Year 8 SQA Advanced Mathematics: Unit Test Mock Paper Analysis | Year 8 进阶数学单元测试模拟卷解析

📚 Year 8 SQA Advanced Mathematics: Unit Test Mock Paper Analysis | Year 8 进阶数学单元测试模拟卷解析

This article provides a detailed walkthrough of a complete unit test mock paper for Year 8 SQA Advanced Mathematics. Each question is carefully selected to cover key topics, including algebra, geometry, ratio, speed, indices, statistics, factorising, inequalities, sequences and probability. You will see full solutions with step-by-step reasoning, helping you to consolidate your understanding and boost your confidence before the real assessment.

本文详细解析了一份完整的 Year 8 SQA 进阶数学单元测试模拟卷。每道题目都经过精心挑选，覆盖了代数、几何、比和比例、速度、指数、统计、因式分解、不等式、数列和概率等核心主题。你将看到完整的解答和逐步推理过程，这有助于巩固理解并在正式评估前提升自信心。

1. Solving Linear Equations | 解一元一次方程

Question: Solve for x: 3(2x - 4) = 18 + 2x.

问题：求解 x：3(2x - 4) = 18 + 2x。

Step 1: Expand the left-hand side. Multiply each term inside the bracket by 3. 3 × 2x gives 6x and 3 × (-4) gives -12. The equation becomes 6x - 12 = 18 + 2x.

第1步：展开左侧。将括号内的每一项乘以 3。3 × 2x 得 6x，3 × (-4) 得 -12。方程变为 6x - 12 = 18 + 2x。

Step 2: Collect like terms. Subtract 2x from both sides to bring all x-terms to the left: 6x - 2x - 12 = 18. This simplifies to 4x - 12 = 18.

第2步：合并同类项。两边同时减去 2x，将所有含 x 的项移到左侧：6x - 2x - 12 = 18，简化为 4x - 12 = 18。

Step 3: Isolate the x-term. Add 12 to both sides: 4x = 30.

第3步：分离含 x 的项。两边加 12：4x = 30。

Step 4: Solve for x. Divide both sides by 4: x = 30 ÷ 4 = 7.5.

第4步：解出 x。两边除以 4：x = 30 ÷ 4 = 7.5。

Therefore, the solution is x = 7.5. Always check your answer by substituting back into the original equation.

因此，解为 x = 7.5。务必通过将解代回原方程进行检验。

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2. Angles in a Triangle | 三角形的内角

Question: In a triangle, two of the interior angles measure 45° and 70°. Calculate the size of the third angle.

问题：在一个三角形中，两个内角分别是 45° 和 70°。求第三个角的度数。

Step 1: Recall the angle sum property. The three interior angles of any triangle always add up to 180°.

第1步：回顾内角和性质。任何三角形的三个内角之和总是 180°。

Step 2: Add the known angles. 45° + 70° = 115°.

第2步：将已知角相加。45° + 70° = 115°。

Step 3: Subtract from 180° to find the missing angle. 180° - 115° = 65°.

第3步：从 180° 中减去这个和，求出未知角。180° - 115° = 65°。

The third angle is 65°. This quick method works for any triangle – just remember the total is always 180°.

第三个角是 65°。这种快速方法适用于任何三角形——只需记住总和永远是 180°。

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3. Ratio and Proportion | 比和比例

Question: The ratio of boys to girls in a class is 3 : 5. If there are 12 boys, how many girls are there?

问题：一个班级里男生和女生的人数比是 3 : 5。如果有 12 名男生，女生有多少人？

Step 1: Set up the proportion. Let the number of girls be g. The ratio can be written as boys/girls = 3/5, so 12/g = 3/5.

第1步：建立比例式。设女生人数为 g。比例可写为男生/女生 = 3/5，因此 12/g = 3/5。

Step 2: Cross-multiply to solve for g. 12 × 5 = 3 × g → 60 = 3g.

第2步：交叉相乘求解 g。12 × 5 = 3 × g → 60 = 3g。

Step 3: Divide both sides by 3. g = 60 ÷ 3 = 20.

第3步：两边除以 3。g = 60 ÷ 3 = 20。

There are 20 girls in the class. Notice that multiplying the ratio 3:5 by 4 gives 12:20, which confirms our answer.

班里有 20 名女生。请注意，将比 3:5 乘以 4 得到 12:20，这验证了我们的答案。

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4. Speed, Distance and Time | 速度、距离和时间

Question: A car travels 240 km in 3 hours. Calculate its average speed in km/h. Then convert this speed into metres per second (m/s).

问题：一辆汽车在 3 小时内行驶了 240 公里。请计算它的平均速度（单位 km/h），然后将该速度转换为米每秒（m/s）。

d in km/h. Average speed = Total distance ÷ Total time. So speed = 240 km ÷ 3 h = 80 km/h.

第1步：求以 km/h 为单位的平均速度。平均速度 = 总距离 ÷ 总时间。因此速度 = 240 km ÷ 3 h = 80 km/h。

m/s. 1 km = 1000 m and 1 hour = 3600 seconds. Therefore, 80 km/h = 80 × (1000 m / 3600 s).

第2步：将 km/h 转换为 m/s。1 km = 1000 m，1 小时 = 3600 秒。因此，80 km/h = 80 × (1000 m / 3600 s)。

0 ÷ 400 = 200, 3600 ÷ 400 = 9. So the speed is 200/9 m/s, which is approximately 22.2 m/s.

第3步：化简分数。80 × 1000 / 3600 = 80000 / 3600。分子分母同时除以 400 进行化简：80000 ÷ 400 = 200，3600 ÷ 400 = 9。所以速度为 200/9 m/s，约等于 22.2 m/s。

The car's average speed is 80 km/h, or 200/9 m/s (≈ 22.2 m/s).

汽车的平均速度是 80 km/h，或 200/9 m/s（约 22.2 m/s）。

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5. Laws of Indices | 指数运算法则

Question: Simplify (2³ × 2⁴) ÷ 2⁵ and give your answer as a number.

问题：化简 (2³ × 2⁴) ÷ 2⁵，并将答案写成一个数。

le for indices. When multiplying powers with the same base, add the exponents: 2³ × 2⁴ = 2³⁺⁴ = 2⁷.

第1步：应用指数乘法法则。同底数幂相乘，指数相加：2³ × 2⁴ = 2³⁺⁴ = 2⁷。

Step 2: Apply the division rule. When dividing, subtract the exponents: 2⁷ ÷ 2⁵ = 2^7-5 = 2².

第2步：应用除法法则。同底数幂相除，指数相减：2⁷ ÷ 2⁵ = 2^7-5 = 2²。

Step 3: Evaluate the resulting power. 2² = 4.

第3步：计算最终的幂。2² = 4。

The simplified result is 4. Using index laws saves time and avoids multiplying out large numbers.

化简后的结果是 4。使用指数法则可以节省时间，避免展开大数相乘。

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6. Median and Range | 中位数和极差

Question: The following set of data represents the scores of seven students: 12, 7, 9, 15, 11, 8, 14. Find the median and the range.

问题：以下数据集表示七名学生的分数：12, 7, 9, 15, 11, 8, 14。求中位数和极差。

Step 1: Order the data from smallest to largest. The ordered list is: 7, 8, 9, 11, 12, 14, 15.

第1步：将数据从小到大排序。排序后的列表为：7, 8, 9, 11, 12, 14, 15。

Step 2: Find the median. Because there are 7 values (an odd number), the median is the middle value, i.e. the 4th term. The 4th term is 11, so the median is 11.

第2步：求中位数。因为有 7 个值（奇数个），中位数是中间的那个值，即第 4 项。第 4 项是 11，所以中位数是 11。

Step 3: Calculate the range. Range = largest value - smallest value = 15 - 7 = 8.

第3步：计算极差。极差 = 最大值 - 最小值 = 15 - 7 = 8。

The median is 11 and the range is 8. The median tells you the centre of the data, while the range shows how spread out the values are.

中位数为 11，极差为 8。中位数反映数据的中心位置，而极差显示数值的分散程度。

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7. Factorising Algebraic Expressions | 代数式的因式分解

Question: Factorise fully: 4x² - 8x.

问题：完全因式分解：4x² - 8x。

Step 1: Identify the greatest common factor (GCF) of the two terms. The coefficients 4 and 8 share a GCF of 4. Both terms also contain at least one factor of x. So the GCF is 4x.

第1步：找出两项的最大公因子（GCF）。系数 4 和 8 的最大公因子是 4。两项都至少包含一个 x 因子。因此最大公因子是 4x。

Step 2: Divide each term by the GCF. 4x² ÷ 4x = x, and -8x ÷ 4x = -2.

第2步：将每一项除以最大公因子。4x² ÷ 4x = x，-8x ÷ 4x = -2。

Step 3: Write the expression as the product of the GCF and the remaining terms inside brackets. This gives 4x(x - 2).

第3步：将表达式写成最大公因子与括号内剩余项的乘积。得到 4x(x - 2)。

The fully factorised form is 4x(x - 2). You can check by expanding: 4x × x = 4x² and 4x × (-2) = -8x.

完全分解后的形式为 4x(x - 2)。你可以通过展开来检验：4x × x = 4x²，4x × (-2) = -8x。

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8. Solving Inequalities | 解不等式

Question: Solve the inequality 5 - 2x > 1 and represent the solution on a number line.

问题：解不等式 5 - 2x > 1，并在数轴上表示解集。

Step 1: Isolate the term containing x. Subtract 5 from both sides: -2x > 1 - 5, which gives -2x > -4.

第1步：分离含 x 的项。两边同时减去 5：-2x > 1 - 5，得到 -2x > -4。

Step 2: Divide both sides by -2. Remember that dividing or multiplying by a negative number reverses the inequality sign. So x < 2 (because -4 ÷ -2 = 2 and the sign flips from > to <).

第2步：两边除以 -2。记住，除以或乘以负数会反转不等号方向。因此 x < 2（因为 -4 ÷ -2 = 2，且符号由 > 变为 <）。

Step 3: Draw a number line. Mark an open circle at 2 (to show that 2 is not included) and draw an arrow to the left, indicating all numbers less than 2.

第3步：绘制数轴。在 2 处画一个空心圆（表示 2 不包含在内），然后向左画一个箭头，表示所有小于 2 的数。

The solution is x < 2. Always check your solution by testing a value, e.g. x = 1 gives 5 - 2 = 3, which is indeed > 1.

解为 x < 2。总是通过代入检验来验证你的解，例如 x = 1 时，5 - 2 = 3，果然 > 1。

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9. Sequences and the nth Term | 数列与第 n 项

Question: The nth term of a sequence is given by 3n + 2. Write down the first three terms and find the 10th term.

问题：一个数列的第 n 项由 3n + 2 给出。写出前三个项，并求出第 10 项。

Step 1: Substitute n = 1 for the first term. 3(1) + 2 = 3 + 2 = 5. So the first term is 5.

第1步：代入 n = 1 得到第一项。3(1) + 2 = 3 + 2 = 5。所以第一项是 5。

Step 2: Substitute n = 2 for the second term. 3(2) + 2 = 6 + 2 = 8.

第2步：代入 n = 2 得到第二项。3(2) + 2 = 6 + 2 = 8。

Step 3: Substitute n = 3 for the third term. 3(3) + 2 = 9 + 2 = 11.

第3步：代入 n = 3 得到第三项。3(3) + 2 = 9 + 2 = 11。

Step 4: Find the 10th term by substituting n = 10. 3(10) + 2 = 30 + 2 = 32.

第4步：代入 n = 10 求出第 10 项。3(10) + 2 = 30 + 2 = 32。

The first three terms are 5, 8, 11, and the 10th term is 32. The difference between consecutive terms is constant (3), so this is an arithmetic sequence.

前三个项是 5, 8, 11，第 10 项是 32。相邻项之间的差是常数 (3)，所以这是一个等差数列。

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10. Basic Probability | 基础概率

Question: A bag contains 3 red, 2 blue and 5 green marbles. One marble is chosen at random. Calculate the probability that the marble is not green.

问题：一个袋子里有 3 个红色、2 个蓝色和 5 个绿色弹珠。随机取出一个弹珠。求取出的弹珠不是绿色的概率。

Step 1: Determine the total number of marbles. Total = 3 + 2 + 5 = 10.

第1步：

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