📚 Year 7 CAIE Further Maths: Unit Test Mock Paper Analysis | Year 7 CAIE 进阶数学：单元测试模拟卷解析
This mock paper walkthrough is carefully crafted for Year 7 students following the CAIE Further Mathematics curriculum. It covers a range of essential topics such as integers, BIDMAS, algebraic manipulation, equations, sequences, angle properties, perimeter, mean, and ratio. Each question is fully explained with step‑by‑step reasoning to help you identify common mistakes and build confidence for the unit test.
本模拟卷解析专为学习 CAIE 进阶数学课程的七年级学生精心设计。内容涵盖整数、运算顺序、代数化简、方程、数列、角的基本性质、周长、平均数以及比例等核心知识点。每道题目均配有完整的逐步解析，帮助你发现常见错误，并为单元测试树立信心。
1. Introduction to the Mock Paper | 模拟卷简介
The mock paper consists of 10 questions, including multiple‑choice, short‑answer and problem‑solving items. It is worth 50 marks and the recommended working time is 45 minutes. The questions reflect the style and level of a typical unit test, so treat this as a real exam rehearsal.
模拟卷共 10 道题目，包含选择题、填空题和解答题，满分 50 分，建议用时 45 分钟。题目在风格和难度上贴近真实单元测试，请将此次练习视为正式考试预演。
Key areas tested: order of operations, directed numbers, simplifying expressions, solving linear equations, sequences, substitution with negative numbers, angle sum in a triangle, perimeter of a rectangle, calculating the mean, and word problems with ratio.
重点考查范围：运算顺序、有向数、代数式化简、一元一次方程、数列、含负数的代入求值、三角形内角和、矩形周长、平均数计算以及比例应用问题。
2. Question 1: Negative Numbers and BIDMAS | 题1：负数与运算顺序
Calculate (−3) × (−4) + 6 ÷ (−2). Choose the correct option: A. 9 B. −9 C. 3 D. −3.
计算 (−3) × (−4) + 6 ÷ (−2)。选择正确的一项：A. 9 B. −9 C. 3 D. −3。
left to right. (−3) × (−4) equals 12 because a negative times a negative gives a positive.
根据运算顺序（括号、指数、乘除、加减），我们先从左到右计算乘除法。 (−3) × (−4) = 12，因为负负得正。
a negative result. Finally, add the two outcomes: 12 + (−3) = 9. The correct answer is A.
接着，6 ÷ (−2) = −3。正数除以负数得负数。最后将两步结果相加：12 + (−3) = 9。正确答案是 A。
A frequent error is to add 6 and (−4) first, or to perform addition before division. Always respect the order of operations to avoid sign mistakes.
常见错误是先加 6 和 (−4)，或在除法前先做加法。务必严格遵循运算顺序，避免符号错误。
3. Question 2: Simplifying Algebraic Expressions | 题2：代数式化简
Simplify 3a − 2b + 4a + 5b − a. Options: A. 6a + 3b B. 7a + 7b C. 6a + 7b D. 4a + 3b.
化简 3a − 2b + 4a + 5b − a。选项：A. 6a + 3b B. 7a + 7b C. 6a + 7b D. 4a + 3b。
Collect like terms: group the a‑terms together and the b‑terms together. The a‑ 3a, 4a, and −a. Adding them gives 3a + 4a − a = 6a. Be careful with the invisible 1 in −a.
合并同类项：将含 a 的项与含 b 的项分别归类。a 项有 3a、4a 和 −a，相加得 3a + 4a − a = 6a。注意 −a 中隐含的系数 1。
The b‑ ing them: −2b + 5b = 3b. Therefore, the simplified expression is 6a + 3b. The answer is A.
b 项有 −2b 和 +5b。合并得 −2b + 5b = 3b。因此化简结果为 6a + 3b。答案是 A。
Learners sometimes treat −2b + 5b as −7b or forget the sign in front of −a. Re‑write the expression with signs clearly if needed.
有同学可能会把 −2b + 5b 算成 −7b，或忽略 −a 前的负号。必要时可以把每项的符号清楚地标出来再合并。
4. Question 3: Solving a Linear Equation | 题3：解一元一次方程
Solve 2x + 5 = 17. Options: A. x = 6 B. x = 11 C. x = 8.5 D. x = 12.
解方程 2x + 5 = 17。选项：A. x = 6 B. x = 11 C. x = 8.5 D. x = 12。
rom both sides to isolate the term containing x: 2x + 5 − 5 = 17 − 5, which gives 2x = 12.
首先，方程两边同时减去 5，以分离含 x 的项：2x + 5 − 5 = 17 − 5，得到 2x = 12。
2, so x = 6. Always check by substituting back: 2 × 6 + 5 = 12 + 5 = 17, which is correct.
然后，两边同时除以 2：2x ÷ 2 = 12 ÷ 2，得 x = 6。务必代入原方程检验：2 × 6 + 5 = 12 + 5 = 17，结果正确。
A common mistake is to divide only one side or to divide before subtracting. Performing inverse operations in the correct sequence is essential.
常见错误是只给一边除以 2，或先除后减。按正确的顺序进行逆运算是解方程的关键。
5. Question 4: Number Sequence | 题4：数字序列
Find the next term in the sequence: 2, 5, 10, 17, 26, ?
找出数列的下一项：2, 5, 10, 17, 26, ？
, 17 − 10 = 7, 26 − 17 = 9. The differences are increasing by 2 each time (3, 5, 7, 9...).
观察相邻项的差：5 − 2 = 3，10 − 5 = 5，17 − 10 = 7，26 − 17 = 9。差值依次增加 2（3, 5, 7, 9 ...）。
ce will be 9 + 2 = 11. Adding this to the last term: 26 + 11 = 37. So the next term is 37.
按照规律，下一个差值为 9 + 2 = 11。将 11 加到上一项：26 + 11 = 37。因此下一项是 37。
1²+1=2, 2²+1=5, 3²+1=10, etc. Recognising this algebraic rule is another powerful method.
该数列也可以用通项公式 n² + 1（n 从 1 开始）表示：1²+1=2，2²+1=5，3²+1=10…… 能够识别这一代数规律是更高级的方法。
6. Question 5: Evaluating Expressions with Negative Values | 题5：含负数值的表达式求值
If a = −2, evaluate 3a² − 2a + 7.
若 a = −2，求 3a² − 2a + 7 的值。
the square applies to the negative sign because of the brackets, giving a positive result.
将表达式中的 a 全部替换为 −2。先计算 a²：(−2)² = 4。注意，因为有括号，平方作用于整个 −2，结果为正。
l: a negative times a negative gives a positive, and the minus sign in front of 2a stays as subtraction of a negative, which is equivalent to addition.
然后逐项求值：3a² = 3 × 4 = 12。接着，−2a 表示 −2 × (−2) = 4。注意：负负得正，此时原表达式中的减号与负值结合，相当于加上 4。
Finally, add the constant 7: 12 + 4 + 7 = 23. The result is 23.
最后加上常数 7：12 + 4 + 7 = 23。最终结果为 23。
dle −2a by writing −2 × −2 = −4. Writing the substitution step clearly avoids such errors.
典型错误包括把 (−2)² 算成 −4（因为忽略了括号），或在计算 −2a 时得出 −4。把代入过程一步步写清楚可以有效避免这类错误。
7. Question 6: Angle Sum in a Triangle | 题6：三角形内角和
The interior angles of a triangle are (2x)°, (3x)° and (4x)°. Find the value of x.
一个三角形的三个内角分别为 (2x)°、(3x)° 和 (4x)°。求 x 的值。
angles in any triangle is 180°. Therefore, we can set up the equation: 2x + 3x + 4x = 180.
任何三角形的内角和均为 180°。因此可以列出方程：2x + 3x + 4x = 180。
Combine like terms on the left: 9x = 180. Divide both sides by 9: x = 180 ÷ 9 = 20.
合并左边同类项：9x = 180。两边除以 9：x = 180 ÷ 9 = 20。
tute x = 20 back: 2×20=40°, 3×20=60°, 4×20=80°. The sum 40+60+80 = 180°, which is correct.
检验：将 x = 20 代入各角，得 40°、60° 和 80°，总和 40+60+80 = 180°，结果正确。
Common errors include forgetting that the sum is 180° or adding the coefficients incorrectly. Always write the full equation before solving.
常见错误包括忘记内角和为 180°，或合并系数时计算错误。解题前先把完整的方程写下来有助于避免失误。
8. Question 7: Perimeter of a Rectangle | 题7：矩形周长
A rectangle has length (x + 3) cm and width 5 cm. Its perimeter is 34 cm. Determine x.
一个矩形的长是 (x + 3) cm，宽是 5 cm，周长为 34 cm。求 x 的值。
rectangle is P = 2(length + width). Substituting the given values: 2( (x + 3) + 5 ) = 34.
矩形周长公式为 P = 2 ×（长 + 宽）。代入已知条件：2( (x + 3) + 5 ) = 34。
Simplify inside the brackets: (x + 3) + 5 = x + 8. Now the equation becomes 2(x + 8) = 34.
化简括号内部分：(x + 3) + 5 = x + 8。方程变为 2(x + 8) = 34。
Divide both sides by 2: x + 8 = 17. Subtract 8 from both sides to find x = 9.
两边同时除以 2：x + 8 = 17。再两边减 8，得 x = 9。
Many mistakes happen when students forget to multiply the sum by 2 or combine the terms incorrectly.
验证：长 = 12 cm，宽 = 5 cm，周长 = 2(12+5)=34 cm。许多同学容易忘记将括号内的和乘以 2，或在合并项时出错。
9. Question 8: Calculating the Mean | 题8：计算平均数
Find the mean of the data set: 8, 12, 15, 20, 10.
求这组数据的平均数：8, 12, 15, 20, 10。
ether and then dividing by the number of values. Sum the data: 8 + 12 + 15 + 20 + 10 = 65.
平均数等于所有数据之和除以数据的个数。求和：8 + 12 + 15 + 20 + 10 = 65。
There are 5 values, so divide the total by 5: 65 ÷ 5 = 13. The mean is 13.
共有 5 个数据，因此用总和除以 5：65 ÷ 5 = 13。平均数为 13。
When a data set is not given in order, it is helpful to double‑check the sum to avoid missing a number. The mean gives a measure of central tendency and is useful for comparing data sets.
当数据未排序时，建议重新核对一次总和，以防遗漏个别数字。平均数反映了数据集的集中趋势，常用于比较不同数据集。
10. Question 9: Vertically Opposite and Supplementary Angles | 题9：对顶角和邻补角
Two straight lines intersect. One of the angles formed is 65°. State the sizes of the other three angles and give reasons for your answers.
两条直线相交，其中一个角为 65°。写出其余三个角的度数，并说明理由。
When two straight lines intersect, vertically opposite angles are equal. Therefore, the angle opposite the 65° angle is also 65°.
当两条直线相交时，对顶角相等。因此，与 65° 角相对的角也是 65°。
65° = 115°. This adjacent angle and its vertically opposite angle are therefore both 115°.
同一直线上的邻补角之和为 180°。与 65° 角相邻的角为 180° − 65° = 115°。该邻角的对顶角同样为 115°。
In summary, the four angles are 65°, 115°, 65°, 115°. Giving clear reasons 'vertically opposite angles are equal' and 'angles on a straight line sum to 180°' earns full marks.
综上所述，四个角分别为 65°, 115°, 65°, 115°。明确写出理由「对顶角相等」和「邻补角和为 180°」就能拿到全部分数。
11. Question 10: Ratio and Proportion Word Problem | 题10：比例应用问题
The ratio of boys to girls in a class is 3 : 4. There are 35 students in total. How many boys and how many girls are there?
某班级男生与女生的人数比为 3 : 4，班级总人数为 35。男生和女生各有多少人？
girls be 4x, where x is a common multiplier. The total number of students is 3x + 4x = 7x.
设男生人数为 3x，女生人数为 4x，其中 x 为一份的数量。总人数为 3x + 4x = 7x。
ividing both sides by 7 gives x = 5. Therefore, boys = 3 × 5 = 15, and girls = 4 × 5 = 20.
列出方程 7x = 35，解得 x = 5。所以男生有 3 × 5 = 15 人，女生有 4 × 5 = 20 人。
simplifies to 3 : 4, which matches the given ratio. The method of using a multiplier x is key to solving ratio problems.
验证：15 + 20 = 35，且 15 : 20 约简后得到 3 : 4，符合已知条件。使用乘数 x 是解决比例问题的核心方法。
12. Common Mistakes and Tips | 常见错误与技巧
Throughout the mock paper, several typical errors appear. With negative numbers, always use brackets when substituting to preserve signs. For BIDMAS, perform Division and Multiplication at the same priority level before Addition and Subtraction.
在整份模拟卷中，有几个典型错误反复出现。处理负数时，代入表达式中一定要加括号，以正确保留符号。对于运算顺序，乘除同级同级运算从左到右进行，做完后再算加减。
When simplifying expressions, double‑check that you have collected like terms correctly and paid attention to the sign in front of each term. In geometry, memorise the standard angle facts: vertically opposite angles are equal, angles on a straight line sum to 180°, and the angles in a triangle add to 180°.
化简代数式时，务必确认同类项合并无误，并留意每项前面的正负号。在几何部分，牢记基本角度性质：对顶角相等、邻补角和为 180°、三角形内角和为 180°。
For ratio word problems, introducing a multiplier such as x is much safer than guesswork. Always check your solution by ensuring the parts add to the total and reduce to the original ratio. Practising these techniques will significantly improve your accuracy in the unit test.
对于比例应用题，引入乘数（如 x）远比猜测更稳妥。每次都要检验：各部分之和是否等于总量，比例是否可约简回原比。多练习这些技巧将大大提高你在单元测试中的准确性。
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