Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

📚 Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

This walkthrough takes you through a typical Year 10 Eduqas Mathematics unit test, covering key topics from number, algebra, geometry, statistics and probability. Each question is broken down step by step so you can see exactly how marks are earned. Use this guide to build confidence and sharpen your exam technique.

本次解析带你完成一份典型的 Year 10 Eduqas 数学单元测试模拟卷，涵盖数、代数、几何、统计与概率等重要主题。每个题目都逐步拆解，让你清楚看到如何得分。请用这份指南建立信心、打磨应试技巧。

1. Question 1: Simplifying Expressions | 问题1：化简表达式

Simplify 3(x + 2) – 2(x – 1). First expand both brackets: 3x + 6 – 2x + 2. Then collect like terms: 3x – 2x gives x, and 6 + 2 gives 8. The final simplified expression is x + 8.

化简 3(x + 2) – 2(x – 1)。首先展开括号：3x + 6 – 2x + 2。接着合并同类项：3x – 2x 得 x，6 + 2 得 8。最终化简结果为 x + 8。

2. Question 2: Solving Linear Equations | 问题2：解一元一次方程

Solve 5x – 3 = 2x + 9. Subtract 2x from both sides to get 3x – 3 = 9. Add 3 to both sides: 3x = 12. Divide by 3: x = 4. Always check by substituting back: 5(4) – 3 = 20 – 3 = 17; 2(4) + 9 = 8 + 9 = 17. Correct.

解方程 5x – 3 = 2x + 9。两边同时减去 2x，得 3x – 3 = 9。两边同时加 3：3x = 12。两边除以 3：x = 4。代回检验：5(4) – 3 = 17，2(4) + 9 = 17，一致。

3. Question 3: Solving Inequalities and Number Line | 问题3：解不等式并在数轴上表示

Solve 4 – 3x > 7. Subtract 4 from both sides: -3x > 3. Divide both sides by -3, remembering to reverse the inequality sign: x < -1. On a number line, draw an open circle at -1 and shade to the left.

解不等式 4 – 3x > 7。两边减4：-3x > 3。两边除以 -3，注意不等号方向反转：x < -1。在数轴上，在 -1 处画空心圆，并向左画阴影线表示 x 小于 -1。

4. Question 4: Plotting a Linear Graph | 问题4：绘制一次函数图像

Draw the graph of y = 2x + 1 for x from -2 to 3. Create a table of values: when x = -2, y = -3; x = -1, y = -1; x = 0, y = 1; x = 1, y = 3; x = 2, y = 5; x = 3, y = 7. Plot these points and draw a straight line. The y-intercept is (0,1) and the line crosses the x-axis when y = 0: 0 = 2x + 1 ⇒ x = -½, so at (-0.5, 0).

绘制 y = 2x + 1 在 x 从 -2 到 3 的图像。先列表：x = -2 时 y = -3；x = -1 时 y = -1；x = 0 时 y = 1；x = 1 时 y = 3；x = 2 时 y = 5；x = 3 时 y = 7。描点连成直线。y 轴截距为 (0,1)；x 轴截距通过令 y = 0 求解：0 = 2x + 1 → x = -½，即 (-0.5, 0)。

5. Question 5: Percentage Decrease | 问题5：百分比减少

An item originally costs £80. It is reduced by 15%. Find the sale price. Method 1: 15% of £80 = 0.15 × 80 = £12. Sale price = 80 – 12 = £68. Method 2: A 15% decrease leaves 85% of the original. 0.85 × 80 = £68. Both methods give the same answer.

一件商品原价 £80，降价 15%，求现价。方法一：£80 的 15% 为 0.15 × 80 = £12。现价 = 80 – 12 = £68。方法二：降价 15% 后剩余原价的 85%。0.85 × 80 = £68。两种方法结果一致。

6. Question 6: Right-Angled Triangle – Pythagoras and Area | 问题6：直角三角形 – 勾股定理与面积

A right-angled triangle has hypotenuse 10 cm and one leg 6 cm. Find the other leg and the area of the triangle. Using Pythagoras’ theorem: a² + b² = c² → 6² + b² = 10² → 36 + b² = 100 → b² = 64 → b = √64 = 8 cm. Area = ½ × base × height = ½ × 6 × 8 = 24 cm².

直角三角形斜边长 10 cm，一直角边长 6 cm。求另一直角边和三角形面积。应用勾股定理：a² + b² = c² → 6² + b² = 10² → 36 + b² = 100 → b² = 64 → b = √64 = 8 cm。面积 = ½ × 底 × 高 = ½ × 6 × 8 = 24 cm²。

7. Question 7: Probability Without Replacement | 问题7：不放回概率

A bag contains 3 red and 5 blue counters. Two counters are drawn at random without replacement. Find the probability that at least one is red. It is easier to find the probability of the complementary event – both blue. P(both blue) = (5/8) × (4/7) = 20/56 = 5/14. So P(at least one red) = 1 – 5/14 = 9/14.

袋中有 3 红、5 蓝共 8 枚筹码。随机抽取两枚，不放回。求至少有一枚红色的概率。先求对立事件 – 两枚都是蓝色的概率。P(双双蓝) = (5/8) × (4/7) = 20/56 = 5/14。因此 P(至少一红) = 1 – 5/14 = 9/14。

8. Question 8: Mean from a Frequency Table | 问题8：频数表求平均值

Frequency table: Score 1 (freq 3), Score 2 (freq 5), Score 3 (freq 7), Score 4 (freq 4), Score 5 (freq 1). To find the mean, multiply each score by its frequency: 1×3 = 3, 2×5 = 10, 3×7 = 21, 4×4 = 16, 5×1 = 5. Sum of fx = 3+10+21+16+5 = 55. Total frequency = 3+5+7+4+1 = 20. Mean = 55 ÷ 20 = 2.75.

频数表：得分 1（频数 3）、2（频数 5）、3（频数 7）、4（频数 4）、5（频数 1）。求平均值：得分×频数：1×3=3，2×5=10，3×7=21，4×4=16，5×1=5。fx 总和 = 55。总频数 = 20。平均值 = 55 ÷ 20 = 2.75。

9. Question 9: Ratio Sharing | 问题9：按比例分配

Share £120 in the ratio 2 : 3 : 5. Total number of parts = 2 + 3 + 5 = 10. Value of one part = £120 ÷ 10 = £12. So the three shares are: 2 × £12 = £24, 3 × £12 = £36, and 5 × £12 = £60. Check: £24 + £36 + £60 = £120.

将 £120 按 2 : 3 : 5 分配。总份数 = 2 + 3 + 5 = 10。每份价值 = £120 ÷ 10 = £12。因此三人所得分别为：2 × £12 = £24，3 × £12 = £36，5 × £12 = £60。验算：24 + 36 + 60 = 120。

10. Question 10: Angles in Parallel Lines | 问题10：平行线中的角度

Two parallel lines are cut by a transversal. One of the alternate interior angles is given as 75°. State the size of the corresponding angle and the co-interior angle on the same side. The corresponding angle is equal to 75°. Co-interior angles are supplementary: 180° – 75° = 105°. Also, vertically opposite angles are equal.

两条平行线被一条截线所截，已知一个内错角为 75°。指出同位角的大小和同旁内角的大小。同位角与已知角相等，为 75°。同旁内角互补：180° – 75° = 105°。对顶角也相等。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析