📚 Year 7 SQA Maths: Core Concepts Overview | Year 7 SQA 数学：核心知识点梳理
The Year 7 SQA Mathematics course, part of Scotland’s Broad General Education (S1), builds a strong foundation in key mathematical areas. This article outlines the essential topics covered, helping students consolidate their understanding and prepare for progression. From place value and fractions to basic algebra and geometry, mastering these concepts is crucial for future success.
Year 7 SQA 数学是苏格兰广义教育（S1）的一部分，旨在为学生打下坚实的数学基础。本文梳理了核心知识模块，帮助学生巩固理解，为后续学习做好准备。从数位值、分数到基础代数和几何，掌握这些概念对未来至关重要。
1. Place Value and Rounding | 数位值与四舍五入
In S1 Maths, you must confidently read, write and understand numbers up to millions. The place value columns include units, tens, hundreds, thousands, ten thousands, hundred thousands, and millions. For example, in 5 678 234, the digit 5 represents 5 millions. Rounding is a key skill: to round 3456 to the nearest 100, look at the tens digit (5) – since it is 5 or more, round up to 3500.
在 S1 数学中，你需要熟练读写并理解直至百万的数字。数位值包括个位、十位、百位、千位、万位、十万位和百万位。例如，在 5 678 234 中，数字 5 代表 5 个百万。四舍五入是一项关键技能：将 3456 四舍五入到最接近的百位时，看十位数字（5）——由于大于等于 5，所以向上舍入为 3500。
When rounding decimals, identify the decimal place required and check the next digit. For instance, 3.764 rounded to 2 decimal places is 3.76 because the thousandths digit is 4. Also, understanding the effect of multiplying and dividing by 10, 100, 1000 helps to grasp decimal place shifts.
在对小数进行四舍五入时，确定所需要的小数位数，然后看下一位数字。例如，3.764 四舍五入到两位小数是 3.76，因为千分位数字是 4。此外，理解乘除以 10、100、1000 对小数点的移动效果有助于掌握数位变化。
2. Adding and Subtracting Whole Numbers and Negative Numbers | 整数及负数的加减法
You should be able to add and subtract large numbers using column methods. For example, 4876 + 3295 = 8171. When subtracting, borrowing may be needed. Mental strategies, such as partitioning numbers, can speed up calculations.
你应该掌握使用竖式进行大数加减法。例如，4876 + 3295 = 8171。减法可能需要借位。心算策略，如拆分数字，可以加快计算。
Negative numbers appear on a number line to the left of zero. Adding a negative number moves left, subtracting a negative moves right. Key rule: a − (−b) = a + b. For example, 3 − (−2) = 3 + 2 = 5. Real-life contexts, like temperatures dropping below zero, reinforce these ideas.
负数在数轴上位于零的左侧。加一个负数相当于向左移动，减一个负数相当于向右移动。关键规则：a − (−b) = a + b。例如，3 − (−2) = 5。生活中的场景，如温度降到零下，有助于加深理解。
3. Multiplying and Dividing Whole Numbers | 整数乘除法
Multiplication and division are fundamental. Know all multiplication tables up to 12 × 12. Use the grid method or column multiplication for larger numbers, e.g., 23 × 45 = 1035. For division, try the bus stop method: 315 ÷ 5 = 63. Understanding factors and multiples is also important.
乘除法是基础。要熟记 12 × 12 以内的乘法表。对于较大数字，可以使用网格法或竖式乘法，如 23 × 45 = 1035。除法可使用短除法（公交车站法）：315 ÷ 5 = 63。理解因数和倍数也很重要。
A prime number has exactly two factors. Prime factorisation expresses a number as a product of primes, e.g., 24 = 2³ × 3. The highest common factor (HCF) and lowest common multiple (LCM) are used when working with fractions and problem solving.
质数是只有两个因数的数。质因数分解将数字表示为质数的乘积，例如 24 = 2³ × 3。在处理分数和解决实际问题时，最高公因数（HCF）和最低公倍数（LCM）会经常用到。
4. Understanding Fractions | 理解分数
Fractions represent parts of a whole. Equivalent fractions have the same value, e.g., ½ = 2/4 = 3/6. To simplify a fraction, divide the numerator and denominator by their highest common factor. For example, 8/12 simplifies to 2/3. Use improper fractions and mixed numbers: 11/4 = 2 ¾.
分数表示整体的一部分。等值分数具有相同的值，如 ½ = 2/4 = 3/6。化简分数时，将分子和分母同时除以它们的最高公因数，如 8/12 化简为 2/3。使用假分数和带分数：11/4 = 2 ¾。
To add ½ + ⅓, convert to equivalent fractions with denominator 6: 3/6 + 2/6 = 5/6. When finding a fraction of an amount, divide by the denominator and multiply by the numerator: 2/5 of 60 = (60 ÷ 5) × 2 = 24.
计算 ½ + ⅓ 时，转换为分母 6 的等值分数：3/6 + 2/6 = 5/6。求一个数的几分之几时，除以分母再乘以分子：60 的 2/5 = (60 ÷ 5) × 2 = 24。
5. Working with Decimals | 小数的运算
Decimals extend the place value system beyond units. For addition and subtraction, align the decimal points: 4.25 + 3.7 = 7.95. When multiplying a decimal by a whole number, perform multiplication normally and then place the decimal point: 0.6 × 4 = 2.4.
小数将数位值系统扩展到个位以下。加减法时对齐小数点：4.25 + 3.7 = 7.95。小数乘整数时，先按整数乘法计算，再点小数点：0.6 × 4 = 2.4。
Multiplying by 10, 100 shifts digits left; dividing shifts right. For example, 3.4 × 10 = 34, and 5.71 ÷ 100 = 0.0571. Dividing a decimal by an integer uses the bus stop method; make sure the decimal point stays in line.
乘 10 或 100 时，数字向左平移；除以 10 或 100 时向右平移。例如，3.4 × 10 = 34，5.71 ÷ 100 = 0.0571。小数除以整数使用短除法，确保小数点对齐。
6. Percentages, Fractions and Decimals | 百分数、分数与小数之间的转换
Percentage means ‘out of 100’. 50% = 50/100 = ½ = 0.5. To convert a fraction to a percentage, make the denominator 100, or divide numerator by denominator and multiply by 100: ¾ = 0.75 = 75%.
百分数表示“每一百”。50% = 50/100 = ½ = 0.5。将分数转换为百分数，可将分母变为 100，或分子除以分母再乘以 100：¾ = 0.75 = 75%。
Non-calculator methods: find 10% then scale. For example, 15% of 60: 10% is 6, 5% is 3, so 15% = 9. Percentage increase: increase 200 by 15% gives 230. Linking percentages, decimals and fractions is crucial for solving real-world problems.
非计算器方法：先求 10% 再缩放。例如，60 的 15%：10% 是 6，5% 是 3，所以 15% = 9。百分比增加：将 200 增加 15% 得到 230。把百分数、小数和分数联系起来对于解决实际问题至关重要。
7. Introduction to Algebraic Expressions | 代数表达式入门
Algebra uses letters to represent unknown numbers or variables. Write expressions like 3x + 2y − 5. The term 3x means 3 times x. Like terms can be collected: 4a + 3b + 2a − b = 6a + 2b.
代数使用字母表示未知数或变量。写出表达式如 3x + 2y − 5。项 3x 表示 3 乘以 x。同类项可以合并：4a + 3b + 2a − b = 6a + 2b。
Substitution means replacing letters with given numbers. For example, if a = 2 and b = 3, then 4a − b = 8 − 3 = 5. Expanding a bracket multiplies each term inside: 2(x + 3) = 2x + 6. These skills lay the groundwork for solving equations.
代入法是用给定数值替换字母。例如，若 a = 2, b = 3，则 4a − b = 5。去括号时，将括号外的因数乘以括号内的每一项：2(x + 3) = 2x + 6。这些技能为解方程打下基础。
8. Solving Simple Equations | 解简单方程
An equation shows two expressions are equal. Solve x + 5 = 12 by subtracting 5 from both sides: x = 7. For 3y = 18, divide both sides by 3: y = 6. Two-step equations: 2p + 1 = 11 → subtract 1: 2p = 10 → divide by 2: p = 5.
方程表示两个表达式相等。解 x + 5 = 12，两边同时减 5 得 x = 7。对于 3y = 18，两边同时除以 3 得 y = 6。两步方程：2p + 1 = 11 → 减 1 得 2
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