📚 Year 8 OCR Further Mathematics: Case Study Practice Sessions | Year 8 OCR 进阶数学：案例分析实战演练
In this article, we tackle real-life problem-solving through eight carefully selected case studies. Each case targets a key topic from the Year 8 OCR Further Mathematics syllabus, helping you build confidence in applying concepts to unfamiliar situations. Read the English description, then use the Chinese translation to check your understanding. Work through each step to turn theory into mastery.
本文通过八个精选案例进行实战演练，每个案例都围绕 Year 8 OCR 进阶数学的核心知识点。我们将实际问题转化为数学模型，逐步展示分析思路与解题过程。中英双语配对的讲解，不仅能帮助你理解数学概念，还能强化逻辑表达能力。请跟随每一步推导，将知识真正内化。
1. Solving Age Problems with Linear Equations | 用线性方程解决年龄问题
Problem: Sarah is 5 years older than Tom. In 3 years, Sarah’s age will be twice Tom’s age. Find their current ages.
问题：莎拉比汤姆大5岁。3年后，莎拉的年龄将是汤姆年龄的两倍。求他们现在的年龄。
Let Tom’s current age be x years. Then Sarah’s current age is x + 5.
设汤姆现在的年龄为 x 岁，则莎拉现在的年龄为 x + 5 岁。
In 3 years, Tom’s age will be x + 3, and Sarah’s age will be (x + 5) + 3 = x + 8.
3年后，汤姆的年龄为 x + 3，莎拉的年龄为 (x + 5) + 3 = x + 8。
The condition says Sarah’s future age will be twice Tom’s future age, so we write the equation:
题目条件说明莎拉未来的年龄是汤姆未来年龄的两倍，因此列出方程：
x + 8 = 2(x + 3)
Solve for x: expand the right side to get x + 8 = 2x + 6. Then bring x terms together: 8 – 6 = 2x – x → 2 = x.
解方程：将右边展开得 x + 8 = 2x + 6。移项合并得 8 – 6 = 2x – x，即 2 = x。
Therefore, Tom is 2 years old and Sarah is 2 + 5 = 7 years old.
因此，汤姆现在2岁，莎拉现在7岁。
Check the answer: in 3 years, Tom will be 5 and Sarah will be 10. 10 is indeed twice 5, so the solution is correct.
检验：3年后汤姆5岁，莎拉10岁，10正好是5的两倍，答案正确。
2. Perimeter and Area of Composite Shapes | 复合图形的周长与面积
Problem: A running track consists of a rectangle of length 80 m and width 50 m, with a semicircle attached to each of the shorter ends. The diameter of each semicircle equals the width of the rectangle. Calculate the total perimeter of the track and the area of the region it encloses. Use π = 3.14.
问题：一个跑道由一个长80米、宽50米的长方形和两个半圆组成，半圆连接在长方形的两条短边上。半圆的直径等于长方形的宽度。求跑道的总周长以及它所围区域的面积。π 取 3.14。
The two semicircles together form a full circle of diameter 50 m, so radius r = 25 m.
两个半圆恰好组成一个直径为50米的整圆，因此半径 r = 25 m。
The perimeter of the track consists of the two straight sides of the rectangle plus the circumference of the full circle. The two straights add 2 × 80 = 160 m. The circumference is 2πr = 2 × 3.14 × 25 = 157 m.
跑道的周长由两条直道和整圆的周长组成。两条直道总长 2 × 80 = 160 m。圆的周长为 2πr = 2 × 3.14 × 25 = 157 m。
Total perimeter = 160 + 157 = 317 m.
总周长 = 160 + 157 = 317 m。
For the area, add the area of the rectangle (length × width = 80 × 50 = 4000 m²) and the area of the full circle (πr² = 3.14 × 25² = 3.14 × 625 = 1962.5 m²).
计算面积时，将长方形面积（长 × 宽 = 80 × 50 = 4000 m²）与整圆面积（πr² = 3.14 × 25² = 3.14 × 625 = 1962.5 m²）相加。
Total area = 4000 + 1962.5 = 5962.5 m²
3. LCM Application: Gear Alignment | 最小公倍数应用：齿轮对齐
Problem: Two meshing gears have 15 teeth and 20 teeth respectively. A marked tooth on each gear is aligned at the start. After how many rotations of the larger gear will the marked teeth align again for the first time?
问题：两个啮合齿轮分别有15个齿和20个齿。开始时两个齿轮上各有一个标记齿恰好对齐。问大齿轮转过多少圈后，这两个标记齿将第一次重新对齐？
For the marked teeth to realign, each gear must have turned through a whole number of teeth that is a common multiple of both gear teeth counts. The smallest such number is the least common multiple (LCM) of 15 and 20.
要使标记齿重新对齐，每个齿轮转过的总齿数必须是两个齿数的公倍数。第一次对齐对应的最小总齿数即为15和20的最小公倍数（LCM）。
Break into prime factors: 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 60.
分解质因数：15 = 3 × 5，20 = 2² × 5。LCM = 2² × 3 × 5 = 60。
So 60 teeth must pass the point of mesh. The larger gear has 20 teeth, so the number of rotations it makes is 60 ÷ 20 = 3.
因此，总共需要有60个齿经过啮合点。大齿轮有20个齿，所以大齿轮的转数为 60 ÷ 20 = 3 圈。
The smaller gear will make 60 ÷ 15 = 4 rotations in the same time, agreeing with the requirement for realignment.
小齿轮在同一时间内转过 60 ÷ 15 = 4 圈，验证了对齐条件。
4. Ratio and Proportion in Mixtures | 混合物中的比例与比率
Problem: A cordial drink is made by mixing concentrate and water in the ratio 1 : 4. (a) How much concentrate is needed to make 2.5 litres of drink? (b) A jug contains 300 ml of concentrate. How much water should be added to use up all the concentrate? Give your answers in millilitres.
问题：一种浓缩果汁饮料按浓缩液与水的比例为 1 : 4 调制。(a) 若要制作2.5升饮料，需要多少浓缩液？(b) 一个壶里有300毫升浓缩液，要全部用完，需要加多少水？答案以毫升为单位。
First, note 2.5 litres = 2500 ml. The ratio 1 : 4 means the total number of parts is 1 + 4 = 5.
首先，2.5升 = 2500毫升。比例 1 : 4 表示总份数为 1 + 4 = 5 份。
One part (concentrate) out of 5 corresponds to 2500/5 = 500 ml.
浓缩液占5份中的1份，因此体积为 2500 ÷ 5 = 500 ml。
For part (b), the concentrate corresponds to 1 part. So 1 part = 300 ml. The water must be 4 parts, i.e., 4 × 300 = 1200 ml.
第(b)问中，浓缩液对应1份，即1份 = 300 ml。水占4份，因此需要 4 × 300 = 1200 ml。
Always check that the final mixture has the intended ratio: 300 : 1200 simplifies to 1 : 4, confirming the answer.
务必检查最终混合物的比例：300 : 1200 化简为 1 : 4，答案正确。
5. Probability with Two Dice | 两个骰子的概率
Problem: Two fair six-sided dice are rolled. Find the probability that the sum of the numbers shown is exactly 9.
问题：同时投掷两个公平的六面骰子，求所显示点数之和恰好为9的概率。
List all possible outcomes that give a sum of 9: (3,6), (4,5), (5,4), (6,3). There are 4 favourable outcomes.
列举所有和为9的可能结果：(3,6)、(4,5)、(5,4)、(6,3)。共有4种有利情况。
The total number of equally likely outcomes when rolling two dice is 6 × 6 = 36.
投掷两个骰子时，所有等可能结果的总数为 6 × 6 = 36。
Probability = 4/36 = 1/9
Therefore, the probability of getting a sum of 9 is one ninth. You can also confirm by drawing a sample space table.
因此，和为9的概率为九分之一。你也可以通过画样本空间表格来验证。
This demonstrates how systematic listing helps to avoid missing combinations in probability problems.
这表明在概率问题中，系统性地列出所有组合可避免遗漏。
6. Finding the nth Term of a Sequence | 求数列的第n项
Problem: A pattern uses matchsticks to form a row of squares. 1 square uses 4 matchsticks, 2 squares use 7, 3 squares use 10, and so on. Find an expression for the number of matchsticks needed for n squares, and calculate how many matchsticks are required for 20 squares.
问题：用火柴棍拼成一排相连的正方形。1个正方形需4根火柴，2个正方形需7根，3个正方形需10根，依此类推。求拼 n 个正方形所需火柴棍数量的表达式，并计算20个正方形需要多少根火柴。
Observe the sequence for the number of matchsticks: 4, 7, 10, … The difference between consecutive terms is constant at 3.
观察火柴棍数的序列：4, 7, 10, … 相邻两项的差恒为3。
For an arithmetic sequence with first term a = 4 and common difference d = 3, the nth term is given by a + (n – 1)d.
对于首项 a = 4，公差 d = 3 的等差数列，第n项的公式为 a + (n – 1)d。
Substitute: nth term = 4 + (n – 1)×3 = 4 + 3n – 3 = 3n + 1.
代入公式：第n项 = 4 + (n – 1)×3 = 4 + 3n – 3 = 3n + 1。
So for n squares, the number of matchsticks is 3n + 1. For n = 20: 3×20 + 1 = 61.
因此，拼 n 个正方形需要 (3n + 1) 根火柴。当 n = 20 时，需 3×20 + 1 = 61 根。
Check small values: n=1 gives 4, n=2 gives 7, confirming the rule works.
验证小数据：n=1 得 4，n=2 得 7，表明公式正确。
7. Statistics: Mean, Median and Range | 统计：平均数、中位数和极差
Problem: The marks of nine students in a test are: 11, 14, 15, 17, 18, 19, 21, 22, 25. Calculate the mean, the median, and the range of these marks.
问题：九名学生的测验分数为：11, 14, 15, 17, 18, 19, 21, 22, 25。计算这些分数的平均数、中位数和极差。
To find the mean, add all values: 11+14+15+17+18+19+21+22+25 = 162. Divide by the number of values, 9.
求平均数：先将所有数值相加：11+14+15+17+18+19+21+22+25 = 162。再除以数据个数9。
Mean = 162 ÷ 9 = 18
The data is already ordered. The median is the middle value when arranged from smallest to largest. With 9 numbers, the 5th value is the median: here it is 18.
数据已按从小到大排列。数据个数为奇数（9个），中位数是第5个值，即18。
The range is the difference between the largest and smallest values: 25 – 11 = 14.
极差为最大值与最小值之差：25 – 11 = 14。
Notice that the mean equals the median in this symmetric set, but this is not always the case; always compute each measure separately.
在这个对称数据集中平均数等于中位数，但并非总是如此，应分别计算每个统计指标。
8. Speed, Time and Distance: Meeting Problem | 速度、时间与距离：相遇问题
Problem: Town A and Town B are 120 km apart. A car leaves Town A travelling towards Town B at a constant speed of 40 km/h. At the same time, a truck leaves Town B travelling towards Town A at 60 km/h. How long will it take for the two vehicles to meet, and how far from Town A will the meeting point be?
问题：A城与B城相距120公里。一辆汽车从A城出发以40 km/h 的速度匀速驶向B城；同时，一辆卡车从B城出发以60 km/h 匀速驶向A城。两车多久后相遇？相遇地点距离A城多远？
Since the vehicles are moving towards each other, their relative speed is the sum of their individual speeds: 40 + 60 = 100 km/h.
两车相向而行，它们的相对速度为各自速度之和：40 + 60 = 100 km/h。
The total distance to be covered is 120 km. Time = Distance ÷ Relative Speed = 120 ÷ 100 = 1.2 hours (which is 1 hour 12 minutes).
需要行驶的总距离为120公里。时间 = 距离 ÷ 相对速度 = 120 ÷ 100 = 1.2 小时（即1小时12分钟）。
To find the distance from Town A, use the car’s travel: Distance = speed × time = 40 km/h × 1.2 h = 48 km.
求相遇点距A城的距离，使用汽车行驶的路程：距离 = 速度 × 时间 = 40 km/h × 1.2 h = 48 km。
Check with the truck: 60 km/h × 1.2 h = 72 km, and 48 + 72 = 120 km, confirming the meeting point divides the journey correctly.
用卡车验证：60 km/h × 1.2 h = 72 km，而 48 + 72 = 120 km，符合总路程。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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