📚 Year 9 CAIE Maths: Case Study Practical Exercises | Year 9 CAIE 数学：案例分析实战演练
In this article, we will work through eight real-world case studies covering key topics from the Year 9 CAIE mathematics syllabus. Each case study presents a problem, followed by step-by-step reasoning to strengthen your problem-solving skills.
在本文中，我们将通过八个真实案例来练习 Year 9 CAIE 数学大纲中的关键主题。每个案例都呈现一个问题，然后逐步推理，以加强你的解题能力。
1. Case Study 1: Discounts and Sales Tax | 案例分析1：折扣与消费税
A store offers a 20% discount on all items, then applies a 5% sales tax on the discounted price. A shirt originally costs £80. What is the final price the customer pays?
一家商店对所有商品提供20%的折扣，然后在折扣价的基础上加收5%的消费税。一件衬衫原价为80英镑。顾客最终需要支付多少？
Step 1: Calculate the discount amount. 20% of £80 is 0.20 × 80 = £16.
步骤1：计算折扣金额。80英镑的20%是 0.20 × 80 = 16英镑。
Step 2: Find the discounted price. £80 − £16 = £64.
步骤2：计算折扣价。80 − 16 = 64英镑。
Step 3: Calculate the sales tax. 5% of £64 is 0.05 × 64 = £3.20.
步骤3：计算消费税。64英镑的5%是 0.05 × 64 = 3.20英镑。
Add the tax to the discounted price. Final price = £64 + £3.20 = £67.20.
步骤4：将消费税加到折扣价上。最终价格 = 64 + 3.20 = 67.20英镑。
Therefore, the customer pays £67.20.
因此，顾客支付67.20英镑。
2. Case Study 2: Ratio and Proportion in Recipes | 案例分析2：配方中的比率和比例
A cake recipe requires flour and sugar in the ratio 3:2. If you use 450 g of flour, how much sugar is needed? If the total mixture must weigh 1 kg, what quantity of each ingredient is required?
一个蛋糕配方需要面粉和糖的比例为3:2。如果你用了450克面粉，需要多少糖？如果混合物总重量必须为1千克，每种原料各需要多少？
iven 3x = 450 g, so x = 450 ÷ 3 = 150 g. Sugar needed = 2x = 2 × 150 = 300 g.
第一部分：设一份为x。则面粉 : 糖 = 3x : 2x。已知3x = 450克，所以x = 450 ÷ 3 = 150克。所需糖 = 2x = 2 × 150 = 300克。
= 200 g. Flour = 3 parts = 3 × 200 = 600 g. Sugar = 2 parts = 2 × 200 = 400 g.
第二部分：总份数 = 3 + 2 = 5份。1千克 = 1000克。一份 = 1000 ÷ 5 = 200克。面粉 = 3份 = 3 × 200 = 600克。糖 = 2份 = 2 × 200 = 400克。
3. Case Study 3: Solving Linear Equations – Ages | 案例分析3：线性方程 – 年龄问题
Ali is 5 years younger than his sister. The sum of their ages is 29. How old is each person?
阿里比他的姐姐小5岁。两人的年龄之和是29岁。他们各是多少岁？
Let Ali's age be a. Then his sister's age is a + 5. The equation: a + (a + 5) = 29.
设阿里的年龄为a。那么他姐姐的年龄为a + 5。方程：a + (a + 5) = 29。
29. Subtract 5 from both sides: 2a = 24. Divide by 2: a = 12. Ali is 12 years old. Sister's age = 12 + 5 = 17.
化简：2a + 5 = 29。两边减去5：2a = 24。除以2：a = 12。阿里12岁。姐姐的年龄 = 12 + 5 = 17岁。
4. Case Study 4: Perimeter and Area – Garden Design | 案例分析4：周长与面积 – 花园设计
A rectangular garden is 12 m long and 8 m wide. A circular flower bed of diameter 3 m is placed in the centre. Find the area of the lawn (the grass area) and the length of fencing required to enclose the garden.
一个矩形花园长12米，宽8米。中心有一个直径3米的圆形花坛。求草坪的面积（草地面积）以及围住花园所需的篱笆长度。
Area of rectangle = length × width = 12 × 8 = 96 m²;.
矩形面积 = 长 × 宽 = 12 × 8 = 96平方米。
; × radius² ≈ 3.14 × 1.5² = 3.14 × 2.25 = 7.065 m²;.
圆的半径 = 3 ÷ 2 = 1.5米。圆的面积 = π × 半径² ≈ 3.14 × 1.5² = 3.14 × 2.25 = 7.065平方米。
Lawn area = 96 − 7.065 = 88.935 m²; (about 88.9 m²).
草坪面积 = 96 − 7.065 = 88.935平方米（约88.9平方米）。
(length + width) = 2 × (12 + 8) = 2 × 20 = 40 m. So 40 m of fencing is needed.
花园周长 = 2 × (长 + 宽) = 2 × (12 + 8) = 2 × 20 = 40米。因此需要40米长的篱笆。
5. Case Study 5: Statistics – Analysing Test Scores | 案例分析5：统计 – 测验分数分析
Twenty students scored the following marks: 55, 60, 65, 70, 75, 80, 85, 90, 90, 95, 100, 100, 55, 60, 70, 75, 80, 85, 90, 95. Find the mean, median, mode, and range.
20名学生获得了如下分数：55, 60, 65, 70, 75, 80, 85, 90, 90, 95, 100, 100, 55, 60, 70, 75, 80, 85, 90, 95。求平均数、中位数、众数和极差。
First, sort the data: 55, 55, 60, 60, 65, 70, 70, 75, 75, 80, 80, 85, 85, 90, 90, 90, 95, 95, 100, 100.
首先，将数据排序：55, 55, 60, 60, 65, 70, 70, 75, 75, 80, 80, 85, 85, 90, 90, 90, 95, 95, 100, 100。
;2 + 100×2 = 110 + 120 + 65 + 140 + 150 + 160 + 170 + 270 + 190 + 200 = 1575. Mean = 1575 ÷ 20 = 78.75.
平均数：将所有数字求和后除以20。总和 95×2 + 100×2 = 110 + 120 + 65 + 140 + 150 + 160 + 170 + 270 + 190 + 200 = 1575。平均数 = 1575 ÷ 20 = 78.75。
Median: The middle two values are the 10th and 11th numbers. Both are 80, so median = 80.
中位数：中间的两个值是第10和第11个数，两者都是80，因此中位数 = 80。
Mode: The most frequent mark is 90 (appears 3 times).
众数：出现最频繁的分数是90（出现3次）。
Range: Maximum − minimum = 100 − 55 = 45.
极差：最大值 − 最小值 = 100 − 55 = 45。
6. Case Study 6: Fractions and Decimals – Adjusting Quantities | 案例分析6：分数与小数 – 调整数量
A model is built to ¾ of the original size. The original length is 2.4 m. Find the new length. If the original object is a cube, by what factor does the volume change?
一个模型按原尺寸的¾ 建造。原长为2.4米。求新长度。如果原物体是一个立方体，体积会按什么比例变化？
× 0.75 = 1.8 m. (Alternatively, 2.4 × 3 ÷ 4 = 7.2 ÷ 4 = 1.8 m.)
新长度 = 原长 × ¾ = 2.4 × 0.75 = 1.8米。（或 2.4 × 3 ÷ 4 = 7.2 ÷ 4 = 1.8米）
= 27/64. The new volume is 27/64 of the original volume, a reduction by a factor of 27/64.
对于立方体，体积缩放因子是线性缩放因子的立方：(¾)³ = 27/64。新体积是原体积的27/64，缩小为27/64。
7. Case Study 7: Coordinate Geometry – Distance Between Points | 案例分析7：坐标几何 – 点间距离
Point A is at (2,3) and point B at (5,7). Calculate the distance between A and B, and find the coordinates of the midpoint.
点A坐标为(2,3)，点B坐标为(5,7)。计算A与B之间的距离，并求中点坐标。
up2; + (y&sub2; − y&sub1;)²]. Here, x&sub1;=2, y&sub1;=3, x&sub2;=5, y&sub2;=7.
距离公式：距离 = √[(x&sub2; − x&sub1;)² + (y&sub2; − y&sub1;)²]。此处 x&sub1;=2, y&sub1;=3, x&sub2;=5, y&sub2;=7。
(7 − 3)²] = √(3² + 4²) = √(9 + 16) = √25 = 5 units.
距离 up2; + (7 − 3)²] = √(3² + 4²) = √(9 + 16) = √25 = 5 个单位。
((x&sub1;+x&sub2;)/2, (y&sub1;+y&sub2;)/2) = ((2+5)/2, (3+7)/2) = (7/2, 10/2) = (3.5, 5).
中点坐标：((x&sub1;+x&sub2;)/2, (y&sub1;+y&sub2;)/2) = ((2+5)/2, (3+7)/2) = (7/2, 10/2) = (3.5, 5)。
8. Case Study 8: Probability – Drawing Cards | 案例分析8：概率 – 抽卡片
A box contains 5 red cards, 3 blue cards, and 2 green cards. One card is drawn at random. Find the probability that it is (a) red, (b) not green.
一个盒子里有5张红色卡片、3张蓝色卡片和2张绿色卡片。随机抽取一张卡片。求抽到 (a) 红色、(b) 非绿色的概率。
Total number of cards = 5 + 3 + 2 = 10.
卡片总数 = 5 + 3 + 2 = 10。
(a) Probability of red = number of red cards / total = 5/10 = ½ = 0.5.
(a) 红色概率 = 红色卡片数 / 总数 = 5/10 = ½ = 0.5。
.8. Alternatively, favorable outcomes: red or blue = 5+3 = 8, so probability = 8/10 = 4/5.
(b) 非绿色的概率 = 1 − 绿色的概率 = 1 − 2/10 = 8/10 = 4/5 = 0.8。或者，有利结果：红色或蓝色 = 5+3 = 8，所以概率 = 8/10 = 4/5。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图