📚 Year 7 WJEC Mathematics: Unit Test Mock Paper Solutions | Year 7 WJEC 数学：单元测试模拟卷解析
This article provides fully worked solutions to a mock unit test designed for Year 7 students following the WJEC mathematics curriculum. Each section focuses on a key topic area and explains the step-by-step methods needed to solve typical questions. Use this guide to check your answers, consolidate your understanding, and build confidence before your real test.
本文提供了为遵循WJEC数学课程的Year 7学生设计的单元测试模拟卷的详细解答。每个小节聚焦一个核心主题，并解释了解答典型题目所需的逐步方法。你可以利用这份指南核对答案、巩固理解，并在真正考试前建立自信。
1. Whole Number and Integer Calculations | 整数与负整数计算
Let's start with basic operations including addition, subtraction, multiplication, division, and simple negative numbers. These questions test your ability to use written methods accurately.
让我们从基本的加法、减法、乘法、除法以及简单的负数运算开始。这些题目考查你准确使用书面方法的能力。
Q1. Calculate 456 + 789.
题目1：计算 456 + 789。
y 1. Hundreds: 4 + 7 + 1 = 12, write 2 and carry 1 to make 1 thousand. The result is 1245.
解答：将数字按列对齐。个位相加：6 + 9 = 15，写5，向十位进1。十位：5 + 8 + 1 = 14，写4，向百位进1。百位：4 + 7 + 1 = 12，写2，进1到千位。结果是1245。
Q2. Work out 1234 − 567.
题目2：计算 1234 − 567。
s). 14 − 7 = 7. Tens: now 2 tens left, 2 − 6 cannot be done, exchange 1 hundred (making 12 tens). 12 − 6 = 6. Hundreds: 1 hundred left (after exchanging), 1 − 5 cannot be done, exchange 1 thousand (making 11 hundreds). 11 − 5 = 6. Thousands: 0. Answer: 667.
解答：按列写出，从右向左相减，需要时借位。个位：4 − 7 不够，从十位借1（变成14个一）。14 − 7 = 7。十位：剩下2个十，2 − 6 不够，从百位借1（变成12个十）。12 − 6 = 6。百位：剩下1，1 − 5 不够，从千位借1（变成11个百）。11 − 5 = 6。千位：0。答案是667。
Q3. Multiply 23 × 15.
题目3：计算 23 × 15。
e two results: 230 + 115 = 345. Alternatively, use the grid method or long multiplication.
解答：把15拆分为10 + 5。23 × 10 = 230，23 × 5 = 115。两部分相加：230 + 115 = 345。也可以使用网格法或竖式乘法。
Q4. Divide 144 ÷ 12.
题目4：计算 144 ÷ 12。
u prefer short division, 12 goes into 14 once (12), remainder 2, bring down 4 to make 24. 12 goes into 24 twice, giving 12.
解答：回想乘法表：12 × 12 = 144，所以144 ÷ 12 = 12。如果你喜欢短除法，12除以14得到1（12），余2，把4移下来变成24。12除24得2，结果是12。
Q5. Evaluate −3 + 7 and 5 − (−2).
题目5：计算 -3 + 7 和 5 − (-2)。
on +4. For 5 − (−2), subtracting a negative is the same as adding the positive: 5 + 2 = 7.
解答：对于 −3 + 7，从0向左移动3格，再向右移动7格，到达+4。对于 5 − (−2)，减去负数等于加上正数：5 + 2 = 7。
2. Fractions: Adding, Subtracting, and Finding a Fraction of an Amount | 分数：加法、减法与求一个数量的几分之几
Fractions appear in many areas of mathematics. You need to be comfortable adding and subtracting fractions with the same or related denominators, and finding a fraction of a whole number.
分数在数学的许多领域都会出现。你需要熟练掌握分母相同或相关时分数的加减法，以及求一个整数数量的几分之几。
Q1. Work out 2/5 + 1/5.
题目1：计算 2/5 + 1/5。
he same, so simply add the numerators: 2 + 1 = 3. The denominator stays as 5. Answer: 3/5.
解答：分母相同，直接将分子相加：2 + 1 = 3。分母保持为5。答案是3/5。
Q2. Calculate 3/4 − 1/2.
题目2：计算 3/4 − 1/2。
est common denominator of 4 and 2 is 4. Convert 1/2 to 2/4. Now subtract: 3/4 − 2/4 = 1/4.
解答：将分母变为相同。4和2的最小公分母是4。把1/2化成2/4。现在相减：3/4 − 2/4 = 1/4。
Q3. Find 2/3 of 24.
题目3：求24的2/3。
1/3 of 24 by dividing: 24 ÷ 3 = 8. Then multiply by 2 to get 2/3: 8 × 2 = 16. Answer: 16.
解答：先用除法求出24的1/3：24 ÷ 3 = 8。再乘以2得到2/3：8 × 2 = 16。答案是16。
3. Decimals and Percentages | 小数与百分比
Decimals and percentages are closely linked to fractions. You need to convert between them and calculate percentages of quantities.
小数和百分比与分数密切相关。你需要掌握它们之间的转换，并计算一个数量的百分比。
Q1. Convert the decimal 0.75 to a percentage.
题目1：把小数0.75化成百分比。
Solution: A percentage means 'out of 100'. Multiply 0.75 by 100 to get 75, so 0.75 = 75%.
解答：百分比的意思是“每一百中的份数”。将0.75乘以100得到75，因此0.75 = 75%。
Q2. What is 20% of 50?
题目2：50的20%是多少？
Solution: 20% means 20/100 or 1/5. To find 1/5 of 50, divide 50 by 5: 50 ÷ 5 = 10. So 20% of 50 is 10.
解答：20%表示20/100，即1/5。求50的1/5，用50除以5：50 ÷ 5 = 10。所以50的20%是10。
4. Simplifying Algebraic Expressions | 化简代数表达式
Algebra uses letters to stand for unknown numbers. Simplifying expressions involves collecting like terms – terms that have the same variable raised to the same power.
代数用字母代表未知数。化简表达式就是合并同类项——即含有相同变量且指数相同的项。
Q1. Simplify 3a + 2b + 5a − b.
题目1：化简 3a + 2b + 5a − b。
a = 8a. Group the b terms: 2b − b = 1b (or simply b). The simplified expression is 8a + b.
解答：将含a的项合并：3a + 5a = 8a。将含b的项合并：2b − b = 1b（或写作b）。化简后的表达式为 8a + b。
Q2. Find the value of 2x + 5 when x = 3.
题目2：当x = 3时，求式子2x + 5的值。
Solution: Substitute 3 in place of x: 2 × 3 + 5. Follow the order of operations – multiplication first: 6 + 5 = 11.
解答：把3代入x的位置：2 × 3 + 5。按照运算顺序，先乘：6 + 5 = 11。
5. Solving Simple Equations | 解简单方程
An equation shows that two expressions are equal. Solving an equation means finding the value of the unknown that makes the equality true.
方程表示两个表达式相等。解方程意味着找到使等式成立的未知数的值。
Q1. Solve x + 7 = 15.
题目1：解方程 x + 7 = 15。
Solution: To isolate x, subtract 7 from both sides: x + 7 − 7 = 15 − 7, so x = 8.
解答：为了把x单独留在一边，两边同时减去7：x + 7 − 7 = 15 − 7，得到 x = 8。
Q2. Solve 4y = 28.
题目2：解方程 4y = 28。
Solution: Divide both sides by 4: 4y ÷ 4 = 28 ÷ 4, so y = 7.
解答：两边同时除以4：4y ÷ 4 = 28 ÷ 4，得到 y = 7。
6. Angles in Triangles and on Straight Lines | 三角形和直线上的角
Angle problems require you to recall basic angle facts, such as angles on a straight line adding to 180° and the sum of the interior angles of a triangle being 180°.
角度问题要求你记住基本的角性质，例如直线上的角之和为180°，三角形内角之和为180°。
Q1. Two angles of a triangle are 50° and 60°. Find the third angle.
题目1：一个三角形的两个角分别是50°和60°。求第三个角。
n angles: 50° + 60° = 110°. Subtract from 180°: 180° − 110° = 70°. The third angle is 70°.
解答：三角形内角之和为180°。先把已知角相加：50° + 60° = 110°。再用180°减去：180° − 110° = 70°。第三个角是70°。
Q2. On a straight line, one angle is 130°. What is the size of the missing adjacent angle?
题目2：在一条直线上，其中一个角是130°。与其相邻的未知角的大小是多少？
to 180°. Subtract the known angle from 180°: 180° − 130° = 50°. The missing angle is 50°.
解答：直线上的角之和为180°。用180°减去已知角：180° − 130° = 50°。未知角是50°。
7. Perimeter and Area of Rectangles | 长方形的周长与面积
Perimeter is the total distance around the outside of a shape. Area is the amount of space inside the shape, measured in square units.
周长是围绕形状外部的总距离。面积是形状内部的空间大小，以平方单位计量。
Q1. A rectangle has length 8 cm and width 5 cm. Find its perimeter and area.
题目1：一个长方形的长是8厘米，宽是5厘米。求它的周长和面积。
of all four sides, or use the formula 2 × (length + width). 2 × (8 + 5) = 2 × 13 = 26 cm.
周长解答：四边长度相加，或使用公式 2 × (长 + 宽)。2 × (8 + 5) = 2 × 13 = 26 厘米。
Solution for area: Multiply length by width.
面积解答：用长乘以宽。
Area = length × width = 8 × 5 = 40 cm²
面积 = 长 × 宽 = 8 × 5 = 40 平方厘米
8. Working with Averages (Mean) | 平均数的计算
The mean (often called the average) is found by adding up all the values and then dividing by the number of values. It is a useful way to summarise a set of data.
平均数（通常简称均值）是通过将所有数值相加再除以数值的个数得到的。它是概括一组数据的有用方法。
Q. Find the mean of the numbers: 5, 8, 12, 9, 6.
题目：求下列数的平均数：5, 8, 12, 9, 6。
9 + 6 = 40. There are 5 numbers in total. Divide the sum by 5: 40 ÷ 5 = 8. The mean is 8.
解答：先将所有数相加：5 + 8 + 12 + 9 + 6 = 40。一共有5个数。用总和除以5：40 ÷ 5 = 8。平均数是8。
9. Interpreting Bar Charts and Tables | 解读柱状图和表格
Data can be presented in charts and tables, and you need to be able to read values directly from them, as well as compare categories.
数据可以用图表和表格呈现，你需要能够直接从中读取数值并进行类别比较。
The bar chart below shows the favourite fruits of a group of Year 7 students. (Imagine a simple bar chart with frequencies: Apple 10, Banana 8, Orange 6, Grape 4.)
下面的柱状图显示了一组Year 7学生最喜欢的水果。（想象一个简单的柱状图，频数分别为：苹果10，香蕉8，橘子6，葡萄4。）
Q1. How many students chose banana?
题目1：有多少名学生选择了香蕉？
Solution: The bar for banana reaches up to 8 on the frequency axis, so 8 students chose banana.
解答：香蕉的柱形在频数轴上达到8，因此有8名学生选择了香蕉。
Q2. How many more students chose apple than grape?
题目2：选择苹果的学生比选择葡萄的多多少人？
requency of 10, grape has 4. The difference is 10 − 4 = 6. So 6 more students chose apple.
解答：苹果的频数是10，葡萄是4。差值是 10 − 4 = 6。所以选择苹果的学生多6人。
10. Word Problems: Speed, Distance, and Time | 应用题：速度、距离与时间
Real-life problems often involve a combination of skills. You may need to apply formulas and decide which operations to use.
现实生活中的问题常常需要综合技能。你可能需要应用公式并决定使用哪些运算。
Q1. A car travels at an average speed of 50 km per hour. How far does it travel in 3 hours?
题目1：一辆车以平均 50 公里/小时 的速度行驶。3小时后它能行驶多远？
Solution: Use the relationship: distance = speed × time.
解答：使用关系式：距离 = 速度 × 时间。
Distance = 50 × 3 = 150 km
距离 = 50 × 3 = 150 公里
Q2. A school buys 15 packs of pens. Each pack contains 6 pens. How many pens are bought in total?
题目2：一所学校买了15包笔，每包有6支。学校一共买了多少支笔？
Solution: Multiply the number of packs by the pens per pack: 15 × 6 = 90 pens.
解答：用包数乘以每包的笔数：15 × 6 = 90 支笔。
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