📚 Year 9 CIE Maths: Case Analysis Practice | Year 9 CIE 数学：案例分析实战演练
In Year 9 CIE Maths, applying concepts to real-world scenarios is essential. This article presents a series of case analysis drills covering algebra, geometry, statistics, and more. Each case encourages you to identify the mathematical tools needed, carry out calculations, and interpret results critically.
在 Year 9 CIE 数学中，将概念应用于实际情境至关重要。本文提供一系列案例分析演练，涵盖代数、几何、统计等内容。每个案例都鼓励你识别所需的数学工具、进行计算并批判性地解读结果。
1. Modelling with Linear Equations | 线性方程建模案例
A mobile phone company offers two plans. Plan A: a fixed fee of $20 per month and $0.05 per minute of calls. Plan B: no fixed fee but $0.08 per minute. We need to find the number of minutes for which the total cost is the same.
一家移动电话公司提供两种套餐。A套餐：每月固定费用20美元，通话每分钟0.05美元。B套餐：无固定费用，但每分钟0.08美元。我们需要找出总费用相同时的通话分钟数。
Let m be the number of minutes. The cost for Plan A is 20 + 0.05m, and for Plan B is 0.08m. Set them equal:
设 m 为分钟数。A套餐的成本为 20 + 0.05m，B套餐为 0.08m。令它们相等：
20 + 0.05m = 0.08m
ide; 0.03 ≈ 666.67 minutes. Therefore, at about 667 minutes the costs are equal. Plan A is cheaper for more minutes, Plan B for fewer.
解方程得 20 = 0.03m，因此 m = 20 ÷ 0.03 ≈ 666.67 分钟。所以约667分钟时费用相同。通话超过此时间选A更便宜，少于则选B更划算。
2. Pythagoras in Real Life | 毕达哥拉斯定理的实际应用
A ladder of length 5 metres leans against a vertical wall. The foot of the ladder is 1.5 metres away from the wall. How high up the wall does the ladder reach?
一架长5米的梯子斜靠在垂直的墙壁上。梯脚离墙1.5米。梯子顶端能达到墙上的多高位置？
Using Pythagoras' √22.75 ≈ 4.77 m (to 3 s.f.). Always sketch a diagram and check your rounding.
运用毕达哥拉斯定理：高² + 1.5² = 5²;。因此高² = 25 − 2.25 = 22.75。高 = √22.75 ≈ 4.77米（保留三位有效数字）。务必画草图并检查舍入。
3. Statistics: Interpreting Test Scores | 统计：考试分数的解读
A class of 20 students scored the following marks in a test: 45, 50, 55, 60, 62, 65, 68, 70, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 95, 98. Find the mean, median, and range, and comment on the distribution.
一个班20名学生的测验成绩如下：45, 50, 55, 60, 62, 65, 68, 70, 70, 72, 75, 78, 80, 82, 85, 88, 90, 92, 95, 98。求平均数、中位数和全距，并评价分布。
− 45 = 53. The distribution is slightly positively skewed because mean > median.
平均数 = 总和 ÷ 20。总和计算为 1520，因此平均数为 76。中位数（第10和第11个值）= (72+75)/2 = 73.5。全距 = 98 − 45 = 53。分布略显正偏态，因为平均数大于中位数。
In a report, we might say the typical score is around 73.5, with a wide spread of 53 marks. Outliers could be investigated.
在报告中，可以说典型分数约为73.5，分数散布较广（53分）。可调查是否存在异常值。
4. Probability in Games | 游戏中的概率
Two fair six-sided dice are rolled. What is the probability that the sum of the numbers is 7 or more? List the outcomes to understand the sample space.
同时抛掷两枚公平的六面骰子。求点数之和为7或以上的概率。列出所有结果以理解样本空间。
, (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6). That's 21 outcomes. So P(sum ≥ 7) = 21/36 = 7/12.
总结果数为36。和为7或以上的组合有：(1,6), (2,5), (2,6) ... 共21种。因此 P(和≥7) = 21/36 = 7/12。
Alternatively, use symmetry: the probability of sum 7 exactly is 6/36, and we can count above. Critical thinking avoids listing all.
或者利用对称性：和为7的概率正好是6/36，再计算大于7的情况。批判性思维可以避免全部列出。
5. Ratio and Map Scales | 比和地图比例尺
On a map with a scale of 1 : 50,000, two towns are 8.4 cm apart. Find the actual distance in kilometres.
在一幅比例尺为 1 : 50 000 的地图上，两城镇相距 8.4 厘米。求实际距离（千米）。
etres by dividing by 100: 4,200 m, then to km by dividing by 1,000: 4.2 km. So the towns are 4.2 km apart.
地图上1厘米代表实际50 000厘米。实际距离 = 8.4 × 50 000 = 420 000 厘米。换算成米除以100得 4 200 米，再换算成千米除以1 000得 4.2 千米。因此两城镇相距4.2公里。
km, divide by 100,000 directly (50,000 cm = 0.5 km per cm). Then 8.4 × 0.5 = 4.2 km.
务必检查单位：由厘米到千米可直接除以100 000（1 cm = 0.5 km）。那么 8.4 × 0.5 = 4.2 km。
6. Sequences: Matchstick Patterns | 数列：火柴棒图形规律
Consider a sequence of squares made from matchsticks. Pattern 1 uses 4 sticks, pattern 2 uses 7 sticks, pattern 3 uses 10 sticks. Find the rule for the nth term and hence the number of sticks for pattern 50.
考虑用火柴棒拼成的正方形序列。图形1用4根，图形2用7根，图形3用10根。求第n项的规则，并由此计算第50个图形所需的火柴棒数。
= 151 sticks. This linear pattern can be justified by adding 3 sticks for each new square.
数列为 4, 7, 10, ... 每次增加3。第n项 = 3n + 1（当n=1：3×1+1=4）。因此第50个图形需 3×50 + 1 = 151 根火柴棒。这个线性规律可以通过每增加一个正方形就添加3根火柴棒来证明。
7. Travel Graphs: Interpreting a Journey | 行程图：解读旅程
A distance–time graph shows a car journey. From 0 to 2 hours it travels 120 km, then stops for 0.5 hours, then returns 80 km in 1.5 hours. Find the speeds for each section and the average speed for the whole trip.
距离-时间图展示了一次汽车旅行。从0到2小时行驶了120千米，然后停留0.5小时，之后用1.5小时返回80千米。求各段的速率以及整个行程的平均速率。
. Total distance = 120 + 80 = 200 km; total time = 2 + 0.5 + 1.5 = 4 h. Average speed = 200 ÷ 4 = 50 km/h.
第一段速率 = 120 km ÷ 2 h = 60 km/h。第二段速率 = 0 km/h（停留）。第三段速率 = 80 km ÷ 1.5 h ≈ 53.3 km/h。总距离 = 120 + 80 = 200 km；总时间 = 2 + 0.5 + 1.5 = 4 h。平均速率 = 200 ÷ 4 = 50 km/h。
Note that average speed is not the average of the speeds, because the time spent at each speed matters.
注意平均速率并非各速率的简单平均，因为各速率所用的时间不同。
8. Measurement and Error Bounds | 测量与误差界限
A rectangle has length measured as 12.0 cm and width as 5.0 cm, both to the nearest 0.1 cm. Determine the upper and lower bounds for its area.
一个矩形的长度测量为12.0厘米，宽度测量为5.0厘米，都精确到0.1厘米。确定其面积的上下界限。
t area = 12.05 × 5.05 = 60.8525 cm². So the area lies between 59.2 and 60.9 cm² (to 3 s.f.).
长度：下界 = 11.95 cm，上界 = 12.05 cm。宽度：下界 = 4.95 cm，上界 = 5.05 cm。最小面积 = 11.95 × 4.95 = 59.1525 cm²;。最大面积 = 12.05 × 5.05 = 60.8525 cm²;。因此面积介于 59.2 cm² 和 60.9 cm² 之间（保留三位有效数字）。
Always state the bounds clearly. Error analysis is crucial in scientific contexts.
务必清晰说明界限。在科学情境中误差分析至关重要。
9. Linear Functions: Modelling Water Drainage | 线性函数：排水建模
A tank initially contains 200 litres of water. A pump drains water at a constant rate of 15 litres per minute. Write an equation for the volume V litres left after t minutes, and find when the tank will be empty.
一个水槽最初装有200升水。一台泵以恒定速率每分钟15升抽水。写出 t 分钟后剩余水量 V 升的方程，并求水槽何时排空。
t = 200/15 ≈ 13.33 minutes. So the tank empties in about 13 minutes and 20 seconds.
方程：V = 200 − 15t。令 V = 0：0 = 200 − 15t → 15t = 200 → t = 200/15 ≈ 13.33 分钟。所以水槽约在13分20秒后排空。
Plotting the graph would show a straight line with negative slope, intercepting
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