📚 Year 8 OCR Further Maths: Case Study Practical Exercises | Year 8 OCR 进阶数学：案例分析实战演练
In Year 8 OCR Further Mathematics, applying concepts to real-life scenarios is one of the most effective ways to deepen understanding. This article presents ten case studies that blend algebra, geometry, data handling, probability and more. Each case is followed by a clear, step-by-step solution that mirrors the reasoning expected at this level.
在Year 8 OCR进阶数学中，将概念应用于真实场景是加深理解最有效的方法之一。本文通过十个案例分析，融合代数、几何、数据处理、概率等主题。每个案例都配有清晰的逐步解题过程，反映这个阶段所期望的推理能力。
1. Patterns in Tile Designs | 瓷砖图案中的序列
A designer arranges square tiles in rows to form a pattern. The first row uses 4 tiles, the second row 7 tiles, the third row 10 tiles, and so on. Find a formula for the number of tiles in the nth row and use it to determine how many tiles are needed for the 20th row.
一位设计师用方形瓷砖排成图案。第一排用了4块瓷砖，第二排7块，第三排10块，依此类推。找出第n排瓷砖数量的公式，并计算第20排需要多少块瓷砖。
quence is given by a + (n - 1)d. Here a = 4 and d = 3, so T(n) = 4 + (n - 1) × 3 = 3n + 1.
注意到序列4, 7, 10,...每次增加3。等差数列的第n项公式为 a + (n - 1)d。这里 a = 4, d = 3，因此 T(n) = 4 + (n - 1) × 3 = 3n + 1。
For the 20th row, substitute n = 20: T(20) = 3 × 20 + 1 = 61 tiles.
对于第20排，代入 n = 20：T(20) = 3 × 20 + 1 = 61 块瓷砖。
number k satisfies 3k + 1 ≤ 100, giving k ≤ 33. So the designer can complete up to row 33.
如果有100块瓷砖可用，最大可能的排数 k 满足 3k + 1 ≤ 100，解得 k ≤ 33。因此设计师最多可以完成到第33排。
2. Scaling Recipes | 食谱比例调整
A cake recipe for 8 people requires 200 g of flour, 150 g of sugar, 100 g of butter and 4 eggs. You need to bake the same cake for 12 people. Calculate the new quantities, keeping the ingredient ratios exactly the same.
一份8人份的蛋糕食谱需要200克面粉、150克糖、100克黄油和4个鸡蛋。你需要为12人烘焙相同的蛋糕。计算新的用量，保持配料比例完全相同。
× 1.5 = 300 g, sugar = 150 × 1.5 = 225 g, butter = 100 × 1.5 = 150 g, eggs = 4 × 1.5 = 6.
从8人到12人的比例因子为 12 ÷ 8 = 3/2 或 1.5。将每种配料乘以1.5：面粉 = 200 × 1.5 = 300克，糖 = 150 × 1.5 = 225克，黄油 = 100 × 1.5 = 150克，鸡蛋 = 4 × 1.5 = 6个。
s confirms that proportional reasoning works for both discrete items (eggs) and continuous quantities.
使用分数计算，结果相同：面粉 = 200 × 3/2 = 300克，等等。这验证了比例推理同样适用于离散物品（鸡蛋）和连续量。
the factor would be 6/8 = 3/4, giving 150 g flour, 112.5 g sugar, 75 g butter and 3 eggs.
如果后来想将食谱减为6人份，比例因子为 6/8 = 3/4，得到150克面粉、112.5克糖、75克黄油和3个鸡蛋。
3. Logo Transformations | Logo 变换
A graphic designer places a rectangular logo with vertices at (1, 2), (3, 2), (3, 4) and (1, 4) on a coordinate grid. First the logo is reflected in the y-axis, and then it is rotated 90° clockwise about the origin. Find the final coordinates of the vertices.
一位平面设计师在坐标网格上放置一个矩形标志，顶点为 (1,2), (3,2), (3,4) 和 (1,4)。首先将标志关于y轴反射，然后绕原点顺时针旋转90°。求最终顶点的坐标。
Reflection in the y-axis changes each (x, y) to (-x, y). The vertices become (-1, 2), (-3, 2), (-3, 4), (-1, 4).
关于y轴反射将每个 (x, y) 变为 (-x, y)。顶点变为 (-1,2), (-3,2), (-3,4), (-1,4)。
x). Applying this: (-1, 2) → (2, 1); (-3, 2) → (2, 3); (-3, 4) → (4, 3); (-1, 4) → (4, 1).
绕原点顺时针旋转90°将 (x, y) 变为 (y, -x)。应用变换：(-1,2) → (2,1); (-3,2) → (2,3); (-3,4) → (4,3); (-1,4) → (4,1)。
The final shape is still a rectangle, now positioned in the first quadrant. Combining transformations helps designers predict how an image moves on screen.
最终形状仍为矩形，现位于第一象限。组合变换帮助设计师预测图像在屏幕上如何移动。
4. Survey Data Analysis | 调查数据分析
Ten students recorded the number of minutes they spend reading each day: 30, 45, 20, 35, 50, 60, 25, 40, 45, 30. As a data analyst, calculate the mean, median, mode and range. Then decide which average best represents the data.
十名学生记录了他们每天阅读的分钟数：30, 45, 20, 35, 50, 60, 25, 40, 45, 30。作为数据分析师，计算平均数、中位数、众数和极差。然后判断哪种平均数最能代表这组数据。
45, 45, 50, 60. The mean is (20+25+30+30+35+40+45+45+50+60) ÷ 10 = 380 ÷ 10 = 38 minutes.
首先将数据排序：20, 25, 30, 30, 35, 40, 45, 45, 50, 60。平均数为 (20+25+30+30+35+40+45+45+50+60) ÷ 10 = 380 ÷ 10 = 38 分钟。
0) ÷ 2 = 37.5 minutes. The mode is 30 and 45 (bimodal). The range is 60 - 20 = 40 minutes.
数据个数为偶数时，中位数是第5和第6个数的平均值：(35 + 40) ÷ 2 = 37.5分钟。众数为30和45（双众数）。极差为 60 - 20 = 40分钟。
ced analysis might include the interquartile range: Q1 = 30, Q3 = 45, so IQR = 15 minutes.
由于存在两个众数，且极端值60将平均数拉高，中位数37.5分钟更能代表典型的阅读时间。更深入的分析可包含四分位距：Q1 = 30, Q3 = 45，因此 IQR = 15分钟。
5. Probability Games | 概率游戏
A stall at a school fair offers a game: you roll two fair dice and win a prize if the sum is exactly 7. It costs 50p to play, and the prize is £2.50. Is the game fair? If not, suggest a prize that would make the expected gain zero.
学校义卖会上有一个游戏摊位：你掷两个公平的骰子，如果点数之和恰好为7就可以赢得奖品。玩一次花费50便士，奖品价值2.50英镑。这个游戏公平吗？如果不公平，请建议一个使期望收益为零的奖品金额。
= 6 × 6 = 36. The combinations that sum to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 6 outcomes. So P(sum = 7) = 6/36 = 1/6.
掷两个骰子总共有 6 × 6 = 36 种可能结果。和为7的组合有 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 共6种结果。因此 P(和为7) = 6/36 = 1/6。
Expected gain for a player = (probability of winning × prize) - cost = (1/6 × £2.50) - £0.50 = £0.4167 - £0.50 = -£0.0833. The negative expectation means the game favours the stall.
玩家的期望收益 = (赢的概率 × 奖品价值) - 花费 = (1/6 × 2.50英镑) - 0.50英镑 = 0.4167英镑 - 0.50英镑 = -0.0833英镑。负的期望值意味着游戏对摊位有利。
For a fair game, expected gain should be 0: (1/6 × P) - 0.50 = 0 ⇒ P = £3.00. So a prize of £ ld make the game fair. Probability ensures games are designed with expected value in mind.
要使游戏公平，期望收益应为0：(1/6 × P) - 0.50 = 0 ⇒ P = 3.00英镑。因此3.00英镑的奖品将使游戏变得公平。概率知识确保游戏在设计时就考虑到期望值。
6. Mobile Phone Plans | 手机套餐
Two mobile networks offer monthly plans: Plan A charges a £10 fixed fee plus 5p per minute of calls. Plan B charges a £20 fixed fee plus 2p per minute. For how many minutes of calls per month do the two plans cost the same? Which plan is cheaper for heavy users?
两家移动网络提供月度套餐：套餐A收取10英镑固定费用，另加每分钟5便士通话费。套餐B收取20英镑固定费用，另加每分钟2便士通话费。每月通话多少分钟时，两种套餐费用相同？对于通话量大的用户，哪种套餐更便宜？
+ 0.05m (in pounds). Cost for B: C₂ = 20 + 0.02m. Set them equal: 10 + 0.05m = 20 + 0.02m.
设通话分钟数为 m。A的费用：C₁ = 10 + 0.05m（英镑）。B的费用：C₂ = 20 + 0.02m。令两者相等：10 + 0.05m = 20 + 0.02m。
Subtract 10 and 0.02m from both sides: 0.03m = 10 ⇒ m = 10 ÷ 0.03 = 333⅓ minutes. At about 333 minutes, the costs are equal.
两边同时减去10和0.02m：0.03m = 10 ⇒ m = 10 ÷ 0.03 = 333⅓ 分钟。在大约333分钟时，费用相等。
er-minute rate outweighs the higher fixed fee. For instance, at 500 minutes: A = 10 + 25 = £35, B = 20 + 10 = £0. Graphical representation shows two straight lines intersecting at the break-even point.
当 m > 333.33 时，套餐B更便宜，因为较低的单分钟费用超过了较高的固定费用。例如在500分钟时：A = 10 + 25 = 35英镑，B = 20 + 10 = 30英镑。用图像表示可以看到两条直线在损益平衡点相交。
7. Garden Design | 花园设计
A rectangular garden is 3 m longer than it is wide. Its area is 54 m². The owner wants to install a fence around the garden, costing £12 per metre. Find the dimensions and the total fencing cost.
一个长方形花园的长比宽多3米。它的面积为54平方米。园主想围绕花园建围栏，每米成本12英镑。求花园的尺寸和围栏总成本。
length = w + 3. Area = w(w + 3) = 54. This gives the quadratic equation w² + 3w - 54 = 0.
设宽为 w 米，则长 = w + 3。面积 = w(w + 3) = 54。得到二次方程 w² + 3w - 54 = 0。
- 6) = 0. The solutions are w = -9 (not possible) and w = 6. So width = 6 m, length = 9 m.
因式分解：(w + 9)(w - 6) = 0。解为 w = -9（不可能）和 w = 6。因此宽为6米，长为9米。
Perimeter = 2 × (6 + 9) = 30 m. Fencing cost = 30 × £12 = £ dimensions even when the relationship between length and width was not initially obvious.
周长 = 2 × (6 + 9) = 30米。围栏成本 = 30 × 12英镑 = 360英镑。运用代数我们能够求出精确尺寸，即使在长宽关系最初不明显的情况下。
8. Bicycle Hire Comparison | 自行车租赁比较
Company X offers bicycle hire at a £5 daily rental plus £0.20 per kilometre cycled. Company Y charges £10 per day plus £0.10 per km. Model the cost for each company using linear equations, draw their graphs, and interpret the point of intersection.
X公司提供自行车租赁服务，每日租金5英镑，另加每公里0.20英镑。Y公司每日收费10英镑，另加每公里0.10英镑。用线性方程表示每家公司的费用，画出图像，并解释交点的含义。
e d km in a day, Cost X = 5 + 0.20d, Cost Y = 10 + 0.10d. Both are of the form y = mx + c.
一天骑行距离为 d 公里时，X的费用 = 5 + 0.20d，Y的费用 = 10 + 0.10d。两者均为 y = mx + c 的形式。
find the intersection: 5 + 0.20d = 10 + 0.10d → 0.10d = 5 → d = 50 km. At 50 km, both cost £ ances below 50 km, Company X is cheaper; above 50 km, Company Y becomes the better choice.
令方程相等求交点：5 + 0.20d = 10 + 0.10d → 0.10d = 5 → d = 50公里。在50公里处，两家费用均为15英镑。当距离短于50公里时，X公司更便宜；超过50公里后，Y公司成为更好的选择。
Plotting the graphs on the same axes shows two lines crossing at (50, 15). The gradient reflects the per-kilometre rate, while the y-intercept shows the fixed daily charge. This visual comparison helps customers decide quickly.
在同一坐标轴上绘图可看到两条直线在 (50,15) 处相交。斜率反映每公里费率，y轴截距显示每日固定费用。这种视觉化比较能帮助顾客快速做出决定。
9. Bacteria Growth | 细菌增长
A laboratory culture starts with 100 bacteria, and the population doubles every hour. Write a formula for the number of bacteria after t hours. How many bacteria are present after 5 hours? After how many hours will the population first exceed 100,000?
一个实验室培养物最初有100个细菌，且数量每小时翻倍。写出t小时后细菌数量的公式。5小时后有多少细菌？经过多少小时细菌数量将首次超过100,000？
s exponential: N(t) = 100 × 2ᵗ. After 5 hours, N(5) = 100 × 2⁵ = 100 × 32 = 3200 bacteria.
增长模式是指数型的：N(t) = 100 × 2ᵗ。5小时后，N(5) = 100 × 2⁵ = 100 × 32 = 3200 个细菌。
12 is too small, so after 10 hours the population reaches 102,400, exceeding the threshold.
要计算何时 N(t) > 100,000，解 100 × 2ᵗ > 100,000 → 2ᵗ > 1000。因为 2¹⁰ = 1024，t 必须至少为10。确实，2⁹ = 512 太小，所以10小时后数量达到102,400，超过阈值。
Exponential growth appears in many real-world scenarios such as compound interest and viral spread. Understanding doubling time is key to making predictions.
指数增长出现在许多真实场景中，如复利和病毒传播。理解翻倍时间对于做出预测至关重要。
10. Password Combinations | 密码组合
A security system requires a 4-digit PIN code using digits 0-9. How many different PINs are possible? If the system demands that the code must contain at least one even digit, how many valid codes exist?
一个安全系统需要一个由数字0-9组成的4位PIN码。有多少种不同的PIN码可能？如果系统要求密码至少包含一个偶数数字，那么有多少个有效密码？
Total number of 4-digit codes = 10⁴ = 10,000 (each position has 10 choices).
4位密码的总数 = 10⁴ = 10,000（每位有10种选择）。
odd digits: 1, 3, 5, 7, 9). That count is 5⁴ = 625. So valid codes = 10,000 - 625 = 9375.
要找出至少包含一个偶数数字的密码，更容易的方法是减去全为奇数数字（1,3,5,7,9）的密码数。全奇数密码的数量为 5⁴ = 625。因此有效密码数 = 10,000 - 625 = 9375。
Using complementary counting simplifies the problem significantly. The same principle can be applied to password strength and probability of guessing correctly.
使用补集计数大大简化了问题。同样的原理可应用于密码强度分析和正确猜测的概率计算。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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