📚 Year 7 OCR Further Mathematics: Unit Test Mock Paper Walkthrough | Year 7 OCR 进阶数学：单元测试模拟卷解析
A unit test in Year 7 Further Mathematics is designed to assess your understanding of core topics and challenge your problem-solving skills. By working through a mock paper and its detailed solutions, you can build confidence and refine your exam technique. This walkthrough provides step-by-step explanations for ten typical questions, following the OCR syllabus. Use it to review key concepts, learn from common mistakes, and ensure you are fully prepared for your upcoming assessment.
七年级进阶数学的单元测试旨在评估你对核心主题的理解，并考验你的解题能力。通过完成一份模拟试卷及其详细解析，你可以建立信心、打磨考试技巧。本篇解析紧扣 OCR 课程大纲，为十道典型题目提供逐步讲解。请利用它回顾关键概念、从常见错误中学习，确保你为即将到来的评估做好充分准备。
1. How the Unit Test Is Structured | 单元测试的结构
Before you attempt any question, it helps to understand the shape of the paper. A Year 7 Further Mathematics unit test usually lasts 45 to 60 minutes and carries around 50 to 60 marks. The first section checks fluency with number and algebra, while the later questions combine two or more topics and reward clear written reasoning. Marks are given for method as well as for the final answer, so a wrong answer with correct working can still earn most of the credit.
在动手做题之前，先了解试卷的结构会很有帮助。七年级进阶数学的单元测试通常时长 45 至 60 分钟，总分约 50 至 60 分。前面部分考查数与代数的熟练度，后面的题目则会综合两个或更多主题，并奖励清晰的书面推理。评分既看方法也看最终答案，因此即使答案算错，只要过程正确仍可拿到大部分分数。
The ten questions below mirror the balance of a real OCR-style paper: four number questions, three algebra questions, two shape and space questions, and one probability and statistics question. Notice how each solution is written in full sentences with the operation shown at every step. That is exactly how you should present your own work in the test.
下面这十道题对应真实 OCR 风格试卷的题型分布：四道数与计算题、三道代数题、两道图形与空间题，以及一道概率与统计题。请注意每个解析都用完整语句书写，并逐步展示运算过程——这正是你在考试中应当采用的书写方式。
- Number and calculation: order of operations, indices, fractions, percentages, ratio | 数与计算：运算顺序、指数、分数、百分数、比
- Algebra: expanding, factorising, substitution, equations, inequalities, sequences | 代数：展开、因式分解、代入、方程、不等式、数列
- Shape and space: angle facts, perimeter, area, volume | 图形与空间：角度性质、周长、面积、体积
- Probability and statistics: probability scale, mean, median, mode, range | 概率与统计：概率刻度、平均数、中位数、众数、极差
2. Question 1: Order of Operations (BIDMAS) | 第1题：运算顺序（BIDMAS）
Question: Evaluate 24 ÷ (3 + 5) × 2 − 3².
题目：计算 24 ÷ (3 + 5) × 2 − 3²。
To solve this, we must follow the order of operations, often remembered by the acronym BIDMAS (Brackets, Indices, Division and Multiplication, Addition and Subtraction). This rule ensures calculations are performed in a consistent sequence.
要解决这道题，我们必须遵循运算顺序，通常用首字母缩写 BIDMAS（括号、指数、乘除、加减）来记忆。这一规则确保计算按一致的顺序进行。
Brackets: work out (3 + 5) first, which gives 8. The expression now reads 24 ÷ 8 × 2 − 3².
第一步——括号：先算 (3 + 5)，得到 8。此时式子变为 24 ÷ 8 × 2 − 3²。
Step 2 - Indices: 3² means 3 × 3 = 9. The expression becomes 24 ÷ 8 × 2 − 9.
第二步——指数：3² 表示 3 × 3 = 9。式子变成 24 ÷ 8 × 2 − 9。
ion, working from left to right: 24 ÷ 8 = 3, then 3 × 2 = 6. The expression becomes 6 − 9.
第三步——乘除，自左向右：24 ÷ 8 = 3，再 3 × 2 = 6。式子变成 6 − 9。
Step 4 - Subtraction: 6 − 9 = −3.
第四步——减法：6 − 9 = −3。
24 ÷ (3 + 5) × 2 − 3² = 24 ÷ 8 × 2 − 9 = 3 × 2 − 9 = 6 − 9 = −3
ity, so you must work strictly from left to right. Another frequent slip is reading 3² as 3 × 2 = 6 rather than 9.
最常见的错误是先乘后除，或误以为除法永远优先于乘法。二者优先级相同，必须严格从左到右计算。另一个常见失误是把 3² 读成 3 × 2 = 6，而不是 9。
3. Question 2: Indices, Roots and Negative Numbers | 第2题：指数、根号与负数
Question: Work out (a) 2⁵, (b) 3⁻², (c) √144, (d) ∛27, (e) (−2)³, (f) (−2)⁴.
题目：计算 (a) 2⁵，(b) 3⁻²，(c) √144，(d) ∛27，(e) (−2)³，(f) (−2)⁴。
number is the index (or power), and it tells you how many copies of the base are multiplied together.
(a) 是重复相乘：2⁵ = 2 × 2 × 2 × 2 × 2 = 32。右上角的小数字叫做指数（幂），它表示底数被相乘的次数。
3² = 1 ÷ 9 = 1/9. A negative index never makes the answer negative; it makes a reciprocal.
(b) 用到负指数法则：a⁻ⁿ = 1 ÷ aⁿ。所以 3⁻² = 1 ÷ 3² = 1 ÷ 9 = 1/9。负指数绝不会让结果变成负数，它只是取倒数。
Part (c): √144 = 12, because 12 × 12 = 144. Part (d): ∛ re numbers up to 15² and the cube numbers up to 5³ by heart, since they appear constantly.
(c)：√144 = 12，因为 12 × 12 = 144。(d)：∛27 = 3，因为 3 × 3 × 3 = 27。请熟记 15² 以内的平方数和 5³ 以内的立方数，它们出现频率极高。
+16. Be careful to distinguish (−2)⁴ from −2⁴; the first is 16 but the second is −(2⁴) = −16.
(e) 和 (f) 考查幂的符号。负数的奇次幂为负，偶次幂为正：(−2)³ = −8，(−2)⁴ = +16。注意区分 (−2)⁴ 和 −2⁴：前者等于 16，后者等于 −(2⁴) = −16。
aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ a⁰ = 1
4. Question 3: Fractions, Decimals and Percentages | 第3题：分数、小数与百分数
Question: (a) Work out 3/4 + 5/6, giving your answer as a mixed number. (b) Find 15% of 240.
题目：(a) 计算 3/4 + 5/6，答案写成带分数。(b) 求 240 的 15%。
For part (a) you cannot add fractions until the denominators match. The lowest common multiple of 4 and 6 is 12, so 3/4 becomes 9/12 and 5/6 becomes 10/12.
做 (a) 时，分母不同不能直接相加。4 和 6 的最小公倍数是 12，所以 3/4 化为 9/12，5/6 化为 10/12。
r as a mixed number is 1 7/12. Always check whether your final fraction can be simplified.
然后把分子相加、分母不变：9/12 + 10/12 = 19/12。因为 19 ÷ 12 = 1 余 7，所以写成带分数是 1 7/12。最后一定要检查分数是否还能约分。
For part (b), remember that "of" means multiply and that 15% = 15 ÷ 100 = 0.15. So 15% of 240 = 0.15 × 240 = 36. A quick mental check: 10% of 240 is 24 and 5% is 12, and 24 + 12 = 36, which confirms the answer.
做 (b) 时记住 "of" 表示相乘，且 15% = 15 ÷ 100 = 0.15。所以 240 的 15% = 0.15 × 240 = 36。快速心算检验：240 的 10% 是 24，5% 是 12，24 + 12 = 36，答案得到确认。
| Fraction 分数 | Decimal 小数 | Percentage 百分数 |
| 1/2 | 0.5 | 50% |
| 1/4 | 0.25 | 25% |
| 1/5 | 0.2 | 20% |
| 1/8 | 0.125 | 12.5% |
| 1/3 | 0.333... | 33⅓% |
5. Question 4: Ratio, Proportion and Scale | 第4题：比、比例与比例尺
Question: A prize of £84 is shared between Ava and Ben in the ratio 3 : 4. How much does each receive?
题目：84 英镑的奖金按 3 : 4 分给 Ava 和 Ben，两人各得多少？
one part. The ratio 3 : 4 has 3 + 4 = 7 parts in total, and the 7 parts together are worth £84.
比例题的核心是求出一份的值。比 3 : 4 总共有 3 + 4 = 7 份，这 7 份合起来等于 84 英镑。
So one part is £84 ÷ 7 = £12. Ava receives 3 parts: 3 × £12 = £36. Ben receives 4 parts: 4 × £12 = £48.
因此一份是 84 ÷ 7 = 12 英镑。Ava 得 3 份：3 × 12 = 36 英镑。Ben 得 4 份：4 × 12 = 48 英镑。
Always check by adding the shares back together: £36 + £48 = £ numbers by 12. This backwards check takes five seconds and catches most arithmetic slips.
务必把两份相加检验：36 + 48 = 84；而 36 : 48 同时除以 12 得 3 : 4，与题意一致。这个反向检验只需五秒，却能抓住大多数计算失误。
Proportion questions use the same unitary method. If 5 identical pens cost £3.50, then one pen costs £3.50 ÷ 5 = £0.70, and 8 pens cost 8 × £0.70 = £ cm on the diagram represents 200 cm in real life, so 6 cm on the plan is 1200 cm, or 12 m.
比例问题用的是同样的"归一法"。如果 5 支相同的笔共 3.50 英镑，那么 1 支为 3.50 ÷ 5 = 0.70 英镑，8 支则为 8 × 0.70 = 5.60 英镑。在比例尺题中，1 : 200 表示图上每 1 cm 代表实际 200 cm，因此图上 6 cm 等于实际 1200 cm，即 12 m。
6. Question 5: Expanding, Factorising and Substitution | 第5题：展开、因式分解与代入求值
Question: simplify (x + 3)(x − 4). (c) Factorise 6x + 9. (d) Find the value of 4x² − 3x when x = −2.
题目：(a) 展开 3(2x − 5)。(b) 展开并化简 (x + 3)(x − 4)。(c) 因式分解 6x + 9。(d) 当 x = −2 时，求 4x² − 3x 的值。
× 2x = 6x and 3 × (−5) = −15, so the answer is 6x − 15. Watch the sign on the second term.
(a) 使用乘法分配律：括号外的项与括号内每一项相乘。3 × 2x = 6x，3 × (−5) = −15，所以答案是 6x − 15。特别注意第二项的符号。
3 × x = 3x, and 3 × (−4) = −12. Collecting the like terms −4x + 3x = −x gives x² − x − 12.
(b) 需要算四次乘积。把第一个括号的每一项分别乘第二个括号的每一项：x × x = x²，x × (−4) = −4x，3 × x = 3x，3 × (−4) = −12。合并同类项 −4x + 3x = −x，得到 x² − x − 12。
3). Check by expanding: 3 × 2x = 6x and 3 × 3 = 9, which matches the original expression.
(c) 是上述过程的逆运算，需要提取最大公因式。6x 与 9 的最大公因数是 3，所以 6x + 9 = 3(2x + 3)。展开检验：3 × 2x = 6x，3 × 3 = 9，与原式一致。
istake is to write (−2)² as −4; squaring a negative number always gives a positive result.
(d) 是代入求值，必须加括号。当 x = −2 时：4x² = 4 × (−2)² = 4 × 4 = 16，−3x = −3 × (−2) = +6，所以 4x² − 3x = 16 + 6 = 22。最常见的错误是把 (−2)² 写成 −4；负数的平方永远是正数。
7. Question 6: Solving Linear Equations and Inequalities | 第6题：解一元一次方程与不等式
Question: = 3x + 9. (b) Solve the inequality 4x + 3 < 19 and show the solution on a number line.
题目：(a) 解方程 5x − 7 = 3x + 9。(b) 解不等式 4x + 3 < 19，并在数轴上表示解集。
cting 3x from both sides gives 2x − 7 = 9. Adding 7 to both sides gives 2x = 16, so x = 8.
做 (a) 时，把含 x 的项集中到一边，常数项集中到另一边。两边同时减 3x，得 2x − 7 = 9；两边同时加 7，得 2x = 16，所以 x = 8。
equal and the solution is correct. This one-line check is worth doing in every equation question.
务必代回检验：5 × 8 − 7 = 33，3 × 8 + 9 = 33，左右两边相等，解正确。这个一行检验值得在每道方程题中都做一遍。
For part (b), treat the inequality like an equation. Subtract 3: 4x < 16. Divide by 4: x < 4. Draw an open circle at 4 (because 4 is not included) with an arrow pointing left. Remember the one special rule: multiplying or dividing both sides by a negative number reverses the inequality sign.
做 (b) 时，把不等式当作方程来处理。两边减 3：4x < 16；两边除以 4：x < 4。在数轴上 4 处画空心圆圈（因为不含 4），箭头指向左侧。记住那条特殊规则：不等式两边同乘或同除一个负数时，不等号方向要改变。
If a < b then a + c < b + c but −2a > −2b
8. Question 7: Sequences and the nth Term | 第7题：数列与第 n 项
Question: The sequence 5, 9, 13, 17, ... continues in the same pattern. (a) Write down the next two terms. (b) Find the nth term. (c) Hence find the 20th term.
题目：数列 5, 9, 13, 17, ... 按同一规律延续。(a) 写出接下来的两项。(b) 求第 n 项。(c) 由此求第 20 项。
s is 9 − 5 = 4, so we add 4 each time. The next two terms are 17 + 4 = 21 and 21 + 4 = 25.
先找出相邻项之间的规律。相邻两项之差为 9 − 5 = 4，因此每次加 4。接下来的两项是 17 + 4 = 21 和 21 + 4 = 25。
For the nth term, the constant difference 4 tells us the term contains 4n. Compare 4n (4, 8, 12, 16, ...) with the given sequence (5, 9, 13, 17, ...): each term is 1 more, so the nth term is 4n + 1.
求第 n 项时，恒定公差 4 说明式中含有 4n。把 4n（4, 8, 12, 16, ...）与原数列（5, 9, 13, 17, ...）比较：每一项都多 1，所以第 n 项是 4n + 1。
n a known term - putting n = 3 should give 4 × 3 + 1 = 13, which is indeed the third term.
做 (c) 时，代入 n = 20：4 × 20 + 1 = 81。注意你还可以用已知项检验公式——代入 n = 3 应得 4 × 3 + 1 = 13，而 13 正是第三项。
Some sequences do not increase by a fixed amount, such as 3, 6, 12, 24, ... where each term is double the previous one. Others are square numbers (1, 4, 9, 16, ...) or triangle numbers (1, 3, 6, 10, ...). The method is the same: find the rule first, then write it algebraically.
有些数列并非按固定差值递增，例如 3, 6, 12, 24, ...，每一项是前一项的 2 倍。还有平方数列（1, 4, 9, 16, ...）和三角数列（1, 3, 6, 10, ...）。方法相同：先找规律，再用代数式表示。
9. Question 8: Angle Facts and Geometric Reasoning | 第8题：角度性质与几何推理
Question: gle C. (b) A regular polygon has interior angles of 150° each. How many sides does it have? (c) Explain your reasoning using correct angle language.
题目：在三角形 ABC 中，∠A = 68°，∠B = 47°。(a) 求 ∠C。(b) 一个正多边形的每个内角都是 150°，它有多少条边？(c) 用正确的角度术语说明你的推理。
180°. So angle C = 180° − 68° − 47° = 65°. In a written answer you should name the reason: "angles in a triangle sum to 180°".
(a) 依据是三角形内角和为 180°。所以 ∠C = 180° − 68° − 47° = 65°。书面作答时应写出理由："三角形内角和为 180°"。
Part (b) uses the exterior angle. If each interior
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