📚 Year 8 AQA Maths: Unit Test Mock Paper Walkthrough | 八年级AQA数学：单元测试模拟卷解析
This walkthrough takes you through a typical Year 8 AQA unit test mock paper, showing step-by-step methods and common pitfalls. Each section focuses on a key topic and includes an example question with a full solution and commentary.
本文详细解析一套典型的八年级AQA单元测试模拟卷，展示逐步解题方法并指出常见错误。每个小节聚焦一个核心知识点，提供例题、完整解答与评注。
1. Number Skills: Order of Operations and Integers | 数字运算：运算顺序与整数
Mock question: Evaluate 3 + 4 × (5 − 2)² ÷ 3 − 7.
模拟题：计算 3 + 4 × (5 − 2)² ÷ 3 − 7。
w the expression becomes 3 + 4 × 9 ÷ 3 − 7. Multiplication and division have equal priority, so work left to right: 4 × 9 = 36, then 36 ÷ 3 = 12. Finally, addition and subtraction: 3 + 12 = 15, then 15 − 7 = 8. The answer is 8. A common mistake is doing 3 + 4 first – always follow the hierarchy.
遵循 BIDMAS 法则（括号、指数、除法/乘法、加法/减法）。先算括号：(5 − 2) = 3。再算指数：3² = 9。此时式子变为 3 + 4 × 9 ÷ 3 − 7。乘除法同级，从左往右计算：4 × 9 = 36，再 36 ÷ 3 = 12。最后加减法：3 + 12 = 15，然后 15 − 7 = 8。答案为 8。常见错误是直接先算 3 + 4 —— 务必按照运算优先顺序计算。
2. Fractions, Decimals and Percentages: Conversions and Calculations | 分数、小数与百分比：转换与运算
Mock question: Convert 0.375 to a fraction in its simplest form, and find 15% of £64.
模拟题：将 0.375 化为最简分数，并求 £64 的 15%。
ator and denominator by 125: 375 ÷ 125 = 3, 1000 ÷ 125 = 8, so the fraction is 3/8. For 15% of £64, remember percent means ‘out of 100’, so 15% = 15/100 = 0.15. Multiply 0.15 × 64. You can do 10% = £6.40 and 5% = £3.20, so total = £9.60. Alternatively, 15/100 × 64 = (15 × 64)/100 = 960/100 = £9.60. Always give units for money answers.
0.375 表示千分之三百七十五：375/1000。分子分母同除以 125 化简：375 ÷ 125 = 3，1000 ÷ 125 = 8，得到 3/8。求 £64 的 15%：百分号表示“每一百”，15% = 15/100 = 0.15。计算 0.15 × 64。可以这样想：10% = £6.40，5% = £3.20，合计 £9.60。或用分数：15/100 × 64 = 960/100 = £9.60。涉及金额时务必带上单位。
3. Ratio and Proportion: Sharing in a Given Ratio | 比与比例：按给定比例分配
Mock question: Share £120 between Alice, Ben and Chloe in the ratio 2 : 3 : 5. How much does each receive?
模拟题：将 £120 按 2 : 3 : 5 分配给 Alice、Ben 和 Chloe，每人各得多少？
First, add the parts of the ratio: 2 + 3 + 5 = 10 parts in total. The total amount is £120, so one part is £120 ÷ 10 = £12. Alice gets 2 parts: 2 × £12 = £24. Ben gets 3 parts: 3 × £12 = £36. Chloe gets 5 parts: 5 × £12 = £60. Always check the sum: £24 + £36 + £60 = £120. Some students forget to add the parts before dividing, so watch out for that.
先求总份数：2 + 3 + 5 = 10 份。总金额 £120，故每份为 £120 ÷ 10 = £12。Alice 得 2 份：2 × £12 = £24；Ben 得 3 份：£36；Chloe 得 5 份：£60。检验总和：£24 + £36 + £60 = £120，无误。常见错误是先除后乘份数时混淆顺序，务必先求每份价值。
4. Algebraic Expressions: Simplifying and Expanding Brackets | 代数表达式：化简与去括号
Mock question: Simplify 3a + 5b − 2a + 7b, and expand 4(2x − 3y + 5).
模拟题：化简 3a + 5b − 2a + 7b，并展开 4(2x − 3y + 5)。
e bracket by 4: 4 × 2x = 8x, 4 × (−3y) = −12y, 4 × 5 = 20. The expanded form is 8x − 12y + 20. Remember the sign rule: a positive multiplied by a negative gives a negative term.
第一部分，合并同类项：3a − 2a = a，5b + 7b = 12b，化简为 a + 12b。展开时，将括号内每项乘以 4：4 × 2x = 8x，4 × (−3y) = −12y，4 × 5 = 20，得 8x − 12y + 20。符号规则：正乘负得负，不能漏掉负号。
5. Solving Linear Equations: One-step and Two-step | 解一元一次方程：一步与两步方程
Mock question: Solve (a) x/5 = 9, and (b) 2y − 7 = 15.
模拟题：解方程 (a) x/5 = 9，以及 (b) 2y − 7 = 15。
1 − 7 = 22 − 7 = 15. Some students try to subtract 2 first, which is incorrect because the 2 is a coefficient, not a separate term. Work in reverse BIDMAS order when undoing operations.
(a) 中，两边同乘 5 解出 x：x = 9 × 5 = 45。(b) 中，先两边加 7：2y = 22，再除以 2 得 y = 11。验算：2 × 11 − 7 = 15。常见错误是先除以 2 再处理 −7，但 2 是系数，应当最后解除。原则是按照逆运算顺序（先加减后乘除）求解。
6. Angles: On a Straight Line, Around a Point and in Triangles | 角度：平角、周角与三角形内角
Mock question: Find the missing angle x in the following scenarios: (a) two angles 120° and x on a straight line; (b) angles 85°, 55° and x in a triangle.
模拟题：求下列图形中的未知角 x：(a) 平线上有 120° 和 x；(b) 三角形内角为 85°、55° 和 x。
o do it in two steps: 85° + 55° = 140°, then 180° − 140°. Always label your reasoning and units.
(a) 平线上的角之和为 180°，所以 x = 180° − 120° = 60°。(b) 三角形内角和为 180°，x = 180° − (85° + 55°) = 180° − 140° = 40°。也可分步：85° + 55° = 140°，180° − 140° = 40°。解答时要注明理由并带上度数符号。
7. Perimeter and Area: Rectangles, Triangles and Compound Shapes | 周长与面积：矩形、三角形与组合图形
Mock question: A rectangle has length 8 cm and width 5 cm. A right-angled triangle sits on top of the rectangle with base 8 cm and height 4 cm. Find the total area of the compound shape and its perimeter.
模拟题：一个矩形长 8 cm，宽 5 cm，矩形上方有一个底边 8 cm、高 4 cm 的直角三角形，形成组合图形。求总面积与周长。
cm, left side = 5 cm, top of rectangle is partly covered by the triangle. The triangle sides: hypotenuse = √4² + 4² = √16+16 = √32 = 4√2 gle must be accounted. Actually, rectangle top is 8 cm; triangle base sits exactly on the top, so the rectangle top is not part of the perimeter. The perimeter runs: bottom 8 cm, right side 5 cm, then up along triangle’s right side? e on top, sharing the top side of rectangle as base. The triangle’s right angle is at left? Let's assume the right triangle' t 8 cm (base), up 5 cm (right side of rectangle), then up along the hypotenuse of triangle? ould be wrong because triangle sits on top; the right side of rectangle meets the triangle's right vertex? o avoid confusion, we can set a simpler perimeter question: rectangle plus a triangle on top, all external edges. Let' the compound shape looks like a house: rectangle 8 by 5, and an isosceles triangle on top? ase 8, height 4. Then the two triangle sides are the height (vertical) of 4 cm on the left? vertical side of rectangle is 5 cm, but the triangle adds a vertical line of 4 cm above it? Actually the left side of rectangle and the triangle's left side would be aligned if triangle' ght side 5, then sloping side (hypotenuse) of triangle, then top of triangle (a horizontal? eter = 8 (bottom) + 5 (right) + 5 (left side of rectangle, but above the rectangle the triangle adds 4 cm vertical? Actually left side of rectangle is 5 cm, above it the triangle' e. So perimeter = 8 (bottom) + 5 (right side of rectangle) + sloping side length + (5+4) left edge. Sloping side = √8² + 4² = √64+16 = √80 = 4√5 ≈ 8.94 cm. Left edge total = 9 cm. Sum = 8 + 5 + 4√5 + 9 = 22 + 4√5 cm (≈ 22+8.94=30.94 cm). This is a bit advanced but Year 8 can handle Pythagoras? gle right-angled with base 6 and height 8 to make hypotenuse 10, a Pythagorean triple. Let' m², triangle 24 cm², total 54 cm². Perimeter: left side = 5+8=13 cm, right side 5 cm, bottom 6 cm, hypotenuse 10 cm. Total = 6+5+10+13 = 34 cm. Perfect. I'll rewrite the question: rectangle 6 cm by 5 cm, triangle base 6 cm, height 8 cm. So I' stion in this planning stated 8 cm and 5 cm with height 4, but I can change for clarity. I' cm (same as rectangle length) and height 8 cm. Find total area and perimeter. This avoids messy surds. I'll write that.
Mock question revised: A rectangle has length 6 cm and width 5 cm. A right-angled triangle with base 6 cm and height 8 cm sits exactly on top of the rectangle. Find the total area and the perimeter of the compound shape.
tline consists of bottom (6 cm), right side (5 cm), sloping side of triangle (hypotenuse = √6²+8² iangle) = 13 cm. Sum = 6 + 5 + 10 + 13 = 34 cm. Always check shared sides are not counted.
面积：矩形 = 6 × 5 = 30 cm²；三角形面积 = ½ × 6 × 8 = 24 cm²；总面积 54 cm²。周长：外轮廓由底边 6 cm、右侧边 5 cm、三角形的斜边（由 6-8-10 勾股数得斜边 10 cm）和左侧边（5 cm 矩形边 + 8 cm 三角形高 = 13 cm）组成。总和 = 6 + 5 + 10 + 13 = 34 cm。注意重合边不计入周长。
8. Statistics: Mean, Median and Bar Charts | 统计：平均数、中位数与条形图
Mock question: The bar chart below shows the number of books read by 8 students in a month: two students read 1 book, three read 2 books, two read 3 books, one read 4 books. Calculate the mean and median number of books read.
模拟题：条形图显示 8 名学生一个月的阅读本数：2 人读 1 本，3 人读 2 本，2 人读 3 本，1 人读 4 本。计算阅读本数的平均数和中位数。
ts = 8. Mean = 18 ÷ 8 = 2.25. For median, the data is already ordered; with 8 values (even), median is the average of the 4th and 5th values: both are 2 (the fourth is 2, fifth is 2), so median = 2. Some students forget to order data or use the wrong position for median.
列出全部数据：1, 1, 2, 2, 2, 3, 3, 4。平均数 = 总和 ÷ 人数。总和 = (2×1) + (3×2) + (2×3) + (1×4) = 2+6+6+4 = 18。8 名学生，平均数为 18 ÷ 8 = 2.25 本。中位数：数据已按顺序排列，共 8 个（偶数），取第 4 和第 5 个数的平均值：第 4 个数是 2，第 5 个也是 2，中位数为 2。常见错误是不排序或中位数定位错误。
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