📚 Year 8 CIE Maths: Mock Unit Test Breakdown | Year 8 CIE 数学：单元测试模拟卷解析
This walkthrough covers a typical Year 8 CIE Mathematics unit test, breaking down ten carefully selected questions that represent the core syllabus areas. Each solution is explained step by step, with common mistakes highlighted so you can refine your technique and boost your confidence before the real assessment.
这份解析针对一份典型的 Year 8 CIE 数学单元测试，拆解了十道精心挑选的试题，覆盖核心考纲范围。每道题的解答都逐步讲解，并点出常见错误，帮助你完善解题技巧，在真正的评估前增强信心。
1. Paper Overview | 试卷概览
This mock test is designed for a 45‑minute session. It includes arithmetic, fractions, decimals, percentages, algebraic manipulation, equation solving, geometry, statistics and probability. The questions mirror the style and difficulty found in CIE Year 8 unit tests, with a mix of straightforward calculations and multi‑step problems. You are expected to show full working where appropriate.
这份模拟卷按照 45 分钟完成来设计。内容包括算术、分数、小数、百分比、代数操作、方程求解、几何、统计和概率。试题在风格和难度上贴近 CIE Year 8 单元测试，兼顾直接计算与多步骤问题。你需要视情况写出完整的演算过程。
2. Question 1: Order of Operations | 问题1：运算顺序
Question: Evaluate 5 + 3 × (8 − 2)² ÷ 9.
题目：计算 5 + 3 × (8 − 2)² ÷ 9。
ply the index: 6² = 36. Then work from left to right with multiplication and division: 3 × 36 = 108, then 108 ÷ 9 = 12. Lastly, perform the addition: 5 + 12 = 17.
解答：按 BIDMAS（括号、指数、乘除、加减）顺序进行。先处理括号：8 − 2 = 6。接着计算指数：6² = 36。然后从左到右进行乘除运算：3 × 36 = 108，接着 108 ÷ 9 = 12。最后做加法：5 + 12 = 17。
Answer: 17
答案：17
Common mistake: Students often rush to add 5 + 3 first, getting 8, which breaks the convention. Always remember that indices are handled before multiplication, and multiplication before addition.
常见错误：学生经常急于先算 5 + 3 得 8，这就打乱了运算法则。务必牢记指数优先于乘法，乘法优先于加法。
3. Question 2: Fractions & Decimals | 问题2：分数与小数
Question: (a) Write 7/8 as a decimal. (b) Write 0.375 as a fraction in its simplest form.
题目：(a) 将 7/8 写作小数。(b) 将 0.375 化成最简分数。
umerator and denominator by 125 (their highest common factor): 375 ÷ 125 = 3, 1000 ÷ 125 = 8. Thus 0.375 = 3/8.
解答：(a) 用 7 除以 8。7 ÷ 8 = 0.875，因此小数为 0.875。(b) 小数 0.375 表示 375 个千分之一，写作 375/1000。分子分母同时除以最大公因数 125：375 ÷ 125 = 3，1000 ÷ 125 = 8。因此 0.375 = 3/8。
Answers: (a) 0.875 (b) 3/8
答案：(a) 0.875 (b) 3/8
speed up conversions. To simplify a fraction, always divide by the greatest common factor.
温馨提示：熟记 1/8 = 0.125 等常见转换能加快速度。化简分数时，一定要除以最大公因数。
4. Question 3: Percentages | 问题3：百分比
Question: A shirt originally priced at $40 is reduced by 15% in a sale. Find the sale price.
题目：一件原价 $40 的衬衫在促销中降价 15%。求促销价。
Solution: Method 1: Find 15% of $ 0.15 × 40 = 6. Subtract this from the original: 40 − 6 = 34. Method 2: Recognise that a 15% reduction leaves 85% of the original price. 85% as a decimal is 0.85, so 0.85 × 40 = 34. Therefore the sale price is $34.
解答：方法1：求出 $40 的 15%：0.15 × 40 = 6。从原价中减去：40 − 6 = 34。方法2：15% 的降价意味着剩余原价的 85%。85% 化为小数是 0.85，因此 0.85 × 40 = 34。促销价为 $34。
Sale price: $34
促销价：$34
Common mistake: Some learners calculate 15% and then add it, or forget to convert the percentage to a decimal correctly. Multiplying by 0.85 directly is often quicker and reduces errors.
常见错误：部分学生会求出 15% 后反而加上去，或忘记把百分数正确化为小数。直接乘以 0.85 通常更快捷且不易出错。
5. Question 4: Algebraic Simplification | 问题4：代数化简
Question: Simplify the expression 4x − 3y + 2x + 7y − 5.
题目：化简代数式 4x − 3y + 2x + 7y − 5。
Solution: Group like terms. The x‑terms: 4x + 2x = 6x. The y‑ 3y + 7y = 4y. The constant term is −5. The simplified expression is therefore 6x + 4y − 5.
解答：合并同类项。含 x 的项：4x + 2x = 6x。含 y 的项：−3y + 7y = 4y。常数项为 −5。因此化简后为 6x + 4y − 5。
Simplified: 6x + 4y − 5
化简结果：6x + 4y − 5
Key point: Only terms with exactly the same variable part can be combined. Watch the signs in front of each term; −3y still stays negative when moved.
关键点：只有变量部分完全相同的项才能合并。注意每项前面的符号；−3y 无论移到哪都依然是负的。
6. Question 5: Solving Equations | 问题5：解方程
Question: Solve 3(y − 4) = 15.
题目：解方程 3(y − 4) = 15。
9. Method 2: Divide both sides by 3: y − 4 = 5. Then add 4: y = 9. The solution is y = 9.
解答：可先展开括号或先将两边除以 3。方法1：展开：3y − 12 = 15。两边加 12：3y = 27。除以 3：y = 9。方法2：两边除以 3：y − 4 = 5。再加 4：y = 9。解为 y = 9。
Answer: y = 9
答案：y = 9
rm by 3 and forget the −4. Always use the distributive property carefully: 3 × (−4) = −12.
常见错误：展开时，学生有时只把 3 乘给第一项而忘记 −4。务必小心运用分配律：3 × (−4) = −12。
7. Question 6: Angles on a Straight Line | 问题6：直线上的角
Question: Two angles lie on a straight line. The angles are (2x)° and (x + 30)°. Find the value of x and state the size of each angle.
题目：两个角位于一条直线上，分别为 (2x)° 和 (x + 30)°。求 x 的值并写出每个角的大小。
ubtract 30: 3x = 150. Divide by 3: x = 50. The angles are 2 × 50 = 100° and 50 + 30 = 80°.
解答：直线上的角之和为 180°。列出方程：2x + x + 30 = 180。化简：3x + 30 = 180。减 30：3x = 150。除以 3：x = 50。角分别为 2 × 50 = 100° 和 50 + 30 = 80°。
Answers: x = 50, angles 100° and 80°
答案：x = 50，角为 100° 和 80°
um of the angles you find equals 180°. 100° + 80° = 180° confirms the solution is correct.
提醒：务必检查所求角度之和是否等于 180°。100° + 80° = 180°，验证了答案正确。
8. Question 7: Area and Perimeter | 问题7：面积与周长
Question: A rectangle has a length of 8 cm and a width of 5 cm. Calculate its perimeter and area.
题目：一个长方形的长为 8 cm，宽为 5 cm。计算它的周长和面积。
× (length + width) = 2 × (8 + 5) = 2 × 13 = 26 cm. Area = length × width = 8 × 5 = 40 cm².
解答：长方形的周长 = 2 × (长 + 宽) = 2 × (8 + 5) = 2 × 13 = 26 cm。面积 = 长 × 宽 = 8 × 5 = 40 cm²。
Perimeter: 26 cm, Area: 40 cm²
周长：26 cm，面积：40 cm²
Common pitfall: Mixing up units – perimeter is a length (cm) while area is measured in square units (cm²). Also, do not forget to double both length and width when finding the perimeter.
常见陷阱：混淆单位——周长是长度 (cm)，而面积以平方单位 (cm²) 计量。此外，求周长时别忘了长和宽都需要乘以 2。
9. Question 8: Mean, Median, Mode | 问题8：平均数、中位数、众数
Question: The set of numbers is: 7, 9, 12, 7, 5. Find the mean, median and mode.
题目：数据集为 7, 9, 12, 7, 5。求平均数、中位数和众数。
er: 5, 7, 7, 9, 12. The middle value is the third one: 7. Mode: The number that appears most often is 7 (appears twice).
解答：平均数：所有数字总和 = 7 + 9 + 12 + 7 + 5 = 40。除以个数：40 ÷ 5 = 8。中位数：按升序排列：5, 7, 7, 9, 12。中间的第三个数值是 7。众数：出现次数最多的数是 7（出现两次）。
Mean = 8, Median = 7, Mode = 7
平均数 = 8，中位数 = 7，众数 = 7
Key point: The median requires ordering the data first. If there is an even number of values, the median is the average of the two middle numbers. The mode can be more than one value, or none if all values are unique.
关键点：求中位数要先排序。如果数据个数为偶数，中位数是中间两个数的平均数。众数可以不止一个，若所有值都唯一则没有众数。
10. Question 9: Probability | 问题9：概率
Question: A bag contains 3 red, 2 blue and 5 green marbles. One marble is picked at random. What is the probability that it is red?
题目：一个袋子里有 3 个红色、2 个蓝色和 5 个绿色弹珠。随机抽取一个，求抽到红色的概率。
d) = 3. Probability = (number of favourable outcomes) / (total number of outcomes) = 3/10.
解答：弹珠总数 = 3 + 2 + 5 = 10。有利结果数（红色）= 3。概率 = (有利结果数) / (总结果数) = 3/10。
P(red) = 3/10
P(红色) = 3/10
Tip: Probability can be expressed as a fraction, decimal or percentage, but the fraction in simplest form is usually preferred. Here 3/10 is already simplest. Always check the total carefully.
提示：概率可用分数、小数或百分数表示，但通常最简分数为佳。这里的 3/10 已是最简。务必仔细核对总数。
11. Question 10: Time Problem | 问题10：时间问题
Question: A train leaves the station at 09:30 and travels for 2 hours and 45 minutes. At what time does it arrive?
题目：一列火车于 09:30 离开车站，行驶 2 小时 45 分钟。它何时到达？
es = 12:15. Alternatively, add all minutes: 2 h 45 min = 165 min; 09:30 + 165 min = 12:15.
解答
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