📚 Year 7 AQA Further Maths Unit Test Mock Paper Walkthrough | Year 7 AQA 进阶数学单元测试模拟卷解析
This article provides a full walkthrough of a mock unit test designed for Year 7 students following the AQA Further Maths progression route. Every question has been carefully chosen to mirror the depth and style of assessment common at this level, blending fluency with reasoning and problem-solving. Use this walkthrough to strengthen your revision and build confidence before your real test.
本文为参加 AQA 进阶数学路径的 Year 7 学生提供一份完整的单元测试模拟卷解析。每道题目都经过精心挑选，反映该阶段测试的深度与风格，融合了运算流畅性、逻辑推理与问题解决能力。请利用这份解析巩固复习内容，在实际考试前建立信心。
1. Mock Test Overview | 模拟卷概览
The mock paper contains ten questions covering number, algebra, geometry, statistics and probability. The total marks are 50, and calculators are allowed for specific questions only. The questions increase in difficulty, starting with straightforward number skills and moving towards multi-step problem-solving. You should aim to complete the paper in 45 minutes.
模拟卷包含十道题，覆盖数、代数、几何、统计与概率。满分 50 分，仅特定题目允许使用计算器。题目难度逐步上升，从直接的数字技能过渡到多步骤问题解决。建议在 45 分钟内完成全卷。
- Question 1–3: Number and operations (10 marks) | 数与运算 (10 分)
- Question 4–6: Algebra and sequences (15 marks) | 代数与序列 (15 分)
- Question 7–8: Geometry and measures (10 marks) | 几何与测量 (10 分)
- Question 9–10: Statistics and probability (15 marks) | 统计与概率 (15 分)
2. Question 1: Adding Fractions with Mixed Numbers | 问题 1：分数相加与带分数
Question: Work out 2/3 + 3/4. Give your answer as a mixed number in its simplest form. (5 marks)
题目：计算 2/3 + 3/4。将结果写成最简带分数。(5 分)
tions: 2/3 = 8/12 and 3/4 = 9/12. Now add the numerators: 8 + 9 = 17, so the sum is 17/12.
首先找出公分母。3 和 4 的最小公倍数是 12。转换两个分数：2/3 = 8/12，3/4 = 9/12。分子相加：8 + 9 = 17，因此和为 17/12。
l part 5/12 is already in simplest form because 5 and 12 have no common factor other than 1.
17/12 是一个假分数。用 17 除以 12：商 1 余 5，所以 17/12 = 1 5/12。分数部分 5/12 已经是最简形式，因为 5 和 12 除了 1 再无公因数。
Answer: 1 5/12. Always check if you can simplify the fraction at the end; rapid checking of common factors avoids losing accuracy marks.
答案：1 5/12。最后一定要检查分数是否可化简；快速核对公因数可以避免丢失准确性分。
3. Question 2: Simplifying Algebraic Expressions | 问题 2：代数式化简
Question: Simplify 4(2a – 3) – 3(a + 2). (4 marks)
题目：化简 4(2a – 3) – 3(a + 2)。(4 分)
ibutive law first. Multiply 4 by each term inside the first bracket: 4 × 2a = 8a and 4 × (–3) = –12. So the first part becomes 8a – 12.
首先应用分配律。将 4 乘以第一个括号内的每一项：4 × 2a = 8a，4 × (–3) = –12。于是第一部分变成 8a – 12。
Now expand the second bracket: –3 multiplied by a gives –3a, and –3 multiplied by 2 gives –6. Since the original operation is subtraction, we must be careful: –3(a + 2) = –3a – 6.
现在展开第二个括号：–3 乘以 a 得到 –3a，–3 乘以 2 得到 –6。由于原先的运算是减法，需格外留意：–3(a + 2) = –3a – 6。
Combine the expressions: 8a – 12 – 3a – 6. Group like terms: 8a – 3a = 5a, and –12 – 6 = –18. The simplified expression is 5a – 18.
合并表达式：8a – 12 – 3a – 6。合并同类项：8a – 3a = 5a，–12 – 6 = –18。化简后的表达式为 5a – 18。
value, e.g. a = 1, into both the original and simplified expressions to ensure they match.
可代入数值（如 a = 1）双重检验原式与化简结果是否一致。
4. Question 3: Order of Operations with Exponents | 问题 3：指数与运算顺序
Question: Evaluate 2³ × 5 – (4² – 6). (3 marks)
题目：计算 2³ × 5 – (4² – 6)。(3 分)
cation and division, finally addition and subtraction. Inside the brackets: 4² = 16, so 16 – 6 = 10. The expression becomes 2³ × 5 – 10.
遵循 BIDMAS/BODMAS 规则：先算括号，再算指数，然后乘除，最后加减。括号内：4² = 16，所以 16 – 6 = 10。表达式变为 2³ × 5 – 10。
handle the index: 2³ = 2 × 2 × 2 = 8. Multiply: 8 × 5 = 40. Finally, subtract 10 to get 40 – 10 = 30.
现在处理指数：2³ = 2 × 2 × 2 = 8。乘法：8 × 5 = 40。最后减去 10 得到 40 – 10 = 30。
A common mistake is to perform 5 – 10 first, but multiplication has higher priority. Writing each step clearly prevents this error.
常见错误是先计算 5 – 10，但乘法优先级更高。逐步写下每一步可避免此类错误。
5. Question 4: nth Term and Sequences | 问题 4：第 n 项与序列
Question: The nth term of a sequence is 3n + 2. Write down the first four terms and find the 10th term. (5 marks)
题目：某个序列的第 n 项为 3n + 2。写出前四项并求出第 10 项。(5 分)
For n = 3: 3(3) + 2 = 11. For n = 4: 3(4) + 2 = 14. The first four terms are 5, 8, 11, 14.
将 n = 1, 2, 3, 4 代入表达式。n = 1: 3(1) + 2 = 5。n = 2: 3(2) + 2 = 8。n = 3: 3(3) + 2 = 11。n = 4: 3(4) + 2 = 14。前四项为 5, 8, 11, 14。
s the coefficient of n. To find the 10th term, substitute n = 10: 3(10) + 2 = 30 + 2 = 32.
注意到序列每次增加 3，这与 n 的系数吻合。求第 10 项时，代入 n = 10：3(10) + 2 = 30 + 2 = 32。
Using nth term rules is much faster than continuing the sequence manually, especially for distant terms like the 100th.
使用第 n 项法则比手动延续序列快得多，尤其是对于像第 100 项这样遥远的项。
6. Question 5: Algebraic Equations with Variables on Both Sides | 问题 5：带两边变量的代数方程
Question: Solve 5x – 3 = 2x + 12. (4 marks)
题目：解方程 5x – 3 = 2x + 12。(4 分)
llect x terms on one side and constant terms on the other. Subtract 2x from both sides: 5x – 3 – 2x = 2x + 12 – 2x, which simplifies to 3x – 3 = 12.
需要将含 x 的项移到一边，常数项移到另一边。方程两边同时减去 2x：5x – 3 – 2x = 2x + 12 – 2x，化简得 3x – 3 = 12。
Next, add 3 to both sides: 3x – 3 + 3 = 12 + 3, giving 3x = 15. Divide both sides by 3: x = 5.
接着，两边加 3：3x – 3 + 3 = 12 + 3，得到 3x = 15。两边除以 3：x = 5。
Check the solution by substituting x = 5 back into the original equation: left side 5(5) – 3 = 25 – 3 = 22; right side 2(5) + 12 = 10 + 12 = 22. Both sides equal, so x = 5 is correct.
将 x = 5 代回原方程检验：左边 5(5) – 3 = 25 – 3 = 22；右边 2(5) + 12 = 10 + 12 = 22。两边相等，故解正确。
7. Question 6: Prime Factorisation, HCF and LCM | 问题 6：质因数分解、HCF 与 LCM
Question: Express 72 and 90 as products of prime factors using index notation. Hence find the highest common factor (HCF) and lowest common multiple (LCM) of 72 and 90. (6 marks)
题目：用指数记法将 72 和 90 表示为质因数乘积。由此求出 72 和 90 的最大公因数 (HCF) 和最小公倍数 (LCM)。(6 分)
2 = 8 × 9 = 2³ × 3². For 90: 90 = 9 × 10 = 3² × 2 × 5, which can be written as 2 × 3² × 5.
画出因数树。对 72：72 = 8 × 9 = 2³ × 3²。对 90：90 = 9 × 10 = 3² × 2 × 5，可写成 2 × 3² × 5。
factor. Both have 2 (lowest power 2¹) and 3 (lowest power 3²). HCF = 2¹ × 3² = 2 × 9 = 18.
求 HCF 时，取每个公共质因数的最小指数。两者都有 2（最低指数 2¹）和 3（最低指数 3²）。HCF = 2¹ × 3² = 2 × 9 = 18。
sent: 2³ from 72, 3² (both have this), and 5 from 90. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.
求 LCM 时，取所有出现过的质因数的最高指数：取自 72 的 2³，两者共有的 3²，以及 90 中的 5。LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360。
This method is reliable and quicker than listing multiples when numbers are large. Always present prime factors in ascending order.
此法可靠，且当数字较大时比罗列倍数更迅速。质因数应始终按升序表示。
8. Question 7: Area and Perimeter of a Triangle | 问题 7：三角形的面积与周长
Question: A triangle has base 12 cm and area 42 cm². Find its height. Then, if the other two sides are 10 cm and 7 cm, calculate the perimeter. (4 marks)
题目：一个三角形底边长 12 cm，面积为 42 cm²。求其高。若另两边长分别为 10 cm 和 7 cm，计算周长。(4 分)
42 = ½ × 12 × h. Simplify: ½ × 12 = 6, so 42 = 6h. Divide both sides by 6 to get h = 7 cm.
使用三角形面积公式：面积 = ½ × 底 × 高。代入已知值：42 = ½ × 12 × h。化简：½ × 12 = 6，因此 42 = 6h。两边除以 6 得 h = 7 cm。
are 12 cm, 10 cm and 7 cm. Perimeter is the sum of all side lengths: 12 + 10 + 7 = 29 cm.
现在三边长度分别为 12 cm, 10 cm, 7 cm。周长为所有边长之和：12 + 10 + 7 = 29 cm。
Remember that the height is perpendicular to the base. Even though the triangle is probably scalene, the area relationship still holds using the given base and calculated height.
记住高是与底边垂直的。尽管三角形可能是不等边三角形，但利用给定的底和算出的高，面积关系依然成立。
9. Question 8: Proportion and Scaling | 问题 8：比例与缩放
Question: A recipe uses 200 g of flour to make 15 biscuits. How much flour is needed to make 36 biscuits? Use a clear unitary method. (4 marks)
题目：一份食谱使用 200 g 面粉制作 15 块饼干。制作 36 块饼干需要多少面粉？请使用清晰的单位法。(4 分)
: 200 g ÷ 15 = 200/15 = 40/3 g per biscuit. (You can leave it as a fraction for accuracy.)
先求出一块饼干所需面粉量：200 g ÷ 15 = 200/15 = 40/3 克每块。（为确保精度可保留分数形式。）
40 × 12 = 480 g. Alternatively, note that 36 ÷ 15 = 2.4, so multiply 200 g × 2.4 = 480 g.
乘以所需饼干数：(40/3) × 36 = 40 × 12 = 480 g。或者，注意到 36 ÷ 15 = 2.4，所以 200 g × 2.4 = 480 g。
Always check the answer makes sense: 36 is more than twice 15, so the flour needed should be more than 400 g, and 480 g is reasonable.
务必检查答案是否合理：36 是 15 的两倍多，故面粉需求量应超过 400 g，480 g 是合理的。
10. Question 9: Mean, Median and Range | 问题 9：平均数、中位数与范围
Question: The stem-and-leaf diagram below shows the marks of 15 students. Find the median, mean and range.
Stem (tens) | Leaf (units)
1 | 2 7 8
2 | 0 1 3 5 5 9
3 | 4 6 7 8
4 | 1 5
(6 marks)
题目：下方的茎叶图显示了 15 名学生的分数。求中位数、平均数和范围。
茎（十位）| 叶（个位）
1 | 2 7 8
2 | 0 1 3 5 5 9
3 | 4 6 7 8
4 | 1 5
(6 分)
List all scores in order: 12, 17, 18, 20, 21, 23, 25, 25, 29, 34, 36, 37, 38, 41, 45. There are 15 values. The median is the 8th value: count to the eighth, which is 25.
按顺序列出所有分数：12, 17, 18, 20, 21, 23, 25, 25, 29, 34, 36, 37, 38, 41, 45。共有 15 个值。中位数是第 8 个值：数到第 8 个，为 25。
34+36+37+38+41+45 = 421. Divide by 15: 421 ÷ 15 = 28.066... which rounds to 28.1 (1 d.p.).
求平均数，将所有分数相加：12+17+18+20+21+23+25+25+29+34+36+37+38+41+45 = 421。除以 15：421 ÷ 15 = 28.066...，四舍五入为 28.1（保留一位小数）。
Range = highest – lowest = 45 – 12 = 33. Stem-and-leaf helps to see the spread quickly.
范围 = 最大值 – 最小值 = 45 – 12 = 33。茎叶图有助于快速观察数据散布情况。
11. Question 10: Probability of Prime Numbers | 问题 10：质数的概率
Question: A fair 10-sided die with faces numbered 1 to 10 is rolled. What is the probability of getting a prime number? Give your answer as a fraction in simplest form. (4 marks)
题目：掷一枚公平的 10 面骰子，面朝上的数字为 1 至 10。得到质数的概率是多少？将答案写成最简分数。(4 分)
List the prime numbers between 1 and 10: 2, 3, 5, 7. Note that 1 is NOT prime. So there are 4 favourable outcomes.
列出 1 到 10 之间的质数：2, 3, 5, 7。注意 1 不是质数。因此有 4 个有利结果。
s / total outcomes = 4/10. Simplify by dividing numerator and denominator by 2 to get 2/5.
可能结果总数 = 10。概率 = 有利结果数 / 总结果数 = 4/10。分子分母同时除以 2 化简得 2/5。
Always check whether 1 is counted; many students mistakenly include it. Remember a prime has exactly two distinct factors.
一定要检查是否将 1 计入；许多学生错误地将其包含。记住质数恰好有两个不同的因数。
12. Final Tips for the Unit Test | 单元测试终极贴士
Before submitting, review your worked solutions for common arithmetic errors. Show all steps clearly—even on calculator-allowed questions, method marks are awarded. Practice writing units (cm, g, cm²) consistently and simplify fractions wherever possible. With this walkthrough, you have seen the key skills: fraction arithmetic, algebraic manipulation, sequences, solving equations, prime factorisation, area, proportion, statistics and probability. Keep this checklist in mind, and approach the test calmly and methodically.
提交前回顾解题过程，排查常见的算术错误。清晰写出所有步骤——即使允许使用计算器的题目，也仍有方法分。规范书写单位（cm, g, cm²），并尽可能化简分数。通过这份解析，你已见识到核心技能：分数运算、代数处理、序列、解方程、质因数分解、面积、比例、统计与概率。牢记这份清单，沉着、有条不紊地应对测试。
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