📚 Year 7 Cambridge Mathematics: Unit Test Mock Paper Walkthrough | Year 7 剑桥数学:单元测试模拟卷解析
This walkthrough takes you through a typical Year 7 Cambridge Mathematics unit test, covering key topics such as number operations, algebra basics, geometry, fractions, and data handling. Every question is explained step by step in both English and Chinese, so you can build confidence, avoid common mistakes, and deepen your understanding before the real test.
本文带你逐题解析一份典型的 Year 7 剑桥数学单元测试卷,涵盖数的运算、代数入门、几何、分数与数据处理等核心内容。每道题均提供中英文同步讲解,帮助你建立信心、避开易错点,并在真正的考试前加深理解。
1. Test Overview and Structure | 试卷结构与整体说明
This mock paper contains three sections: A – short‑answer arithmetic (15 marks), B – problem solving with diagrams (20 marks), and C – reasoning and data interpretation (15 marks). The total time allowed is 45 minutes, and calculators are not permitted throughout the test.
本模拟卷分为三节:A 节——简短算术作答(15 分),B 节——带图形的应用题(20 分),C 节——推理与数据解读(15 分)。考试总时间为 45 分钟,全程禁止使用计算器。
You must show all working in the spaces provided. Marks are awarded for method, not only for the final answer. Even if your answer is wrong, a correct step can still earn part of the marks.
所有解题过程须写在指定区域。评分看重解题方法,而不仅仅是最终答案。即使最终答案错误,正确的解题步骤仍可获得部分分数。
2. Question 1 – Adding and Subtracting Negative Numbers | 第 1 题:负数的加减
The first question asks: Work out -7 + 4 and 3 – (-2). Remember that subtracting a negative is the same as adding a positive: 3 – (-2) = 3 + 2 = 5.
第一道题要求计算 -7 + 4 和 3 – (-2)。请记住:减去一个负数等于加上它的相反数,即 3 – (-2) = 3 + 2 = 5。
For -7 + 4, imagine moving 7 units left on a number line, then 4 units right. You land at -3. Many students confuse signs; writing a number line quickly can prevent sign errors.
对于 -7 + 4,想像在数轴上向左移动 7 个单位,再向右移动 4 个单位,最终停在 -3。很多学生容易搞混符号,快速画出数轴可以有效避免符号错误。
3. Question 2 – Order of Operations (BIDMAS) | 第 2 题:运算顺序(BIDMAS 法则)
Evaluate 4 + 3 × 5 – 2. According to BIDMAS (Brackets, Indices, Division/Multiplication, Addition/Subtraction), multiplication comes before addition and subtraction. So do 3 × 5 = 15 first, then 4 + 15 – 2 = 17.
计算 4 + 3 × 5 – 2。根据 BIDMAS 法则(括号、指数、除/乘、加/减),乘法优先于加减。因此先算 3 × 5 = 15,再算 4 + 15 – 2 = 17。
A common mistake is to work left to right: 4 + 3 = 7, then 7 × 5 = 35, then 35 – 2 = 33, which is incorrect. Always scan the expression for multiplication or division first.
常见错误是从左往右依次计算:4 + 3 = 7,接着 7 × 5 = 35,再 35 – 2 = 33,这是错误的。务必先扫描整个表达式,优先处理乘除。
4. Question 3 – Finding Factors and Multiples | 第 3 题:因数与倍数的寻找
List all the factors of 36. Factors come in pairs: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6. So the complete list is 1, 2, 3, 4, 6, 9, 12, 18, 36. Writing them in order prevents missing any.
列出 36 的所有因数。因数成对出现:1 × 36、2 × 18、3 × 12、4 × 9、6 × 6。因此完整因数为 1、2、3、4、6、9、12、18、36。按顺序书写可避免遗漏。
The question also asks for the first three common multiples of 4 and 6. Multiples of 4: 4, 8, 12, 16, 20, 24, … Multiples of 6: 6, 12, 18, 24, … The first three common ones are 12, 24, and 36.
题目还要求写出 4 和 6 的前三个公倍数。4 的倍数:4、8、12、16、20、24…… 6 的倍数:6、12、18、24…… 前三个公倍数是 12、24 和 36。
5. Question 4 – Equivalent Fractions and Simplification | 第 4 题:等值分数与化简
Fill in the missing number: 3/5 = ?/20. Since 5 × 4 = 20, multiply the numerator by the same factor: 3 × 4 = 12. So 3/5 = 12/20. This checks understanding of equivalent fractions.
填空:3/5 = ?/20。因为分母 5 × 4 = 20,所以分子也要乘以相同的倍数:3 × 4 = 12。因此 3/5 = 12/20。这考查对等值分数的理解。
Simplify 18/24 to its simplest form. Find the highest common factor (HCF) of 18 and 24, which is 6. Divide both numerator and denominator by 6: 18 ÷ 6 = 3, 24 ÷ 6 = 4. Answer: 3/4.
将 18/24 化为最简分数。先找出 18 和 24 的最大公因数(HCF)为 6,分子分母同时除以 6:18 ÷ 6 = 3,24 ÷ 6 = 4,答案为 3/4。
6. Question 5 – Algebraic Expressions and Substitution | 第 5 题:代数表达式与代入求值
If a = 4 and b = -2, find the value of 3a + 2b. Substitute carefully: 3 × 4 + 2 × (-2) = 12 + (-4) = 8. Always use brackets around negative numbers when substituting to avoid sign mistakes.
已知 a = 4,b = -2,求 3a + 2b 的值。仔细代入:3 × 4 + 2 × (-2) = 12 + (-4) = 8。代入负数时务必使用括号,以免符号出错。
Another part asks to collect like terms: 5x + 3y – 2x + y. Combine x terms: 5x – 2x = 3x. Combine y terms: 3y + y = 4y. Final expression: 3x + 4y.
另一部分要求合并同类项:5x + 3y – 2x + y。合并 x 项:5x – 2x = 3x;合并 y 项:3y + y = 4y。最终表达式为 3x + 4y。
7. Question 6 – Angles on a Straight Line and Around a Point | 第 6 题:直线上的角与围绕一点的角
A diagram shows three angles on a straight line: two are 65° and 45°, find the third. Angles on a straight line sum to 180°. 65 + 45 = 110, so the missing angle is 180 – 110 = 70°.
图中显示一条直线上有三个角,其中两个分别为 65° 和 45°,求第三个角。直线上的角之和为 180°。65 + 45 = 110,因此未知角为 180 – 110 = 70°。
Another diagram shows angles around a point: 130°, 90°, and 70°, find angle x. Angles around a point sum to 360°. 130 + 90 + 70 = 290, so x = 360 – 290 = 70°.
另一幅图展示围绕一点的角:130°、90° 和 70°,求角 x。围绕一点的角之和为 360°。130 + 90 + 70 = 290,因此 x = 360 – 290 = 70°。
8. Question 7 – Perimeter and Area of Rectangles | 第 7 题:长方形的周长与面积
A rectangle has length 8 cm and width 5 cm. Find its perimeter. P = 2 × (length + width) = 2 × (8 + 5) = 2 × 13 = 26 cm. Be sure to include the unit.
一个长方形的长为 8 cm,宽为 5 cm,求其周长。周长 = 2 × (长 + 宽) = 2 × (8 + 5) = 2 × 13 = 26 cm。务必带上单位。
Find the area of the same rectangle. Area = length × width = 8 × 5 = 40 cm². Note the square unit for area. Knowing the difference between perimeter (linear) and area (square) is essential.
求同一个长方形的面积。面积 = 长 × 宽 = 8 × 5 = 40 cm²。注意面积单位是平方单位。区分周长(线性单位)与面积(平方单位)至关重要。
9. Question 8 – Interpreting a Bar Chart | 第 8 题:条形统计图的解读
A bar chart shows the number of books read by five students. The task: find the range, which is the difference between the highest and lowest values. If the highest bar is at 14 books and the lowest at 5, range = 14 – 5 = 9 books.
条形统计图显示了五名学生阅读的书籍数量。要求计算极差,即最大值与最小值的差。若最高柱形对应 14 本书,最低柱形对应 5 本书,则极差 = 14 – 5 = 9 本书。
Also calculate the mean number of books read. Add all values: 14 + 10 + 8 + 5 + 13 = 50. Divide by 5: mean = 10 books. Show the addition step clearly to gain full method marks.
同时计算阅读本书的平均数。将所有值相加:14 + 10 + 8 + 5 + 13 = 50。除以 5:平均数 = 10 本书。清晰地展示加法步骤,以获得完整的方法分。
10. Question 9 – Fraction of an Amount and Word Problem | 第 9 题:求一个数的几分之几与应用题
In a school of 540 students, 2/5 are in Year 7. Find 1/5 first: 540 ÷ 5 = 108. Then multiply by 2: 108 × 2 = 216 students. This two‑step method works for any unit fraction.
一所学校有 540 名学生,其中 2/5 就读于 Year 7。先求 1/5:540 ÷ 5 = 108。再乘以 2:108 × 2 = 216 名学生。这种两步法适用于任何单位分数。
The question extends: Of these Year 7 students, 1/3 walk to school. How many walk? Calculate 1/3 of 216: 216 ÷ 3 = 72. Always reread the problem to ensure you are taking the fraction of the correct quantity.
题目延伸:在这些 Year 7 学生中,有 1/3 步行上学。问步行人数是多少?计算 216 的 1/3:216 ÷ 3 = 72。务必再次读题,确保求的是正确数量的几分之几。
11. Question 10 – Solving One‑Step Equations | 第 10 题:解一步方程
Solve x + 7 = 15. Subtract 7 from both sides: x = 8. For y – 3 = 10, add 3 to both sides: y = 13. Always do the inverse operation to isolate the variable.
解方程 x + 7 = 15。两边同时减去 7,得 x = 8。对于 y – 3 = 10,两边同时加上 3,得 y = 13。始终使用逆运算来分离未知数。
For 5p = 45, divide both sides by 5: p = 9. For m/4 = 6, multiply both sides by 4: m = 24. Writing the operation on both sides balances the equation, which is a fundamental concept.
对于 5p = 45,两边同时除以 5,得 p = 9。对于 m/4 = 6,两边同时乘以 4,得 m = 24。在等式两边书写相同的运算以保持平衡,这是一个基本概念。
12. Common Mistakes and Final Advice | 常见错误与最后建议
Many errors come from rushing through signs, forgetting BIDMAS, or mixing up area and perimeter. Always double‑check negative number calculations and ensure fractions are in simplest form before writing the final answer.
许多错误源自匆忙导致符号错漏、忘记 BIDMAS 法则,或混淆面积与周长。务必复查负数计算,并确保分数化为最简形式后再书写最终答案。
Use the last few minutes of the test to review your answers. If a question asks for units, make sure you have written them. A well‑presented solution with clear working is your best tool for scoring high marks.
利用考试最后几分钟检查答案。如果题目要求写出单位,确保已经标上。清晰展示解题步骤是你获取高分的最佳工具。
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