📚 Year 8 CIE Advanced Mathematics: Mock Unit Test Analysis | CIE Year 8 进阶数学单元测试模拟卷解析
Welcome to the detailed walkthrough of a mock unit test for Year 8 CIE Advanced Mathematics. This mock paper covers key topics such as algebraic manipulation, equations, inequalities, coordinate geometry, mensuration, sequences, probability, and more. By working through each question and studying the solutions, you will reinforce essential concepts and improve your problem-solving skills for the actual test.
欢迎阅读 Year 8 CIE 进阶数学单元测试模拟卷的详细解析。本模拟试卷涵盖了代数运算、方程、不等式、坐标几何、测量、数列、概率等重要知识点。通过逐题演练并学习解题步骤，你能够巩固核心概念，提升应对真实考试的解题能力。
1. Simplifying Algebraic Expressions | 化简代数表达式
Simplify 2(3x - 4) - (x + 5) + 3x.
化简：2(3x - 4) - (x + 5) + 3x。
6x - 8. The minus sign in front of (x + 5) changes the signs inside, so -(x + 5) = -x - 5.
步骤1：去括号。2(3x - 4) 得到 6x - 8。(x + 5) 前的负号会改变括号内符号，因此 -(x + 5) = -x - 5。
Step 2: Rewrite the whole expression: 6x - 8 - x - 5 + 3x.
步骤2：将整个表达式重写为：6x - 8 - x - 5 + 3x。
up and combine the x terms: 6x - x + 3x = 8x. Then combine the number terms: -8 - 5 = -13.
步骤3：合并 x 项：6x - x + 3x = 8x。再合并常数项：-8 - 5 = -13。
Therefore, the simplified expression is 8x - 13.
因此，化简后的表达式为 8x - 13。
2. Solving Linear Equations with Fractions | 解含分数的线性方程
Solve the equation: (2x + 1)/3 - (x - 2)/2 = 1.
解方程：(2x + 1)/3 - (x - 2)/2 = 1。
Step 1: Find the lowest common multiple of the denominators 3 and 2, which is 6. Multiply every term by 6 to clear the fractions.
步骤1：找出分母 3 和 2 的最小公倍数 6。将每项乘以 6 以消去分母。
6 × [(2x + 1)/3] - 6 × [(x - 2)/2] = 6 × 1 simplifies to 2(2x + 1) - 3(x - 2) = 6.
6 × [(2x + 1)/3] - 6 × [(x - 2)/2] = 6 × 1 化简为 2(2x + 1) - 3(x - 2) = 6。
Step 2: Expand the brackets: 4x + 2 - 3x + 6 = 6.
步骤2：去括号：4x + 2 - 3x + 6 = 6。
Step 3: Combine like terms on the left side: (4x - 3x) + (2 + 6) = x + 8. So x + 8 = 6.
步骤3：合并左边同类项：(4x - 3x) + (2 + 6) = x + 8。得到 x + 8 = 6。
Step 4: Subtract 8 from both sides: x = -2.
步骤4：两边同时减去 8：x = -2。
3. Solving Inequalities and the Number Line | 解不等式与数轴表示
Solve 4 - 3x ≤ 7 and represent the solution on a number line.
解不等式 4 - 3x ≤ 7，并将解集表示在数轴上。
p 1: Subtract 4 from both sides to isolate the term containing x: -3x ≤ 7 - 4, so -3x ≤ 3.
步骤1：两边同时减去 4，使含 x 的项独立：-3x ≤ 7 - 4，得 -3x ≤ 3。
y -3. Remember that dividing by a negative number reverses the inequality sign, so x ≥ -1.
步骤2：两边同时除以 -3。注意除以负数时不等号方向要改变，因此 x ≥ -1。
Step 3: To draw the number line representation, place a closed dot at -1 (since -1 is included) and shade the line to the right, indicating all numbers greater than or equal to -1.
步骤3：在数轴上表示时，在 -1 处画实心圆点（代表包含 -1），向右画射线并涂阴影，表示所有大于或等于 -1 的数。
4. Equation of a Straight Line | 直线的方程
Find the equation of the line passing through the points (2, 3) and (4, 7).
求经过 (2, 3) 和 (4, 7) 两点的直线方程。
pe) m using the formula m = (y2 - y1) / (x2 - x1). Here m = (7 - 3) / (4 - 2) = 4 / 2 = 2.
步骤1：利用公式 m = (y2 - y1) / (x2 - x1) 计算斜率。这里 m = (7 - 3) / (4 - 2) = 4 / 2 = 2。
nt-slope form y - y1 = m(x - x1) with one point, say (2, 3). Substitute: y - 3 = 2(x - 2).
步骤2：使用点斜式方程 y - y1 = m(x - x1)，任选一点，例如 (2, 3)。代入得 y - 3 = 2(x - 2)。
Step 3: Expand and simplify: y - 3 = 2x - 4, so y = 2x - 1.
步骤3：展开并化简：y - 3 = 2x - 4，因此 y = 2x - 1。
The equation of the line is y = 2x - 1.
该直线方程为 y = 2x - 1。
5. Angles in a Triangle Using Ratios | 用比例求三角形内角
The angles of a triangle are in the ratio 2 : 3 : 4. Find the size of each angle.
一个三角形的三个内角之比为 2 : 3 : 4。求每个角的度数。
Since the sum of angles in a triangle is 180°, we write the equation 2x + 3x + 4x = 180°.
步骤1：设三个角分别为 2x, 3x 和 4x。三角形内角和为 180°，列出方程 2x + 3x + 4x = 180°。
Step 2: Combine like terms: 9x = 180°.
步骤2：合并同类项：9x = 180°。
Step 3: Divide by 9: x = 20°.
步骤3：两边除以 9：x = 20°。
Step 4: Find each angle: 2x = 40°, 3x = 60°, 4x = 80°.
步骤4：计算各角度：2x = 40°，3x = 60°，4x = 80°。
6. Volume and Surface Area of a Cylinder | 圆柱体的体积与表面积
5 cm and a height of 12 cm. Taking π ≈ 3.14, calculate its volume and total surface area.
一个圆柱体底面半径为 5 cm，高为 12 cm，π 取 3.14。计算它的体积和总表面积。
Step 1: Volume formula V = πr²h. Substitute: V = 3.14 × 5² × 12 = 3.14 × 25 × 12.
步骤1：体积公式 V = πr²h。代入：V = 3.14 × 5² × 12 = 3.14 × 25 × 12。
ep 2: Multiply stepwise: 3.14 × 25 = 78.5, then 78.5 × 12 = 942. So the volume is 942 cm³.
步骤2：逐步计算：3.14 × 25 = 78.5，再乘以 12 得 942。因此体积为 942 cm³。
2πr² + 2πrh. This means the sum of the areas of the two circular ends and the curved side.
步骤3：总表面积公式 A = 2πr² + 2πrh，即两个底圆面积加侧面面积之和。
: Substitute values: 2πr² = 2 × 3.14 × 25 = 157 cm². 2πrh = 2 × 3.14 × 5 × 12 = 376.8 cm².
步骤4：代入数值：2πr² = 2 × 3.14 × 25 = 157 cm²。2πrh = 2 × 3.14 × 5 × 12 = 376.8 cm²。
Step 5: Total surface area = 157 + 376.8 = 533.8 cm².
步骤5：总表面积 = 157 + 376.8 = 533.8 cm²。
7. Application of the Pythagorean Theorem | 勾股定理的实际应用
A 13 m ladder leans against a vertical wall. The foot of the ladder is 5 m away from the wall. How high up the wall does the ladder reach?
一个 13 m 长的梯子靠在竖直墙面上，梯脚离墙 5 m。梯子顶端能达到多高的墙壁位置？
tance from the wall is one leg (a = 5 m), and the height up the wall is the other leg (b).
步骤1：梯子、墙面和地面构成直角三角形。梯子为斜边 (c = 13 m)，梯脚到墙的距离为一条直角边 (a = 5 m)，墙上的高度为另一条直角边 (b)。
Step 2: Apply Pythagoras' theorem: a² + b² = c², so 5² + b² = 13².
步骤2：应用勾股定理：a² + b² = c²，即 5² + b² = 13²。
Step 3: Work out the squares: 25 + b² = 169. Subtract 25 from both sides: b² = 144.
步骤3：计算平方：25 + b² = 169。两边同时减去 25：b² = 144。
Step 4: Take the positive square root (height is positive): b = √144 = 12 m.
步骤4：取正平方根（高度为正数）：b = √144 = 12 m。
8. Finding the nth Term of a Linear Sequence | 求线性数列的第n项
Consider the sequence: 3, 10, 17, 24, ... Find the nth term formula and hence the 20th term.
已知数列：3, 10, 17, 24, ... 找出第 n 项的通项公式，并求第 20 项的值。
on difference d between consecutive terms: 10 - 3 = 7, 17 - 10 = 7, 24 - 17 = 7. So d = 7.
步骤1：求相邻两项的公差：10 - 3 = 7，17 - 10 = 7，24 - 17 = 7。因此 d = 7。
nth term of an arithmetic sequence is given by aₙ = a₁ + (n - 1)d. Here a₁ = 3 and d = 7.
步骤2：等差数列的第 n 项公式为 aₙ = a₁ + (n - 1)d。此处 a₁ = 3，d = 7。
Step 3: Substitute the values: aₙ = 3 + (n - 1) × 7 = 3 + 7n - 7 = 7n - 4.
步骤3：代入数值：aₙ = 3 + (n - 1) × 7 = 3 + 7n - 7 = 7n - 4。
Step 4: For the 20th term, set n = 20: a₂₀ = 7 × 20 - 4 = 140 - 4 = 136.
步骤4：求第 20 项，令 n = 20：a₂₀ = 7 × 20 - 4 = 140 - 4 = 136。
9. Probability without Replacement | 无放回概率
A bag contains 4 red, 3 blue and 5 green marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are the same colour.
一个袋子中有 4 个红色、3 个蓝色和 5 个绿色弹珠。随机不放回地抽取两个。求两颗弹珠颜色相同的概率。
Step 1: Total number of marbles = 4 + 3 + 5 = 12.
步骤1：弹珠总数 = 4 + 3 + 5 = 12。
Step 2: The favourite outcomes are both red (RR), both blue (BB), or both green (GG). Since the draws are without replacement, we multiply the probabilities along each branch.
步骤2：有利结果包括同为红色 (RR)、同为蓝色 (BB) 或同为绿色 (GG)。由于是不放回抽取，沿各分支将概率相乘。
P(RR) = (4/12) × (3/11) = 12/132.
P(RR) = (4/12) × (3/11) = 12/132。
P(BB) = (3/12) × (2/11) = 6/132.
P(BB) = (3/12) × (2/11) = 6/132。
P(GG) = (5/12) × (4/11) = 20/132.
P(GG) = (5/12) × (4/11) = 20/132。
re mutually exclusive events, add them: P(same colour) = 12/132 + 6/132 + 20/132 = 38/132.
步骤3：因为这些事件互斥，将概率相加：P(同色) = 12/132 + 6/132 + 20/132 = 38/132。
Step 4: Simplify the fraction by dividing numerator and denominator by 2: 38/132 = 19/66.
步骤4：将分数化简，分子分母同时除以 2：38/132 = 19/66。
10. Average Speed over a Journey with Two Speeds | 两段路程中的平均速度问题
A car completes a journey in two halves. It travels the first half of the distance at 60 km/h and the second half at 80 km/h. What is the average speed for the whole journey?
一辆车完成一段旅程，前半段路程以 60 km/h 行驶，后半段路程以 80 km/h 行驶。求全程的平均速度。
Step 1: Let the total distance be 2d, so each half is d km.
步骤1：设总路程为 2d，则每半段为 d km。
: Time for the first half: t1 = d / 60 hours. Time for the second half: t2 = d / 80 hours.
步骤2：前半段时间：t1 = d / 60 小时。后半段时间：t2 = d / 80 小时。
d these fractions, find a common denominator: 240. T = (4d/240) + (3d/240) = 7d/240 hours.
步骤3：总时间 T = d/60 + d/80。通分相加，公分母为 240。T = (4d/240) + (3d/240) = 7d/240 小时。
Step 4: Average speed = total distance / total time = 2d
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