Year 10 Cambridge Maths: Unit Test Mock Paper Walkthrough | 剑桥10年级数学：单元测试模拟卷解析

📚 Year 10 Cambridge Maths: Unit Test Mock Paper Walkthrough | 剑桥10年级数学：单元测试模拟卷解析

This article walks you through a full mock paper designed for a typical Year 10 Cambridge Mathematics unit test, covering algebra, equations, sequences, ratio, coordinate geometry and statistics. Each question is presented together with a step-by-step solution, modelling the thinking and working needed to secure top marks. Use this as a revision tool to sharpen your skills and build confidence before the real test.

本文完整解析一份为剑桥10年级数学单元测试设计的模拟试卷，涵盖代数、方程、数列、比、坐标几何和统计。每道题目都配有逐步求解过程，展示了获取高分所需的思考方式和书写步骤。将此解析用作复习工具，帮助你打磨技巧并在真正的测试前建立信心。

1. Solving a Linear Equation | 解一元一次方程

Question: Solve 3x + 5 = 20.

题目：解方程 3x + 5 = 20。

ct 5 from both sides of the equation to isolate the term containing x. This gives 3x = 15.

步骤1：方程两边同时减去5，使含x的项单独在一边，得到 3x = 15。

Step 2: Divide both sides by 3 to solve for x. 3x ÷ 3 = 15 ÷ 3, therefore x = 5.

步骤2：两边同时除以3，解得 x。3x ÷ 3 = 15 ÷ 3，因此 x = 5。

inal equation: 3×5 + 5 = 15 + 5 = 20, which satisfies the equation. The solution is x = 5.

检验：将 x = 5 代回原方程：3×5 + 5 = 15 + 5 = 20，等式成立。解为 x = 5。

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2. Solving a Quadratic Equation by Factorising | 用因式分解解二次方程

Question: Solve x² − 5x + 6 = 0.

题目：解方程 x² − 5x + 6 = 0。

6 and add to −5. The numbers are −2 and −3. So we factorise: x² − 5x + 6 = (x − 2)(x − 3).

步骤1：寻找乘积为 +6、和为 −5 的两个数。这两个数是 −2 和 −3。于是因式分解：x² − 5x + 6 = (x − 2)(x − 3)。

Step 2: Set each factor equal to zero. x − 2 = 0 gives x = 2. x − 3 = 0 gives x = 3.

步骤2：令每个因式等于零。x − 2 = 0 得 x = 2。x − 3 = 0 得 x = 3。

Both values satisfy the original equation. The solution set is x = 2 or x = 3.

两个值均满足原方程。解集为 x = 2 或 x = 3。

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3. Expanding Brackets | 去括号展开

Question: Expand and simplify (2x + 3)(x − 4).

题目：展开并化简 (2x + 3)(x − 4)。

ket. First, 2x × x = 2x². Then 2x × (−4) = −8x. Next, 3 × x = 3x. Finally, 3 × (−4) = −12.

步骤1：将第一个括号中的每一项分别乘以第二个括号中的每一项。首先 2x × x = 2x²。然后 2x × (−4) = −8x。接着 3 × x = 3x。最后 3 × (−4) = −12。

Step 2: Combine like terms. −8x + 3x = −5x. Write the result: 2x² − 5x − 12.

步骤2：合并同类项。−8x + 3x = −5x。写出结果：2x² − 5x − 12。

The expanded and simplified expression is 2x² − 5x − 12.

展开化简后的表达式为 2x² − 5x − 12。

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4. Factorising the Difference of Two Squares | 平方差因式分解

Question: Factorise completely x² − 9.

题目：对 x² − 9 进行完全因式分解。

so x² − 9 is a difference of two squares. It follows the pattern a² − b² = (a − b)(a + b).

步骤1：识别出9是3的平方，因此 x² − 9 是平方差，符合公式 a² − b² = (a − b)(a + b)。

Step 2: Apply the pattern with a = x and b = 3. x² − 9 = (x − 3)(x + 3).

步骤2：套用公式，取 a = x, b = 3。x² − 9 = (x − 3)(x + 3)。

No further common factors exist, so this is the complete factorisation.

没有其他公因子，这就是完全因式分解。

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5. Solving Simultaneous Equations | 解联立方程组

Question: Solve the simultaneous equations: y = 2x + 1 and 3x + y = 14.

题目：解联立方程组：y = 2x + 1 与 3x + y = 14。

he second equation. Replace y in 3x + y = 14 with (2x + 1). This gives 3x + (2x + 1) = 14.

步骤1：将第一个方程中y的表达式代入第二个方程。把 3x + y = 14 中的 y 换成 (2x + 1)，得到 3x + (2x + 1) = 14。

p 2: Simplify and solve for x. 3x + 2x + 1 = 14 → 5x + 1 = 14 → 5x = 13 → x = 13/5 or 2.6.

步骤2：化简并解出 x。3x + 2x + 1 = 14 → 5x + 1 = 14 → 5x = 13 → x = 13/5 或 2.6。

tute x back into the first equation to find y. y = 2(13/5) + 1 = 26/5 + 5/5 = 31/5 or 6.2.

步骤3：将 x 代回第一个方程求 y。y = 2(13/5) + 1 = 26/5 + 5/5 = 31/5 或 6.2。

The solution is x = 13/5 (2.6), y = 31/5 (6.2).

解为 x = 13/5（或2.6），y = 31/5（或6.2）。

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6. Solving a Linear Inequality | 解一元一次不等式

Question: Solve 4x − 7 ≤ 5 and represent the solution on a number line.

题目：解不等式 4x − 7 ≤ 5，并在数轴上表示解集。

Step 1: Add 7 to both sides: 4x ≤ 12.

步骤1：两边加7：4x ≤ 12。

Step 2: Divide both sides by 4: x ≤ 3.

步骤2：两边除以4：x ≤ 3。

Step 3: On the number line, draw a solid circle at 3 to show it is included, and shade the region to the left extending to negative infinity. The inequality is read as 'x is less than or equal to 3'.

步骤3：在数轴上，用实心圆点标出3表示包含该点，并将左侧区域涂上阴影，延伸至负无穷。此不等式读作“x小于或等于3”。

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7. Finding Terms of a Linear Sequence | 求线性数列的项

Question: a sequence is given by uₙ = 2n + 3. Find (a) the 10th term, (b) which term is equal to 29.

题目：某数列的第n项公式为 uₙ = 2n + 3。求 (a) 第10项；(b) 哪一项等于29。

For part (a): Substitute n = 10 into the formula. u₁₀ = 2(10) + 3 = 20 + 3 = 23.

第(a)部分：将 n = 10 代入公式。u₁₀ = 2(10) + 3 = 20 + 3 = 23。

For part (b): Set 2n + 3 equal to 29 and solve. 2n + 3 = 29 → 2n = 26 → n = 13.

第(b)部分：令 2n + 3 = 29 并求解。2n + 3 = 29 → 2n = 26 → n = 13。

Therefore, the 10th term is 23 and 29 is the 13th term in the sequence.

因此，第10项是23，29是该数列的第13项。

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8. Combining Ratios | 组合比

Question: Given a : b = 3 : 4 and b : c = 2 : 5, find the ratio a : b : c in its simplest form.

题目：已知 a : b = 3 : 4 且 b : c = 2 : 5，求 a : b : c 的最简整数比。

4 and 2 is 4. Multiply the second ratio by 2 to make b = 4: b : c = 2 : 5 becomes 4 : 10.

步骤1：写出比，使共同项b连接两者。让两个比中的b值相同。当前b值为4和2，最小公倍数为4。将第二个比乘以2使b = 4：b : c = 2 : 5 变为 4 : 10。

Now we have a : b = 3 : 4 and b : c = 4 : 10. Combine them to get a : b : c = 3 : 4 : 10.

步骤2：现在有 a : b = 3 : 4 和 b : c = 4 : 10。合并得到 a : b : c = 3 : 4 : 10。

The ratio is already in its simplest form, as 3, 4 and 10 share no common factor other than 1.

该比已是最简形式，因为3、4和10除了1外没有其他公因子。

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9. Equation of a Straight Line from Two Points | 由两点求直线方程

Question: ne : Find the equation of the straight line passing through the points (1, 3) and (3, 7). Write your answer in the form y = mx + c.

题目：求经过点(1, 3)和(3, 7)的直线方程，并以 y = mx + c 的形式表示。

) / (x₂ − x₁). Here, y₂ = 7, y₁ = 3, x₂ = 3, x₁ = 1. So m = (7 − 3) / (3 − 1) = 4 / 2 = 2.

步骤1：用公式 (y₂ − y₁) / (x₂ − x₁) 计算斜率 m。这里 y₂ = 7, y₁ = 3, x₂ = 3, x₁ = 1。则 m = (7 − 3) / (3 − 1) = 4 / 2 = 2。

, say (1, 3), into y = mx + c to find the y-intercept c. 3 = 2(1) + c → 3 = 2 + c → c = 1.

步骤2：将斜率 m = 2 和其中一个点（如(1, 3)）代入 y = mx + c，求截距 c。3 = 2(1) + c → 3 = 2 + c → c = 1。

Step 3: Write the full equation: y = 2x + 1.

步骤3：写出完整方程：y = 2x + 1。

You can verify with the second point: when x = 3, y = 2(3) + 1 = 7, which is correct.

可用第二个点验证：当 x = 3 时，y = 2(3) + 1 = 7，结果正确。

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10. Mean and Mode of a Data Set | 一组数据的平均数与众数

Question: Eleven students recorded their test scores: 3, 7, 5, 7, 9, 6, 7, 8, 5, 7, 4. Find (a) the mean, (b) the mode.

题目：11名学生的考试成绩为：3, 7, 5, 7, 9, 6, 7, 8, 5, 7, 4。求 (a) 平均数；(b) 众数。

3+7+5+7+9+6+7+8+5+7+4 = 68. There are 11 scores, so mean = 68 ÷ 11 = 68/11 or about 6.18.

平均数：将所有分数相加。总和 = 3+7+5+7+9+6+7+8+5+7+4 = 68。共有11个分数，因此平均数 = 68 ÷ 11 = 68/11（约6.18）。

For the mode: Look for the score that appears most frequently. The score 7 occurs four times (more than any other). Therefore, the mode is 7.

众数：找出出现频率最高的分数。分数7出现了四次（多于其他任何分数）。因此众数为7。

The mean gives an average performance, while the mode indicates the most common score in the class.

平均数体现了整体水平，而众数则指出了班级中最常见的成绩。

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