📚 Mastering Cambridge Lower Secondary Mathematics Learner’s Book 7: Core Skills and Study Guide | 掌握剑桥初中数学学生用书第7册：核心技能与学习指南
Cambridge Lower Secondary Mathematics Learner’s Book 7 builds the essential foundations for secondary mathematics by linking primary number work to formal algebra, geometry and statistics. This guide walks through the key topics, common misconceptions and revision strategies you need to succeed in Stage 7.
剑桥初中数学学生用书第7册将小学的数的运算与正式的代数、几何和统计联系起来，打下中学数学的坚实基础。本指南将带你梳理关键主题、常见误区以及在第7阶段取得好成绩所需的复习策略。
1. Number and Place Value | 数与位值
At Stage 7 you must read, write and order integers up to one billion, including negative numbers. Place value tells you the value of each digit according to its position, so in 6 732 405 the digit 3 means 30 000.
在第7阶段，你需要读写并排序大到十亿的整数，包括负数。位值告诉你每个数字因位置不同而具有的值，例如在 6 732 405 中，数字 3 表示 30 000。
Negative numbers appear in temperature, bank balances and height above or below sea level. Remember that on a number line, a smaller negative number is further left, so -7 is less than -3.
负数出现在温度、银行余额以及海平面以上或以下的高度中。请记住，在数轴上，较小的负数更靠左，因此 -7 小于 -3。
A useful skill is rounding to the nearest 10, 100 or 1000, and using estimation to check answers. For example, 4 782 rounded to the nearest 100 is 4 800.
一个有用的技能是四舍五入到最近的十、百或千，并用估算来检查答案。例如，4 782 精确到最近的百是 4 800。
2. Factors, Multiples and Primes | 因数、倍数与质数
A factor divides exactly into a number without leaving a remainder. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. A multiple is the result of multiplying by an integer, so the first five multiples of 6 are 6, 12, 18, 24, 30.
因数是能整除某个数且没有余数的数。24 的因数有 1、2、3、4、6、8、12 和 24。倍数是与整数相乘得到的结果，因此 6 的前五个倍数是 6、12、18、24、30。
A prime number has exactly two distinct factors: 1 and itself. The primes below 20 are 2, 3, 5, 7, 11, 13, 17 and 19. Note that 1 is not prime because it has only one factor.
质数只有两个不同的因数：1 和它本身。20 以下的质数是 2、3、5、7、11、13、17 和 19。注意 1 不是质数，因为它只有一个因数。
Prime factorisation breaks a number into its prime building blocks. For example, 60 = 2² × 3 × 5. You can use a factor tree to find the prime factors.
质因数分解将一个数拆成它的质数组成部分。例如，60 = 2² × 3 × 5。你可以使用因数树来找出质因数。
3. Fractions and Decimals | 分数与小数
Equivalent fractions represent the same value: ½ = 2/4 = 4/8. To simplify a fraction, divide the numerator and denominator by their highest common factor.
等值分数表示相同的值：½ = 2/4 = 4/8。要化简分数，将分子和分母同时除以它们的最大公因数。
Adding and subtracting fractions requires a common denominator. For example, ⅓ + ¼ = 4/12 + 3/12 = 7/12. Multiplying fractions is simpler: multiply numerators and denominators, so ⅖ × ¾ = 6/20 = 3/10.
分数的加减需要公分母。例如，⅓ + ¼ = 4/12 + 3/12 = 7/12。分数的乘法更简单：分子乘分子，分母乘分母，所以 ⅖ × ¾ = 6/20 = 3/10。
Decimals are fractions with denominators of powers of 10. The decimal 0.47 means 47/100. To convert a fraction to a decimal, divide the numerator by the denominator.
小数是分母为 10 的幂的分数。小数 0.47 表示 47/100。要把分数转换为小数，用分子除以分母。
4. Percentages | 百分数
A percentage is a fraction out of 100. The word ‘per cent’ means ‘out of one hundred’, so 35% = 35/100 = 0.35.
百分数是分母为 100 的分数。‘percent’一词意为“每一百”，因此 35% = 35/100 = 0.35。
To find a percentage of a quantity, convert the percentage to a decimal and multiply. For example, 20% of 80 = 0.20 × 80 = 16. Alternatively, find 10% first and scale up.
要求一个数量的百分数，先将百分数化为小数再相乘。例如，80 的 20% = 0.20 × 80 = 16。也可以先求 10%，再按比例放大。
Percentage increase and decrease problems often involve a multiplier. A 15% increase uses a multiplier of 1.15, and a 15% decrease uses 0.85.
百分数增加和减少的问题通常涉及乘数。增加 15% 使用的乘数是 1.15，减少 15% 使用的乘数是 0.85。
5. Introduction to Algebra | 代数初步
Algebra uses letters to represent unknown or changing numbers. An expression such as 3n + 5 means ‘multiply n by 3, then add 5’. Letters and numbers multiplied together are written without the multiplication sign: 4 × y = 4y.
代数用字母表示未知或变化的数。像 3n + 5 这样的表达式表示“将 n 乘以 3，再加 5”。字母和数字相乘时省略乘号：4 × y = 4y。
Like terms have exactly the same letter part, so 2a and 5a can be combined to give 7a, but 2a and 3b cannot. This is called collecting like terms.
同类项具有完全相同的字母部分，因此 2a 和 5a 可以合并为 7a，但 2a 和 3b 不能。这叫做合并同类项。
Substitution means replacing letters with numbers. If p = 4 and q = 3, then 2p + q = 2(4) + 3 = 11. Always use brackets when substituting negative numbers.
代入是指用数字替换字母。如果 p = 4 且 q = 3，那么 2p + q = 2(4) + 3 = 11。代入负数时始终要使用括号。
6. Equations and Expressions | 方程与表达式
An equation states that two expressions are equal, and solving it means finding the value of the unknown. A one-step equation like x + 7 = 12 is solved by subtracting 7 from both sides: x = 5.
方程表示两个表达式相等，解方程就是求出未知数的值。像 x + 7 = 12 这样的一步方程，可以通过在两边同时减去 7 来解：x = 5。
For two-step equations, reverse the order of operations. To solve 2x − 3 = 9, first add 3 to both sides to get 2x = 12, then divide by 2: x = 6.
对于两步方程，要逆着运算顺序进行。解 2x − 3 = 9 时，先在两边加上 3 得到 2x = 12，再除以 2：x = 6。
Always check your answer by substituting it back into the original equation. If the left side equals the right side, the solution is correct.
一定要把答案代回原方程进行检验。如果左边等于右边，解就是正确的。
7. Geometry: Angles and Shapes | 几何：角与图形
Angles are measured in degrees. A right angle is 90°, an acute angle is less than 90°, an obtuse angle is between 90° and 180°, and a reflex angle is greater than 180°.
角以度为单位来测量。直角是 90°，锐角小于 90°，钝角在 90° 到 180° 之间，优角大于 180°。
Angles on a straight line add up to 180°, and angles around a point add up to 360°. If three angles on a straight line are 70°, 40° and x°, then x = 180 − 70 − 40 = 70°.
直线上的角之和为 180°，一个点周围的角之和为 360°。如果直线上的三个角分别是 70°、40° 和 x°，那么 x = 180 − 70 − 40 = 70°。
Regular polygons have equal sides and equal angles. The interior angles of a regular pentagon each measure 108°, because the sum of interior angles is (5 − 2) × 180° = 540°.
正多边形有相等的边和相等的角。正五边形的每个内角是 108°，因为内角和为 (5 − 2) × 180° = 540°。
8. Perimeter, Area and Volume | 周长、面积与体积
Perimeter is the total distance around a shape. For a rectangle, perimeter = 2(length + width). A rectangle with length 8 cm and width 5 cm has perimeter 2(8 + 5) = 26 cm.
周长是图形外轮廓的总长度。对于矩形，周长 = 2(长 + 宽)。长 8 厘米、宽 5 厘米的矩形周长为 2(8 + 5) = 26 厘米。
Area measures the surface inside a shape. The area of a rectangle is length × width, so the same rectangle has area 8 × 5 = 40 cm². Triangle area is ½ × base × height.
面积测量图形内部的表面。矩形的面积是长 × 宽，因此同一个矩形的面积是 8 × 5 = 40 平方厘米。三角形的面积是 ½ × 底 × 高。
Volume measures space inside a 3D object. The volume of a cuboid is length × width × height. A cuboid with dimensions 3 cm, 4 cm and 5 cm has volume 60 cm³.
体积测量三维物体内部的空间。长方体的体积是长 × 宽 × 高。尺寸为 3 厘米、4 厘米、5 厘米的长方体体积为 60 立方厘米。
9. Data Handling and Averages | 数据处理与平均数
The mean is the sum of all values divided by the number of values. For the data set 5, 7, 8, 10, 15, the mean is (5 + 7 + 8 + 10 + 15) ÷ 5 = 45 ÷ 5 = 9.
平均数是所有数据之和除以数据的个数。对于数据集 5、7、8、10、15，平均数是 (5 + 7 + 8 + 10 + 15) ÷ 5 = 45 ÷ 5 = 9。
The median is the middle value when data are ordered. For even numbers of data, take the mean of the two middle values. The mode is the value that occurs most often.
中位数是将数据排序后位于中间的值。数据个数为偶数时，取中间两个值的平均数。众数是出现次数最多的值。
Bar charts, pictograms and line graphs display data clearly. Always label axes, use equal intervals and give the chart a title.
条形图、象形图和折线图能清晰地展示数据。始终要标注坐标轴、使用相等的间隔，并为图表加上标题。
10. Ratio and Proportion | 比与比例
A ratio compares two or more quantities. The ratio 3 : 5 means that for every 3 of one item there are 5 of another. Ratios can be simplified like fractions: 12 : 18 = 2 : 3.
比用来比较两个或更多数量。比 3 : 5 表示每有 3 个一种物品，就有 5 个另一种物品。比可以像分数一样化简：12 : 18 = 2 : 3。
To divide a quantity in a given ratio, add the parts first. Divide 60 in the ratio 2 : 3: total parts = 5, one part = 60 ÷ 5 = 12, so the shares are 24 and 36.
按给定比例分配一个数量，先求出总份数。把 60 按 2 : 3 的比例分配：总份数 = 5，每份 = 60 ÷ 5 = 12，因此两份分别是 24 和 36。
Proportion problems often involve direct scaling. If 5 pens cost 15 dollars, then 1 pen costs 3 dollars and 8 pens cost 24 dollars.
比例问题通常涉及直接缩放。如果 5 支笔花 15 美元，那么 1 支笔花 3 美元，8 支笔花 24 美元。
11. Coordinates and Graphs | 坐标与图像
Coordinates locate points on a grid. The x-coordinate gives the horizontal position and the y-coordinate gives the vertical position. Point (3, -2) means 3 units right and 2 units down from the origin.
坐标用来在网格上确定点的位置。x 坐标表示水平位置，y 坐标表示垂直位置。点 (3, -2) 表示从原点向右 3 个单位、向下 2 个单位。
Plotting straight-line graphs from tables of values helps link algebra and geometry. For y = 2x + 1, when x = 0, y = 1; when x = 2, y = 5; plot these points and draw a line.
根据数值表绘制直线图像有助于把代数和几何联系起来。对于 y = 2x + 1，当 x = 0 时 y = 1；当 x = 2 时 y = 5；描出这些点并画一条直线。
The point where two lines cross is the solution to the system of equations. This visual method supports algebraic solving skills.
两条直线相交的点就是方程组的解。这种图形方法有助于提升代数解题能力。
12. Problem-Solving Strategies | 解题策略
Successful problem solving in Stage 7 often requires identifying the operation, writing a number sentence and checking the answer. Read the question twice and underline key numbers and words.
第7阶段成功解题通常需要确定运算、列出数字表达式并检查答案。把题目读两遍，并在关键数字和词语下面画线。
Use the RUDE method: Read, Underline, Draw, Estimate. Drawing a bar model or number line can turn a word problem into a clear visual.
使用 RUDE 方法：阅读、划线、画图、估算。画条形模型或数轴可以把应用题变成清晰的直观图形。
Regular practice with mixed problems from the Learner’s Book, especially the end-of-unit review exercises, builds confidence and speed for the Cambridge Checkpoint assessment.
经常练习学生用书中的混合题，尤其是单元末复习题，可以为剑桥 Checkpoint 评估建立信心并提高速度。
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