IGCSE Mathematics · ⁨IGCSE الرياضيات⁩

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

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. 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 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.

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:

• 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: $2^{3} \times 3^{2} \times 5 = 360$.

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:

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.
• $n(A \cup B) = 7$.

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 $1 \leqslant x \leqslant 5$".

• 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 $5^{2} \times \sqrt[3]{8}$.

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

Some special powers:

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

Worked examples.

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

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:

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:

Simplify a surd by taking out the largest square factor.

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

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 $\dfrac{1}{-1+\sqrt{3}}$.

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

• 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

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 $100x$:

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: $18 + 14 = 32$.

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 $15$:

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

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

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

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.

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.

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 $60 \div 1.2 = \$50$.

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

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 $5 \times 6 = \$30$.

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 $7 \times 0.60 = \$4.20$.

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

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

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

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.

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 $\dfrac{41.3}{9.79 \times 0.765}$.

A rounded value could really be anything that rounds to it. The smallest possible value is the lower bound 下界; the largest is the upper bound 上界. 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.

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 largest possible 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.

• $60$ 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.

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.

• 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.

• 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 $7.5 \le x < 8.5$).

A point on a graph is described by its coordinates 坐标, sometimes called Cartesian coordinates, written $(x, y)$. The first number is the across value and the second is the up value.

• The two number lines are the axes 坐标轴: the horizontal 水平 $x$-axis and the vertical 竖直 $y$-axis.
• They cross at the origin 原点, the point $(0, 0)$.
• The axes split the grid into four quadrants 象限.

So the point $(3, -2)$ is found by going $3$ to the right and $2$ down.

Most straight lines can be written as

where $m$ is the gradient 斜率 (steepness) and $c$ is the intercept 截距 — the $y$-value where the line crosses the $y$-axis.

• A line like $x = k$ (for example $x = 3$) is vertical.
• A line like $y = k$ (for example $y = 3$) is horizontal.

A line may also be given as $ax + by = c$. Rearrange it into $y = mx + c$ to read off the gradient and intercept.

Worked example. Find the gradient and $y$-intercept of $5x + 4y = 8$.

So the gradient is $-\tfrac{5}{4}$ and the $y$-intercept is $2$.

The gradient measures how steep a line is:

A positive gradient goes up to the right; a negative gradient goes down to the right.

Worked example. Find the gradient of the line through $(1, 2)$ and $(4, 11)$.

To draw $y = mx + c$, the quickest way is:

1. Mark the intercept $c$ on the $y$-axis.
2. Use the gradient to step to more points (for $m = 3$, go $1$ right and $3$ up).
3. Join the points with a straight line.

You can also make a small table of values 数值表: choose two or three $x$-values, work out $y$, and plot the points.

If you know the gradient $m$ and one point on the line, put the point into $y = mx + c$ to find $c$.

Worked example. A line has gradient $3$ and passes through $(1, 2)$. Find its equation.

So the equation is $y = 3x - 1$. (If you are given two points, first find the gradient, then do this.)

A line segment 线段 is the straight piece between two points.

To find the length 长度 of the segment between $(x_1, y_1)$ and $(x_2, y_2)$, use Pythagoras' theorem 勾股定理 on the horizontal and vertical gaps:

To find the midpoint 中点 (the point exactly in the middle), average the coordinates:

Worked example. Find the length and midpoint of the segment from $(1, 2)$ to $(4, 6)$.

Parallel 平行 lines never meet, so they have the same gradient.

Worked example. Find the equation of the line parallel to $y = 4x - 1$ that passes through $(1, -3)$.

The gradient is also $4$. Put the point in:

so the line is $y = 4x - 7$.

Two lines are perpendicular 垂直 if they meet at a right angle 直角. Their gradients multiply to $-1$:

In words: flip the fraction and change the sign.

Worked example. Find the gradient of a line perpendicular to $2y = 3x + 1$.

Rearrange: $y = \tfrac{3}{2}x + \tfrac{1}{2}$, so the gradient is $\tfrac{3}{2}$. The perpendicular gradient is $-\tfrac{2}{3}$.

Perpendicular bisector

The perpendicular bisector 垂直平分线 of a segment cuts it in half at a right angle. To find its equation: get the midpoint, then use the perpendicular gradient through that midpoint.

Worked example. Find the perpendicular bisector of the segment joining $(-3, 8)$ and $(9, -2)$.

• Midpoint: $\left( \frac{-3 + 9}{2}, \frac{8 + (-2)}{2} \right) = (3, 3)$.
• Gradient of the segment: $\frac{-2 - 8}{9 - (-3)} = \frac{-10}{12} = -\tfrac{5}{6}$.
• Perpendicular gradient: $\frac{6}{5}$.

Through $(3, 3)$: $\; 3 = \tfrac{6}{5}(3) + c \Rightarrow c = 3 - \tfrac{18}{5} = -\tfrac{3}{5}$. So the bisector is

• The straight line is $y = mx + c$: $m$ is the gradient and $c$ is where the line crosses the $y$-axis.
• Gradient = (change in $y$) ÷ (change in $x$). Keep the two coordinates in the same order on the top and the bottom.
• Parallel lines have the same gradient; perpendicular lines have gradients that multiply to $-1$ (the negative reciprocal).
• The midpoint is the average of the coordinates; the distance between two points comes from Pythagoras on the differences.