📚 KS3 Maths: Past Paper Analysis | KS3 数学：历年真题解析
Mastering KS3 Maths requires consistent practice with past paper questions. This article walks you through ten typical exam-style problems, providing clear step-by-step solutions and bilingual explanations to strengthen your understanding of key topics from number operations to geometry and statistics.
掌握 KS3 数学需要不断练习历年真题。本文通过十道典型考题，提供逐步清晰的解答和中英双语解析，帮助巩固从数字运算到几何与统计的核心知识点。
1. Fractions, Decimals, and Percentages | 分数、小数与百分比
Question: Write 0.375 as a fraction in its simplest form, and convert 5/8 to a decimal. Then calculate 2/5 of 60 and express the result as a percentage of 80.
题目： 将 0.375 写为最简分数，将 5/8 化为小数。然后计算 60 的 2/5，并将结果表示为 80 的百分之几。
÷ 25 = 15, 1000 ÷ 25 = 40, so 15/40. Then divide by 5: 15 ÷ 5 = 3, 40 ÷ 5 = 8, giving 3/8.
步骤1：将 0.375 化为分数。0.375 = 375/1000。分子分母同除以 25 化简：375 ÷ 25 = 15，1000 ÷ 25 = 40，得到 15/40。再同除以 5：15 ÷ 5 = 3，40 ÷ 5 = 8，得 3/8。
Step 2: To write 5/8 as a decimal, divide 5 by 8. 5 ÷ 8 = 0.625. So 5/8 = 0.625.
步骤2：将 5/8 化为小数，计算 5 ÷ 8 = 0.625。所以 5/8 = 0.625。
/5 × 60 = (2 × 60) / 5 = 120 / 5 = 24. Now express 24 as a percentage of 80: (24/80) × 100% = 0.3 × 100% = 30%.
步骤3：计算 60 的 2/5。2/5 × 60 = (2 × 60)/5 = 120/5 = 24。然后将其表示为 80 的百分比：(24/80) × 100% = 0.3 × 100% = 30%。
2. Ratio and Proportion | 比与比例
Question: A recipe for 10 biscuits uses 200 g of flour, 100 g of sugar, and 50 g of butter. How much of each ingredient is needed for 25 biscuits? The price of flour is £1.20 per 500 g. How much will the flour for 25 biscuits cost?
题目： 一份 10 块饼干的食谱需要面粉 200 g、糖 100 g、黄油 50 g。制作 25 块饼干各需食材多少？面粉价格为每 500 g £1.20，25 块饼干所需的面粉将花费多少？
y 2.5. Flour: 200 g × 2.5 = 500 g. Sugar: 100 g × 2.5 = 250 g. Butter: 50 g × 2.5 = 125 g.
步骤1：求倍数。25 块饼干的比例因子为 25 ÷ 10 = 2.5。将每种食材乘以 2.5。面粉：200 g × 2.5 = 500 g。糖：100 g × 2.5 = 250 g。黄油：50 g × 2.5 = 125 g。
Step 2: Flour cost. 500 g is exactly one bag, so cost is £1.20. If bought in proportion, 500 g costs £1.20, so it remains £1.20.
步骤2：面粉成本。500 g 刚好是一袋，因此价格为 £1.20。
3. Algebraic Simplification | 代数化简
Question: plify the expression 3(2a − 4) + 5a + 7. Then find the value of the expression when a = 3.
题目： 化简表达式 3(2a − 4) + 5a + 7。然后求出 a = 3 时表达式的值。
pand the bracket. 3 × 2a = 6a, 3 × (−4) = −12. So the expression becomes 6a − 12 + 5a + 7.
步骤1：展开括号。3 × 2a = 6a，3 × (−4) = −12。表达式变为 6a − 12 + 5a + 7。
2: Collect like terms. 6a + 5a = 11a. −12 + 7 = −5. The simplified expression is 11a − 5.
步骤2：合并同类项。6a + 5a = 11a。−12 + 7 = −5。化简后的表达式为 11a − 5。
Step 3: Substitute a = 3. 11 × 3 − 5 = 33 − 5 = 28.
步骤3：代入 a = 3。11 × 3 − 5 = 33 − 5 = 28。
4. Solving Equations | 解方程
Question: Solve the equation 4(x + 3) = 2x + 16. Check your answer by substitution.
题目： 解方程 4(x + 3) = 2x + 16。用代入法检验答案。
Step 1: Expand the left side. 4 × x + 4 × 3 = 4x + 12. Equation: 4x + 12 = 2x + 16.
步骤1：左边展开。4 × x + 4 × 3 = 4x + 12。方程：4x + 12 = 2x + 16。
Step 2: Subtract 2x from both sides. 4x − 2x + 12 = 16 → 2x + 12 = 16.
步骤2：两边减去 2x。4x − 2x + 12 = 16 → 2x + 12 = 16。
Step 3: Subtract 12 from both sides. 2x = 4. Divide by 2: x = 2.
步骤3：两边减 12。2x = 4。除以 2：x = 2。
Step 4: Check: left side 4(2 + 3) = 4×5 = 20. Right side 2×2 + 16 = 4+16=20. Both equal.
步骤4：检验：左边 4(2+3)=4×5=20，右边 2×2+16=4+16=20，相等。
5. Linear Sequences | 线性数列
Question: The first four terms of a sequence are 7, 12, 17, 22. Write down the nth term rule. Use it to find the 50th term.
题目： 某数列前四项为 7, 12, 17, 22。写出第 n 项公式，并用它求第 50 项。
Step 1: Find the common difference. 12 − 7 = 5, 17 − 12 = 5, 22 − 17 = 5. So it's an arithmetic sequence with common difference d = 5.
步骤1：找公差。12 − 7 = 5，17 − 12 = 5，22 − 17 = 5。所以是公差 d = 5 的等差数列。
m formula is: first term + (n − 1) × d. So nth term = 7 + (n − 1)×5 = 7 + 5n − 5 = 5n + 2.
步骤2：第 n 项公式：首项 + (n − 1) × 公差。所以第 n 项 = 7 + (n−1)×5 = 7 + 5n − 5 = 5n + 2。
Step 3: For the 50th term, substitute n = 50. 5×50 + 2 = 250 + 2 = 252.
步骤3：求第 50 项，代入 n=50。5×50 + 2 = 250 + 2 = 252。
6. Angle Rules | 角度规则
Question: 48°, angle B = 3x + 10°, and angle C = 2x + 20°. Find x and the size of all three angles.
题目： 在三角形 ABC 中，∠A = 48°，∠B = 3x + 10°，∠C = 2x + 20°。求 x 和三个角的度数。
Step 1: Use the angle sum of a triangle: 48 + (3x + 10) + (2x + 20) = 180.
步骤1：利用三角形内角和：48 + (3x+10) + (2x+20) = 180。
0 + 20 = 78, 3x + 2x = 5x. So 5x + 78 = 180. Subtract 78: 5x = 102. Divide by 5: x = 20.4.
步骤2：合并同类项。48+10+20=78，3x+2x=5x。得 5x + 78 = 180。减 78：5x=102。除以 5：x=20.4。
0 = 61.2 + 10 = 71.2°, C = 2×20.4 + 20 = 40.8 + 20 = 60.8°. Check: 48 + 71.2 + 60.8 = 180.
步骤3：角度：∠B = 3×20.4+10=61.2+10=71.2°，∠C=2×20.4+20=40.8+20=60.8°。检验：48+71.2+60.8=180。
7. Area and Perimeter of Compound Shapes | 复合图形的面积与周长
Question: The shape consists of a rectangle 8 cm by 5 cm with a right-angled triangle cut from one corner. The triangle has legs 2 cm and 3 cm. Calculate the perimeter and area of the remaining shape.
题目： 一个图形由 8 cm × 5 cm 的长方形切去一个直角三角形组成，三角形两直角边为 2 cm 和 3 cm。求剩余图形的周长和面积。
= 8 × 5 = 40 cm². Area of triangle = ½ × 2 × 3 = 3 cm². Remaining area = 40 − 3 = 37 cm².
步骤1：长方形面积 = 8×5=40 cm²。三角形面积 = ½×2×3=3 cm²。剩余面积 = 40−3=37 cm²。
remaining top and right. Better to sketch: rectangle 8×5 with a corner cut from top right? h the hypotenuse. Original top 8 cm becomes (8 − 3) = 5 cm, right side (5 − 2) = 3 cm, plus the two legs of triangle? ndary: starting bottom-left, go right 8 cm, up 3 cm (since right side reduced), then slant? Let' angle from corner B with legs along BA and BC: take point E on AB such that AE=3 cm from A? (top remaining) + (5−2)=3 cm (right remaining) + the hypotenuse of triangle. The hypotenuse = √2²+3² = √4+9=√13 Total perimeter = 8 + 5 + 5 + 3 + 3.61 = 24.61 cm. But we need exact: perimeter = 8+5+5+3+√13 = 21+√13 cm. Provide both.
步骤2：周长。假设切去右上角。原长方形底边 8 cm，左边 5 cm。切后，底边仍为 8 cm，左边 5 cm。上边剩余 (8−3)=5 cm，右边剩余 (5−2)=3 cm，加上三角形斜边。斜边长 = √2²+3² = √13 cm。周长 = 8+5+5+3+√13 = 21+√13 cm，约 24.61 cm。
8. Pythagoras' Theorem | 勾股定理
Question: A ladder leans against a vertical wall. The ladder is 5 m long and its foot is 1.5 m away from the wall. How high up the wall does the ladder reach? Give your answer to 2 decimal places.
题目： 一架梯子斜靠在竖直墙上。梯长 5 m，梯脚距墙 1.5 m。梯子能达到墙多高？答案保留两位小数。
Step 1: Use Pythagoras' theorem. Let height be h. Then h² + 1.5² = 5².
步骤1：利用勾股定理。设高度为 h，则 h² + 1.5² = 5²。
h² + 1.5² = 5²
Step 2: 1.5² = 2.25, 5² = 25. So h² = 25 − 2.25 = 22.75.
步骤2：1.5² = 2.25，5² = 25。故 h² = 25 − 2.25 = 22.75。
Step 3: h = √22.75 ≈ 4.77 m (to 2 decimal places).
步骤3：h = √22.75 ≈ 4.77 m（保留两位小数）。
9. Statistics: Mean, Median, Mode, and Range | 统计：平均数、中位数、众数和极差
Question: The numbers of books read by 9 students in a month are: 3, 7, 2, 7, 5, 4, 7, 1, 8. Find the mean, median, mode, and range.
题目： 9 名学生一个月读书的本数分别为：3, 7, 2, 7, 5, 4, 7, 1, 8。求平均数、中位数、众数和极差。
Step 1: Order the data: 1, 2, 3, 4, 5, 7, 7, 7, 8.
步骤1：数据排序：1, 2, 3, 4, 5, 7, 7, 7, 8。
Step 2: Mean = sum ÷ count. Sum = 1+2+3+4+5+7+7+7+8 = 44. Mean = 44 ÷ 9 ≈ 4.89.
步骤2：平均数 = 总和 ÷ 个数。总和 = 44，平均数 = 44 ÷ 9 ≈ 4.89。
Step 3: Median is the middle value. 9 values, so 5th value = 5. Median = 5.
步骤3：中位数是中间值。9 个数，第 5 个为 5，中位数 = 5。
Mode is the most frequent: 7 appears three times, mode = 7. Range = max − min = 8 − 1 = 7.
步骤4：众数为出现最多次的 7（三次）。极差 = 最大值 − 最小值 = 8 − 1 = 7。
10. Probability Experiments | 概率实验
Question: A bag contains 3 red balls, 2 blue balls, and 5 green balls. One ball is taken at random. Calculate the probability that it is (a) red, (b) not blue, (c) either red or green. If the ball is replaced and another is drawn, what is the probability both are green?
题目： 一个袋子里有 3 个红球、2 个蓝球和 5 个绿球。随机抽取一个球，求 (a) 抽到红色的概率，(b) 不是蓝色的概率，(c) 红色或绿色的概率。如果放回后再抽一个，两个都是绿色的概率是多少？
Step 1: Total balls = 3+2+5 = 10.
步骤1：总球数 = 3+2+5 = 10。
Step 2: (a) P(red) = 3/10.
步骤2：(a) P(红) = 3/10。
: (b) P(not blue) = 1 − P(blue) = 1 − 2/10 = 8/10 = 4/5. Or direct: (3+5)/10 = 8/10 = 4/5.
步骤3：(b) P(不是蓝) = 1 − P(蓝) = 1 − 2/10 = 8/10 = 4/5。或直接 (3+5)/10 = 8/10 = 4/5。
Step 4: (c) P(red or green) = (3+5)/10 = 8/10 = 4/5.
步骤4：(c) P(红或绿) = (3+5)/10 = 8/10 = 4/5。
replacement, P(both green) = P(green) × P(green) = (5/10) × (5/10) = (1/2) × (1/2) = 1/4.
步骤5：放回情况下，P(两个绿) = 5/10 × 5/10 = 1/2 × 1/2 = 1/4。
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