📚 Case Study Practice in Year 9 CIE Advanced Mathematics | CIE 9 年级进阶数学：案例分析实战演练
Case studies form a pivotal part of the CIE Year 9 Advanced Mathematics curriculum, bridging the gap between abstract theory and real-world problem-solving. This article presents a series of carefully structured practical exercises that challenge students to apply algebraic reasoning, geometric insight, statistical analysis, and probabilistic thinking in authentic scenarios. Each case study is designed not merely to test knowledge, but to cultivate the analytical mindset required for higher-level mathematical study.
案例分析是 CIE 9 年级进阶数学课程中的关键环节，它连接了抽象理论与现实问题解决之间的桥梁。本文提供一系列精心设计的实战演练，要求学生在真实情境中运用代数推理、几何直觉、统计分析和概率思维。每个案例不仅旨在检验知识，更致力于培养高阶数学学习所需的分析思维模式。
1. Maximising Profit at the School Fair | 校园义卖利润最大化
A student-run tuck shop at a school fair sells homemade lemonade. The cost to produce one bottle is £0.40, and the selling price is set at £1.20. The fixed costs for the stall rental and equipment amount to £15. The team aims to understand how many bottles they must sell to break even, and how to maximise their profit if they consider adjusting the price.
学校义卖会上，学生经营的小食摊出售自制柠檬水。每瓶生产成本为 0.40 英镑，售价定为 1.20 英镑。摊位租金和设备固定成本总计 15 英镑。团队希望了解必须售出多少瓶才能实现盈亏平衡，以及在考虑调整价格时如何实现利润最大化。
Let x represent the number of bottles sold. The revenue function R(x) and total cost function C(x) can be expressed as follows. Revenue is simply the price per bottle multiplied by quantity sold, giving R(x) = 1.20x. The total cost includes both variable and fixed components: C(x) = 0.40x + 15. The profit function P(x) is the difference between revenue and cost, yielding P(x) = 1.20x − (0.40x + 15) = 0.80x − 15.
设 x 代表售出的瓶数。收入函数 R(x) 和总成本函数 C(x) 可表示如下。收入即每瓶价格乘以售出数量，得出 R(x) = 1.20x。总成本包含可变成本和固定成本两部分：C(x) = 0.40x + 15。利润函数 P(x) 为收入与成本之差，即 P(x) = 1.20x − (0.40x + 15) = 0.80x − 15。
To find the break-even point, we set profit equal to zero and solve for x. This gives 0.80x − 15 = 0, so 0.80x = 15, and therefore x = 18.75. Since a fractional bottle cannot be sold, the team must sell at least 19 bottles to begin making a profit. For every additional bottle beyond the break-even point, the profit increases by £0.80.
为求盈亏平衡点，设利润为零并求解 x。得到 0.80x − 15 = 0，因此 0.80x = 15，x = 18.75。由于无法售出分数瓶，团队至少需要售出 19 瓶才能开始盈利。盈亏平衡点之后，每多售出一瓶，利润增加 0.80 英镑。
Now suppose the team experiments with a higher selling price of £1.50 per bottle, believing demand might still be strong. The new revenue function becomes R(x) = 1.50x, while costs remain unchanged. The profit function updates to P(x) = 1.50x − (0.40x + 15) = 1.10x − 15. The break-even quantity shifts to approximately 14 bottles (since 1.10x = 15 implies x ≈ 13.64). This sensitivity analysis demonstrates how small pricing adjustments can substantially alter profit margins.
现在假设团队尝试将售价提高到每瓶 1.50 英镑，认为需求可能仍然强劲。新的收入函数变为 R(x) = 1.50x，而成本保持不变。利润函数更新为 P(x) = 1.50x − (0.40x + 15) = 1.10x − 15。盈亏平衡数量下移至约 14 瓶（因为 1.10x = 15 意味着 x ≈ 13.64）。这种敏感性分析展示了微小的定价调整如何显著改变利润率。
2. Designing the Optimal Garden Pathway | 最优花园小径设计
A rectangular school garden measures 24 metres in length and 15 metres in width. The grounds team plans to construct a uniform gravel pathway around the entire perimeter, with the remaining central area reserved for planting. The total available budget for gravel allows covering exactly 156 m². The challenge is to determine the width of the pathway that uses the full budget.
学校一个矩形花园长 24 米，宽 15 米。园艺团队计划围绕整个周边修建一条等宽的砾石小径，剩余中央区域保留用于种植。砾石的总预算恰好能覆盖 156 平方米。挑战在于确定用尽全部预算的小径宽度。
Let the pathway width be w metres, applied uniformly on all four sides. The overall dimensions of the garden including the pathway become (24 + 2w) by (15 + 2w). The area of the pathway alone is the difference between the outer rectangle area and the inner planting area. The inner planting rectangle measures 24 by 15, while the outer rectangle measures (24 + 2w) by (15 + 2w). Therefore, the pathway area equation is (24 + 2w)(15 + 2w) − 24 × 15 = 156.
设小径宽度为 w 米，四边等宽。包含小径的花园总尺寸为 (24 + 2w) × (15 + 2w)。小径的面积是外矩形面积与内种植区域面积之差。内种植矩形尺寸为 24 × 15，外矩形尺寸为 (24 + 2w) × (15 + 2w)。因此，小径面积方程为 (24 + 2w)(15 + 2w) − 24 × 15 = 156。
Expanding the product and simplifying yields a quadratic equation. First, (24 + 2w)(15 + 2w) = 360 + 48w + 30w + 4w² = 360 + 78w + 4w². Subtract the inner area of 360 to obtain the pathway expression: 4w² + 78w. Setting this equal to 156 gives 4w² + 78w − 156 = 0. Dividing through by 2 simplifies to 2w² + 39w − 78 = 0.
展开乘积并简化得到二次方程。首先，(24 + 2w)(15 + 2w) = 360 + 48w + 30w + 4w² = 360 + 78w + 4w²。减去内部面积 360 得到小径表达式：4w² + 78w。将其设为 156 得 4w² + 78w − 156 = 0。全式除以 2 简化为 2w² + 39w − 78 = 0。
Applying the quadratic formula with a = 2, b = 39, c = −78 produces two candidate solutions. The discriminant is b² − 4ac = 39² − 4 × 2 × (−78) = 1521 + 624 = 2145. Taking the square root gives √2145 ≈ 46.31. The solutions are w = (−39 ± 46.31) / 4. The positive root is w ≈ 1.83, while the negative root is discarded as geometrically meaningless. Thus, a pathway width of approximately 1.83 metres exhausts the gravel budget.
应用二次方程求根公式，其中 a = 2，b = 39，c = −78，得到两个候选解。判别式为 b² − 4ac = 39² − 4 × 2 × (−78) = 1521 + 624 = 2145。开平方得 √2145 ≈ 46.31。解为 w = (−39 ± 46.31) / 4。正根为 w ≈ 1.83，负根因无几何意义而被舍弃。因此，约 1.83 米宽的小径正好用尽砾石预算。
3. Analysing Student Heights Across Year Groups | 跨年级学生身高分析
A school conducts a survey measuring the heights of 60 students, with equal samples drawn from Year 7, Year 9, and Year 11. The raw data is organised into a grouped frequency table to facilitate comparison. The task is to calculate measures of central tendency and dispersion for each year group, and to draw comparative conclusions about growth patterns between ages 11–12 and 15–16.
某学校进行了一项调查，测量了 60 名学生的身高，样本等量来自 7 年级、9 年级和 11 年级。原始数据被整理成分组频率表以便进行比较。任务是计算每个年级的集中趋势和离散程度指标，并就 11–12 岁与 15–16 岁之间的生长模式得出比较性结论。
Below is the grouped frequency table for the three year groups, showing height intervals in centimetres and the number of students falling within each interval.
以下是三个年级的分组频率表，显示以厘米为单位的身高区间及各区间内的学生人数。
| Height Interval (cm) | 身高区间（厘米） | Year 7 | 7年级 | Year 9 | 9年级 | Year 11 | 11年级 |
|---|---|---|---|
| 140 ≤ h < 150 | 8 | 1 | 0 |
| 150 ≤ h < 160 | 9 | 6 | 1 |
| 160 ≤ h < 170 | 3 | 8 | 7 |
| 170 ≤ h < 180 | 0 | 4 | 9 |
| 180 ≤ h < 190 | 0 | 1 | 3 |
To estimate the mean height for each year group, we use the midpoint of each interval as the representative value. The midpoints are 145, 155, 165, 175, and 185 cm respectively. For Year 7, the estimated mean is calculated as follows: (8 × 145 + 9 × 155 + 3 × 165) ÷ 20 = (1160 + 1395 + 495) ÷ 20 = 3050 ÷ 20 = 152.5 cm. This represents the typical height of an 11–12 year old in the sample.
为估算每个年级的平均身高，我们以每个区间的中点作为代表值。中点分别为 145、155、165、175 和 185 厘米。对于 7 年级，估算均值计算如下：(8 × 145 + 9 × 155 + 3 × 165) ÷ 20 = (1160 + 1395 + 495) ÷ 20 = 3050 ÷ 20 = 152.5 厘米。这代表了样本中 11–12 岁学生的典型身高。
Repeating the process for Year 11 yields an estimated mean of (1 × 155 + 7 × 165 + 9 × 175 + 3 × 185) ÷ 20 = (155 + 1155 + 1575 + 555) ÷ 20 = 3440 ÷ 20 = 172.0 cm. The difference between Year 7 and Year 11 means is 19.5 cm, reflecting the significant growth occurring during adolescence. The modal class for Year 7 is 150–159 cm, while for Year 11 it shifts to 170–179 cm, providing additional evidence of the upward shift in central tendency.
对 11 年级重复此过程得到估算均值：(1 × 155 + 7 × 165 + 9 × 175 + 3 × 185) ÷ 20 = (155 + 1155 + 1575 + 555) ÷ 20 = 3440 ÷ 20 = 172.0 厘米。7 年级与 11 年级均值之差为 19.5 厘米，反映了青春期发生的显著生长。7 年级的众数区间为 150–159 厘米，而 11 年级则移至 170–179 厘米，进一步证明了集中趋势上移。
4. Probability in a Two-Stage Game | 两阶段游戏中的概率
In a classroom activity, students design a probability game involving two fair spinners. The first spinner is divided into four equal sectors labelled A, B, C, and D. The second spinner has three equal sectors numbered 1, 2, and 3. A player spins both spinners simultaneously and records the ordered pair of outcomes. Certain combinations trigger bonus points, and students are asked to calculate various probabilities associated with the game.
在课堂活动中，学生设计了一个涉及两个均匀转盘的概率游戏。第一个转盘分为四个等份扇形，标有 A、B、C 和 D。第二个转盘分为三个等份扇形，标有数字 1、2 和 3。玩家同时旋转两个转盘并记录有序结果对。特定组合会触发奖励积分，学生需要计算与游戏相关的各种概率。
The sample space for this two-stage experiment contains 4 × 3 = 12 equally likely outcomes, since the spinners are independent and fair. These outcomes can be listed systematically as (A,1), (A,2), (A,3), (B,1), (B,2), (B,3), (C,1), (C,2), (C,3), (D,1), (D,2), (D,3). The sample space diagram provides a clear visual representation, aiding in the calculation of event probabilities.
该两阶段实验的样本空间包含 4 × 3 = 12 个等可能结果，因为两个转盘是独立且均匀的。这些结果可系统地列为 (A,1), (A,2), (A,3), (B,1), (B,2), (B,3), (C,1), (C,2), (C,3), (D,1), (D,2), (D,3)。样本空间图提供了清晰的视觉呈现，有助于事件概率的计算。
Suppose bonus points are awarded when the letter is a vowel (A only) and the number is even (2 only). The favourable outcome is solely (A,2), so the probability of earning bonus points is 1/12. If instead the condition is that the letter is a consonant (B, C, or D) or the number is odd (1 or 3), we use the addition rule carefully to avoid double-counting. The consonants occupy 3 × 3 = 9 outcomes, the odd numbers occupy 4 × 2 = 8 outcomes, but the intersection (consonant and odd) accounts for 3 × 2 = 6 outcomes. The union probability is therefore (9 + 8 − 6) / 12 = 11/12. This single case illustrates the importance of the inclusion-exclusion principle in compound probability calculations.
假设当字母为元音（仅 A）且数字为偶数（仅 2）时获得奖励积分。有利结果仅为 (A,2)，因此获得奖励积分的概率为 1/12。如果条件改为字母为辅音（B、C 或 D）或数字为奇数（1 或 3），我们需要仔细使用加法法则以避免重复计数。辅音占 3 × 3 = 9 个结果，奇数占 4 × 2 = 8 个结果，但交集（辅音且奇数）占 3 × 2 = 6 个结果。因此并集概率为 (9 + 8 − 6) / 12 = 11/12。这个案例说明了容斥原理在复合概率计算中的重要性。
5. Exploring the Fibonacci Sequence in Nature | 探索自然界中的斐波那契数列
The Fibonacci sequence, defined by the recurrence relation Fₙ = Fₙ₋₁ + Fₙ₋₂ with seed values F₁ = 1 and F₂ = 1, appears in numerous natural phenomena, from the arrangement of leaves on a stem to the spiral patterns of pinecones. This case study invites students to generate terms, examine ratios of consecutive terms, and observe the convergence towards the golden ratio φ, approximately equal to 1.618.
斐波那契数列由递推关系 Fₙ = Fₙ₋₁ + Fₙ₋₂ 定义，初始值为 F₁ = 1 和 F₂ = 1，它出现在众多自然现象中，从茎上叶片的排列到松果的螺旋图案。本案例研究要求学生生成数列项、检查连续项的比率，并观察其向黄金比例 φ（约等于 1.618）的收敛。
Generating the first twelve terms of the Fibonacci sequence is straightforward using the recurrence. Starting with F₁ = 1, F₂ = 1, we obtain F₃ = 1 + 1 = 2, F₄ = 1 + 2 = 3, F₅ = 2 + 3 = 5, F₆ = 3 + 5 = 8, F₇ = 5 + 8 = 13, F₈ = 8 + 13 = 21, F₉ = 13 + 21 = 34, F₁₀ = 21 + 34 = 55, F₁₁ = 34 + 55 = 89, and F₁₂ = 55 + 89 = 144.
使用递推关系生成斐波那契数列的前十二项非常简单。从 F₁ = 1，F₂ = 1 开始，得到 F₃ = 1 + 1 = 2，F₄ = 1 + 2 = 3，F₅ = 2 + 3 = 5，F₆ = 3 + 5 = 8，F₇ = 5 + 8 = 13，F₈ = 8 + 13 = 21，F₉ = 13 + 21 = 34，F₁₀ = 21 + 34 = 55，F₁₁ = 34 + 55 = 89，F₁₂ = 55 + 89 = 144。
Now consider the ratios of consecutive terms: F₂/F₁ = 1/1 = 1.000, F₃/F₂ = 2/1 = 2.000, F₄/F₃ = 3/2 = 1.500, F₅/F₄ = 5/3 ≈ 1.667, F₆/F₅ = 8/5 = 1.600, F₇/F₆ = 13/8 = 1.625, F₈/F₇ = 21/13 ≈ 1.615, F₉/F₈ = 34/21 ≈ 1.619, F₁₀/F₉ = 55/34 ≈ 1.618, F₁₁/F₁₀ = 89/55 ≈ 1.618, F₁₂/F₁₁ = 144/89 ≈ 1.618. The sequence of ratios oscillates around the golden ratio, converging rapidly after the eighth term. This limit can be derived algebraically by assuming Fₙ/Fₙ₋₁ → L as n → ∞, leading to the characteristic equation L = 1 + 1/L, which rearranges to L² − L − 1 = 0, yielding L = (1 + √5)/2.
现在考虑连续项的比率：F₂/F₁ = 1/1 = 1.000，F₃/F₂ = 2/1 = 2.000，F₄/F₃ = 3/2 = 1.500，F₅/F₄ = 5/3 ≈ 1.667，F₆/F₅ = 8/5 = 1.600，F₇/F₆ = 13/8 = 1.625，F₈/F₇ = 21/13 ≈ 1.615，F₉/F₈ = 34/21 ≈ 1.619，F₁₀/F₉ = 55/34 ≈ 1.618，F₁₁/F₁₀ = 89/55 ≈ 1.618，F₁₂/F₁₁ = 144/89 ≈ 1.618。比率数列围绕黄金比例振荡，在第八项之后迅速收敛。这个极限可以通过假设 n → ∞ 时 Fₙ/Fₙ₋₁ → L 来代数推导，得到特征方程 L = 1 + 1/L，重新排列为 L² − L − 1 = 0，解得 L = (1 + √5)/2。
6. Modelling Mobile Phone Battery Drain | 手机电池耗电建模
A student notices that their mobile phone battery decreases at a roughly constant rate when streaming video. Starting from a full charge of 100% at 08:00, the battery level drops to 72% by 09:30. The student wishes to construct a linear model to predict when the battery will reach a critically low level of 10%, allowing them to plan charging breaks effectively throughout the day.
一名学生注意到，在流媒体播放视频时，手机电池以大致恒定的速率下降。从 08:00 充满电 100% 开始，到 09:30 电池电量降至 72%。该学生希望构建一个线性模型来预测电池何时会达到 10% 的临界低电量水平，从而能够有效地规划全天的充电休息时间。
Let t represent the time in hours elapsed since 08:00, and let B(t) denote the battery percentage as a function of time. The linear model takes the form B(t) = mt + c, where m is the constant rate of change (drain rate) and c is the initial battery level. From the given information, we know that at t = 0, B(0) = 100, so c = 100. At t = 1.5 hours (09:30), B(1.5) = 72. Substituting these values into the linear equation gives 72 = m(1.5) + 100.
设 t 表示自 08:00 起经过的小时数，B(t) 表示作为时间函数的电池百分比。线性模型形式为 B(t) = mt + c，其中 m 是恒定变化率（耗电速率），c 是初始电池电量。根据已知信息，t = 0 时 B(0) = 100，因此 c = 100。在 t = 1.5 小时（09:30）时，B(1.5) = 72。将这些值代入线性方程得到 72 = m(1.5) + 100。
Solving for m, we rearrange: 1.5m = 72 − 100 = −28, so m = −28 / 1.5 = −18.67 (rounded to two decimal places). The negative sign confirms that the battery is draining. The complete linear model is B(t) = −18.67t + 100. This model allows the student to predict the battery level at any time t within a reasonable domain, assuming the constant drain rate persists.
求解 m，重新排列：1.5m = 72 − 100 = −28，因此 m = −28 / 1.5 = −18.67（四舍五入至两位小数）。负号确认电池正在耗电。完整的线性模型为 B(t) = −18.67t + 100。该模型允许学生在恒定耗电速率持续的合理范围内预测任何时间 t 的电池电量。
To find when the battery reaches 10%, we set B(t) = 10 and solve: −18.67t + 100 = 10 ⇒ −18.67t = −90 ⇒ t ≈ 4.82 hours. Adding 4.82 hours to 08:00 gives a time of approximately 12:49. Therefore, if streaming continues uninterrupted, the phone battery will drop to the critical 10% threshold just before 1 PM. This analysis highlights the power of linear modelling in everyday technology management.
为求电池何时降至 10%，设 B(t) = 10 并求解：−18.67t + 100 = 10 ⇒ −18.67t = −90 ⇒ t ≈ 4.82 小时。将 4.82 小时加到 08:00 得到时间约 12:49。因此，如果不间断地继续流媒体播放，手机电池将在下午 1 点前降至 10% 的临界阈值。这一分析突显了线性建模在日常技术管理中的强大功能。
7. Optimising the Volume of an Open Box | 无盖盒子的体积优化
A classic optimisation problem in CIE Advanced Mathematics involves constructing an open-topped box by cutting equal squares from the corners of a flat rectangular sheet and folding up the sides. Consider a sheet of card measuring 30 cm by 20 cm. Squares of side length x cm are removed from each corner, and the resulting flaps are folded to form a tray. The objective is to determine the value of x that maximises the volume of the box.
CIE 进阶数学中一个经典的优化问题涉及通过从平展矩形纸板的四角切去相等正方形并向上折叠各边来构造无盖盒子。考虑一张 30 厘米 × 20 厘米的卡纸。从每个角切去边长为 x 厘米的正方形，然后将产生的边翼折叠以形成一个托盘。目标是确定使盒子体积最大化的 x 值。
After removing the corner squares and folding, the dimensions of the resulting box are: length = 30 − 2x, width = 20 − 2x, and height = x. The volume V(x), measured in cubic centimetres, is the product of these three dimensions: V(x) = x(30 − 2x)(20 − 2x). Expanding this expression yields V(x) = x(600 − 60x − 40x + 4x²) = x(600 − 100x + 4x²) = 4x³ − 100x² + 600x.
切去四角正方形并折叠后，所得盒子的尺寸为：长 = 30 − 2x，宽 = 20 − 2x，高 = x。体积 V(x) 以立方厘米为单位，是这三个维度的乘积：V(x) = x(30 − 2x)(20 − 2x)。展开该表达式得 V(x) = x(600 − 60x − 40x + 4x²) = x(600 − 100x + 4x²) = 4x³ − 100x² + 600x。
The domain for x is constrained by the physical dimensions: x must be positive and less than half the smaller side length, so 0 < x < 10. To find the maximum volume within this interval, we examine the turning points of the cubic function. Setting the derivative dV/dx equal to zero gives 12x² − 200x + 600 = 0. Dividing by 4 simplifies to 3x² − 50x + 150 = 0. Using the quadratic formula produces x = (50 ± √(2500 − 1800)) / 6 = (50 ± √700) / 6 = (50 ± 10√7) / 6. The two critical values are approximately x ≈ 12.74 (outside the domain) and x ≈ 3.93 (within the domain).
x 的取值范围受物理尺寸限制：x 必须为正且小于较短边长的一半，因此 0 < x < 10。为求该区间内的最大体积，我们考察三次函数的驻点。令导数 dV/dx 等于零得到 12x² − 200x + 600 = 0。除以 4 简化为 3x² − 50x + 150 = 0。使用二次求根公式得 x = (50 ± √(2500 − 1800)) / 6 = (50 ± √700) / 6 = (50 ± 10√7) / 6。两个临界值约为 x ≈ 12.74（在取值范围内）和 x ≈ 3.93（在取值范围内）。
Evaluating the volume at x ≈ 3.93 gives V ≈ 3.93 × (30 − 7.86) × (20 − 7.86) = 3.93 × 22.14 × 12.14 ≈ 1056 cm³. Checking the endpoints x → 0 and x → 10 confirms that V approaches zero, establishing x ≈ 3.93 as the global maximum within the feasible domain. This application of calculus to geometric design demonstrates optimisation in a tangible context.
将 x ≈ 3.93 代入体积计算得 V ≈ 3.93 × (30 − 7.86) × (20 − 7.86) = 3.93 × 22.14 × 12.14 ≈ 1056 cm³。检查端点 x → 0 和 x → 10 确认 V 趋近于零，从而确定 x ≈ 3.93 为可行范围内的全局最大值。这种将微积分应用于几何设计的方法展示了具体情境下的优化过程。
8. Compound Interest and Exponential Growth | 复利与指数增长
A young entrepreneur invests £800 in a savings account that offers an annual compound interest rate of 3.5%, compounded monthly. They plan to leave the money untouched for five years. This case study explores the difference between simple and compound interest, the effect of compounding frequency, and the exponential nature of the growth.
一位年轻企业家将 800 英镑投资于一个年复利利率为 3.5%、按月复利的储蓄账户。他们计划将资金不动用存放五年。本案例研究探讨简单利息与复利之间的差异、复利频率的影响以及增长指数性质。
The compound interest formula is A = P(1 + r/n)^(nt), where A is the final amount, P is the principal, r is the annual interest rate expressed as a decimal, n is the number of compounding periods per year, and t is the time in years. Substituting the given values, P = 800, r = 0.035, n = 12, and t = 5, yields A = 800(1 + 0.035/12)^(12 × 5) = 800(1 + 0.0029167)^60 = 800(1.0029167)^60.
复利公式为 A = P(1 + r/n)^(nt)，其中 A 为最终金额，P 为本金，r 为以小数表示的年利率，n 为每年复利次数，t 为年数。代入给定值，P = 800，r = 0.035，n = 12，t = 5，得 A = 800(1 + 0.035/12)^(12 × 5) = 800(1 + 0.0029167)^60 = 800(1.0029167)^60。
Computing the power using a calculator gives (1.0029167)^60 ≈ 1.1909. Multiplying by the principal, A ≈ 800 × 1.1909 = 952.72. The total interest earned over five years is approximately £152.72. In contrast, simple interest calculated as I = Prt would yield only 800 × 0.035 × 5 = £140. The extra £12.72 arises from the compounding effect, where interest earned in earlier periods itself earns interest in subsequent periods.
使用计算器计算幂值得 (1.0029167)^60 ≈ 1.1909。乘以本金得 A ≈ 800 × 1.1909 = 952.72。五年内获得的总利息约为 152.72 英镑。相比之下，按 I = Prt 计算的简单利息仅产生 800 ×
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