📚 Year 9 OCR Further Mathematics Unit Test Mock Paper Walkthrough | Year 9 OCR 进阶数学单元测试模拟卷解析
This article provides a step-by-step walkthrough of a typical Year 9 OCR Further Mathematics unit test mock paper. Each question is broken down with clear solutions, common pitfalls and key concepts to help you master the techniques needed for the exam. Use this resource to check your understanding and refine your problem-solving skills.
本文逐步解析了一份典型的 Year 9 OCR 进阶数学单元测试模拟卷。每道题目均配有清晰的解答、常见错误提示和关键概念讲解，旨在帮助同学们掌握考试所需的技巧。请利用这份资料检验自己的理解，并提升解题能力。
1. Question 1 – Linear Equations with Fractions | 第1题 — 含分数的线性方程
Solve: (2x+1)/3 = (x-2)/2
First, identify the type of equation: it is a linear equation involving algebraic fractions. To clear the denominators, we can cross-multiply because both sides are single fractions.
首先判断方程类型：这是一个含有代数分式的线性方程。由于两边各是一个分式，可以通过交叉相乘来去分母。
ives: 2(2x+1) = 3(x-2). Remember to place the numerators in brackets to avoid sign errors.
交叉相乘得：2(2x+1) = 3(x-2)。注意分子要加括号，以免出现符号错误。
ommon mistake here is to write 3x - 2 instead of 3x - 6, so double-check the distribution.
展开两边：4x + 2 = 3x - 6。常见的错误是误写成 3x - 2 而不是 3x - 6，因此务必仔细检查乘法分配。
act 3x from both sides to collect x terms on the left: 4x - 3x + 2 = -6 → x + 2 = -6.
接下来，两边同时减去 3x，将含 x 的项移到左边：x + 2 = -6。
LHS = (2(-8)+1)/3 = (-15)/3 = -5; RHS = (-8-2)/2 = (-10)/2 = -5. The solution checks out.
两边再减去 2，得 x = -8。务必代回原方程检验：左边 = (2(-8)+1)/3 = -15/3 = -5，右边 = (-8-2)/2 = -10/2 = -5，成立。
2. Question 2 – Factorising Quadratics | 第2题 — 二次方程因式分解
Solve: x² - 5x + 6 = 0
We solve this quadratic by factorisation. Look for two numbers that multiply to +6 and add to -5.
我们通过因式分解来解这个二次方程。寻找两个数，其乘积为 +6，和为 -5。
2) × (-3) = 6 and (-2) + (-3) = -5. Thus, the factorised form is (x - 2)(x - 3) = 0.
这两个数是 -2 和 -3，因为 (-2)×(-3)=6 且 (-2)+(-3)=-5。因此因式分解为 (x - 2)(x - 3) = 0。
t each factor equal to zero: x - 2 = 0 or x - 3 = 0, giving the solutions x = 2 and x = 3.
令每个因式等于零：x - 2 = 0 或 x - 3 = 0，解得 x = 2 和 x = 3。
A frequent error is to write factors as (x+2)(x+3), which would expand to x²+5x+6. Always check the signs by expanding mentally.
常见的错误是写成 (x+2)(x+3)，展开后为 x²+5x+6。应在心中快速展开检验符号是否正确。
3. Question 3 – Inequalities and Number Line | 第3题 — 不等式与数轴表示
Solve: 3x - 5 > 2x + 1, and represent the solution on a number line.
Treat the inequality like an equation initially, but be careful when multiplying or dividing by a negative number – here we do not need that step.
初步可将不等式当作方程处理，但当乘除负数时需格外小心——本题没有这一步。
Subtract 2x from both sides: 3x - 2x - 5 > 1 → x - 5 > 1.
两边同时减去 2x：x - 5 > 1。
Add 5 to both sides: x > 6. This means any number strictly greater than 6 satisfies the inequality.
两边加 5：x > 6。这表示所有严格大于 6 的数都满足不等式。
On a number line, place an open circle at 6 (since 6 is not included) and draw an arrow extending to the right. Label the line clearly.
在数轴上，于 6 处画一个空心圆（因为 6 不包含在内），并向右画一条射线。务必清晰标注数轴。
4. Question 4 – Simultaneous Equations | 第4题 — 联立方程组
Solve: y = 2x + 1, 3x - y = 6.
Since the first equation is already solved for y, substitution is the most straightforward method.
由于第一个方程已经用 y 表示，使用代入法最为直接。
Substitute y = 2x + 1 into the second equation: 3x - (2x + 1) = 6.
将 y = 2x + 1 代入第二个方程：3x - (2x + 1) = 6。
lify: 3x - 2x - 1 = 6 → x - 1 = 6, so x = 7. Then substitute back: y = 2(7) + 1 = 15.
化简：3x - 2x - 1 = 6 得 x - 1 = 6，所以 x = 7。再代回得 y = 2(7)+1 = 15。
, 15) is correct. Some students forget to distribute the minus sign over the bracket; always write the bracket to avoid sign errors.
用原方程检验：3(7)-15=6，解 (7,15) 正确。有学生会忘记将负号分配到括号内，代入时一定要先保留括号以免符号出错。
5. Question 5 – Functions and Inverses | 第5题 — 函数与反函数
Given f(x) = 2x + 3, find the inverse function f⁻¹(x).
find the inverse, we set y = f(x), so y = 2x + 3, and then swap x and y to get x = 2y + 3.
求反函数时，设 y = f(x)，即 y = 2x + 3，然后交换 x 和 y 得到 x = 2y + 3。
Solve for y: x - 3 = 2y, then divide both sides by 2: y = (x - 3)/2.
解出 y：x - 3 = 2y，两边除以 2 得 y = (x - 3)/2。
Therefore, f⁻¹(x) = (x - 3)/2. In function notation, this means 'take the input, subtract 3, then divide by 2'.
因此 f⁻¹(x) = (x - 3)/2。用函数语言描述就是“输入减去3后除以2”。
ommon error is to write f⁻¹(x)= x/2 - 3; always apply inverse operations in reverse order.
可验证 f(f⁻¹(x)) = 2[(x-3)/2] + 3 = x。常见的错误是写成 x/2 - 3；务必按逆向运算的顺序进行。
6. Question 6 – Pythagoras and Trigonometry | 第6题 — 勾股定理与三角比
In a right-angled triangle, one leg is 5 cm and the hypotenuse is 13 cm. Find the other leg and sin θ, where θ is the angle opposite the 5 cm leg.
Let the missing leg be b. By Pythagoras' eorem: 5² + b² = 13² → 25 + b² = 169, so b² = 144 → b = 12 cm (positive length).
设未知直角边为 b。根据勾股定理：5² + b² = 13²，得 25 + b² = 169，b² = 144，b = 12 cm（取正值）。
fied further. Many students mistakenly use adjacent instead of opposite; always label the sides relative to the chosen angle.
于是 sin θ = 对边/斜边 = 5/13，不能再约分。许多学生会错用邻边而非对边，解题时务必根据所选角度标清各边。
so find the other trigonometric ratios if needed: cos θ = 12/13, tan θ = 5/12.
必要时也可求出其他三角比：cos θ = 12/13，tan θ = 5/12。
7. Question 7 – Volume of a Cone | 第7题 — 圆锥的体积
A cone has base radius 3 cm and perpendicular height 4 cm. Calculate its volume. Leave your answer in terms of π.
Recall the formula for the volume of a cone: V = 1/3 π r² h.
回忆圆锥体积公式：V = 1/3 π r² h。
d h = 4: V = 1/3 × π × 3² × 4 = 1/3 × π × 9 × 4.
代入 r=3，h=4：V = 1/3 × π × 9 × 4。
. Therefore V = 12π cm³. Note that the answer is exact when left as a multiple of π.
化简：9×4=36，36的1/3是12，因此 V = 12π cm³。保留 π 的倍数可得到精确值。
A slip often seen is using the diameter instead of the radius, or squaring the height by mistake. Always identify r and h carefully from the diagram.
常见的失误是将直径当作半径，或错误地对高进行平方。一定要依据图示仔细辨别 r 和 h。
8. Question 8 – Arithmetic Sequences | 第8题 — 等差数列
In an arithmetic sequence, the 3rd term is 7 and the 6th term is 16. Find the nth term formula and the 20th term.
Represent the nth term as a + (n-1)d, where a is the first term and d is the common difference.
将第 n 项表示为 a + (n-1)d，其中 a 为首项，d 为公差。
m = a + 5d = 16. Subtract the equations: (a+5d) - (a+2d) = 16 - 7 → 3d = 9, so d = 3.
根据已知条件：第3项 a+2d=7，第6项 a+5d=16。两式相减得 3d=9，故 d=3。
=3 into a+2d=7: a + 6 = 7, thus a = 1. The nth term formula is 1 + (n-1)×3 = 3n - 2.
将 d=3 代入 a+2d=7 得 a=1。第 n 项公式为 1+(n-1)×3 = 3n - 2。
20) - 2 = 60 - 2 = 58. Checking the pattern: 1, 4, 7, 10, 13, 16, ... confirms the values.
求第 20 项，令 n=20：T₂₀ = 3(20)-2 = 58。序列 1,4,7,10,13,16,... 验证了结果。
9. Question 9 – Direct and Inverse Proportion | 第9题 — 正比例与反比例
If 3 machines take 4 hours to produce 120 parts, how many parts can 5 machines produce in 6 hours, assuming the machines work at the same rate?
s ÷ (3 machines × 4 hours) = 120 ÷ 12 = 10 parts per machine per hour.
先求出每台机器每小时的生产量：120 个 ÷ (3 台 × 4 小时) = 10 个/台·小时。
s. This is a direct proportion problem in which the total production scales with both the number of machines and the time.
那么 5 台机器 6 小时的产量 = 5 × 6 × 10 = 300 个。这是一个正比例问题，总产量与机器数量和工时同时成正比。
Some learners mistakenly treat this as an inverse proportion and set up an incorrect ratio. Always check the context: more machines or more time should increase total output.
部分学生误以为这是反比例而列错比例式。一定要根据题意判断：机器越多或时间越长，总产量应该越多。
10. Question 10 – Probability without Replacement | 第10题 — 无放回概率
From a standard 52-card deck, one card is drawn at random. Find the probability that it is a Jack, Queen or King. Then two cards are drawn successively without replacement; find the probability both are face cards (J, Q or K).
Kings). The probability of drawing a face card on a single draw is P(face) = 12/52 = 3/13.
一副牌中共有 12 张花牌（4张J、4张Q、4张K）。单张抽到花牌的概率为 P = 12/52 = 3/13。
rds is calculated using conditional probability: P(first face) = 12/52, then P(second face | first face) = 11/51.
无放回连抽两张均为花牌的概率需使用条件概率：P(第一张花牌)=12/52，P(第二张花牌|第一张花牌)=11/51。
denominator by 12: 132÷12 = 11, 2652÷12 = 221, so the probability is 11/221.
两者相乘得 132/2652。分子分母同除以 12 化简：132÷12=11，2652÷12=221，所以概率为 11/221。
A tree diagram can help visualise the scenarios. A common mistake is to forget to reduce the denominator for the second draw, making the probability too high.
画树状图有助于呈现各种情形。常见错误是忘记第二次抽取时分母应减小，导致概率值偏高。
Published by TutorHao | Further Mathematics Revision Series | aleveler
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