📚 Essential Maths Book 8C Answers.compressed: Key Topic Revision | Essential Maths Book 8C 答案压缩知识点精讲
Essential Maths Book 8C is a core resource for Key Stage 3 students tackling more advanced topics. This compressed revision guide extracts the most important concepts from the book’s exercises and answer keys, breaking them down into clear, digestible explanations. Whether you’re checking your homework or preparing for a test, these key points will help you understand the ‘why’ behind every answer.
《Essential Maths 8C》是Key Stage 3阶段学生攻克更深入数学内容的重要教材。这份精炼的复习指南提取了书中习题和答案的核心知识点，将其拆解为清晰易懂的讲解。无论你是在核对作业还是在备考，这些要点都能帮你理解每个答案背后的原理。
1. Simplifying Algebraic Expressions | 化简代数表达式
In Book 8C, you often need to collect like terms. An expression such as 5a + 3b − 2a + 4b can be simplified by adding or subtracting the coefficients of the same letter. Here, 5a minus 2a gives 3a, and 3b plus 4b gives 7b. The final expression is 3a + 7b. Remember that the sign in front of a term belongs to it.
在8C教材中，经常需要合并同类项。比如表达式5a + 3b − 2a + 4b，可以通过对相同字母的系数进行加减来化简。这里5a减去2a得到3a，3b加上4b得到7b。最终表达式为3a + 7b。切记，项前面的符号是属于该项的。
When you see brackets and a number outside, like 3(x + 2y), multiply every term inside by the number outside: 3 times x is 3x, and 3 times 2y is 6y. So the answer is 3x + 6y. Never just multiply the first term.
当你看到括号外有数字，比如3(x + 2y)，要用外面的数乘以括号内的每一项：3乘x得3x，3乘2y得6y。答案为3x + 6y。千万不要只乘第一项。
2. Solving Two-Step Equations | 解两步方程
A two-step equation often looks like 2x + 5 = 13. The goal is to isolate x. First, undo the addition: subtract 5 from both sides to get 2x = 8. Then undo the multiplication: divide both sides by 2 to obtain x = 4. Always check by substituting 4 back into the original equation: 2(4) + 5 = 13, which is true.
两步方程通常形如2x + 5 = 13。目标是隔离x。首先消除加法：两边同时减去5，得到2x = 8。然后消除乘法：两边同时除以2，得到x = 4。一定要将4代回原方程检验：2(4) + 5 = 13，等式成立。
If the variable appears on both sides, like 5x − 3 = 2x + 9, first collect the x terms on one side. Subtract 2x from both sides: 3x − 3 = 9. Then add 3 to both sides: 3x = 12, so x = 4.
如果变量出现在两边，比如5x − 3 = 2x + 9，首先把含x的项移到一边。两边同减2x：3x − 3 = 9。然后两边加3：3x = 12，因此x = 4。
3. Ratio and Proportion | 比和比例
In Book 8C, ratio problems often involve sharing a quantity in a given ratio. For example, divide £60 between two people in the ratio 2 : 3. First, find the total number of parts: 2 + 3 = 5. One part is worth £60 ÷ 5 = £12. The first person gets 2 × £12 = £24, and the second gets 3 × £12 = £36. Always check that the amounts add back up to the original total.
在8C中，比例问题常涉及按给定比例分配数量。例如，将60英镑按2:3的比例分给两人。先求总份数：2 + 3 = 5。一份是£60 ÷ 5 = £12。第一个人得2 × £12 = £24，第二个人得3 × £12 = £36。务必核实两个金额加起来等于原始总数。
For equivalent ratios, scale up or down by multiplying or dividing both sides by the same number. The ratio 15 : 10 can be simplified to 3 : 2 by dividing both numbers by 5. This skill is essential when comparing mixtures or recipes.
对于等值比例，可以通过同时乘或除以相同的数来放大或缩小。比例15:10可同时除以5简化为3:2。在比较混合物或配方时，这一技能至关重要。
4. Percentage Increase and Decrease | 百分比增减
To increase an amount by 20%, you can find 20% and add it on, or use a multiplier directly. The multiplier for a 20% increase is 1.20 (since 100% + 20% = 120% = 1.20). So increasing £80 by 20% gives £80 × 1.20 = £96. Using multipliers is faster and less error‑prone for multi‑step problems.
要将一个金额增加20%，可以先求出20%再加上，或直接使用乘数。增加20%的乘数为1.20（因为100% + 20% = 120% = 1.20）。因此，£80增加20%为£80 × 1.20 = £96。对于多步问题，使用乘数更快速且不易出错。
For a decrease of 15%, the multiplier is 0.85 (100% − 15% = 85% = 0.85). So reducing £200 by 15% yields £200 × 0.85 = £170. For reverse percentage problems (e.g., a price after a 10% discount is £45, find the original), divide by 0.90 (since the sale price is 90% of the original).
减少15%的乘数是0.85（100% − 15% = 85% = 0.85）。因此£200减少15%为£200 × 0.85 = £170。对于逆向百分数问题（比如打九折后价格为£45，求原价），需要除以0.90（因为售价是原价的90%）。
5. Angles in Triangles and Quadrilaterals | 三角形和四边形的角
KS3 students must know that the angles in any triangle sum to 180°. If two angles are given, say 45° and 75°, the third angle is 180° − (45° + 75°) = 60°. In isosceles triangles, the base angles are equal, which helps when only one angle is known.
KS3学生必须知道任何三角形的内角和为180°。如果已知两个角是45°和75°，第三个角就是180° − (45° + 75°) = 60°。在等腰三角形中，底角相等，这有助于在只知道一个角时求解。
Quadrilaterals have an angle sum of 360°. For a kite, a trapezium, or a standard quadrilateral, missing angles are found by subtracting the sum of the known angles from 360°. In a parallelogram, opposite angles are equal and adjacent angles add to 180°.
四边形的内角和为360°。无论是风筝形、梯形还是一般的四边形，未知角都可以用360°减去已知角之和来求得。在平行四边形中，对角相等，邻角之和为180°。
6. Area of Compound Shapes | 组合图形的面积
Compound shapes made of rectangles and triangles appear frequently in Book 8C. Split the shape into distinct parts whose areas you know how to calculate. For an L-shape, divide it into two rectangles, work out each area (length × width), and add them together. Always label the missing side lengths first.
由长方形和三角形组成的组合图形在8C中频繁出现。将图形分割成你知道如何计算面积的几个部分。对于L形，可分成两个长方形，分别计算面积（长×宽）再相加。务必先标出所有缺失的边长。
If the shape includes a triangle, remember the area is ½ × base × height. For a shape that combines a rectangle and a right-angled triangle on top, find the rectangle’s area and add the triangle’s area. Pay attention to perpendicular heights; never use a slanted side.
如果图形包含三角形，牢记面积是½ × 底 × 高。对于顶部有一个直角三角形的长方形组合体，求出长方形面积再加上三角形面积。注意要用垂直高度，绝不能使用斜边。
7. Circles: Circumference and Area | 圆的周长和面积
The circumference (distance around) a circle is calculated using C = πd or C = 2πr. The value of π is approximately 3.14, though answers are often left in terms of π. For a circle with diameter 10 cm, circumference = π × 10 ≈ 31.4 cm. When the radius is 7 cm, C = 2 × π × 7 = 14π cm.
圆的周长（周边距离）计算公式为C = πd或C = 2πr。π约等于3.14，但答案常保留π。对于直径为10 cm的圆，周长 = π × 10 ≈ 31.4 cm。当半径为7 cm时，C = 2 × π × 7 = 14π cm。
Area is given by A = πr². The radius must be squared first. For a circle with radius 5 m, area = π × 5² = 25π m², which is about 78.5 m². Do not confuse the two formulas; a common mistake is using diameter in the area formula or forgetting to square the radius.
面积公式为A = πr²。半径必须先平方。半径为5 m的圆，面积= π × 5² = 25π m²，约78.5 m²。不要混淆两个公式；常见错误是在面积公式中使用直径或者忘记将半径平方。
8. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
These four values summarise a data set. The mean is calculated by adding all numbers and dividing by how many there are. The median is the middle value when data is ordered; if there are two middle numbers, take their mean. The mode is the most frequent value, and the range is the largest minus the smallest.
这四个数值可以概括一组数据。平均数由所有数之和除以数据个数得到。中位数是将数据排序后位于中间的数；若有两个中间数，则取它们的平均数。众数是出现次数最多的值，极差是最大值与最小值之差。
Example: for the set 4, 7, 2, 7, 10, 3, 7, the ordered list is 2, 3, 4, 7, 7, 7, 10. The mean is (2+3+4+7+7+7+10) ÷ 7 = 40 ÷ 7 ≈ 5.71. The median is 7. The mode is 7. The range is 10 − 2 = 8. Outliers affect the mean but not the median.
示例：对于数据集4, 7, 2, 7, 10, 3, 7，排序后为2, 3, 4, 7, 7, 7, 10。平均数 = (2+3+4+7+7+7+10) ÷ 7 = 40 ÷ 7 ≈ 5.71。中位数是7。众数是7。极差是10 − 2 = 8。异常值会影响平均数，但不会影响中位数。
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