1到15的平方及其对应的平方根,以及1、2、3、4、5和10的立方及其对应的立方根等,例如: • 写出 $\sqrt{169}$ 的值。 • 计算 $5^2 \times \sqrt[3]{8}$ 。
学科内容 注释与示例
进行以下运算: • 平方 • 平方根 • 立方 • 立方根 • 数的其他次幂和开方。 包括回忆1至15的平方及其对应的平方根,以及1、2、3、4、5和10的立方及其对应的立方根等知识,例如: • 写出 $\sqrt{169}$ 的值。 • 计算 $5^2 \times \sqrt[3]{8}$。

来源:Cambridge International 教学大纲

English
  • A power 幂 (also called an index 指数, plural indices) shows how many times to multiply a number by itself: $2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$.
  • A square root 平方根 of a number gives that number when squared: $\sqrt{169} = 13$ because $13^{2} = 169$.
  • A cube root 立方根 works the same way for cubes: $\sqrt[3]{8} = 2$ because $2^{3} = 8$.

You should be able to recall the squares from $1^2$ to $15^2$ (and their roots), and the cubes of $1, 2, 3, 4, 5$ and $10$.

Worked example. Work out $5^{2} \times \sqrt[3]{8}$.

$$5^{2} \times \sqrt[3]{8} = 25 \times 2 = 50.$$
探索

幂与根的练习

平方 = x^2

改变底数观察幂的增长以及根如何抵消幂运算。

词汇 训练
English 中文 拼音
power/ˈpaʊə/ 幂 mì
index/ˈɪndeks/ 指数 zhǐ shù
square root/skweə ruːt/ 平方根 píng fāng gēn
cube root/kjuːb ruːt/ 立方根 lì fāng gēn
1.7

指数 I

教学大纲
学科内容 注释与示例
1 理解并使用指数(正整数、零和负整数)。 例如:求 $7^{-2}$ 的值。
2 理解并使用指数的运算法则。 例如:求 $2^{-3} \times 2^4$,$(2^3)^2$,$2^3 \div 2^4$ 的值。
学科内容 注释与示例
1 理解并使用指数(正整数、零、负数及分数)。 示例包括:• $6^{\frac{1}{2}} = \sqrt{6}$ • $16^{\frac{1}{4}} = \sqrt[4]{16}$ • 求 $7^{-2}$,$81^{\frac{1}{2}}$,$8^{-\frac{2}{3}}$ 的值。
2 理解并使用指数法则。 例如:求 $2^{-3} \times 2^4$,$(2^3)^2$,$2^3 \div 2^4$ 的值。

来源:Cambridge International 教学大纲

English

When you multiply or divide powers of the same base 底数, use these rules:

$$a^{m} \times a^{n} = a^{m+n}, \qquad a^{m} \div a^{n} = a^{m-n}, \qquad (a^{m})^{n} = a^{mn}.$$

Some special powers:

$$a^{0} = 1, \qquad a^{-n} = \frac{1}{a^{n}}, \qquad a^{\frac{1}{n}} = \sqrt[n]{a}, \qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}.$$

(Fractional powers like $a^{\frac{m}{n}}$ are Extended.)

Worked examples.

  • $2^{-3} \times 2^{4} = 2^{-3+4} = 2^{1} = 2.$
  • $(2^{3})^{2} = 2^{6} = 64.$
  • $2^{3} \div 2^{4} = 2^{3-4} = 2^{-1} = \dfrac{1}{2}.$
  • $7^{-2} = \dfrac{1}{7^{2}} = \dfrac{1}{49}.$
  • $81^{\frac{1}{2}} = \sqrt{81} = 9.$
  • $8^{-\frac{2}{3}} = \dfrac{1}{8^{\frac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{8}\right)^{2}} = \dfrac{1}{2^{2}} = \dfrac{1}{4}.$

Common mistakes. Multiplying the indices when you multiply powers ($2^{3} \times 2^{4}$ is $2^{7}$, not $2^{12}$), thinking a negative power makes a negative number ($7^{-2}$ is positive), and writing $a^{0} = 0$ (it is $1$).

Three cards for the index laws: multiplying powers adds the indices, dividing subtracts them, and a power of a power multiplies them, each with a worked example
The three index laws: multiply adds the powers, divide subtracts them, a power of a power multiplies them
词汇 训练
English 中文 拼音
base/beɪs/ 底数 dǐ shù
1.8

标准形式(科学记数法)

教学大纲
学科内容 注释与示例
1 使用标准形式 $A \times 10^n$,其中 $n$ 为正整数或负整数,且 $1 \leqslant A < 10$。
2 将数字转换为标准形式及从标准形式转换出来。
3 对以标准形式表示的数值进行计算。 核心层考生仅被要求在试卷3中使用标准形式进行计算。
学科内容 注释与示例
1 使用科学记数法 $A \times 10^n$,其中 $n$ 为正整数或负整数,且 $1 \leqslant A < 10$。
2 将数字转换为科学记数法以及从科学记数法转换回普通数字。
3 对科学记数法表示的数值进行计算。

来源:Cambridge International 教学大纲

English
The Andromeda galaxy
A galaxy: huge distances are written compactly in standard form.

Standard form 科学记数法 writes a number as $A \times 10^{n}$, where $1 \leqslant A < 10$ and $n$ is an integer. It is used for very large or very small numbers.

To convert, count how many places the decimal point moves:

  • $4\,500\,000 = 4.5 \times 10^{6}$ — the point moves $6$ places left, so the power is positive.
  • $0.00072 = 7.2 \times 10^{-4}$ — the point moves $4$ places right, so the power is negative.

Worked example. Work out $(3 \times 10^{5}) \times (2 \times 10^{-2})$.

Multiply the front numbers and add the powers:

$$3 \times 2 = 6, \qquad 10^{5} \times 10^{-2} = 10^{3}, \qquad \text{so the answer is } 6 \times 10^{3}.$$

Common mistake. Leaving an answer such as $15 \times 10^{4}$. The front number must be between $1$ and $10$, so write $1.5 \times 10^{5}$.

Standard form: a number written as a × 10^n with 1 ≤ a < 10
Standard form: a number written as a × 10^n with 1 ≤ a < 10
探索

科学记数法路径

将大数或小数转换为 A × 10^n 形式。

词汇 训练
English 中文 拼音
standard form/ˈstændəd fɔːm/ 科学记数法 kē xué jì shù fǎ
1.18

无理根式

教学大纲
学科内容 注释与示例
1 理解并使用根式,包括化简表达式。 示例包括:• $\sqrt{20} = 2\sqrt{5}$ • $\sqrt{200} - \sqrt{32} = 6\sqrt{2}$。
2 分母有理化。 示例包括:• $\frac{10}{\sqrt{5}} = 2\sqrt{5}$ • $\frac{1}{-1 + \sqrt{3}} = \frac{1 + \sqrt{3}}{2}$。

来源:Cambridge International 教学大纲

English

A surd 根式 is a root that is irrational, such as $\sqrt{5}$. Leave it in exact form instead of rounding. Two useful rules:

$$\sqrt{a} \times \sqrt{b} = \sqrt{ab}, \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}.$$

Simplify a surd by taking out the largest square factor.

Worked example. Simplify $\sqrt{20}$ and $\sqrt{200} - \sqrt{32}$.

$$\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4}\,\sqrt{5} = 2\sqrt{5}.$$
$$\sqrt{200} - \sqrt{32} = \sqrt{100 \times 2} - \sqrt{16 \times 2} = 10\sqrt{2} - 4\sqrt{2} = 6\sqrt{2}.$$
Simplifying the square root of 20 in three steps: find the largest square factor, split the root, and take the square root out to get 2 root 5
Simplifying a surd: take out the largest square factor

To rationalise the denominator 分母有理化 means to remove a surd from the bottom of a fraction (the denominator 分母). Multiply the top and bottom by a value that clears the surd.

Worked example. Rationalise $\dfrac{10}{\sqrt{5}}$ and $\dfrac{1}{-1+\sqrt{3}}$.

$$\frac{10}{\sqrt{5}} = \frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}.$$
$$\frac{1}{-1+\sqrt{3}} = \frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{3}+1}{3-1} = \frac{1+\sqrt{3}}{2}.$$
探索

根式化简路径

将根式分解为平方因子并化简。

词汇 训练
English 中文 拼音
surd/sɜːd/ 根式 gēn shì
rationalise the denominator/ˈræʃənəlaɪz ðə dɪˈnɒmɪneɪtə/ 分母有理化 fēn mǔ yǒu lǐ huà
denominator/dɪˈnɒmɪneɪtə/ 分母 fēn mǔ
1.4

分数、小数与百分数

教学大纲
学科内容 注释与示例
1 在适当的语境中使用以下内容的语言和符号: • 真分数 • 假分数 • 带分数 • 小数 • 百分数。 考生应能将分数化为最简形式。不要求考生使用循环小数的记号。
2 识别等价关系并在这些形式之间进行转换。 不要求考生演示将循环小数转换为分数,以及将分数转换为循环小数的过程。
学科内容 注释与示例
1 在适当的语境中使用以下内容的语言和符号: • 真分数 • 假分数 • 带分数 • 小数 • 百分数。 考生应能将分数化为最简形式。需要掌握循环小数的记法,例如: • $0.1\dot{7} = 0.1777...$ • $0.1\dot{2}\dot{3} = 0.1232323...$ • $0.\dot{1}2\dot{3} = 0.123123...$
2 识别等价关系并在这些形式之间进行转换。 包括在循环小数与分数之间互相转换,例如:将 $0.1\dot{7}$ 写成分数形式。

来源:Cambridge International 教学大纲

English

A fraction has a numerator 分子 (the top) and a denominator (the bottom).

One half written three ways: the fraction 1/2, the decimal 0.5, and the percentage 50%
One value three ways: a fraction, a decimal and a percentage
  • proper fraction 真分数: numerator smaller than denominator, e.g. $\frac{3}{4}$.
  • improper fraction 假分数: numerator the same or larger, e.g. $\frac{7}{4}$.
  • mixed number 带分数: a whole number plus a fraction, e.g. $1\frac{3}{4}$.

Change between improper and mixed: $\frac{7}{4} = 1\frac{3}{4}$ because $7 \div 4 = 1$ remainder $3$.

A decimal 小数 uses place value after a point. A percentage 百分比 means "out of $100$", so $37\% = \frac{37}{100} = 0.37$.

Converting between forms

To change Method Example
fraction → decimal divide top by bottom $\frac{3}{8} = 3 \div 8 = 0.375$
decimal → percentage multiply by $100$ $0.07 = 7\%$
percentage → fraction put over $100$, then simplify $7\% = \frac{7}{100}$
percentage → decimal divide by $100$ $34\% = 0.34$

Write a fraction in its simplest form 最简形式 by dividing the top and bottom by their HCF: $\frac{18}{24} = \frac{3}{4}$ (both divided by $6$).

Recurring decimals (Extended)

A recurring decimal 循环小数 repeats the same digits forever. Dots mark the repeating part: $0.1\dot{7} = 0.1777\ldots$ and $0.\dot{1}2\dot{3} = 0.123123\ldots$

To turn a recurring decimal into a fraction, multiply so the repeating parts line up, then subtract.

Worked example. Write $0.1\dot{7}$ as a fraction.

Let $x = 0.1777\ldots$ Only the $7$ repeats, so use $10x$ and $100x$:

$$100x = 17.777\ldots, \qquad 10x = 1.777\ldots$$
$$100x - 10x = 16, \qquad 90x = 16, \qquad x = \frac{16}{90} = \frac{8}{45}.$$
The two lines 100x and 10x written with their repeating tails highlighted and lined up; subtracting cancels the tails and leaves 90x equals 16
Line up the repeating tails, subtract, and the tails cancel
探索

数值形式练习

对等价的数值形式和运算进行分类。

词汇 训练
English 中文 拼音
numerator/ˈnjuːməreɪtə/ 分子 fèn zǐ
proper fraction/ˈprɒpə ˈfrækʃn/ 真分数 zhēn fēn shù
improper fraction/ɪmˈprɒpə ˈfrækʃn/ 假分数 jiǎ fēn shù
mixed number/mɪkst ˈnʌmbə/ 带分数 dài fēn shù
decimal/ˈdesɪml/ 小数 xiǎo shù
percentage/pəˈsentɪdʒ/ 百分比 bǎi fēn bǐ
simplest form/ˈsɪmpəlɪst fɔːm/ 最简形式 zuì jiǎn xíng shì
recurring decimal/rɪˈkɜːrɪŋ ˈdesɪml/ 循环小数 xún huán xiǎo shù
1.6

四则运算

教学大纲
学科内容 注释与示例
使用四则运算对整数、分数和小数进行计算,包括正确的运算顺序及括号的使用。 包括:• 负数 • 假分数 • 带分数 • 实际情境,例如温度变化。
学科内容 注释与示例
使用四则运算对整数、分数和小数进行计算,包括正确的运算顺序和括号的使用。 包括:• 负数 • 假分数 • 带分数 • 实际情境,例如温度变化。

来源:Cambridge International 教学大纲

English

Order of operations

Work in this order — the order of operations 运算顺序: brackets first, then indices (powers and roots), then multiply and divide (left to right), then add and subtract (left to right).

Worked example. Work out $-6 \times -3 + 7 \times 2$.

Do the multiplications first: $-6 \times -3 = 18$ and $7 \times 2 = 14$. Then add: $18 + 14 = 32$.

A ladder with four rungs, brackets, indices, multiply and divide, add and subtract, next to the worked example done one level at a time
The order of operations, applied to the worked example

Negative numbers

  • Adding a negative: $5 + (-3) = 5 - 3 = 2$.
  • Subtracting a negative: $5 - (-3) = 5 + 3 = 8$.
  • Multiplying or dividing: same signs give a positive; different signs give a negative. So $-6 \times -3 = 18$ but $-12 \div 4 = -3$.

A change in temperature from $-5\,{}^{\circ}\text{C}$ to $3\,{}^{\circ}\text{C}$ is a rise of $8\,{}^{\circ}\text{C}$.

Calculating with fractions

  • Multiply: multiply the tops, multiply the bottoms: $\frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$.
  • Divide: multiply by the reciprocal of the second fraction: $\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$.
  • Add or subtract: use a common denominator (the LCM of the bottoms).

Worked example. Work out $1\frac{7}{15} - \frac{4}{5}$, giving the answer in its simplest form.

Change the mixed number to an improper fraction, then use denominator $15$:

$$1\frac{7}{15} = \frac{22}{15}, \qquad \frac{4}{5} = \frac{12}{15}, \qquad \frac{22}{15} - \frac{12}{15} = \frac{10}{15} = \frac{2}{3}.$$
词汇 训练
English 中文 拼音
order of operations/ˈɔːdə ɒv ˌɒpəˈreɪʃnz/ 运算顺序 yùn suàn shùn xù
1.5

排序

教学大纲
学科内容 注释与示例
按大小顺序排列数量,并展示对符号 $=, \ne, >, <, \geqslant$ 和 $\leqslant$ 的熟悉程度。
学科内容 注释和示例
按大小顺序排列数量,并展示对符号 $=, \neq, >, <, \geqslant$ 和 $\leqslant$ 的熟悉程度。

来源:Cambridge International 教学大纲

English

Use these symbols to compare numbers by magnitude 大小 (size):

Symbol Meaning
$=$ is equal to
$\neq$ is not equal to
$>$ is greater than
$<$ is less than
$\geqslant$ is greater than or equal to
$\leqslant$ is less than or equal to

To put a mixed list in order, change every value to a decimal first.

Worked example. Put $34\%$, $\frac{1}{3}$ and $\frac{3}{10}$ in order, smallest first.

As decimals: $34\% = 0.34$, $\frac{1}{3} = 0.333\ldots$, $\frac{3}{10} = 0.3$. So the order is

$$\frac{3}{10} < \frac{1}{3} < 34\%.$$
The fraction three tenths, the fraction one third and 34 percent each converted to a decimal and dropped onto a zoomed number line between 0.30 and 0.35
Convert every value to a decimal, then place them on a number line
词汇 训练
English 中文 拼音
magnitude/ˈmæɡnɪtjuːd/ 大小 dà xiǎo
1.13

百分数

教学大纲
学科内容 注释与示例
1 计算给定数量的指定百分比。
2 将一个数量表示为另一个数量的百分比。
3 计算百分比的增减。
4 进行简单利息和复利计算。 不提供公式。百分数计算可能包括:• 存款 • 折扣 • 利润和亏损(以金额或百分比形式)• 收益 • 超过100%的百分数。
学科内容 注释和示例
1 计算给定数量的指定百分比。
2 将一种数量表示为另一种数量的百分比。
3 计算百分比的增减。
4 进行单利和复利的计算。 问题可能包含重复的百分比变化。不提供公式。
5 利用逆运算进行百分比计算。 例如:已知售价和利润率,求进价。百分比计算可能包括: • 定金 • 折扣 • 利润和亏损(以金额或百分比表示) • 收入 • 超过100%的百分比。

来源:Cambridge International 教学大纲

English
Shops along a high street
Shops use percentages for discounts, sales tax and profit margins.

Find a percentage of an amount. $15\%$ of $\$80 = 0.15 \times 80 = \$12$.

Write one amount as a percentage of another. A score of $18$ out of $25$ is $\frac{18}{25} \times 100\% = 72\%$.

Percentage increase or decrease uses

$$\text{percentage change} = \frac{\text{change}}{\text{original amount}} \times 100\%.$$

Worked example. A price rises from $\$40$ to $\$50$. Find the percentage increase.

The change is $\$10$, so $\frac{10}{40} \times 100\% = 25\%$ increase.

Common mistake. Dividing by the new amount ($\frac{10}{50} = 20\%$). A percentage change is always worked out from the original amount.

A quick way is a multiplier 乘数. To increase by $15\%$, multiply by $1.15$; to decrease by $15\%$, multiply by $0.85$.

Simple and compound interest

Interest 利息 is money paid for borrowing or for saving. The principal 本金 is the starting amount.

  • Simple interest 单利 pays the same amount each year, worked out on the principal only:
    $$I = \frac{P \times r \times t}{100},$$
    where $P$ is the principal, $r$ is the rate per year (as a percentage) and $t$ is the number of years.
  • Compound interest 复利 adds the interest on each year, so the next year earns interest on a larger total:
    $$\text{final value} = P\left(1 + \frac{r}{100}\right)^{t}.$$

Worked example. Find the value of $\$500$ saved at $4\%$ compound interest for $3$ years.

$$500 \times 1.04^{3} = 500 \times 1.124864 = \$562.43 \ (\text{to the nearest cent}).$$
A graph of 500 dollars saved at 4 percent: simple interest is a straight line, compound interest curves above it and the gap keeps growing
Simple interest grows in a straight line; compound interest grows faster every year

Reverse percentages (Extended)

A reverse percentage 逆百分比 problem gives the amount after a change and asks for the original. To solve it, divide by the multiplier — do not just take the percentage off.

Worked example. A coat costs $\$60$ after a $20\%$ increase. Find the original price.

$\$60$ is $120\%$ of the original, so the original price is $60 \div 1.2 = \$50$.

Common mistake. Taking $20\%$ off the new price: $60 \times 0.8 = \$48$ is wrong, because the $20\%$ was a percentage of the original price, not of $\$60$. Check: $50 \times 1.2 = 60$.

探索

百分比变化练习

新值 = 旧值 × 乘数

改变乘数并观察最终值的变化。

词汇 训练
English 中文 拼音
multiplier/ˌmʌltɪˈplaɪə/ 乘数 chéng shù
interest/ˈɪntrest/ 利息 lì xī
principal/ˈprɪnsɪpl/ 本金 běn jīn
simple interest/ˈsɪmpl ˈɪntrest/ 单利 dān lì
compound interest/ˈkɒmpaʊnd ˈɪntrest/ 复利 fù lì
reverse percentage/rɪˈvɜːs pəˈsentɪdʒ/ 逆百分比 nì bǎi fēn bǐ
1.17

指数增长与衰减

教学大纲
学科内容 说明与示例
使用指数增长与衰减。 例如:折旧、人口变化。不要求掌握 e 的知识。

来源:Cambridge International 教学大纲

English

When a quantity changes by the same percentage in each time period, it shows exponential growth 指数增长 (it gets bigger) or exponential decay 指数衰减 (it gets smaller). Use the compound formula. Depreciation 折旧, where something like a car loses value each year, is decay.

Worked example. A car worth $\$20\,000$ loses $15\%$ of its value each year. Find its value after $4$ years.

The multiplier is $0.85$, so

$$20\,000 \times 0.85^{4} = 20\,000 \times 0.522\ldots = \$10\,440 \ (\text{to the nearest dollar}).$$
Bars showing the car value each year from 20000 dollars downwards; every bar is 0.85 times the one before, and the bar after 4 years is about 10440 dollars
Exponential decay: the car loses 15 percent of its current value every year
探索

复利

钱每年按 (1 + r) 增长,所以复利曲线高于单利。拖动利率和年数。

探索

指数增长与衰减

y = a·bˣ

改变底数 b:b > 1 增长,0 < b < 1 衰减——对利息和人口有用。

词汇 训练
English 中文 拼音
exponential growth/ˌekspəˈnenʃl ɡrəʊθ/ 指数增长 zhǐ shù zēng zhǎng
exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ 指数衰减 zhǐ shù shuāi jiǎn
depreciation/dɪˌpriːʃɪˈeɪʃn/ 折旧 zhé jiù
1.11

比与比例

教学大纲
学科内容 注释与示例
理解并运用比和比例来解决以下问题:
• 将比化为最简形式 例如,20:30:40 的最简形式为 2:3:4。
• 按给定比例分配数量
• 在实际情境中运用成比例推理和比。 例如:调整食谱;使用地图比例尺;确定最佳性价比。
学科内容 注释与示例
理解并运用比和比例进行:
• 将比化为最简形式 例如,20:30:40 的最简形式为 2:3:4。
• 按给定比例分配数量
• 在实际情境中运用比例推理和比。 例如,调整食谱;使用地图比例尺;确定最佳性价比。

来源:Cambridge International 教学大纲

English

A ratio 比 compares quantities, written like $a:b$. Simplify it like a fraction by dividing by the HCF: $20:30:40 = 2:3:4$.

Dividing in a ratio. Share $\$48$ in the ratio $3:5$.

The total number of parts is $3 + 5 = 8$. One part is $48 \div 8 = \$6$. So the shares are $3 \times 6 = \$18$ and $5 \times 6 = \$30$.

A bar of 8 equal parts worth 6 dollars each: 3 blue parts make 18 dollars and 5 orange parts make 30 dollars
A bar model for sharing 48 dollars in the ratio 3 to 5

Proportion 比例 means two ratios are equal. Use it for recipes, map scales 比例尺 and finding best value.

Worked example. $3$ pens cost $\$1.80$. Find the cost of $7$ pens.

One pen costs $1.80 \div 3 = \$0.60$. So $7$ pens cost $7 \times 0.60 = \$4.20$.

探索

正比例

y = ax

正比例是一条通过原点的直线——x 翻倍,y 也翻倍。

词汇 训练
English 中文 拼音
ratio/ˈreɪʃɪəʊ/ 比 bǐ
proportion/prəˈpɔːʃn/ 比例 bǐ lì
scale/skeɪl/ 比例尺 bǐ lì chǐ
1.12

比率

教学大纲
学科内容 注释与示例
1 使用常见的比率度量。 例如:计算以下内容的比率:
• 时薪
• 货币汇率
• 流量
• 燃油消耗率。
2 应用其他比率度量。 例如:计算以下内容的比率:
• 压强
• 密度
• 人口密度。
所需公式将在题目中给出。
3 解决涉及平均速度的问题。 需要掌握速度/距离/时间公式的知识。例如:一名骑行者在3小时45分钟内骑行了45公里。他们的平均速度是多少?使用的记号示例包括 m/s(米每秒)、$\text{g/cm}^3$(克每立方厘米)。
学科内容 注释与示例
1 使用常见的比率度量。 例如计算:• 时薪 • 货币兑换率 • 流量 • 燃料消耗量。
2 应用其他比率度量。 例如计算:• 压强 • 密度 • 人口密度。所需公式将在题目中给出。
3 解决涉及平均速度的问题。 需要掌握速度/距离/时间公式的知识。例如:一名骑自行车者行驶了 45 km,用时 3 小时 45 分钟。求其平均速度?将使用的记号包括,例如 m/s(米每秒),g/cm$^{3}$(克每立方厘米)。

来源:Cambridge International 教学大纲

English

A rate 比率 compares two quantities measured in different units, such as price per kilogram, or distance per hour.

Average speed 平均速度 uses

$$\text{average speed} = \frac{\text{total distance}}{\text{total time}}.$$

Worked example. A cyclist travels $45\text{ km}$ in $3$ hours $45$ minutes. Find the average speed.

First change the time to hours: $3$ h $45$ min $= 3.75$ h. Then

$$\text{average speed} = \frac{45}{3.75} = 12\text{ km/h}.$$
The distance-speed-time triangle with D on top and S and T below, the three formulas you can read from it, and the cyclist worked example
The distance-speed-time triangle: cover the one you want

Other rates work the same way. Density 密度 is found from mass 质量 and volume 体积:

$$\text{density} = \frac{\text{mass}}{\text{volume}}.$$

Other examples are flow rate, fuel consumption and population density 人口密度. If a rate needs a special formula (such as pressure 压强), the question will give it to you.

词汇 训练
English 中文 拼音
rate/reɪt/ 比率 bǐ lǜ
average speed/ˈævrɪdʒ spiːd/ 平均速度 píng jūn sù dù
density/ˈdensɪti/ 密度 mì dù
mass/mæs/ 质量 zhì liàng
volume/ˈvɒljuːm/ 体积 tǐ jī
population density/ˌpɒpjʊˈleɪʃn ˈdensɪti/ 人口密度 rén kǒu mì dù
pressure/ˈpreʃə/ 压强 yā qiáng
1.9

估算

教学大纲
学科内容 注释与示例
1 将数值四舍五入到指定的精确度。 包括小数位数和有效数字。
2 对涉及数字、数量和测量的计算进行估算。 例如:将 5764 四舍五入到最接近的千位。例如:通过将每个数保留 1 位有效数字,估算
$$\frac{41.3}{9.79 \times 0.765}$$
的值。
3 在给定问题的背景下,将答案四舍五入到合理的精确度。
学科内容 注释与示例
1 将数值四舍五入至指定的精确度。 包括小数位数和有效数字。例如:将 5764 四舍五入到最接近的千位。
2 对涉及数值、数量和测量的计算进行估算。 例如:通过将每个数四舍五入到 1 位有效数字,估算
$$\frac{41.3}{9.79 \times 0.765}$$
的值。
3 根据给定问题的背景,将答案四舍五入到合理的精确度。

来源:Cambridge International 教学大纲

English

Rounding

  • decimal places (d.p.) 小数位: digits counted after the point. $3.14159$ to $2$ d.p. is $3.14$.
  • significant figures (s.f.) 有效数字: digits counted from the first non-zero digit. $5764$ to $1$ s.f. is $6000$; $0.004067$ to $2$ s.f. is $0.0041$.

Rule: look at the next digit. If it is $5$ or more, round up; if it is less, round down.

Estimation

To estimate 估算 an answer, round every number to $1$ s.f., then calculate.

Worked example. Estimate $\dfrac{41.3}{9.79 \times 0.765}$.

$$\frac{41.3}{9.79 \times 0.765} \approx \frac{40}{10 \times 0.8} = \frac{40}{8} = 5.$$
探索

四舍五入与界限练习

根据确定精度所需的决策对数字进行分类。

词汇 训练
English 中文 拼音
decimal place/ˈdesɪml pleɪs/ 小数位 xiǎo shù wèi
significant figure/sɪɡˈnɪfɪkənt ˈfɪɡə/ 有效数字 yǒu xiào shù zì
estimate/ˈestɪmət/ 估算 gū suàn
1.10

精度范围

教学大纲
学科内容 注释与示例
为四舍五入到指定精度的数据给出上界和下界。 例如:写出一个精确到最近一米的长度的上界。考生不需要计算使用过四舍五入到指定精度的数据所得结果的上界和下界。
学科内容 注释与示例
1 给出四舍五入到指定精度的数据的上限和下界。 例如:写出长度测量值精确到最接近米时的上限。
2 求使用四舍五入到指定精度的数据计算出的结果的上限和下界。 示例计算包括:
• 已知矩形尺寸精确到最接近厘米,计算周长或面积的上限
• 已知距离和时间的四舍五入值,求速度的下限。

来源:Cambridge International 教学大纲

English

A rounded value could really be anything that rounds to it. The lower bound 下界 and upper bound 上界 are the limiting values. An upper bound need not itself be allowed. For a value rounded to the nearest unit, the bounds lie half a unit on each side.

Worked example. A height $h$ is $635\text{ m}$, correct to the nearest metre. Give the bounds.

$$634.5 \leqslant h < 635.5.$$
A number line with the band from 634.5 to 635.5 shaded; 634.5 has a filled dot because it is included and 635.5 has an open dot because it is not
Everything in the band rounds to 635: the lower bound is included, the upper bound is not

So the lower bound is $634.5\text{ m}$ and the upper bound is $635.5\text{ m}$.

Bounds in calculations (Extended)

Combine the bounds to get the bound you want.

Worked example. A rectangle is $8\text{ cm}$ by $5\text{ cm}$, each side to the nearest cm. Find the upper bound for its area 面积.

Use the upper bounds of both sides: $8.5 \times 5.5 = 46.75\text{ cm}^{2}$. (The smallest area uses the lower bounds: $7.5 \times 4.5 = 33.75\text{ cm}^{2}$.)

For a divided quantity such as $\text{speed} = \dfrac{\text{distance}}{\text{time}}$, the largest speed comes from the largest distance divided by the smallest time.

Common mistakes. Writing the upper bound as $635.4$ or $635.49$ (it is $635.5$, even though $635.5$ itself would round up), and using the upper bound of the time to get the largest speed. Dividing by a smaller number gives a bigger answer.

词汇 训练
English 中文 拼音
lower bound/ˈləʊə baʊnd/ 下界 xià jiè
upper bound/ˈʌpə baʊnd/ 上界 shàng jiè
area/ˈeərɪə/ 面积 miàn jī
1.15

时间

教学大纲
学科内容 注释与示例
1 进行时间计算:秒(s)、分(min)、小时(h)、天、周、月、年,包括单位间的换算关系。 1年 = 365天。
2 使用24小时制和12小时制表示时间。 在24小时制中,例如上午3点15分记作03:15,下午3点15分记作15:15。
3 识读钟表和时刻表。 包括涉及时区、地方时和时间差的问题。
学科内容 注释与示例
1 进行时间计算:秒(s)、分(min)、时(h)、天、周、月、年,包括单位之间的换算关系。 1年 = 365天。
2 使用24小时制和12小时制进行时间计算。 在24小时制中,例如凌晨3点15分记作0315,下午3点15分记作1515。
3 读取钟表和时刻表。 包括涉及时区、本地时间和时差的问题。

来源:Cambridge International 教学大纲

English
  • $60$ seconds $= 1$ minute, $60$ minutes $= 1$ hour, $24$ hours $= 1$ day, and $1$ year $= 365$ days.
  • The 24-hour clock writes a time as four digits: $3.15$ p.m. is $15\,15$.

Worked example. A film starts at $19\,35$ and lasts $70$ minutes. Find the time it finishes.

$70$ min $= 1$ h $10$ min. Adding $1$ hour gives $20\,35$; adding $10$ minutes gives $20\,45$.

For timetables and time zone 时区 problems, add or subtract the time difference between the places.

探索

时间、金钱和计算器练习

选择与真实测量问题相匹配的运算。

词汇 训练
English 中文 拼音
time zone/taɪm zəʊn/ 时区 shí qū
1.16

货币

教学大纲
学科内容 注释与示例
1 进行货币计算。
2 将一种货币兑换为另一种货币。

来源:Cambridge International 教学大纲

English

Work with money as ordinary decimals, but give answers to $2$ d.p. (so $\$4.8$ is written $\$4.80$).

Currency conversion. Use the exchange rate 汇率 as a multiplier.

Worked example. The exchange rate is $\$1 = €0.92$. Convert $\$150$ to euros, and convert $€138$ back to dollars.

$$150 \times 0.92 = €138, \qquad 138 \div 0.92 = \$150.$$
词汇 训练
English 中文 拼音
exchange rate/eksˈtʃeɪndʒ reɪt/ 汇率 huì lǜ
1.14

使用计算器

教学大纲
学科内容 注释与示例
1 高效使用计算器。 例如:知道不应在计算过程中对数值进行四舍五入,而应仅对最终答案进行四舍五入。
2 在计算器上正确输入数值。 例如:将2小时30分钟输入为2.5小时或2° 30’ 0’’。
3 正确解读计算器显示结果。 例如:金额中4.8表示$4.80;时间中3.25表示3小时15分钟。
学科内容 注释与示例
1 高效使用计算器。 例如:知道不应在计算过程中对数值进行四舍五入,而应仅对最终答案进行四舍五入。
2 在计算器上正确输入数值。 例如:将2小时30分钟输入为2.5小时或2° 30' 0''。
3 正确解读计算器显示结果。 例如:金额中4.8表示$4.80;时间中3.25表示3小时15分钟。

来源:Cambridge International 教学大纲

English
  • Do not round part-way through a calculation. Keep the full value and round only the final answer.
  • Enter time as a decimal of an hour: $2$ hours $30$ minutes is $2.5$ hours, not $2.30$.
  • Read the display in context: in money, $4.8$ means $\$4.80$; in time, $3.25$ hours means $3$ hours $15$ minutes.
1.14

考试技巧

English
  • Follow BIDMAS (brackets, indices, division/multiplication, addition/subtraction) in order, and remember a negative times a negative is positive.
  • In standard form the number in front is between 1 and 10; a small number (like $0.0004$) has a negative power of 10.
  • A percentage change is worked out on the original amount. For a reverse percentage, divide by the multiplier (e.g. $\div 1.2$ undoes a $20\%$ rise).
  • Do not round part-way through — keep the full value and round only at the end, to the accuracy the question asks for (decimal places or significant figures).
  • For limits of accuracy, a value rounded to the nearest whole number can be up to $0.5$ either side (so $8$ means $7.5 \le x < 8.5$).

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