📚 Year 7 CCEA Further Mathematics: Unit Test Mock Paper Analysis | 北爱尔兰CCEA七年级进阶数学：单元测试模拟卷解析
This article walks through a complete mock unit test designed for Year 7 CCEA Further Mathematics. We break down each question step‑by‑step, highlighting key techniques and common pitfalls. Use this as a revision tool to consolidate your understanding of numbers, algebra, geometry, and data handling.
本文完整解析一套专为CCEA七年级进阶数学设计的单元测试模拟卷。我们将逐题分解，详解解题技巧与易错点，帮助你巩固数、代数、几何与数据处理等核心知识，是考前复习的绝佳材料。
1. Evaluating Expressions with Indices and Fractions | 带指数和分数的表达式求值
Question 1: Evaluate 2³ × (4 + 5) – 12 ÷ 3.
问题1：计算 2³ × (4 + 5) – 12 ÷ 3。
We must follow the order of operations – in CCEA this is often remembered as BIDMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction).
解题时必须遵循运算顺序——CCEA课程常记为 BIDMAS（括号、指数、除/乘、加/减）。
Step 1: Calculate inside the brackets first. (4 + 5) gives 9.
步骤1：先算括号内，4 + 5 得 9。
The expression now reads 2³ × 9 – 12 ÷ 3.
此时表达式变为 2³ × 9 – 12 ÷ 3。
Step 2: Deal with the index. 2³ = 2 × 2 × 2 = 8.
步骤2：处理指数，2³ = 2 × 2 × 2 = 8。
We have 8 × 9 – 12 ÷ 3.
得到 8 × 9 – 12 ÷ 3。
vision and multiplication are next, working from left to right. 8 × 9 = 72 and 12 ÷ 3 = 4.
步骤3：接下来做乘除，从左往右进行。8 × 9 = 72，12 ÷ 3 = 4。
The expression simplifies to 72 – 4.
表达式简化为 72 – 4。
Step 4: Finally, subtract to get 68.
步骤4：最后相减得 68。
Always double‑check that you haven’t mixed up the order – a common error is to add before multiplying.
务必复查是否混淆了运算顺序——常见错误是先加后乘。
2. Converting Fractions to Decimals and Percentages | 分数转换为小数和百分数
Question 2: Write 5/8 as a decimal and a percentage.
问题2：将 5/8 写成小数和百分数形式。
To change a fraction to a decimal, divide the numerator by the denominator. Here, 5 ÷ 8.
将分数化为小数，用分子除以分母。此处为 5 ÷ 8。
Bring down 0: 20 ÷ 8 = 2 (16), remainder 4. Bring down 0: 40 ÷ 8 = 5. The result is 0.625.
8 除 5 不够，点上小数点补零：5.000 ÷ 8。8 × 6 = 48，50 – 48 余 2；落下 0 得 20，8 × 2 = 16，余 4；再落下 0 得 40，8 × 5 = 40。结果是 0.625。
So 5/8 = 0.625 as a decimal.
因此 5/8 = 0.625（小数）。
To convert a decimal to a percentage, multiply by 100%. 0.625 × 100% = 62.5%.
将小数化为百分数，乘以 100%。0.625 × 100% = 62.5%。
Remember: fractions, decimals, and percentages are just different ways of showing the same proportion.
切记：分数、小数和百分数只是同一比例的不同表现形式。
3. Simplifying Algebraic Expressions by Collecting Like Terms | 通过合并同类项化简代数式
Question 3: Simplify 3a + 5b – a + 2b.
问题3：化简 3a + 5b – a + 2b。
Identify like terms: the a‑terms are 3a and –a; the b‑terms are 5b and +2b.
识别同类项：a 项为 3a 和 –a；b 项为 5b 和 +2b。
Combine the a‑terms: 3a – a = 2a.
合并 a 项：3a – a = 2a。
Combine the b‑terms: 5b + 2b = 7b.
合并 b 项：5b + 2b = 7b。
The simplified expression is 2a + 7b.
化简后的式子为 2a + 7b。
Never try to add unlike terms such as 2a + 7b – they must stay separate.
千万不要试图合并不同类的项（如 2a + 7b），它们必须保持独立。
4. Solving Two‑Step Linear Equations | 解两步线性方程
Question 4: Solve 4x – 7 = 13.
问题4：解方程 4x – 7 = 13。
Our goal is to isolate x. Start by undoing the subtraction: add 7 to both sides.
目标是分离出 x。先消除减法：方程两边同时加 7。
4x – 7 + 7 = 13 + 7 → 4x = 20.
4x – 7 + 7 = 13 + 7 → 4x = 20。
Now divide both sides by 4: 4x ÷ 4 = 20 ÷ 4 → x = 5.
现在两边除以 4：4x ÷ 4 = 20 ÷ 4 → x = 5。
Check by substituting back: 4(5) – 7 = 20 – 7 = 13, which is correct.
代入检验：4(5) – 7 = 20 – 7 = 13，正确。
Always perform the same operation on both sides to keep the equation balanced.
务必在等号两边同时进行相同运算以保持平衡。
5. Perimeter with Algebraic Expressions | 含代数式的周长问题
Question 5: A rectangle has length (2x + 3) cm and width 4 cm. Find an expression for the perimeter.
问题5：长方形长为 (2x + 3) cm，宽为 4 cm。求周长的表达式。
Perimeter of a rectangle = 2 × (length + width). Substitute the given expressions.
长方形周长 = 2 × (长 + 宽)。代入已知表达式。
So, P = 2[(2x + 3) + 4] = 2(2x + 7).
得 P = 2[(2x + 3) + 4] = 2(2x + 7)。
Expand the bracket: 2 × 2x = 4x, and 2 × 7 = 14. So P = 4x + 14 cm.
展开括号：2 × 2x = 4x，2 × 7 = 14。因此 P = 4x + 14 cm。
Useful perimeter formulas are summarised below:
以下汇总了常用周长公式：
| Shape | Perimeter formula |
| Rectangle | P = 2(l + w) |
| Square | P = 4s |
| Triangle | P = a + b + c |
Expanding brackets carefully is a vital skill for later algebra topics.
仔细展开括号是后续代数学习的关键技能。
6. Angle Sum in a Triangle | 三角形内角和
Question 6: The angles in a triangle are x°, 2x°, and 3x°. Find x.
问题6：三角形三个内角分别为 x°、2x° 和 3x°。求 x。
In any triangle, the sum of the interior angles is 180°.
任意三角形的内角和均为 180°。
Set up an equation: x + 2x + 3x = 180.
列出方程：x + 2x + 3x = 180。
Combine like terms: 6x = 180.
合并同类项：6x = 180。
Divide both sides by 6: x = 30.
两边除以 6：x = 30。
Thus the angles are 30°, 60°, and 90° – this is a right‑angled triangle.
因此三个角分别为 30°、60° 和 90°——这是一个直角三角形。
Knowing the 180° rule for triangles and quadrilaterals (360°) is essential for geometry problems.
掌握三角形（180°）和四边形（360°）的内角和规则对解几何题至关重要。
7. Area of a Circle Using π ≈ 22/7 | 使用 π ≈ 22/7 计算圆面积
Question 7: Find the area of a circle with radius 7 cm. Use π ≈ 22/7.
问题7：半径为 7 cm 的圆，求面积。取 π ≈ 22/7。
The area A of a circle is given by A = πr².
圆面积公式为 A = πr²。
Substitute r = 7 and π = 22/7: A = (22/7) × 7².
代入 r = 7，π = 22/7：A = (22/7) × 7²。
7² = 49, so A = (22/7) × 49.
7² = 49，故 A = (22/7) × 49。
Cancel the 7 into 49: 49 ÷ 7 = 7. Then A = 22 × 7 = 154.
用 7 约分 49：49 ÷ 7 = 7。于是 A = 22 × 7 = 154。
The area is therefore 154 cm². Always include the square units.
因此面积为 154 cm²。务必带上平方单位。
Using 22/7 avoids messy decimals and is common in CCEA non‑calculator papers.
使用 22/7 可避免复杂小数，这在 CCEA 非计算器试卷中很常见。
8. Reverse Mean Problems | 逆向平均数问题
Question 8: The mean of five numbers is 12. Four numbers are 9, 14, 11, and 16. Find the fifth number.
问题8：五个数的平均数是 12，其中四个数为 9、14、11 和 16。求第五个数。
Mean = Total sum ÷ Number of values. So Total sum = Mean × Number of values.
平均数 = 总和 ÷ 个数。因此，总和 = 平均数 × 个数。
Here, Total sum = 12 × 5 = 60.
此处总和 = 12 × 5 = 60。
Add the four known numbers: 9 + 14 + 11 + 16 = 50.
将已知的四个数相加：9 + 14 + 11 + 16 = 50。
Subtract from the total sum to find the missing number: 60 – 50 = 10.
从总和中减去已知之和以求得缺失数：60 – 50 = 10。
The fifth number is 10.
第五个数是 10。
Reverse mean questions test your understanding of the relationship between sum, mean, and number of items.
逆向平均数问题旨在考查你对总和、平均数与项数之间关系的理解。
9. Ratio and Proportion with Counters | 比例与筹码问题
Question 9: A bag contains red, blue, and green counters in the ratio 3:2:1. There are 24 red counters. Find the total number of counters.
问题9：袋中红、蓝、绿筹码的比例为 3:2:1。红色筹码有 24 个。求筹码总数。
or every 3 reds there are 2 blues and 1 green. The total number of parts is 3 + 2 + 1 = 6.
比例 3:2:1 表示每 3 个红筹码对应 2 个蓝筹码和 1 个绿筹码。总份数为 3 + 2 + 1 = 6。
Red represents 3 parts, which equal 24 counters. So 1 part = 24 ÷ 3 = 8 counters.
红色占 3 份，等于 24 个。因此 1 份 = 24 ÷ 3 = 8 个筹码。
Since there are 6 parts in total, the total number of counters = 6 × 8 = 48.
共有 6 份，所以筹码总数 = 6 × 8 = 48。
A quick check: blues = 2 × 8 = 16, greens = 1 × 8 = 8. Total = 24 + 16 + 8 = 48.
快速验证：蓝筹码 = 2 × 8 = 16，绿筹码 = 1 × 8 = 8。总数 = 24 + 16 + 8 = 48。
Identifying the value of one part is the key to solving most ratio problems.
找出“一份”对应的数量是解决大多数比例问题的关键。
10. Expanding and Simplifying Bracketed Expressions | 去括号并化简表达式
Question 10: Simplify 2(3y – 1) – (y + 4).
问题10：化简 2(3y – 1) – (y + 4)。
Start by expanding the brackets. For 2(3y – 1), multiply each term inside by 2: 2 × 3y = 6y and 2 × (–1) = –2.
首先去括号。2(3y – 1) 中，括号内每一项乘以 2：2 × 3y = 6y，2 × (–1) = –2。
So 2(3y – 1) becomes 6y – 2.
因此 2(3y – 1) 变为 6y – 2。
Next, the – (y + 4) means multiply by –1: –1 × y = –y and –1 × 4 = –4.
接着，– (y + 4) 表示乘以 –1：–1 × y = –y，–1 × 4 = –4。
Now write the expression: 6y – 2 – y – 4.
现在写出表达式：6y – 2 – y – 4。
Collect like terms: 6y – y = 5y, and –2 – 4 = –6.
合并同类项：6y – y = 5y；–2 – 4 = –6。
The simplified expression is 5y – 6.
化简结果为 5y – 6。
Notice how the minus sign in front of a bracket changes the sign of every term inside – don’t lose those negatives!
注意括号前的负号会改变括号内每一项的正负号——千万别漏掉负号！
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