📚 GCSE OCR Maths: Typical Example Questions Explained in Detail | GCSE OCR 数学：典型例题详解
This article walks you through a carefully selected set of typical GCSE OCR Mathematics questions, covering the key topics you are most likely to encounter. Each example is broken down step by step with clear reasoning, showing exactly how to secure full marks in the exam. Use these detailed solutions to build your confidence and sharpen your problem‑solving skills.
本文带你逐一攻克一组精心挑选的 GCSE OCR 数学典型考题，覆盖考试中最常见的关键主题。每道例题都配有逐步解析与清晰思路，准确展示如何在考试中稳拿满分。通过这些详解，你可以建立信心并提升解题能力。
1. Solving Linear Equations | 解一元一次方程
Example: Solve 4(2x − 3) = 3x + 7.
例题：解方程 4(2x − 3) = 3x + 7。
First expand the bracket on the left: 8x − 12 = 3x + 7.
首先把左边的括号展开：8x − 12 = 3x + 7。
Collect like terms by subtracting 3x from both sides: 5x − 12 = 7.
将含 x 的项移到一边，两边同时减去 3x：5x − 12 = 7。
Add 12 to both sides: 5x = 19, then divide by 5 to obtain x = 19/5 or 3.8.
两边加 12：5x = 19，再除以 5 得到 x = 19/5 或 3.8。
Always check by substituting back: 4(2×3.8 − 3) = 4(7.6 − 3) = 4×4.6 = 18.4, and 3×3.8 + 7 = 11.4 + 7 = 18.4. Both sides match, so the solution is correct.
务必代回原方程检验：4(2×3.8 − 3) = 4(7.6 − 3) = 4×4.6 = 18.4，而 3×3.8 + 7 = 11.4 + 7 = 18.4。两边相等，因此解正确。
2. Factorising Quadratic Expressions | 分解二次三项式
Example: Factorise x² + 7x + 12.
例题：分解因式 x² + 7x + 12。
We look for two numbers that multiply to give 12 and add to give 7. The pair 3 and 4 works because 3 × 4 = 12 and 3 + 4 = 7.
我们需要找到两个数，乘积为 12，和为 7。数字 3 和 4 符合，因为 3 × 4 = 12，3 + 4 = 7。
Therefore, x² + 7x + 12 = (x + 3)(x + 4).
因此，x² + 7x + 12 = (x + 3)(x + 4)。
You can verify by expanding: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12. Notice that the constant term is positive and the coefficient of x is positive, so both brackets have plus signs.
你可以通过展开来验证：(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12。注意常数项为正且 x 系数为正，因此两个括号内都是加号。
Example with a negative constant: Factorise x² − 5x − 24. Find two numbers with product −24 and sum −5. The numbers −8 and 3 work because (−8) × 3 = −24 and (−8) + 3 = −5. So it factorises to (x − 8)(x + 3).
带有负常数项的例题：分解 x² − 5x − 24。寻找乘积为 −24 且和为 −5 的两个数。−8 和 3 满足，因为 (−8) × 3 = −24，(−8) + 3 = −5。因此分解为 (x − 8)(x + 3)。
3. Simultaneous Equations | 联立方程组
Example: Solve the simultaneous equations 2x + y = 10 and x − y = 2.
例题：解联立方程组 2x + y = 10 与 x − y = 2。
Label the equations: (1) 2x + y = 10, (2) x − y = 2. Adding them eliminates y: (2x + y) + (x − y) = 10 + 2 → 3x = 12, so x = 4.
将方程标号：(1) 2x + y = 10，(2) x − y = 2。两式相加消去 y：(2x + y) + (x − y) = 10 + 2 → 3x = 12，因此 x = 4。
Substitute x = 4 into equation (2): 4 − y = 2 → y = 2.
把 x = 4 代入方程 (2)：4 − y = 2 → y = 2。
Check using the other equation: 2×4 + 2 = 8 + 2 = 10, which is correct. The solution is x = 4, y = 2.
用另一个方程检验：2×4 + 2 = 8 + 2 = 10，正确。解为 x = 4，y = 2。
If the coefficients do not match for easy elimination, you multiply one or both equations first. For example, solve 3x + 2y = 8 and 2x − y = 3. Multiply the second equation by 2 to get 4x − 2y = 6, then add to the first: 7x = 14, giving x = 2, then y = 2×2 − 3 = 1.
若系数不匹配、不容易直接消元，可以先将一个或两个方程乘以适当的数。例如，解 3x + 2y = 8 与 2x − y = 3。将第二个方程乘以 2 得到 4x − 2y = 6，再与第一个方程相加：7x = 14，得 x = 2，随后 y = 2×2 − 3 = 1。
4. Pythagoras’ Theorem | 勾股定理
Example: In a right‑angled triangle, the two shorter sides are 5 cm and 12 cm. Find the length of the hypotenuse.
例题：在一个直角三角形中，两条直角边分别为 5 cm 和 12 cm。求斜边的长度。
Pythagoras’ theorem states that a² + b² = c², where c is the hypotenuse. Substitute the known values: 5² + 12² = c² → 25 + 144 = c² → c² = 169.
勾股定理指出 a² + b² = c²，其中 c 为斜边。代入已知值：5² + 12² = c² → 25 + 144 = c² → c² = 169。
Take the square root: c = √169 = 13 cm. The hypotenuse is 13 cm.
开平方：c = √169 = 13 cm。斜边长为 13 cm。
If you need to find a shorter side, rearrange: a² = c² − b². For instance, a hypotenuse of 15 cm and one leg of 9 cm give a² = 15² − 9² = 225 − 81 = 144, so a = 12 cm. Always show the square‑root step clearly and include units.
若要求一条直角边，重新整理公式：a² = c² − b²。例如，斜边长 15 cm，一条直角边为 9 cm，则 a² = 15² − 9² = 225 − 81 = 144，因此 a = 12 cm。计算时务必清晰写出开方步骤并标明单位。
5. Trigonometry in Right‑Angled Triangles | 直角三角形三角学
Example: In a right‑angled triangle, an acute angle is 35°, and the side opposite this angle is 8 cm. Find the hypotenuse, labelled h.
例题：在一个直角三角形中，一个锐角为 35°，该角的对边长为 8 cm。求斜边（记为 h）。
The relationship between opposite and hypotenuse is given by sine: sin θ = opposite / hypotenuse. Substitute: sin 35° = 8 / h.
对边与斜边的关系由正弦给出：sin θ = 对边 / 斜边。代入：sin 35° = 8 / h。
Rearrange: h = 8 / sin 35°. Using a calculator, sin 35° ≈ 0.5736, so h ≈ 8 ÷ 0.5736 ≈ 13.9 cm (to 3 significant figures).
整理得：h = 8 / sin 35°。使用计算器，sin 35° ≈ 0.5736，因此 h ≈ 8 ÷ 0.5736 ≈ 13.9 cm（保留三位有效数字）。
For a question requiring cosine, e.g. adjacent = 10 cm, hypotenuse = 15 cm, find the angle: cos θ = 10 / 15 = 2/3, so θ = cos⁻¹(2/3) ≈ 48.2°. Remember to use the correct inverse function and check your calculator mode (degrees).
若题目需要使用余弦，例如邻边 = 10 cm，斜边 = 15 cm，求角度：cos θ = 10 / 15 = 2/3，因此 θ = cos⁻¹(2/3) ≈ 48.2°。注意使用正确的反函数并检查计算器是否处于角度模式。
6. Percentages and Compound Interest | 百分数与复利
Example: £800 is invested at 3% compound interest per annum for 4 years. Calculate the total amount after 4 years.
例题：800 英镑以年利率 3% 的复利投资 4 年。计算 4 年后的总金额。
The compound interest formula is A = P (1 + r/100)ⁿ, where P = 800, r = 3, n = 4. Substitute: A = 800 × (1.03)⁴.
复利公式为 A = P (1 + r/100)ⁿ，其中 P = 800，r = 3，n = 4。代入：A = 800 × (1.03)⁴。
Calculate stepwise: 1.03⁴ = (1.03²)². 1.03² = 1.0609, then (1.0609)² = 1.12550881. Multiply by 800: 800 × 1.12550881 = 900.407048, so A ≈ £900.41 (to the nearest penny).
逐步计算：1.03⁴ = (1.03²)²。1.03² = 1.0609，再平方 (1.0609)² = 1.12550881。乘以 800：800 × 1.12550881 = 900.407048，因此 A ≈ 900.41 英镑（精确到便士）。
For depreciation, use a minus sign: value = original × (1 − rate/100)ⁿ. If a car loses 15% of its value each year, the multiplier is 0.85. After 3 years, £12000 becomes £12000 × 0.85³ ≈ £7369.50.
对于折旧，使用减号：价值 = 原值 × (1 − 折旧率/100)ⁿ。若一辆车每年贬值 15%，则乘数为 0.85。3 年后，12000 英镑变为 12000 × 0.85³ ≈ 7369.50 英镑。
7. Probability Tree Diagrams | 概率树状图
Example: A bag contains 4 red and 6 blue counters. Two counters are drawn at random without replacement. Draw a tree diagram and find the probability that both counters are the same colour.
例题：一个袋子里有 4 个红色和 6 个蓝色筹码。随机抽取两个且不放回。画出树状图，并求两个筹码颜色相同的概率。
First pick: P(Red) = 4/10 = 2/5, P(Blue) = 6/10 = 3/5. If the first is red, 3 red and 6 blue remain, so P(Red|Red) = 3/9 = 1/3, P(Blue|Red) = 6/9 = 2/3. If the first is blue, 4 red and 5 blue remain, so P(Red|Blue) = 4/9, P(Blue|Blue) = 5/9.
第一次抽取：P(红) = 4/10 = 2/5，P(蓝) = 6/10 = 3/5。若第一次抽到红色，剩下 3 红 6 蓝，因此 P(红|红) = 3/9 = 1/3，P(蓝|红) = 6/9 = 2/3。若第一次抽到蓝色，剩下 4 红 5 蓝，因此 P(红|蓝) = 4/9，P(蓝|蓝) = 5/9。
Same colour outcomes: both red or both blue. Probability(both red) = (2/5) × (1/3) = 2/15. Probability(both blue) = (3/5) × (5/9) = 15/45 = 1/3. Total = 2/15 + 1/3 = 2/15 + 5/15 = 7/15.
颜色相同的结果：两个红色或两个蓝色。概率（两个红色）= (2/5) × (1/3) = 2/15。概率（两个蓝色）= (3/5) × (5/9) = 15/45 = 1/3。总和 = 2/15 + 1/3 = 2/15 + 5/15 = 7/15。
A tree diagram helps you check that all branch probabilities add up to 1, reducing careless errors.
树状图有助于检查所有分支概率之和是否为 1，从而减少粗心错误。
8. Scatter Graphs and Correlation | 散点图与相关性
Example: A scatter graph shows the number of hours of revision (x) and the test score (y) for 10 students. The points rise from lower left to upper right. Describe the correlation and draw a line of best fit. Use the line to estimate the score for a student who revises for 4.5 hours.
例题：一幅散点图显示了 10 名学生的复习小时数（x）与测试分数（y）之间的关系。数据点由左下方向右上方分布。描述相关性并画出最佳拟合线。利用此线估计一名复习 4.5 小时学生的分数。
The points show a positive correlation: as revision hours increase, test scores tend to increase. The correlation appears strong because the points lie close to a straight line.
这些点呈现出正相关：复习时间增加时，测试分数也倾向于提高。相关性看起来很强，因为这些点紧密分布在一条直线附近。
Draw a line of best fit that passes through the middle of the points, with roughly equal numbers of points above and below. Then, from x = 4.5 on the horizontal axis, read up to the line and across to the vertical axis to find an estimated score, e.g. 62 marks.
画一条最佳拟合线，使其穿过数据点的中间，并使得线上方和下方的点数大致相等。然后，在横轴上找到 x = 4.5，向上读到直线，再水平读到纵轴，得到估计分数，例如 62 分。
Avoid forcing the line through the origin unless it makes sense. Remember that extrapolation beyond the data range is unreliable, so limit predictions to the given interval.
除非合理，否则不要强迫直线经过原点。请记住，对数据范围之外的推断不可靠，因此预测应限制在给定区间内。
9. Cumulative Frequency and Box Plots | 累积频数与箱线图
Example: The grouped frequency table shows the masses (kg) of 50 parcels: 0 < m ≤ 2 (frequency 6), 2 < m ≤ 4 (10), 4 < m ≤ 6 (15), 6 < m ≤ 8 (12), 8 < m ≤ 10 (7). Construct a cumulative frequency table and use it to estimate the median and interquartile range. Then draw a box plot.
例题：分组频率表显示 50 个包裹的质量（kg）：0 < m ≤ 2（频数 6），2 < m ≤ 4（10），4 < m ≤ 6（15），6 < m ≤ 8（12），8 < m ≤ 10（7）。建立累积频数表，并据此估计中位数和四分位距。然后绘制箱线图。
Cumulative frequencies: ≤2: 6; ≤4: 16; ≤6: 31; ≤8: 43; ≤10: 50. The median position is at (50+1)/2 = 25.5th value, so we can use 25.5 or the 25th and 26th. From the table, the 25.5th value lies in the 4 < m ≤ 6 interval. Using linear interpolation: median ≈ 4 + (25.5 − 16) / (31 − 16) × 2 = 4 + (9.5/15) × 2 ≈ 4 + 1.267 = 5.27 kg.
累积频数：≤2：6；≤4：16；≤6：31；≤8：43；≤10：50。中位数的位置在 (50+1)/2 = 25.5 个值，我们可以用 25.5 或第 25 和 26 个值。由表可知，第 25.5 个值落在 4 < m ≤ 6 区间。使用线性插值：中位数 ≈ 4 + (25.5 − 16) / (31 − 16) × 2 = 4 + (9.5/15) × 2 ≈ 4 + 1.267 = 5.27 kg。
Lower quartile (Q1) at 25% of 50 = 12.5th value: falls in 2 < m ≤ 4. Q1 ≈ 2 + (12.5 − 6)/(16 − 6) × 2 = 2 + 6.5/10 × 2 = 2 + 1.3 = 3.3 kg. Upper quartile (Q3) at 37.5th value: falls in 6 < m ≤ 8. Q3 ≈ 6 + (37.5 − 31)/(43 − 31) × 2 = 6 + 6.5/12 × 2 ≈ 6 + 1.083 = 7.08 kg. IQR = 7.08 − 3.3 = 3.78 kg.
下四分位数（Q1）位于 50 的 25% = 第 12.5 个值：落在 2 < m ≤ 4 区间。Q1 ≈ 2 + (12.5 − 6)/(16 − 6) × 2 = 2 + 6.5/10 × 2 = 2 + 1.3 = 3.3 kg。上四分位数（Q3）位于第 37.5 个值：落在 6 < m ≤ 8 区间。Q3 ≈ 6 + (37.5 − 31)/(43 − 31) × 2 = 6 + 6.5/12 × 2 ≈ 6 + 1.083 = 7.08 kg。四分位距 IQR = 7.08 − 3.3 = 3.78 kg。
Box plot: draw a scale, plot points for minimum (0), Q1 (3.3), median (5.27), Q3 (7.08), maximum (10), and connect them to form the box and whiskers.
箱线图：画出刻度，标出最小值 (0)、Q1 (3.3)、中位数 (5.27)、Q3 (7.08)、最大值 (10) 各点，并将它们连接成箱体和须线。
10. Ratio and Proportion | 比和比例
Example: The ingredients for 8 pancakes are 200 g flour, 2 eggs and 300 ml milk. How much of each ingredient is needed for 20 pancakes?
例题：制作 8 个煎饼需要 200 g 面粉、2 个鸡蛋和 300 ml 牛奶。制作 20 个煎饼需要每种配料各多少？
The scaling factor is 20 ÷ 8 = 2.5. Multiply each quantity by 2.5: flour = 200 × 2.5 = 500 g, eggs = 2 × 2.5 = 5, milk = 300 × 2.5 = 750 ml.
比例缩放因子为 20 ÷ 8 = 2.5。每种配料的量乘以 2.5：面粉 = 200 × 2.5 = 500 g，鸡蛋 = 2 × 2.5 = 5 个，牛奶 = 300 × 2.5 = 750 ml。
Another common OCR question is dividing a quantity in a given ratio. Divide £180 between two people in the ratio 2:3. Total parts = 2 + 3 = 5. One part is £180 ÷ 5 = £36. Therefore the shares are 2 × £36 = £72 and 3 × £36 = £108.
另一种常见的 OCR 题型是按给定比例分配数量。按 2:3 的比例把 180 英镑分给两个人。总份数 = 2 + 3 = 5。一份为 £180 ÷ 5 = £36。因此所得分别为 2 × £36 = £72 和 3 × £36 = £108。
For map scales, e.g. a scale of 1:25000, 1 cm on the map represents 25000 cm = 250 m in real life. A road measuring 7.5 cm on the map would cover 7.5 × 250 = 1875 m, or 1.875 km. Always convert to sensible units.
对于地图比例尺，例如 1:25000，图上 1 cm 代表实际 25000 cm = 250 m。图上一条 7.5 cm 的道路实际长度为 7.5 × 250 = 1875 m，即 1.875 km。注意转换为合理的单位。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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