📚 Year 7 CAIE Maths: Unit Test Mock Paper Analysis | Year 7 CAIE 数学：单元测试模拟卷解析
This article walks you through a complete mock paper for the Year 7 CAIE Mathematics unit test. Every question is carefully chosen to reflect the Cambridge Lower Secondary Stage 7 curriculum, covering integers, fractions, decimals, percentages, algebra, geometry, data handling, ratio and probability. Detailed, step‑by‑step solutions are provided in both English and Chinese, helping you identify common pitfalls and build strong exam techniques. Use this analysis to check your understanding and boost your confidence before the real test.
本文为你详细解析一份完整的 Year 7 CAIE 数学单元测试模拟卷。所有题目均紧扣剑桥初中阶段7年级内容，涵盖整数、分数、小数、百分数、代数、几何、数据处理、比与比例以及概率。每道题都配有中英双语的逐步解答，帮助你发现常见错误、掌握解题技巧。请借助这份解析检验自己的学习成果，在正式考试前增强自信。
1. Integer Operations: Orders and Negatives | 整数运算：运算顺序与负数
Calculate: (−25) + 18 − (−7) × 2.
计算：(−25) + 18 − (−7) × 2。
First, handle the multiplication. (−7) × 2 = −14. The expression becomes (−25) + 18 − (−14). Subtracting a negative means adding a positive, so − (−14) = +14. Now we have −25 + 18 + 14. Work left to right: −25 + 18 = −7, then −7 + 14 = 7. The final answer is 7.
先处理乘法部分：(−7) × 2 = −14。原式变为 (−25) + 18 − (−14)。减去一个负数相当于加上它的相反数，所以 − (−14) = +14。接着从左到右计算：−25 + 18 = −7，然后 −7 + 14 = 7。最终答案是 7。
Many students forget that ‘−(−14)’ turns into addition. Always rewrite subtraction of a negative as addition before moving ahead. Also, remember BIDMAS: Brackets, Indices, Division/Multiplication (left to right), Addition/Subtraction (left to right).
很多学生会忘记“−(−14)”要变成加法。在做题时，先把减去负数改写成加法，才不容易出错。同时要记住运算顺序 BIDMAS：括号、指数、乘除（从左到右）、加减（从左到右）。
2. Fractions: Adding and Subtracting Unlike Denominators | 分数：异分母加减
Work out: 3/5 + 2/3 − 1/6. Give your answer as a mixed number in simplest form.
计算：3/5 + 2/3 − 1/6。结果用最简带分数表示。
Find the lowest common denominator (LCD) of 5, 3 and 6. The multiples show 30 is the smallest number all three divide into. Convert each fraction: 3/5 = 18/30, 2/3 = 20/30, 1/6 = 5/30. Now compute: 18/30 + 20/30 − 5/30 = (18+20−5)/30 = 33/30. Simplify 33/30: divide numerator and denominator by 3 to get 11/10. As a mixed number, 11/10 = 1 1/10.
先找到 5、3、6 的最小公分母。列出倍数可知 30 是它们的最小公倍数。改写各分数：3/5 = 18/30，2/3 = 20/30，1/6 = 5/30。然后计算：18/30 + 20/30 − 5/30 = (18+20−5)/30 = 33/30。约分：分子分母同除以 3 得到 11/10。化为带分数：11/10 = 1 1/10。
A common mistake is adding denominators directly. Always find a common denominator first. Also, remember to simplify your final answer. Leaving 33/30 without simplifying to 11/10 would lose marks.
常见的错误是直接将分母相加。一定要先通分。另外，最后的答案必须化成最简形式，如果保留 33/30 而没有约分成 11/10 是会扣分的。
3. Decimals and Percentages: Conversion and Finding a Percentage | 小数与百分数：互化与求百分数
(a) Convert 0.05 to a percentage. (b) Find 20% of 150 kg.
(a) 把 0.05 化成百分数。 (b) 求 150 千克的 20%。
To change a decimal to a percentage, multiply by 100%. So 0.05 × 100% = 5%. For part (b), 20% means 20/100 = 0.2. Multiply 0.2 by 150 kg: 0.2 × 150 = 30. So 20% of 150 kg is 30 kg. Always include the unit in the answer.
将小数化成百分数，乘以 100% 即可：0.05 × 100% = 5%。第 (b) 题中，20% 可写作 20/100 = 0.2。用 0.2 乘以 150 千克：0.2 × 150 = 30。因此 150 千克的 20% 是 30 千克。答题时务必带上单位。
Students sometimes confuse 0.05 with 5% and write 0.05% by mistake. Remember: 0.05 as a percentage is 5%. Another useful check: ‘of’ in percentage problems means multiply. Using the decimal form 0.2 makes the calculation straightforward.
学生有时会把 0.05 和 5% 弄混，错误地写成 0.05%。请记住：0.05 化成百分数是 5%。一个有用的检查点：在百分数应用题中，“of”意味着乘法运算。使用小数形式 0.2 能让计算更直接。
4. Sequences: Finding the nth Term Pattern | 数列：寻找第 n 项规律
Find the next two terms in the sequence: 4, 9, 16, 25, … . Then write the rule for the nth term.
找出数列 4, 9, 16, 25, … 的后两项，并写出第 n 项的规律。
Observe the pattern: 4 = 2², 9 = 3², 16 = 4², 25 = 5². The terms are perfect squares starting from 2². Position 1 is 2², position 2 is 3², position 3 is 4², so term n = (n+1)². The next two terms are for n=5 and n=6: (5+1)² = 6² = 36, and (6+1)² = 7² = 49. So the next two terms are 36 and 49. The nth term rule is (n+1)².
观察规律：4 = 2²，9 = 3²，16 = 4²，25 = 5²。各项是从 2² 开始的完全平方数。第 1 项是 2²，第 2 项是 3²，第 3 项是 4²，因此第 n 项为 (n+1)²。后两项分别对应 n=5 和 n=6：(5+1)² = 6² = 36，(6+1)² = 7² = 49。所以后两项是 36 和 49。第 n 项规律为 (n+1)²。
Always link the term number (n) to the value. With square sequences, check whether they begin at 1² or 2². Writing the position number above each term helps: n=1 → 4, n=2 → 9, etc. This makes it easier to spot (n+1)² rather than n².
务必把项数 n 与相应数值联系起来。对于平方数列，要检查它是从 1² 还是 2² 开始。可以在每一项的上方标注位置数：n=1 → 4，n=2 → 9，……这样更容易看出规律是 (n+1)²，而不是 n²。
5. Solving Simple Equations: Balancing Method | 解简单方程：平衡法
Solve 3x − 7 = 14.
解方程：3x − 7 = 14。
To isolate x, first add 7 to both sides: 3x − 7 + 7 = 14 + 7, so 3x = 21. Next, divide both sides by 3: 3x/3 = 21/3, giving x = 7. Check by substituting back: 3(7) − 7 = 21 − 7 = 14, which matches the right‑hand side.
为了将 x 单独放在一边，首先两边同时加 7：3x − 7 + 7 = 14 + 7，得 3x = 21。然后两边同时除以 3：3x/3 = 21/3，得到 x = 7。代入检验：3(7) − 7 = 21 − 7 = 14，与等式右边一致。
Always perform the same operation on both sides. Common errors include adding 7 only to the right or forgetting to divide the entire left side by 3. Showing each step clearly prevents these mistakes.
务必对方程两边同时进行相同的运算。常见错误包括只在右边加 7，或者忘记将左边的整体除以 3。清晰地写出每一步能有效避免这类错误。
6. Geometry: Angles in a Triangle and on a Straight Line | 几何：三角形内角与平角
(a) Two angles of a triangle are 45° and 65°. Find the third angle.
(b) On a straight line, one angle is 3x and the adjacent angle is 2x. Calculate the value of x.
(a) 一个三角形的两个内角分别是 45° 和 65°，求第三个角。
(b) 在一条直线上，一个角为 3x，相邻角为 2x，求 x 的值。
For part (a), the angles in a triangle sum to 180°. Add the known angles: 45° + 65° = 110°. Subtract from 180°: 180° − 110° = 70°. The third angle is 70°. For part (b), angles on a straight line add up to 180°. So 3x + 2x = 180°, giving 5x = 180°. Divide by 5: x = 36°.
对于 (a)，三角形内角和为 180°。将已知两角相加：45° + 65° = 110°。用 180° 减去这个和：180° − 110° = 70°，第三个角就是 70°。对于 (b)，平角上的角之和为 180°，因此 3x + 2x = 180°，即 5x = 180°，两边除以 5 得 x = 36°。
Many students confuse the 180° rule for triangles with the 360° rule for quadrilaterals. Stick to clear working: write ‘Sum of angles = 180°’ before substituting. For the straight line question, always add the angles and set the total to 180°.
很多学生会把三角形的 180° 规则与四边形的 360° 规则混淆。做题时在代入前先写出“角和 = 180°”。处理平角问题时，牢记将所有角相加，并令总和等于 180°。
7. Perimeter and Area: Rectangles and Composite Shapes | 周长与面积：矩形和组合图形
A rectangular garden measures 8 m by 5 m. (a) Calculate its perimeter. (b) Calculate its area. The garden is extended by attaching a 3 m by 2 m rectangular shed area along one of the 5 m sides. Find the new total area.
一个长方形花园长 8 米、宽 5 米。(a) 求周长。(b) 求面积。沿其中一条 5 米的边附加一个 3 米×2 米的长方形工具房区域，求新的总面积。
Perimeter of rectangle = 2 × (length + width) = 2 × (8 + 5) = 2 × 13 = 26 m. Area = length × width = 8 × 5 = 40 m². For the extension, the shed area = 3 × 2 = 6 m². The two areas simply add together because they do not overlap. Total area = 40 + 6 = 46 m². The perimeter would change, but the question only asks for total area.
矩形周长 = 2 × (长 + 宽) = 2 × (8 + 5) = 2 × 13 = 26 米。面积 = 长 × 宽 = 8 × 5 = 40 平方米。附加的工具房区域面积为 3 × 2 = 6 平方米。由于两部分不重叠，总面积可以直接相加：40 + 6 = 46 平方米。此时周长会发生变化，但题目只要求总面积。
Always write the formula first. Common errors involve mixing up units or using the wrong pair of dimensions for composite shapes. Draw a quick sketch to check whether areas add or overlap.
做题时先写出公式。常见错误包括单位混淆，或在组合图形中用错尺寸。画一个简单的示意图可以帮助判断面积是直接相加还是有重叠部分。
8. Data Handling: Mean from a Bar Chart | 数据处理：由柱状图求平均数
A bar chart shows the number of books read by students in a week:
0 books: 4 students,
1 book: 6 students,
2 books: 7 students,
3 books: 3 students.
Calculate the mean number of books read.
某柱状图展示了一周内学生读书本数的情况：
0 本：4 名学生，
1 本：6 名学生，
2 本：7 名学生，
3 本：3 名学生。
请计算平均每人读书的本数。
First, find the total number of students: 4 + 6 + 7 + 3 = 20 students. Next, find the total number of books read: (0 × 4) + (1 × 6) + (2 × 7) + (3 × 3) = 0 + 6 + 14 + 9 = 29 books. Mean = total books ÷ total students = 29 ÷ 20 = 1.45 books. The mean number of books read is 1.45.
先求学生总人数：4 + 6 + 7 + 3 = 20 人。再求读书的总本数：(0×4) + (1×6) + (2×7) + (3×3) = 0 + 6 + 14 + 9 = 29 本。平均数 = 总本数 ÷ 总人数 = 29 ÷ 20 = 1.45 本。平均每人读书 1.45 本。
Make sure to multiply each number of books by its frequency. A typical mistake is to add the books ignoring frequency (0+1+2+3) and divide by 4, which gives 1.5, not the correct 1.45. Always use the frequency table carefully.
一定要将每类读书本数乘以其对应的频数。典型错误是把读书本数简单相加 (0+1+2+3) 再除以 4，得到 1.5，而不是正确的 1.45。务必仔细使用频数表。
9. Ratio and Proportion: Scaling a Recipe | 比与比例：食谱的按比例调整
A recipe for 4 people uses 200 g of flour and 50 g of sugar. How much flour and sugar are needed for 10 people?
一份供 4 人食用的食谱需要 200 克面粉和 50 克糖。若要供 10 人食用，需要面粉和糖各多少？
Find the amount needed for 1 person first. Flour per person: 200 g ÷ 4 = 50 g. Sugar per person: 50 g ÷ 4 = 12.5 g. For 10 people, multiply each by 10: flour = 50 × 10 = 500 g; sugar = 12.5 × 10 = 125 g. Alternatively, use the scale factor from 4 to 10, which is 10/4 = 2.5. Then flour = 200 × 2.5 = 500 g, sugar = 50 × 2.5 = 125 g. Both methods give the same answers.
先求每人所需的量。面粉：200 g ÷ 4 = 50 g；糖：50 g ÷ 4 = 12.5 g。10 人的量则各乘以 10：面粉 = 50 × 10 = 500 g；糖 = 12.5 × 10 = 125 g。也可以使用比例因子：4 人到 10 人的放大系数为 10/4 = 2.5。于是面粉 = 200 × 2.5 = 500 g，糖 = 50 × 2.5 = 125 g。两种方法结果相同。
The unitary method (finding for 1 unit) is very reliable and reduces scaling errors. When using a multiplier, always write the ratio new/old to avoid inverting. Clearly state the final amounts with units.
先算出单一量（每份量）的方法非常可靠，能减少比例转换中的错误。使用倍数时，注意用新人数除以原人数，避免颠倒分子分母。最终答案一定要写明单位。
10. Probability: Simple Events and Complementary Outcomes | 概率：简单事件与互补事件
A bag contains 5 red, 3 blue and 2 green marbles. One marble is taken at random. Find: (a) P(red), (b) P(not blue). Give your answers as fractions in simplest form.
一个袋子里有 5 颗红色、3 颗蓝色和 2 颗绿色弹珠。随机取出 1 颗。求：(a) P(红)，(b) P(非蓝)。答案请用最简分数表示。
Total number of marbles = 5 + 3 + 2 = 10. For (a), number of red marbles = 5, so P(red) = 5/10 = 1/2. For (b), ‘not blue’ means red or green. Number of not blue = 5 red + 2 green = 7. So P(not blue) = 7/10. Alternatively, P(blue) = 3/10, therefore P(not blue) = 1 − 3/10 = 7/10. Both ways are correct.
弹珠总数 = 5 + 3 + 2 = 10 颗。(a) 红色弹珠有 5 颗，所以 P(红) = 5/10 = 1/2。(b) “非蓝”即红色或绿色。非蓝弹珠数 = 5 + 2 = 7 颗，因此 P(非蓝) = 7/10。也可以先求 P(蓝) = 3/10，再用 1 − 3/10 = 7/10。两种做法都对。
Always simplify probabilities when possible. 5/10 reduces to 1/2. Many students mistakenly write P(not blue) as 3/10 by only considering the blue marbles. Understanding ‘not’ as the complement helps avoid this: P(not A) = 1 − P(A).
只要有可能，概率分数一定要约分，5/10 要约成 1/2。很多学生错误地认为 P(非蓝) 就是 3/10，因为他们只考虑蓝色部分。理解“非某事件”就是互补事件这一概念有助于避免错误：P(非 A) = 1 − P(A)。
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