Year 7 CIE Advanced Mathematics Unit Test Mock Paper Analysis | Year 7 CIE 进阶数学单元测试模拟卷解析

📚 Year 7 CIE Advanced Mathematics Unit Test Mock Paper Analysis | Year 7 CIE 进阶数学单元测试模拟卷解析

This mock paper is designed to help Year 7 students preparing for a CIE Advanced Mathematics unit test. It covers key topics including number theory, algebra, geometry, measurement, and data handling. Each question is followed by a detailed step-by-step solution.

这份模拟卷旨在帮助 Year 7 学生备考 CIE 进阶数学单元测试。内容涵盖数论、代数、几何、测量和数据处理等重要主题。每道题后均附有详细的逐步解析。

1. Highest Common Factor and Lowest Common Multiple | 最高公因数和最小公倍数

Question: Find the HCF and LCM of 24 and 36.

问题：求 24 和 36 的最高公因数 (HCF) 和最小公倍数 (LCM)。

The highest common factor is the largest whole number that divides both numbers exactly. The lowest common multiple is the smallest number that is a multiple of both. Write each number as a product of prime factors.

最高公因数是能同时整除这两个数的最大整数。最小公倍数是这两个数的最小公倍数。将每个数写成质因数乘积形式。

24 = 2³ × 3    and    36 = 2² × 3²

24 = 2³ × 3    和    36 = 2² × 3²

4 × 3 = 12. For LCM, take the highest powers of all primes present: 2³ × 3² = 8 × 9 = 72.

计算 HCF 时取公共质因数的最低次幂：2² × 3 = 4 × 3 = 12。计算 LCM 时取所有质因数的最高次幂：2³ × 3² = 8 × 9 = 72。

Therefore, HCF(24, 36) = 12 and LCM(24, 36) = 72.

因此，HCF(24, 36) = 12，LCM(24, 36) = 72。

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2. Fractions, Decimals and Percentages | 分数、小数和百分比

Question: Convert 3/5 into a decimal and a percentage, and then calculate 3/5 of 40.

问题：将 3/5 转换为小数和百分数，并计算 40 的 3/5。

e division gives 0.6. To change a decimal to a percentage, multiply by 100: 0.6 × 100 = 60%.

分数可通过除法转换为小数。3/5 即 3 除以 5，计算结果为 0.6。将小数转换为百分数需乘以 100：0.6 × 100 = 60%。

a quantity, multiply the fraction by the quantity. 3/5 of 40 = (3/5) × 40 = 0.6 × 40 = 24.

求某数量的几分之几，将分数乘以该数量。40 的 3/5 = (3/5) × 40 = 0.6 × 40 = 24。

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3. Simplifying Algebraic Expressions | 化简代数表达式

Question: Simplify 5a + 3b - 2a + 4b - a.

问题：化简 5a + 3b - 2a + 4b - a。

An algebraic expression can be simplified by collecting like terms. Like terms have exactly the same variable part. Identify a-terms and b-terms separately.

代数表达式可通过合并同类项化简。同类项是指变量部分完全相同的项。分别找出含 a 的项和含 b 的项。

terms: 3b + 4b = (3 + 4)b = 7b. Combine the results: the simplified expression is 2a + 7b.

含 a 的项：5a - 2a - a = (5 - 2 - 1)a = 2a。含 b 的项：3b + 4b = (3 + 4)b = 7b。合并结果：最简式为 2a + 7b。

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4. Solving Linear Equations | 解线性方程

Question: Solve 3x - 4 = 11.

问题：解方程 3x - 4 = 11。

To isolate x, perform the same operation on both sides of the equation. First, add 4 to both sides to remove the constant term from the left side.

为了分离出 x，需在方程两边同时进行相同运算。首先，两边同时加 4，消除左边的常数项。

3x - 4 + 4 = 11 + 4   ⇒   3x = 15

Then divide both sides by 3, the coefficient of x. x = 15 ÷ 3 = 5.

然后将两边同时除以 x 的系数 3。x = 15 ÷ 3 = 5。

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5. Angles in a Triangle | 三角形内角

Question: Two angles of a triangle are 42° and 86°. Find the size of the third angle.

问题：三角形中两个角分别为 42° 和 86°，求第三个角的度数。

The interior angles of any triangle always sum to 180°. Use this property to find the missing angle.

任何三角形的三个内角之和总是 180°。利用这一性质可求出未知角。

Third angle = 180° - (42° + 86°)

86° = 128°. Then subtract from 180°: 180° - 128° = 52°. Therefore, the third angle is 52°.

先将已知两角相加：42° + 86° = 128°。再从 180° 中减去：180° - 128° = 52°。因此第三个角是 52°。

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6. Perimeter and Area | 周长与面积

Question: A rectangle has length 12 cm and width 7 cm. Find its perimeter and area.

问题：一个矩形的长为 12 cm，宽为 7 cm。求它的周长和面积。

l distance around: P = 2 × (length + width). Area is the space inside: A = length × width.

矩形的周长是四条边总长：P = 2 × (长 + 宽)。面积是内部空间：A = 长 × 宽。

Perimeter = 2 × (12 + 7) = 2 × 19 = 38 cm

周长 = 2 × (12 + 7) = 2 × 19 = 38 cm

Area = 12 × 7 = 84 cm²

面积 = 12 × 7 = 84 cm²

Always include units in the final answer: cm for perimeter, cm² for area.

答案要带单位：周长用 cm，面积用 cm²。

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7. Ratio and Proportion | 比率与比例

Question: Share £150 between Amy and Ben in the ratio 2 : 3. How much does each receive?

问题：将 150 英镑按 2 : 3 分给 Amy 和 Ben，他们各得多少？

at for every 2 parts Amy gets, Ben gets 3 parts. Total number of equal parts is 2 + 3 = 5.

比率表示整体中各部分的关系。2 : 3 表示 Amy 得 2 份，Ben 得 3 份。相等的总份数为 2 + 3 = 5。

Value of one part = £150 ÷ 5 = £30. Amy's share = 2 parts = 2 × £30 = £60. Ben's share = 3 parts = 3 × £30 = £90.

每份金额 = £150 ÷ 5 = £30。Amy 的份额 = 2 份 = 2 × £30 = £60。Ben 的份额 = 3 份 = 3 × £30 = £90。

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8. Mean Value in Statistics | 统计中的平均数

Question: The mean of four numbers is 15. Three of the numbers are 12, 18, and 10. Find the fourth number.

问题：四个数的平均数为 15。已知其中三个数为 12、18 和 10，求第四个数。

ll values by the number of values. Rearranging gives: Total sum = mean × number of values.

平均数由所有数值之和除以数值个数得出。变形可得：总和 = 平均数 × 个数。

wn numbers = 12 + 18 + 10 = 40. The missing number = total sum - known sum = 60 - 40 = 20.

总和 = 15 × 4 = 60。已知三个数的和 = 12 + 18 + 10 = 40。未知数 = 总和 - 已知和 = 60 - 40 = 20。

he numbers 12, 18, 10, 20 have a sum of 60, and 60 ÷ 4 = 15, which matches the given mean.

检验：12、18、10、20 之和为 60，60 ÷ 4 = 15，符合给定的平均数。

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9. Sequences and the nth Term | 数列与第 n 项

Question: The nth term of a sequence is given as 4n - 1. Write down the first four terms of the sequence.

问题：数列的第 n 项公式为 4n - 1。写出该数列的前四项。

, 3, 4 into the formula. A sequence rule describes how to find any term from its position number n.

要生成各项，将 n = 1、2、3、4 代入公式。数列通项公式描述了如何根据位置序号 n 求出对应项。

n = 1: 4(1) - 1 = 3

n = 2: 4(2) - 1 = 7

n = 3: 4(3) - 1 = 11

n = 4: 4(4) - 1 = 15

The first four terms are 3, 7, 11, 15. Notice that the terms increase by 4 each time, which matches the coefficient of n.

前四项为 3、7、11、15。注意到相邻项之差为 4，这与 n 的系数一致。

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10. Basic Probability | 基础概率

Question: A box contains 4 white chocolates and 6 milk chocolates. One chocolate is picked at random. What is the probability it is a white chocolate? Give your answer as a fraction in its simplest form.

问题：一个盒子中有 4 颗白巧克力和 6 颗牛奶巧克力。随机取出一颗，取出白巧克力的概率是多少？答案以最简分数形式给出。

re, favourable outcomes are white chocolates, so there are 4. Total outcomes = 4 + 6 = 10.

概率计算公式为：有利结果数 / 可能结果总数。有利结果是取出白巧克力，有利结果数为 4。总结果数 = 4 + 6 = 10。

P(white) = 4/10

r by their highest common factor, 2. 4 ÷ 2 = 2, 10 ÷ 2 = 5. Thus the simplest form is 2/5.

将分子分母同时除以它们的最大公因数 2 以化简。4 ÷ 2 = 2，10 ÷ 2 = 5。因此最简分数为 2/5。

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