📚 Year 11 OCR Maths: Unit Test Mock Paper Walkthrough | 11年级OCR数学：单元测试模拟卷解析
Mock unit tests are an excellent way to consolidate your understanding of key GCSE Maths topics and build exam confidence. This walkthrough will guide you through a typical Year 11 OCR unit test paper covering algebra, geometry, trigonometry, statistics, and probability. Each question is broken down step by step, with detailed explanations so you can master the methods required for the real exam.
模拟单元测试是巩固GCSE数学关键知识、建立考试信心的绝佳方式。本文将通过一份典型的11年级OCR单元测试卷，带你梳理代数、几何、三角学、统计和概率等重点内容。每道题都配有逐步拆解和详细解析，帮助你掌握考试必备的解题方法。
1. Solving a Linear Equation | 解一元一次方程
The first question tests your ability to solve a linear equation with brackets. Solve: 4(2x – 3) = 5x + 12.
第一题考查解含有括号的一元一次方程。解方程：4(2x – 3) = 5x + 12。
4(2x – 3) = 5x + 12
Expand the left-hand side by multiplying each term inside the bracket by 4: 4 × 2x = 8x and 4 × -3 = -12. The equation becomes 8x – 12 = 5x + 12.
展开左边，用4乘以括号内每一项：4 × 2x = 8x，4 × -3 = -12。方程变为 8x – 12 = 5x + 12。
Collect like terms by subtracting 5x from both sides: 8x – 5x – 12 = 5x – 5x + 12, which simplifies to 3x – 12 = 12.
移项合并同类项，两边同时减去5x：8x – 5x – 12 = 5x – 5x + 12，化简得 3x – 12 = 12。
Add 12 to both sides: 3x – 12 + 12 = 12 + 12, giving 3x = 24.
两边同时加12：3x – 12 + 12 = 12 + 12，得 3x = 24。
Finally, divide by 3: x = 24 ÷ 3 = 8. The solution is x = 8.
最后，两边除以3：x = 24 ÷ 3 = 8。解为 x = 8。
2. Factorising a Quadratic Expression | 因式分解二次三项式
Factorising quadratics is a core algebra skill. Factorise x² – 5x – 14.
因式分解二次式是代数核心技能。请分解 x² – 5x – 14。
x² – 5x – 14
We need two numbers that multiply to -14 and add to -5. The numbers -7 and 2 satisfy this because -7 × 2 = -14 and -7 + 2 = -5.
我们需要两个数，乘积为 -14，和为 -5。数字 -7 和 2 满足条件，因为 -7 × 2 = -14，-7 + 2 = -5。
Write the quadratic as two brackets: (x – 7)(x + 2). You can check by expanding: x(x + 2) – 7(x + 2) = x² + 2x – 7x – 14 = x² – 5x – 14, which matches the original expression.
将二次式写成两个括号乘积：(x – 7)(x + 2)。可以通过展开验证：x(x + 2) – 7(x + 2) = x² + 2x – 7x – 14 = x² – 5x – 14，与原式一致。
3. Using Trigonometry to Find a Missing Side | 利用三角函数求未知边长
In a right-angled triangle, the angle is 37°, the opposite side is 5 cm, and the hypotenuse is unknown. Use trigonometry to find the hypotenuse length to 1 decimal place.
在直角三角形中，已知一个角为37°，对边长为5 cm，求斜边长，结果保留1位小数。
sin θ = opposite / hypotenuse
Since we have the opposite side and need the hypotenuse, we use the sine ratio: sin 37° = 5 / h.
已知对边求斜边，使用正弦函数：sin 37° = 5 / h。
Rearrange to make h the subject: h = 5 / sin 37°. Using a calculator, sin 37° ≈ 0.6018, so h ≈ 5 / 0.6018 ≈ 8.308… Rounded to 1 decimal place, h ≈ 8.3 cm.
将h作为未知量表示：h = 5 / sin 37°。用计算器算出 sin 37° ≈ 0.6018，因此 h ≈ 5 / 0.6018 ≈ 8.308…。四舍五入到1位小数，h ≈ 8.3 cm。
4. Area and Circumference of a Circle | 圆的面积与周长
A circle has a radius of 7 cm. Find its area and circumference, giving your answers in terms of π.
一个圆的半径为7 cm。求它的面积和周长，答案用π表示。
C = 2πr, A = πr²
Circumference formula: C = 2πr = 2 × π × 7 = 14π cm.
周长公式：C = 2πr = 2 × π × 7 = 14π cm。
Area formula: A = πr² = π × 7² = 49π cm². Always include the correct units.
面积公式：A = πr² = π × 7² = 49π cm²。务必写对单位。
5. Direct Proportion | 正比例关系
y is directly proportional to x. When x = 8, y = 20. Find y when x = 14.
y 与 x 成正比。已知 x = 8 时 y = 20，求 x = 14 时 y 的值。
y = kx
Write the direct proportion equation: y = kx, where k is the constant of proportionality. Substitute the known pair: 20 = k × 8, so k = 20 / 8 = 2.5.
写出正比例关系式：y = kx，其中 k 为比例常数。代入已知值：20 = k × 8，得 k = 20 / 8 = 2.5。
Now use the constant to find y when x = 14: y = 2.5 × 14 = 35.
用比例常数求 x = 14 时的 y：y = 2.5 × 14 = 35。
6. Cumulative Frequency and the Median | 累积频率与中位数
The table shows the heights of 80 students. Use it to construct a cumulative frequency graph and estimate the median height.
表格显示了80名学生的身高数据。请据此绘制累积频率图并估算身高中位数。
| Height, h (cm) | Frequency | Cumulative Frequency |
|---|---|---|
| 150 ≤ h < 155 | 8 | 8 |
| 155 ≤ h < 160 | 14 | 22 |
| 160 ≤ h < 165 | 22 | 44 |
| 165 ≤ h < 170 | 20 | 64 |
| 170 ≤ h < 175 | 16 | 更多咨询请联系16621398022（同微信）
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