Essential Maths 9H Compressed Question Types Analysis | KS3 数学：Essential Maths 9H 压缩题型解析

📚 Essential Maths 9H Compressed Question Types Analysis | KS3 数学：Essential Maths 9H 压缩题型解析

The Essential Maths 9H workbook is widely used in Key Stage 3 to challenge higher‑ability students with compressed, multi‑step problems. These questions often combine several skills into one compact task, testing fluency, reasoning, and problem‑solving. In this article we break down the most common question types, show you how to unpack them, and share strategies for tackling even the trickiest 9H compression exercises.

Essential Maths 9H 练习册在 KS3 阶段广泛用于挑战高能力学生，其中的 “压缩题型” 将多种技能融合在一个紧凑的任务中，同步考察计算流畅度、逻辑推理与问题解决。本文我们将拆解最常见的题型，演示如何拆题、梳理解题思路，并分享应对 9H 压缩难题的实用策略。

1. Fractions and Decimals in Context | 情景化的分数与小数

9H compression tasks often embed fractions and decimals within real‑world scenarios – from sharing bills to calculating discounts. A typical question might ask: “Three friends share a restaurant bill of £84.50. A pays 2/5 of the total, B pays 0.3 of the total, and C pays the rest. How much does C pay?” You need to convert between fractions and decimals fluently, then subtract from the total.

9H 压缩题常将分数与小数嵌入真实场景，比如分摊账单或计算折扣。典型题目如：“三人分摊 84.50 英镑的餐费，A 付总额的 2/5，B 付 0.3，C 付余额。C 付了多少？” 你需要灵活地在分数与小数之间转换，再从总量中减去。

Convert 2/5 to 0.4 by division (2 ÷ 5 = 0.4).
将 2/5 转换为 0.4（2 ÷ 5 = 0.4）。
Calculate 0.4 × 84.50 = 33.80, and 0.3 × 84.50 = 25.35.
计算 0.4 × 84.50 = 33.80，以及 0.3 × 84.50 = 25.35。
Then find C’s share: 84.50 − (33.80 + 25.35) = 25.35.
然后求 C 的份额：84.50 − (33.80 + 25.35) = 25.35。

A common 9H twist is to present the information in a table with missing values, forcing you to reverse‑engineer the part‑to‑whole relationship.

9H 常见的变式是用表格给出部分数据并留空，逼迫你逆向推导部分与整体的关系。

Fraction → Decimal: a/b = a ÷ b
Decimal → Percentage: multiply by 100

2. Ratio, Proportion, and Best Buys | 比例与最佳购买方案

Ratio problems in 9H often combine sharing ratios with value‑for‑money comparisons. For example: “Crisps are sold in packs of 150 g for £1.20 and 200 g for £1.55. Which is the better buy? Then, if two friends share a 200 g pack in the ratio 3:2, how much does each get?” This single compressed task tests both proportional reasoning and ratio division.

9H 中的比例问题常将分配比与性价比比较相结合。例如：“薯片有 150 克/1.20 英镑与 200 克/1.55 英镑两种包装，哪个更划算？若两人按 3 : 2 分享一包 200 克，各得多少克？” 一个压缩任务同时考察比例推理和按比分物体的能力。

Find unit price: 120/150 = 0.80p per gram, and 155/200 = 0.775p per gram, so 200 g pack is better value.
计算单价：120 ÷ 150 = 0.80 便士/克，155 ÷ 200 = 0.775 便士/克，因此 200 克装更划算。
Divide 200 g in ratio 3:2: 3+2=5 parts, one part = 200 ÷ 5 = 40 g; friend 1 gets 3 × 40 = 120 g, friend 2 gets 2 × 40 = 80 g.
按 3:2 分配 200 克：3+2=5 份，每份 200 ÷ 5 = 40 克；第一位得 3×40=120 克，第二位得 2×40=80 克。

The 9H pack often presents ratio and proportion inside word‑heavy contexts, so underline the numbers and the key words ‘shares’, ‘ratio’, ‘total’ to stay organised.

9H 练习常在长叙述中隐藏比例信息，建议圈出数字和关键词 “share”、“ratio”、“total” 以保持思路清晰。

3. Algebraic Simplification and Expanding Brackets | 代数化简与去括号

Compressed algebra questions in 9H require you to expand, simplify, and sometimes factorise within the same line of working. Example: “Simplify 3(2x − 4) + 4(x + 1) − 2x.” You must expand both sets of brackets, collect like terms, and watch for negative signs.

9H 压缩代数题要求你在同一步骤中完成去括号、合并和因式分解。例如：“化简 3(2x − 4) + 4(x + 1) − 2x。” 你需要展开两组括号，合并同类项，并小心负号。

Expand: 3(2x − 4) = 6x − 12, and 4(x + 1) = 4x + 4.
展开：3(2x − 4) = 6x − 12，4(x + 1) = 4x + 4。
Combine: 6x − 12 + 4x + 4 − 2x = (6x + 4x − 2x) + (−12 + 4) = 8x − 8.
合并：6x − 12 + 4x + 4 − 2x = (6x + 4x − 2x) + (−12 + 4) = 8x − 8。

Watch out for 9H questions that include a bracket with a minus sign in front, e.g. −(3x − 2). The sign flips every term inside: −3x + 2. Always write the intermediate step to avoid sign errors.

注意 9H 中括号前带负号的情形，如 −(3x − 2)，每一项都要变号：−3x + 2。务必写出中间步骤以规避符号错误。

4. Solving Linear Equations with Unknowns on Both Sides | 解含未知数在两边的一次方程

9H compressed equations often involve brackets, fractions, or variables on both sides. Example: “Solve 5(y − 2) = 3(y + 4).” You need to expand, move variable terms to one side, constants to the other, and solve.

9H 压缩方程常含有括号、分数或两边都有未知数。例如：“解 5(y − 2) = 3(y + 4)。” 你需要展开，将变量项移向一边，常数项移向另一边，然后求解。

Expand: 5y − 10 = 3y + 12.
展开：5y − 10 = 3y + 12。
Subtract 3y: 2y − 10 = 12.
两边减 3y：2y − 10 = 12。
Add 10: 2y = 22, so y = 11.
两边加 10：2y = 22，所以 y = 11。

When fractions appear, multiply every term by the lowest common denominator to clear the fractions. For example, x/3 + 1 = x/4 + 2 becomes 4x + 12 = 3x + 24 after multiplying by 12.

当出现分数时，用最小公分母乘以每一项去分母。例如，x/3 + 1 = x/4 + 2，乘以 12 后得到 4x + 12 = 3x + 24。

Checking solution: 5(11 − 2) = 5 × 9 = 45; 3(11 + 4) = 3 × 15 = 45 ✓

5. Coordinates and Straight Line Graphs | 坐标与直线图像

9H graph questions compress table completion, coordinate plotting, and gradient calculation into a single exercise. You might be given y = 2x − 1 and asked to fill a table for x = −2, 0, 3, then plot the line and find where it crosses the axes.

9H 的图像题将填表、描点和斜率计算压缩为一个练习。你可能会遇到 y = 2x − 1，要求填写 x = −2, 0, 3 时的 y 值，然后描点画线并找出与坐标轴的交点。

Complete table: when x = −2, y = 2(−2) − 1 = −5; x = 0, y = −1; x = 3, y = 5.
填表：x = −2 时 y = 2(−2) − 1 = −5；x = 0 时 y = −1；x = 3 时 y = 5。
Plot points and draw a straight line through them.
描点并用直尺画直线。
To find the axes intercepts: y‑intercept (where x = 0) is −1; x‑intercept set y = 0 → 2x − 1 = 0 → x = 0.5.
求截距：y 截距（x = 0）为 −1；x 截距令 y = 0 → 2x − 1 = 0 → x = 0.5。

The gradient is the coefficient of x, here 2. A 9H extension might ask: “What is the equation of a line parallel to y = 2x − 1 but passing through (0, 3)?” The answer is y = 2x + 3 because parallel lines share the same gradient.

斜率是 x 的系数，此处为 2。9H 的拓展可能会问：“与 y = 2x − 1 平行且经过 (0, 3) 的直线方程是什么？” 答案是 y = 2x + 3，因为平行直线斜率相同。

6. Area and Perimeter of Composite Shapes | 组合图形的面积与周长

A classic 9H compression is a compound shape made of rectangles and triangles, with some side lengths missing. You must use given lengths to deduce the unknown sides, then calculate both perimeter and area. Example: an L‑shape formed by two rectangles.

经典的 9H 压缩题是给出由矩形和三角形组成的组合图形，部分边长未知。你需要用已知长度推导未知边，再分别计算周长和面积。例如：由两个矩形组成的 L 形。

Split the shape into two simpler rectangles, A and B.
将图形分割为两个简单矩形 A 和 B。
Find missing sides using given lengths: if a vertical total is 10 cm and the top part is 4 cm, the missing vertical is 6 cm.
用已知长度求缺失边：若总高度 10 cm，上部高 4 cm，则缺失的竖边为 6 cm。
Area = (length₁ × width₁) + (length₂ × width₂); Perimeter = sum of all outer edges – be careful not to double‑count internal lines.
面积 = (长₁ × 宽₁) + (长₂ × 宽₂)；周长 = 所有外围边长之和 —— 注意内部线段不计算在内。

Sometimes the perimeter question uses the same lengths but the student must realise that an internal edge does not contribute to the perimeter. The 9H pack deliberately blends these to test attention to detail.

有时周长问题使用同样的边长，但学生必须意识到内部边不计入周长。9H 练习有意将两者混合以考察细节意识。

7. Volume and Surface Area of Prisms | 棱柱的体积与表面积

9H often gives a triangular prism with its dimensions shown, asking for volume and surface area in one task. A typical question: “The cross‑section is a triangle with base 5 cm, height 4 cm, length of prism 8 cm. Find the volume and total surface area.”

9H 常给出一个三棱柱并标注尺寸，要求在一次任务中计算体积和表面积。典型题目：“截面是底 5 cm、高 4 cm 的三角形，棱柱长 8 cm。求体积和总表面积。”

Volume = area of cross‑section × length = (½ × 5 × 4) × 8 = 10 × 8 = 80 cm³.
体积 = 截面积 × 长 = (½ × 5 × 4) × 8 = 10 × 8 = 80 cm³。
Surface area = 2 × area of triangle + area of three rectangular faces. The three rectangular dimensions: 5 × 8, 4 × 8, and hypotenuse (√(5² + 4²) = √41 ≈ 6.4) × 8.
表面积 = 2 × 三角形面积 + 三个矩形面面积。矩形尺寸：5×8、4×8，以及斜边 (√(5² + 4²) = √41 ≈ 6.4) × 8。
Total SA ≈ 2×10 + (40 + 32 + 51.2) = 20 + 123.2 = 143.2 cm².
总表面积 ≈ 2×10 + (40 + 32 + 51.2) = 20 + 123.2 = 143.2 cm²。

The compression comes from needing to correctly identify the cross‑section, apply Pythagoras for the slant edge, and sum the faces without missing any. 9H will sometimes swap the prism’s orientation to confuse which face is the base.

压缩的难点在于准确识别截面、运用勾股定理求斜边、以及不遗漏任何面积。9H 有时会刻意旋转棱柱的方向以混淆底面。

8. Angles on Parallel Lines and in Polygons | 平行线上的角度与多边形内角

A single 9H diagram may show two parallel lines with a transversal, plus a triangle attached. The question compresses angle facts: alternate, corresponding, vertically opposite, and interior angles of a triangle. You must chain several angle calculations to find the final answer.

一张 9H 图形可能展示两条平行线与一条截线，还附加一个三角形。此类题目压缩了多种角度关系：内错角、同位角、对顶角以及三角形内角和。你需要连锁计算多个角度才能得到最终结果。

Identify alternate angles (Z‑shape) and corresponding angles (F‑shape) to transfer angles from one line to the parallel line.
利用内错角（Z 形）和同位角（F 形）将角度从一条线传递到平行线。
Use the fact that angles on a straight line sum to 180° and vertically opposite angles are equal.
运用平角 180° 和对顶角相等的事实。
Then apply the triangle angle sum (180°) to find the missing interior angle.
再用三角形内角和 180° 求缺失的内角。

Example: Given angle 1 = 55° and lines are parallel, then its corresponding angle at another intersection is also 55°; the adjacent angle on the straight line is 125°. That 125° might be an exterior angle of a triangle, allowing you to find the interior angles.

例如：已知 ∠1 = 55° 且两线平行，则同位角 ∠2 也是 55°；其邻补角为 125°。该 125° 可能是三角形的一个外角，从而推算出内角。

9H often asks for a reasoning chain: ‘Angle a = … because …’ so practise writing short angle reasons.

9H 经常要求写出推理链：“∠a = … 因为 …”，所以请练习书写简短的角度理由。

9. Probability and Expected Outcomes | 概率与期望结果

Compressed probability tasks in 9H typically combine experimental and theoretical probability, or merge two independent events. You might be given: “A spinner has sections: red, blue, green, yellow in ratio 1:2:3:4. Spun 200 times, how many times do you expect blue? What is the probability of getting red or green?”

9H 中的压缩概率题常将实验概率与理论概率结合，或合并两个独立事件。你会遇到：“一个转盘有红、蓝、绿、黄四色，面积比为 1:2:3:4。旋转 200 次，期望出现蓝色的次数是多少？得到红色或绿色的概率是多少？”

Total parts = 1+2+3+4 = 10.
总份数 = 1+2+3+4 = 10。
Probability of blue = 2/10 = 1/5. Expected frequency = 1/5 × 200 = 40.
蓝色的概率 = 2/10 = 1/5。期望次数 = 1/5 × 200 = 40。
Probability of red or green = (1+3)/10 = 4/10 = 2/5.
红色或绿色的概率 = (1+3)/10 = 4/10 = 2/5。

If a second spinner with numbers 1 to 4 is spun simultaneously, a 9H question might ask for the probability that the colour is blue AND the number is 2. Since events are independent, multiply: 1/5 × 1/4 = 1/20.

如果同时旋转另一个标有 1–4 的数字转盘，9H 题目可能问颜色为蓝色且数字为 2 的概率。因事件独立，相乘：1/5 × 1/4 = 1/20。

Always check if events are independent or if the question asks for ‘at least’ probabilities, which often require a tree diagram or a complementary approach in 9H.

务必检查事件是否独立，以及问题是否要求 “至少” 的概率，这在 9H 中常需树形图或补集法。

10. Averages and Data Comparisons | 平均数与数据比较

9H data handling compressions provide two small data sets and ask for the mean, median, mode, and range – then a comparative sentence. Example: “Set A: 5, 7, 8, 9, 11; Set B: 2, 4, 10, 10, 14. Compare their averages and spreads.”

9H 数据处理压缩题会给出两组小数据，要求计算平均数、中位数、众数和极差，然后写一句比较性结论。例如：“数据集 A：5, 7, 8, 9, 11；数据集 B：2, 4, 10, 10, 14。比较它们的平均数和离散程度。”

Set A: mean = (5+7+8+9+11)÷5 = 8; median = 8; range = 11−5 = 6.
数据集 A：平均数 (5+7+8+9+11)÷5 = 8；中位数 8；极差 11−5 = 6。
Set B: mean = (2+4+10+10+14)÷5 = 8; median = 10; range = 14−2 = 12.
数据集 B：平均数 (2+4+10+10+14)÷5 = 8；中位数 10；极差 14−2 = 12。
Comparison: Both sets have the same mean, but B has a larger median and a wider spread, indicating more variability.
比较：两者平均数相同，但 B 的中位数较大且极差更大，说明数据更分散。

9H questions may include an outlier and ask how it affects the mean. A single extreme value pulls the mean up or down much more than the median, which is why the median is often the better measure in such cases.

9H 问题可能包含异常值并询问其对平均数的影响。一个极端值对平均数的拉动远大于对中位数的影响，因此这种情况下中位数往往是更合适的中心度量。

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Essential Maths 9H Compressed Question Types Analysis | KS3 数学：Essential Maths 9H 压缩题型解析