Year 10 CIE Maths: Case Study Practical Drills | Year 10 CIE 数学：案例分析实战演练

📚 Year 10 CIE Maths: Case Study Practical Drills | Year 10 CIE 数学：案例分析实战演练

Welcome to this practical drill session designed to sharpen your problem-solving skills for the CIE IGCSE Mathematics syllabus (Year 10). Instead of working through typical textbook exercises, we will dive into real-style case studies that blend multiple topics, just as you will see in your final examinations. Each case study is followed by a step-by-step bilingual breakdown so you can follow the logic in both English and Chinese, ensuring deep understanding.

欢迎参加本次实战演练，旨在提升你在 CIE IGCSE 数学（Year 10）课程中的解题能力。我们不再拘泥于传统课本练习，而是深入贴近真题风格的综合案例分析，就像你在最终考试中会遇到的那样。每个案例都配有逐步的双语解析，帮助你用中英文两种思维理清逻辑，确保真正掌握。

1. Case Study 1: Optimising a Rectangular Garden | 案例一：矩形花园的优化

A landscaper has 40 metres of fencing to enclose a rectangular vegetable garden against an existing wall. Only three sides need fencing. What dimensions give the maximum area, and what is that area? This is a classic optimisation problem combining algebra and quadratic functions.

一位园艺师有 40 米长的围栏，想靠着一面已有的墙壁围出一个矩形菜园，只需围三面。求能获得最大面积的尺寸，以及该最大面积是多少？这是一道结合代数与二次函数的经典优化题。

e wall is needed, that side will be (40 − 2x) metres. The area A = x(40 − 2x) = 40x − 2x².

设垂直于墙的边长为 x 米。因为平行于墙的边只需一条，其长度为 (40 − 2x) 米。面积 A = x(40 − 2x) = 40x − 2x²。

) = 10 metres. Then the length is 40 − 2(10) = 20 metres. Maximum area = 10 × 20 = 200 m².

为求最大值，可以配方或使用抛物线顶点 x = −b/(2a)。这里 a = −2，b = 40，得 x = −40/(2 × −2) = 10 米。那么另一边长为 40 − 2×10 = 20 米。最大面积 = 10 × 20 = 200 平方米。

valid. Quadratic optimisation appears frequently in CIE papers, often disguised as a fence, a picture frame, or a profit scenario.

务必检查尺寸的合理性：x > 0 且 40 − 2x > 0 → x < 20，因此 x = 10 合适。二次函数优化题在 CIE 试卷中很常见，往往伪装成围栏、相框或利润场景。

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2. Case Study 2: Exchange Rates and Percentage Profit | 案例二：汇率与百分比利润

Amira travels from London to Tokyo. She exchanges £800 into Japanese yen at a rate of £1 = 162.5 yen. The bureau charges a 2% rling amount. She spends 95,000 yen and converts the remainder back to pounds at a rate of £ 8.0 yen with no commission. Calculate her overall percentage loss or profit on the initial £800.

Amira 从伦敦去东京旅游。她按 £1 = 162.5 日元的汇率将 800 英镑兑换成日元，兑换所按英镑金额收取 2% 的手续费。她花掉 95,000 日元后，将余款按 £1 = 168.0 日元的汇率换回英镑（无手续费）。计算她相对于初始 800 英镑的总体盈亏百分比。

Step 1: Commission reduces the effective sterling amount to 98% of £800 = £784. Yen received = 784 × 162.5 = 127,400 yen.

步骤一：手续费使实际兑换的英镑金额变为 800 的 98%，即 £784。获得日元 = 784 × 162.5 = 127,400 日元。

er spending, remaining yen = 127,400 − 95,000 = 32,400 yen. Convert back: 32,400 ÷ 168.0 = £192.857... ≈ £192.86.

步骤二：消费后剩余日元 = 127,400 − 95,000 = 32,400 日元。换回英镑：32,400 ÷ 168.0 = £192.857... ≈ £192.86。

Step 3: Overall change = £192.86 − £800 = −£607.14. Percentage loss = (607.14/800) × 100% = 75.89% She has effectively lost over three-quarters of her money due to spending and commission.

步骤三：总体变化 = £192.86 − £800 = −£607.14。亏损百分比 = (607.14/800) × 100% = 75.89%。由于消费和手续费，她实际上损失了超过四分之三的钱。

This case reminds us to read carefully: commission on the sterling amount, not the yen. CIE often tests two-stage currency conversions with fees, requiring a clear step-by-step approach.

这个案例提醒我们要仔细审题：手续费是基于英镑金额，而非日元。CIE 常考带手续费的两阶段货币兑换，需要有清晰的逐步解题思路。

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3. Case Study 3: Interpreting Distance-Time Graphs | 案例三：解读距离-时间图

-minute stop. From t=40 to t=70 minutes, the distance decreases linearly back to 0. Find the average speed for the whole journey, and the speed on the return leg.

下图展示了一位骑行者从家到公园再返回的行程。t=0 到 t=30 分钟，距离从 0 线性增加到 12 km。然后停留 10 分钟。从 t=40 到 t=70 分钟，距离线性减少至 0。求全程的平均速度，以及返程段的速度。

time = 70 minutes = 70/60 = 7/6 hours. Average speed = 24 ÷ (7/6) = 24 × 6/7 ≈ 20.57 km/h.

平均速度 = 总距离 ÷ 总时间。总距离 = 12 km 去程 + 12 km 回程 = 24 km。总时间 = 70 分钟 = 70/60 = 7/6 小时。平均速度 = 24 ÷ (7/6) = 24 × 6/7 ≈ 20.57 km/h。

30 minutes = 0.5 h), distance changes from 12 km to 0, so speed = 12 km / 0.5 h = 24 km/h.

返程速度：从 t=40 到 70 分钟（30 分钟 = 0.5 小时），距离从 12 km 降至 0，因此速度 = 12 km ÷ 0.5 h = 24 km/h。

Note: the outward speed was 12 km / 0.5 h = 24 km/h as well? , 30 minutes = 0.5 h, so outward speed = 12/0.5 = 24 km/h. So both legs had the same speed! s. Average speed for the whole trip is not simply the arithmetic mean because of the stop.

注意：去程速度也是 12 km / 0.5 h = 24 km/h。等等，30 分钟 = 0.5 小时，所以去程速度 = 12/0.5 = 24 km/h。因此两段速度相同！图形斜率大小相等，表明来回速度恒定。全程平均速度并非简单的算术平均，因为中途有停留。

CIE questions often mash up distance-time and speed-time interpretations, so always check the axes labels and account for stationary periods.

CIE 考题常常把距离-时间图和速度-时间图的解读混在一起，因此一定要检查坐标轴标签，并考虑静止时段。

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4. Case Study 4: Probability Tree without Replacement | 案例四：无放回概率树

A bag contains 5 red and 3 blue marbles. Two marbles are drawn at random without replacement. Calculate the probability that both are the same colour, and the probability of at least one red.

一个袋子里有 5 个红球和 3 个蓝球。随机抽出两个球且不放回。计算两个球颜色相同的概率，以及至少一个红球的概率。

t draw: P(R) = 5/8, P(B) = 3/8. Second draw probabilities change: if first was R, then P(R|R) = 4/7, P(B|R) = 3/7; if first was B, then P(R|B) = 5/7, P(B|B) = 2/7.

画出树状图。第一次抽取：P(R) = 5/8，P(B) = 3/8。第二次概率相应变化：若第一次是 R，则 P(R|R) = 4/7，P(B|R) = 3/7；若第一次是 B，则 P(R|B) = 5/7，P(B|B) = 2/7。

Same colour: P(RR) + P(BB) = (5/8 × 4/7) + (3/8 × 2/7) = 20/56 + 6/56 = 26/56 = 13/28.

颜色相同：P(RR) + P(BB) = (5/8 × 4/7) + (3/8 × 2/7) = 20/56 + 6/56 = 26/56 = 13/28。

P(RB) + P(BR) = 20/56 + (5/8 × 3/7) + (3/8 × 5/7) = 20/56 + 15/56 + 15/56 = 50/56 = 25/28.

至少一个红球：1 − P(无红球) = 1 − P(BB) = 1 − 6/56 = 50/56 = 25/28。或者相加 P(RB) + P(BR) = 20/56 + (5/8 × 3/7) + (3/8 × 5/7) = 20/56 + 15/56 + 15/56 = 50/56 = 25/28。

Without replacement problems require careful updating of probabilities. CIE often embeds these within larger data handling tasks or combines with conditional probability language.

无放回问题需要谨慎地更新概率。CIE 常将其嵌入更复杂的数据处理题中，或与条件概率的表述结合。

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5. Case Study 5: Simultaneous Equations in Context | 案例五：实际情境中的联立方程

A cinema sells adult tickets for $12 and child tickets for $7. On Saturday, 320 tickets were sold, giving a total revenue of $3,060. How many of each type were sold? Model the situation with two linear equations and solve.

一家电影院成人票售价 $12，儿童票售价 $7。周六共售出 320 张票，总收入为 $3,060。问两种票各售出多少张？用两个线性方程建模并求解。

from (2): (12a − 7a) + (7c − 7c) = 3060 − 2240 → 5a = 820 → a = 164. Then c = 320 − 164 = 156.

设 a 为成人票数量，c 为儿童票数量。方程 (1)：a + c = 320。方程 (2)：12a + 7c = 3060。将 (1) 式乘以 7：7a + 7c = 2240。用 (2) 式相减：(12a − 7a) + (7c − 7c) = 3060 − 2240 → 5a = 820 → a = 164。则 c = 320 − 164 = 156。

ethod is efficient, but you could also use substitution. CIE expects you to form equations from word descriptions and solve accurately.

验证：164 × 12 = 1968，156 × 7 = 1092，合计 3060。正确。这种消元法很高效，你也可以用代入法。CIE 希望你从文字描述中建立方程并准确求解。

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6. Case Study 6: Trigonometry in 3D | 案例六：三维三角学

A vertical flagpole stands on horizontal ground. From a point A due south of the pole, the angle of elevation to the top is 35°. From a point B due east of the pole, the angle of elevation is 22°. The distance between A and B is 50 m. Find the height of the flagpole.

一根垂直旗杆立在水平地面上。从旗杆正南方的 A 点测得顶部仰角为 35°。从旗杆正东方的 B 点测得仰角为 22°。A 和 B 之间的距离为 50 米。求旗杆高度。

tan 35° = h / OA → OA = h / tan 35°. In triangle BOT, tan 22° = h / OB → OB = h / tan 22°.

设旗杆底部为 O，高度为 h。在直角三角形 AOT（T 为顶端）中，tan 35° = h / OA → OA = h / tan 35°。在三角形 BOT 中，tan 22° = h / OB → OB = h / tan 22°。

e A is south and B is east, the angle AOB is a right angle (90°). Therefore, by Pythagoras' theorem, OA² + OB² = AB². Substitute: (h / tan 35°)² + (h / tan 22°)² = 50².

因为 A 在正南、B 在正东，角 AOB 是直角 (90°)。因此，根据勾股定理，OA² + OB² = AB²。代入：(h / tan 35°)² + (h / tan 22°)² = 50²。

+ (2.475h)² = 2500 → (2.039 + 6.126)h² = 2500 → 8.165h² = 2500 → h² ≈ 306.2 → h ≈ 17.5 m.

计算：tan 35° ≈ 0.7002，tan 22° ≈ 0.4040。因此 OA ≈ h/0.7002 ≈ 1.428h，OB ≈ h/0.4040 ≈ 2.475h。方程：² + (2.475h)² = 2500 → (2.039 + 6.126)h² = 2500 → 8.165h² = 2500 → h² ≈ 306.2 → h ≈ 17.5 m。

This 3D problem tests your ability to visualise right-angled triangles in different vertical planes and link them with a horizontal right triangle. Always sketch and label clearly.

这道三维题考查你在不同的竖直平面内想象直角三角形的能力，并用一个水平直角三角形将它们联系起来。一定要清晰绘制草图并标注。

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7. Case Study 7: Cumulative Frequency and IQR | 案例七：累积频数与四分位距

The table gives the marks of 80 students in a test: 0-10: 5, 10-20: 8, 20-30: 14, 30-40: 20, 40-50: 18, 50-60: 10, 60-70: 5. Construct a cumulative frequency table, draw the curve, and estimate the median and interquartile range.

表格给出了 80 名学生的考试分数：0-10：5，10-20：8，20-30：14，30-40：20，40-50：18，50-60：10，60-70：5。建立累积频数表，绘制曲线，并估算中位数和四分位距。

rpolation, median ≈ lower boundary + (40.5 − 27)/(20) × 10 ≈ 30 + (13.5/20)×10 = 36.75 marks.

累积频数：≤10：5，≤20：13，≤30：27，≤40：47，≤50：65，≤60：75，≤70：80。中位数位置：(80+1)/2 = 第 40.5 个值。从图或插值可得，中位数 ≈ 下限 + (40.5 − 27)/20 × 10 ≈ 30 + (13.5/20)×10 = 36.75 分。

Lower quartile Q1 at 20.25th value: in 10-20 interval? 0 + (20.25 − 13)/14 × 10 ≈ 20 + 5.18 = 25.18. Upper quartile Q3 at 60.75th value: in 40-50? 0 + (60.75 − 47)/18 × 10 ≈ 40 + 7.64 = 47.64. IQR = Q3 − Q1 ≈ 47.64 − 25.18 = 22.46 marks.

下四分位数 Q1 位于第 20.25 个值：在 10-20区间？20.25 > 13，所以在 20-30 区间：Q1 ≈ 20 + (20.25 − 13)/14 × 10 ≈ 20 + 5.18 = 25.18。上四分位数 Q3 位于第 60.75 个值：在 40-50？60.75 > 47，所以 Q3 ≈ 40 + (60.75 − 47)/18 × 10 ≈ 40 + 7.64 = 47.64。四分位距 IQR = Q3 − Q1 ≈ 47.64 − 25.18 = 22.46 分。

Cumulative frequency curves are a staple in CIE Statistics. You must be able to plot, smooth, and read off quartiles and percentiles with accurate interpolation.

累积频数曲线是 CIE 统计学的核心内容。你必须能够绘图、描出平滑曲线，并通过精确的插值法读取四分位数和百分位数。

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8. Case Study 8: Transformations of Graphs | 案例八：图形变换

f y = f(x − 1), y = 2f(x), and y = f(½x). Clearly mark the images of the three key points.

已知 y = f(x) 的图像经过点 (−2, 0)、(0, 3)、(2, 0)。请画出 y = f(x − 1)、y = 2f(x) 以及 y = f(½x) 的图像，并清晰标记这三个关键点的对应点。

it to the right. Points become: (−2+1, 0) = (−1, 0), (0+1, 3) = (1, 3), (2+1, 0) = (3, 0).

对于 y = f(x − 1)：向右平移 1 个单位。点变为：(−2+1, 0) = (−1, 0)，(0+1, 3) = (1, 3)，(2+1, 0) = (3, 0)。

al stretch by factor 2. Points: (−2, 0×2) = (−2, 0), (0, 3×2) = (0, 6), (2, 0×2) = (2, 0).

对于 y = 2f(x)：垂直拉伸为原来的 2 倍。点：(−2, 0×2) = (−2, 0)，(0, 3×2) = (0, 6)，(2, 0×2) = (2, 0)。

ply x-coordinates by 2. Points: (−2×2, 0) = (−4, 0), (0×2, 3) = (0, 3), (2×2, 0) = (4, 0).

对于 y = f(½x)：水平拉伸为原来的 2 倍（因为括号内是 ½x）。将 x 坐标乘以 2。点：(−2×2, 0) = (−4, 0)，(0×2, 3) = (0, 3)，(2×2, 0) = (4, 0)。

Remember the order of transformations matters when combined. CIE frequently asks for sketches of transformed functions from a given graph, testing if you understand inside/outside changes.

记住，当多种变换组合时，变换的顺序很重要。CIE 经常要求根据给定图形画出变换后的函数草图，考查你是否理解括号内外的变化。

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9. Case Study 9: Bounds and Accuracy | 案例九：界限与精确度

m (to 1 decimal place). Calculate the upper and lower bounds for the perimeter and area. Then find the maximum possible percentage error for the area if calculated with the given measurements.

一个矩形的长 L = 8.4 cm（精确到 1 位小数），宽 W = 5.3 cm（精确到 1 位小数）。计算周长和面积的上界与下界。然后求出若用所给测量值计算面积时的最大可能百分比误差。

Lower bound of L: 8.35 cm, upper bound: 8.45 cm. For W: 8.35? No, W lower bound = 5.25 cm, upper bound = 5.35 cm.

L 的下界：8.35 cm，上界：8.45 cm。W 的下界：5.25 cm，上界：5.35 cm。

(8.35 + 5.25) = 2 × 13.6 = 27.2 cm. Upper bound = 2 × (8.45 + 5.35) = 2 × 13.8 = 27.6 cm.

周长下界 = 2 × (8.35 + 5.25) = 2 × 13.6 = 27.2 cm。上界 = 2 × (8.45 + 5.35) = 2 × 13.8 = 27.6 cm。

m absolute error = max of (44.52 − 43.8375) = 0.6825 or (45.2075 − 44.52) = 0.6875 → 0.6875. Percentage error = (0.6875 / 44.52) × 100% ≈ 1.54%.

面积下界 = 8.35 × 5.25 = 43.8375 cm²。上界 = 8.45 × 5.35 = 45.2075 cm²。用所给值计算的面积 = 8.4 × 5.3 = 44.52 cm²。最大绝对误差 = (44.52 − 43.8375) = 0.6825 或 (45.2075 − 44.52) = 0.6875 的最大值 → 0.6875。百分比误差 = (0.6875 / 44.52) × 100% ≈ 1.54%。

Bounds problems demand careful attention to the degree of accuracy. Always halve the unit of the last significant figure to find limits. CIE often mixes bounds with formula rearrangements.

界限问题需要仔细关注精确度。通常将最后一位有效数字的单位减半来找到极限。CIE 经常将界限与公式变形混合考查。

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10. Case Study 10: Algebraic Fractions and Solving | 案例十：代数分式与求解

s requires forming a common denominator and solving a quadratic, then checking for extraneous solutions.

求解 x：3/(x + 2) − 1/(x − 1) = 1。需要通分并求解二次方程，然后排除增根。

² + x − 2). Simplify LHS: 2x − 5 = x² + x − 2. Rearrange: 0 = x² − x + 3 → x² − x + 3 = 0.

两边同乘 (x+2)(x−1)，注意 x ≠ −2, 1。得到 3(x−1) − 1(x+2) = 1(x+2)(x−1) → 3x − 3 − x − 2 = (x² + x − 2)。简化左边：2x − 5 = x² + x − 2。移项：0 = x² − x + 3 → x² − x + 3 = 0。

11 < 0, so no real solutions. Thus the equation has no real roots. However, if the discriminant had been positive, we would still need to check the excluded values.

判别式 = (−1)² − 4 × 1 × 3 = 1 − 12 = −11 < 0，因此无实数解。所以该方程没有实数根。不过，如果判别式为正，我们仍需检查排除的值。

Many students forget to state the domain restrictions or to verify solutions. CIE rewards explicit statements like x ≠ −2, 1.

许多学生忘记声明定义域限制或验证解。CIE 乐于看到诸如 x ≠ −2, 1 的明确陈述。

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