📚 Year 9 OCR Maths: Unit Test Mock Paper Walkthrough | Year 9 OCR 数学：单元测试模拟卷解析
Preparing for a Year 9 OCR unit test can feel like a steep climb, but working through a mock paper with detailed solutions is one of the most effective ways to build confidence. In this walkthrough, we will tackle a carefully selected set of questions covering number, algebra, ratio, geometry, statistics, probability and sequences — every answer is explained step by step so that you can see exactly what the examiner expects.
准备 Year 9 OCR 单元测试可能像一次艰难的攀登，但通过带有详细解析的模拟卷进行练习是建立信心的最有效方法之一。在这次串讲中，我们将攻关一套精心挑选的题目，涵盖数字、代数、比、几何、统计、概率和数列——每一步解答都被拆解出来，让你清楚看出考官到底需要什么。
1. Fractions and Decimals Operations | 分数与小数运算
Question: Calculate 2/3 + 1/4, giving your answer as a mixed number.
题目：计算 2/3 + 1/4，并将答案写成带分数形式。
To add fractions, you must first find a common denominator. The denominators are 3 and 4, and their least common multiple is 12.
要相加分数，必须先找到公分母。分母是 3 和 4，它们的最小公倍数是 12。
Convert each fraction to an equivalent fraction with denominator 12: 2/3 becomes (2×4)/(3×4) = 8/12, and 1/4 becomes (1×3)/(4×3) = 3/12.
把每个分数转化为以 12 为分母的等价分数：2/3 变成 (2×4)/(3×4) = 8/12，1/4 变成 (1×3)/(4×3) = 3/12。
2/3 + 1/4 = 8/12 + 3/12 = 11/12
The sum is 11/12. Since 11/12 is a proper fraction (the numerator is less than the denominator), it cannot be expressed as a mixed number. The answer is simply 11/12.
结果是 11/12。因为 11/12 是一个真分数（分子小于分母），它无法写成带分数的形式。答案就是 11/12。
A common mistake is to add numerators and denominators directly — that gives 3/7, which is wrong. Always find a common denominator first and only add the numerators.
一个常见错误是把分子和分母分别直接相加——那会得到 3/7，这是错的。一定要先找公分母，只相加分子。
2. Solving Linear Equations | 解一元一次方程
Question: Solve 3x + 5 = 2x + 9.
题目：解方程 3x + 5 = 2x + 9。
The aim is to collect like terms and isolate x on one side. Start by subtracting 2x from both sides to eliminate x from the right-hand side.
目标是合并同类项，让 x 单独待在一边。首先两边同时减去 2x，消去右边的 x 项。
3x – 2x + 5 = 2x – 2x + 9 → x + 5 = 9
Now the equation is much simpler: x + 5 = 9. Subtract 5 from both sides to find x.
现在方程简化很多：x + 5 = 9。两边再同时减去 5，求出 x。
x + 5 – 5 = 9 – 5 → x = 4
The solution is x = 4. You can check by substituting back: left side 3(4) + 5 = 12 + 5 = 17, right side 2(4) + 9 = 8 + 9 = 17. Both sides are equal, so the answer is correct.
解为 x = 4。你可以代入检验：左边 3(4) + 5 = 12 + 5 = 17，右边 2(4) + 9 = 8 + 9 = 17。两边相等，答案正确。
Always perform the same operation on both sides. A frequent error is forgetting to change signs when moving terms, like writing x + 5 = 9 and then adding 5, which would give x = 14.
始终要对两边执行相同的运算。一个常见的错误是移项时忘记变号，比如写成 x + 5 = 9 之后还加 5，得到 x = 14。
3. Ratio and Best Buys | 比与最划算购买
Question: A 300 g pack of biscuits costs £1.20. A 500 g pack of the same biscuits costs £1.90. Which pack is better value? Show your working.
题目：一包 300 g 的饼干售价 £1.20。同一款饼干的 500 g 包装售价 £1.90。哪一个包装更划算？写出你的计算过程。
To compare value, work out the cost per 100 g for each pack. For the 300 g pack, divide the price by 3 to find the cost of 100 g.
要比较性价比，先计算每种包装每 100 g 的价格。对于 300 g 包装，把价格除以 3 得到 100 g 的费用。
300 g pack: £1.20 ÷ 3 = £0.40 per 100 g
For the 500 g pack, divide the price by 5.
对于 500 g 包装，把价格除以 5。
500 g pack: £1.90 ÷ 5 = £0.38 per 100 g
The 500 g pack costs £0.38 per 100 g, while the 300 g pack costs £0.40 per 100 g. So, the larger pack gives you more biscuits for your money — it is better value.
500 g 包装每 100 g 价格为 £0.38，而 300 g 包装每 100 g 为 £0.40。因此，大包装能让你用同样的钱买到更多饼干——它更划算。
Always use identical units when comparing. Some students mistakenly compare total prices alone, but a fair comparison must use a unit rate like cost per 100 g or cost per gram.
比较时始终使用相同的单位。有些同学错误地只比较总价，但公平的比较必须使用单位价格，比如每 100 g 或每克的价格。
4. Angles in Parallel Lines | 平行线中的角
Question: In the diagram, AB is parallel to CD. The transversal EF intersects AB at B and CD at D. Angle EFB = 65°. Find the size of angle EGD, where G is a point on CD, and E, F, G are collinear. Justify your answer.
题目：如图所示，AB 平行于 CD。横截线 EF 与 AB 交于点 B，与 CD 交于点 D。∠EFB = 65°。点 G 在 CD 上，且 E、F、G 共线。求 ∠EGD 的大小并说明理由。
When two parallel lines are cut by a transversal, corresponding angles are equal. Here, angle EFB and angle EGD are in corresponding positions: EFB is at intersection point B on AB, while EGD is at intersection point D on CD, both on the same side of the transversal EF and above the parallel lines.
当两条平行线被一条横截线所截时，同位角相等。在这里，∠EFB 与 ∠EGD 处于同位角的位置：∠EFB 位于 AB 上的交点 B 处，∠EGD 位于 CD 上的交点 D 处，它们都在横截线 EF 的同侧且位于平行线的上方。
Therefore, angle EGD = angle EFB = 65°. The reason is ‘corresponding angles on parallel lines are equal’.
因此，∠EGD = ∠EFB = 65°。理由是“平行线上的同位角相等”。
It’s easy to confuse corresponding angles with alternate angles. Always check their positions: if one angle is outside the ‘F’ shape and the other is inside but on the same side, they are corresponding.
很容易把同位角和内错角搞混。一定要检查它们的位置：如果一个角在“F”形的外部，另一个在内部且位于同一侧，它们就是同位角。
5. Perimeter and Area of Circles | 圆的周长与面积
Question: A circle has a radius of 7 cm. Use π = 3.14 to calculate its circumference and area. Give your answers to 1 decimal place.
题目：一个圆的半径为 7 cm。使用 π = 3.14 计算它的周长和面积。答案保留一位小数。
The formula for circumference is C = 2πr. Substitute r = 7 cm and π = 3.14.
周长公式为 C = 2πr。代入 r = 7 cm 和 π = 3.14。
C = 2 × 3.14 × 7 = 6.28 × 7 = 43.96 cm → 44.0 cm (to 1 d.p.)
The circumference is 44.0 cm. Next, the area: A = πr². Square the radius first.
周长为 44.0 cm。接下来算面积：A = πr²。先算半径的平方。
r² = 7² = 49
A = 3.14 × 49 = 153.86 cm² → 153.9 cm² (to 1 d.p.)
The area is 153.9 cm². Remember the units: circumference is a length (cm), area is in square units (cm²). Rounding only at the final step helps avoid errors.
面积为 153.9 cm²。记住单位：周长是长度（cm），面积是平方单位（cm²）。只在最后一步四舍五入可以减少错误。
6. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Question: Here are the scores from a maths quiz: 12, 15, 14, 12, 18, 16, 12. Find the mean, median, mode and range.
题目：以下是数学小测验的分数：12, 15, 14, 12, 18, 16, 12。求平均数、中位数、众数和极差。
First, arrange the data in order: 12, 12, 12, 14, 15, 16, 18.
首先，把数据从小到大排列：12, 12, 12, 14, 15, 16, 18。
The mean is the sum divided by the count. Sum = 12+15+14+12+18+16+12 = 99. There are 7 values.
平均数是总和除以个数。总和 = 12+15+14+12+18+16+12 = 99。共 7 个数。
Mean = 99 ÷ 7 ≈ 14.1 (to 1 d.p.)
The median is the middle value. With 7 items, the 4th value is the median: 14.
中位数是中间的值。7 个数据中，第 4 个值为中位数：14。
The mode is the most frequent value — 12 appears three times, so the mode is 12.
众数是出现次数最多的值——12 出现了三次，所以众数是 12。
The range is the difference between the largest and smallest values: 18 – 12 = 6.
极差是最大值与最小值的差：18 – 12 = 6。
Summary: mean ≈ 14.1, median = 14, mode = 12, range = 6. The mean is pulled down slightly by the three 12s, showing the effect of a low mode.
总结：平均数 ≈ 14.1，中位数 = 14，众数 = 12，极差 = 6。平均数被三个 12 稍往下拉，体现出低众数的影响。
7. Probability: Tree Diagrams | 概率：树状图
Question: A bag contains 3 red balls and 2 blue balls. Two balls are drawn at random without replacement. Using a tree diagram, find the probability that both balls are red.
题目：一个袋子里有 3 个红球和 2 个蓝球。随机抽取两个球且不放回。利用树状图求两个球都是红色的概率。
Draw the first branch: P(Red) = 3/5, P(Blue) = 2/5. Without replacement means the total balls decrease by 1 for the second pick.
画出第一层分支：P(红) = 3/5，P(蓝) = 2/5。不放回意味着在抽第二个球时总球数减少 1。
If the first ball is red, 2 red and 2 blue remain, so P(Red | Red) = 2/4 = 1/2. If the first is blue, 3 red and 1 blue remain, so P(Red | Blue) = 3/4.
如果第一个球是红色，则剩 2 红 2 蓝，所以 P(红|红) = 2/4 = 1/2。如果第一个是蓝色，则剩 3 红 1 蓝，所以 P(红|蓝) = 3/4。
To get both red, follow the Red–Red path. Multiply probabilities along the branch.
要得到两个都是红色，顺着红–红路径。沿着分支相乘概率。
P(Both Red) = (3/5) × (2/4) = 6/20 = 3/10
The probability is 3/10. The tree diagram visually confirms all possible outcomes and helps avoid missing the ‘without replacement’ condition — a typical mistake is to treat the second draw as independent.
概率为 3/10。树状图能直观地展示所有可能结果，有助于避免忽略“不放回”条件——一个典型错误是把第二次抽取当作独立事件处理。
8. Sequences: Finding the nth term | 数列：求第 n 项
Question: The first four terms of a sequence are 5, 9, 13, 17. Write down an expression for the nth term, and use it to find the 10th term.
题目：数列的前四项为 5, 9, 13, 17。写出第 n 项的表达式，并用以求出第 10 项。
Find the common difference: 9 – 5 = 4, 13 – 9 = 4, so it’s an arithmetic sequence with difference 4. The nth term of an arithmetic sequence has the form an + b, where a is the common difference.
先找公差：9 – 5 = 4，13 – 9 = 4，所以这是一个公差为 4 的等差数列。等差数列第 n 项的形式为 an + b，其中 a 是公差。
Here, a = 4, so the expression begins with 4n. To find b, plug in n = 1: 4(1) + b = 5, so b = 1. The nth term is 4n + 1.
这里 a = 4，所以表达式以 4n 开头。为求 b，代入 n = 1：4(1) + b = 5，得 b = 1。第 n 项为 4n + 1。
nth term = 4n + 1
To find the 10th term, substitute n = 10: 4(10) + 1 = 40 + 1 = 41.
要求第 10 项，代入 n = 10：4(10) + 1 = 40 + 1 = 41。
Always test the formula with the given terms: n=2 gives 4(2)+1=9, n=3 gives 13 — it works. A quick check like this can catch arithmetic errors.
一定要用已知项检验公式：n=2 时得 4(2)+1=9，n=3 得 13——没问题。这样快速验证能发现计算错误。
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