📚 Year 7 CAIE Maths: Mock Unit Test Walkthrough | 七年级 CAIE 数学：单元测试模拟卷解析
Unit tests are a powerful way to check your understanding before a major exam. In this walkthrough, we break down a full CAIE Year 7 mock paper, question by question, explaining the key methods and common pitfalls. Working through these worked examples will help you build confidence and speed.
单元测试是检验知识掌握程度的有力工具。本篇文章将完整解析一套 CAIE 七年级模拟试卷，逐题讲解关键方法与常见错误。通过这些详尽的例题解析，你将增强信心并提升解题速度。
1. Integer Operations and BIDMAS | 整数运算与运算顺序
Question: Work out 23 + 5 × 4 − 6.
题目：计算 23 + 5 × 4 − 6。
First, follow the order of operations: multiplication before addition and subtraction. Here, 5 × 4 = 20.
首先，遵循运算顺序：先乘除后加减。此处 5 × 4 = 20。
Replace the expression: 23 + 20 − 6. Then work left to right: 23 + 20 = 43, and 43 − 6 = 37. So the answer is 37.
将原式替换为 23 + 20 − 6。从左到右计算：23 + 20 = 43，43 − 6 = 37。答案为 37。
Watch out: a common mistake is to add first (23 + 5 = 28, then 28 × 4 = 112, then − 6 = 106). That ignores BIDMAS and will lose marks.
注意：常见错误是先加（23 + 5 = 28，再 28 × 4 = 112，然后 − 6 = 106）。这忽略了运算法则，会失分。
2. Fractions and Decimals | 分数和小数
Question: Convert 0.875 to a fraction in its simplest form.
题目：将小数 0.875 化为最简分数。
0.875 means 875 thousandths, so write it as 875/1000.
0.875 表示千分之 875，因此写成分数 875/1000。
Now cancel common factors. Both 875 and 1000 end in 5 or 0, so divide by 25: 875 ÷ 25 = 35, 1000 ÷ 25 = 40, giving 35/40.
现在约分。875 和 1000 都能被 25 整除：875 ÷ 25 = 35，1000 ÷ 25 = 40，得到 35/40。
Divide again by 5: 35 ÷ 5 = 7, 40 ÷ 5 = 8, so the simplest form is 7/8.
再除以 5：35 ÷ 5 = 7，40 ÷ 5 = 8，最简分数为 7/8。
You can check your answer: 7 ÷ 8 = 0.875.
你可以验证：7 ÷ 8 = 0.875。
3. Negative Numbers | 负数的运算
Question: Find the value of −7 − (−3) + 5.
题目：计算 −7 − (−3) + 5。
Start with −7. Subtracting a negative is the same as adding the positive: −7 − (−3) = −7 + 3 = −4.
从 −7 开始。减去一个负数等于加上它的正数：−7 − (−3) = −7 + 3 = −4。
Now add 5: −4 + 5 = 1. The answer is 1.
然后加上 5：−4 + 5 = 1。答案为 1。
Remember, double negatives become a plus. You can think of it on a number line: starting at −7, take away a negative step moves you right by 3, landing on −4, then move right 5 to reach 1.
记住，两个负号变成正号。可以在数轴上想象：从 −7 开始，减去一个负步长相当于向右移动 3 个单位，到达 −4，再向右移 5 个单位到 1。
4. Algebraic Substitution | 代数代入求值
Question: If a = 3 and b = −2, work out the value of 2a² + 3b.
题目：已知 a = 3，b = −2，求代数式 2a² + 3b 的值。
Replace a with 3 and b with −2: 2 × (3)² + 3 × (−2).
把 a = 3 和 b = −2 代入：2 × (3)² + 3 × (−2)。
Calculate the index first: (3)² = 9. Then multiply: 2 × 9 = 18, and 3 × (−2) = −6.
先算指数：(3)² = 9。再做乘法：2 × 9 = 18，3 × (−2) = −6。
Finally, add: 18 + (−6) = 12. The answer is 12.
最后相加：18 + (−6) = 12。答案为 12。
Be careful with the signs: 3 × (−2) gives −6, not +6.
注意符号：3 × (−2) 得 −6，而不是 +6。
5. Solving Linear Equations | 解一元一次方程
Question: Solve 4x − 7 = 13.
题目：解方程 4x − 7 = 13。
We want to isolate x. First, add 7 to both sides: 4x − 7 + 7 = 13 + 7, so 4x = 20.
目标是求出 x。首先，两边同时加上 7：4x − 7 + 7 = 13 + 7，得 4x = 20。
Now divide both sides by 4: 4x ÷ 4 = 20 ÷ 4, so x = 5.
然后两边同时除以 4：4x ÷ 4 = 20 ÷ 4，得 x = 5。
Check: 4 × 5 − 7 = 20 − 7 = 13, which matches the right-hand side.
检验：4 × 5 − 7 = 20 − 7 = 13，符合等式右边。
Always perform the inverse operation in reverse order: undo subtraction/addition first, then undo multiplication/division.
始终按照相反的顺序运用逆运算：先处理加减，再处理乘除。
6. Angles on a Straight Line | 直线上的角
Question: Three angles lie on a straight line. Two of them are 72° and 43°. Find the size of the third angle.
题目：三个角共同形成一条直线，其中两个角分别为 72° 和 43°。求第三个角的度数。
Angles on a straight line add up to 180°. Let the missing angle be y. Then 72 + 43 + y = 180.
直线上的角之和为 180°。设未知角为 y，则 72 + 43 + y = 180。
Add the known angles: 72 + 43 = 115. So the equation becomes 115 + y = 180.
将已知角相加：72 + 43 = 115。方程变为 115 + y = 180。
Subtract 115 from both sides: y = 180 − 115 = 65°. The third angle is 65°.
两边减去 115：y = 180 − 115 = 65°。第三个角为 65°。
Check: 72 + 43 + 65 = 180.
验证：72 + 43 + 65 = 180。
7. Area of a Triangle | 三角形的面积
Question: A triangle has a base of 8 cm and a perpendicular height of 5 cm. Calculate its area.
题目：三角形底长 8 cm，对应高为 5 cm。求其面积。
Area of a triangle = ½ × base × height. Substitute the values: ½ × 8 × 5.
三角形面积 = ½ × 底 × 高。代入数值：½ × 8 × 5。
Multiply 8 by 5 to get 40, then take half: 40 ÷ 2 = 20. The area is 20 cm². Remember to include the square units.
先计算 8 × 5 = 40，再取一半：40 ÷ 2 = 20。面积为 20 cm²。务必带上平方单位。
If the question gives different units, make sure they are consistent before calculating.
如果题目给出的单位不同，计算前要统一单位。
8. Mean and Range | 平均数与极差
Question: Here are five game scores: 8, 12, 7, 15, 10. Find the mean and the range.
题目：五个游戏得分如下：8, 12, 7, 15, 10。求平均数和极差。
Mean: add all values: 8 + 12 + 7 + 15 + 10 = 52. Divide by the number of values (5): 52 ÷ 5 = 10.4.
平均数：将所有数值相加：8 + 12 + 7 + 15 + 10 = 52。除以数据个数 5：52 ÷ 5 = 10.4。
Range: subtract the smallest value from the largest. Largest = 15, smallest = 7. Range = 15 − 7 = 8.
极差：最大值减最小值。最大值为 15，最小值为 7。极差 = 15 − 7 = 8。
Mean is a measure of average, while range shows spread. Both are often tested together.
平均数用来衡量集中趋势，极差反映离散程度。两者常同时出现。
9. Ratio Sharing | 按比例分配
Question: Share £120 between Alice and Ben in the ratio 3 : 5. How much does Alice receive?
题目：将 £120 按 3:5 的比例分给 Alice 和 Ben。Alice 得到多少钱？
Total number of parts = 3 + 5 = 8 parts. Each part is worth £120 ÷ 8 = £15.
总份数 = 3 + 5 = 8 份。每份为 £120 ÷ 8 = £15。
Alice gets 3 parts, so her amount = 3 × £15 = £45.
Alice 得到 3 份，金额 = 3 × £15 = £45。
You can check: Ben gets 5 × £15 = £75, and £45 + £75 = £120.
验证：Ben 得到 5 × £15 = £75，而 £45 + £75 = £120。
10. Word Problem: Speed, Distance, Time | 应用问题：速度、距离、时间
Question: A cyclist travels 48 km at a constant speed of 16 km/h. How long does the journey take? Give the answer in hours and minutes.
题目：一名骑行者以 16 km/h 的恒定速度行驶 48 km。行程用时多少？答案用小时和分钟表示。
Time = Distance ÷ Speed. So time = 48 ÷ 16 = 3 hours. That is exactly 3 hours, or 3 hours 0 minutes.
时间 = 距离 ÷ 速度。时间 = 48 ÷ 16 = 3 小时。即恰好 3 小时，或 3 小时 0 分钟。
If the calculation gave a decimal like 2.5 hours, you would convert: 0.5 hours = 30 minutes, so 2 hours 30 minutes. Always present time clearly in the required format.
若算得小数，如 2.5 小时，需换算：0.5 小时 = 30 分钟，即 2 小时 30 分钟。始终按照题目要求明确呈现时间格式。
This simple formula triangle (Speed = D/T) is key for CAIE motion problems.
这个简单的公式关系（速度 = 距离/时间）是 CAIE 运动问题的关键。
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