SQA Year 10 Maths Unit Test Mock Paper Walkthrough | SQA 十年级数学单元测试模拟卷解析

📚 SQA Year 10 Maths Unit Test Mock Paper Walkthrough | SQA 十年级数学单元测试模拟卷解析

Welcome to our step-by-step walkthrough of a SQA Year 10 Mathematics unit test mock paper. This revision resource covers essential topics from algebra, geometry, trigonometry and statistics that frequently appear in school assessments. Each question is broken down into clear, manageable steps so you can build your confidence and exam technique.

欢迎阅读我们为SQA十年级数学单元测试模拟卷提供的逐步解析。这份复习资料涵盖了代数、几何、三角学和统计学中经常出现在校内评估中的核心主题。我们将每道题目拆解成清晰、易于掌握的步骤，帮助你建立信心，提升应试技巧。

1. Solving Linear Equations | 解一元一次方程

Question: Solve 3x + 7 = 22.

7 = 22, start by subtracting 7 from both sides to undo the addition. This leaves 3x = 15.

解方程 3x + 7 = 22，首先两边同时减去 7，消去加数，得到 3x = 15。

Next, divide both sides by 3 to isolate x, giving x = 5.

接下来，两边除以 3，分离出 x，得 x = 5。

substituting it back into the original equation: 3(5) + 7 = 15 + 7 = 22, which is correct.

务必通过代回原方程检验答案：3(5) + 7 = 15 + 7 = 22，结果正确。

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2. Factorising Quadratic Expressions and Solving | 因式分解二次表达式并求解

Question: Solve x² + 5x + 6 = 0.

To factorise x² + 5x + 6, find two numbers that multiply to 6 and add to 5. These are 2 and 3.

对 x² + 5x + 6 进行因式分解，找到两个相乘得 6、相加得 5 的数，即 2 和 3。

the expression as (x + 2)(x + 3). Set each bracket equal to zero: x + 2 = 0 or x + 3 = 0.

将表达式写成 (x + 2)(x + 3)。令每个括号等于零：x + 2 = 0 或 x + 3 = 0。

This gives the solutions x = -2 and x = -3.

由此得到解 x = -2 和 x = -3。

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3. Linear Inequalities | 一元一次不等式

Question: Solve 4x - 3 ≤ 13.

sign if you multiply or divide by a negative number. First, add 3 to both sides: 4x ≤ 16.

解不等式与解方程类似，但如果乘以或除以负数，要记得反转不等号。首先，两边同时加 3：4x ≤ 16。

by the positive number 4: x ≤ 4. The solution is all real numbers less than or equal to 4.

两边同时除以正数 4：x ≤ 4。解集为所有小于或等于 4 的实数。

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4. Pythagoras' Theorem | 勾股定理

Question: A right-angled triangle has legs of length 5 cm and 12 cm. Find the length of the hypotenuse.

Pythagoras' ², where c is the hypotenuse. Substitute a = 5 and b = 12: c² = 5² + 12² = 25 + 144 = 169.

勾股定理公式为 c² = a² + b²，其中 c 代表斜边。代入 a = 5，b = 12：c² = 5² + 12² = 25 + 144 = 169。

Take the square root of both sides: c = √169 = 13. The hypotenuse is 13 cm long.

两边开平方：c = √169 = 13。斜边长为 13 厘米。

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5. Volume of a Cylinder | 圆柱体积计算

stion: A cylinder has radius 3 cm and height 10 cm. Calculate its volume, taking π ≈ 3.14.

formula is V = πr²h. Substitute r = 3 and h = 10: V = π × 3² × 10 = π × 9 × 10 = 90π cm³.

体积公式为 V = πr²h。代入 r = 3，h = 10：V = π × 3² × 10 = π × 9 × 10 = 90π cm³。

Using π ≈ 3.14, the approximate volume is 90 × 3.14 = 282.6 cm³ (to one decimal place).

取 π ≈ 3.14，计算近似体积：90 × 3.14 = 282.6 cm³（保留一位小数）。

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6. Basic Trigonometry – Finding a Side | 基础三角学 – 求边长

Question: In a right-angled triangle, the hypotenuse is 10 cm and one acute angle is 37°. Find the length of the side opposite the 37° angle.

opposite / hypotenuse. Set up the equation sin 37° = x / 10, where x is the opposite side.

运用正弦比：sin(θ) = 对边/斜边。列出方程 sin 37° = x / 10，其中 x 是对边长度。

by 10: x = 10 × sin 37°. Using a calculator, sin 37° ≈ 0.6018, so x ≈ 6.02 cm (to 2 d.p.).

两边乘以 10：x = 10 × sin 37°。使用计算器求得 sin 37° ≈ 0.6018，因此 x ≈ 6.02 cm（保留两位小数）。

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7. Similar Triangles – Scale Factor | 相似三角形 – 比例因子

Question: Two triangles are similar. The smaller triangle has a base of 4 cm, and the larger has a corresponding base of 10 cm. If the height of the smaller triangle is 3 cm, find the height of the larger.

er base by the smaller base: 10 / 4 = 2.5. All corresponding lengths scale by this factor.

从小三角形到大三角形的比例因子用大三角形底边除以小三角形底边求得：10 / 4 = 2.5。所有对应边长都按此因子缩放。

Multiply the smaller height by the scale factor: height = 3 cm × 2.5 = 7.5 cm.

将小三角形的高乘以比例因子：高 = 3 cm × 2.5 = 7.5 cm。

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8. Mean from a Frequency Table | 从频数表求平均数

Question: The table shows the number of pets owned by students. 0 pets: 3 students, 1 pet: 7 students, 2 pets: 5 students, 3 pets: 2 students. Find the mean number of pets.

(value × frequency): 0×3 = 0, 1×7 = 7, 2×5 = 10, 3×2 = 6. Sum of fx = 0 + 7 + 10 + 6 = 23.

增设 fx 栏（数值 × 频数）：0×3 = 0，1×7 = 7，2×5 = 10，3×2 = 6。fx 总和 = 0 + 7 + 10 + 6 = 23。

otal frequency Σf = 3 + 7 + 5 + 2 = 17. Mean = Σfx / Σf = 23 / 17 ≈ 1.35 pets (to 2 d.p.).

总频数 Σf = 3 + 7 + 5 + 2 = 17。平均数 = Σfx / Σf = 23 / 17 ≈ 1.35 只宠物（保留两位小数）。

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9. Basic Probability | 基础概率

Question: A bag contains 4 red, 3 blue and 2 green balls. One ball is drawn at random. Find the probability that it is not blue.

ber of balls = 4 + 3 + 2 = 9. The number of balls that are not blue = 4 red + 2 green = 6.

球的总数 = 4 + 3 + 2 = 9。非蓝色球的数量 = 4 个红球 + 2 个绿球 = 6。

Probability(not blue) = 6/9 = 2/3 after simplifying.

概率（非蓝色）= 6/9，约分为 2/3。

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10. Equation of a Straight Line | 直线方程

hrough the points (1, 3) and (3, 7). Find the equation of the line in the form y = mx + c.

d the slope m using the formula m = (y₂ - y₁) / (x₂ - x₁) = (7 - 3) / (3 - 1) = 4 / 2 = 2.

使用公式求斜率 m：m = (y₂ - y₁) / (x₂ - x₁) = (7 - 3) / (3 - 1) = 4 / 2 = 2。

Use the point-slope form with (1, 3): y - 3 = 2(x - 1). Expand to get y - 3 = 2x - 2.

运用点 (1, 3) 和点斜式：y - 3 = 2(x - 1)。展开得 y - 3 = 2x - 2。

dd 3 to both sides to obtain the slope-intercept form: y = 2x + 1. The y-intercept c is 1.

两边同时加 3，化为斜截式：y = 2x + 1。y 轴截距 c 为 1。

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