📚 Case Studies in Action: Real-Life Math Problem Solving | 案例分析实战演练：生活中的数学解题
Mathematics is not just about numbers in a textbook – it is a powerful tool for solving everyday problems. In this revision guide, we will work through a series of real-life case studies designed for Year 7 Cambridge learners. Each scenario invites you to apply your knowledge of arithmetic, fractions, percentages, geometry, statistics, and algebra in practical contexts. By carefully reading the questions, breaking them into steps, and using clear calculations, you will strengthen your problem-solving skills and build confidence for your exams.
数学不仅仅是课本上的数字——它是解决日常问题的强大工具。在这份复习指南中，我们将通过一系列为剑桥7年级学生设计的真实案例，把知识运用到实践中。每个场景都要求你运用算术、分数、百分比、几何、统计和代数知识。通过仔细审题、分步思考和清晰计算，你将提升解题能力，并为考试树立信心。
1. Budgeting a Birthday Party | 生日聚会预算
Imagine you have a budget of £150 to organize a birthday party for 12 friends. You need to buy snacks, drinks, decorations, and a cake. The costs are: snacks £3.50 per person, drinks £2.20 per person, decorations £25 in total, and cake £30. Can you stay within budget?
假设你有150英镑预算，为12位朋友组织一场生日聚会。你需要购买零食、饮料、装饰品和一个蛋糕。费用如下：零食每人3.50英镑，饮料每人2.20英镑，装饰品总计25英镑，蛋糕30英镑。你能控制在预算内吗？
Let’s calculate step by step. First, find the cost of snacks: 12 × £3.50 = £42.00. Next, drinks: 12 × £2.20 = £26.40. Add these to find the total per-person items cost: £42.00 + £26.40 = £68.40. Then add the fixed costs for decorations (£25) and cake (£30): £68.40 + £25 + £30 = £123.40.
我们分步计算。首先，求零食费用：12 × 3.50 英镑 = 42.00 英镑。然后，饮料：12 × 2.20 英镑 = 26.40 英镑。将两者相加，得到按人计算的物品总费用：42.00 + 26.40 = 68.40 英镑。再加上固定费用装饰品（25英镑）和蛋糕（30英镑）：68.40 + 25 + 30 = 123.40 英镑。
Compare the total with the budget: £123.40 < £150. You have £26.60 remaining. So the answer is yes, you can afford the party and even have money left over.
将总费用与预算比较：123.40 英镑 < 150 英镑。你还剩下 26.60 英镑。因此答案是肯定的，你能负担这场聚会，还有剩余的钱。
| Item | Calculation | Cost |
|---|---|---|
| Snacks | 12 × £3.50 | £42.00 |
| Drinks | 12 × £2.20 | £26.40 |
| Decorations | Fixed | £25.00 |
| Cake | Fixed | £30.00 |
| Total | £123.40 |
This case shows how addition and multiplication with decimals work together in financial planning.
这个案例展示了小数加法和乘法如何在财务规划中协同工作。
2. Planning a Road Trip | 规划公路旅行
A family drives from Cambridge to Edinburgh, a distance of 330 miles. Their car travels 45 miles on one gallon of petrol. Petrol costs £1.45 per gallon. The family also stops for a 45-minute lunch break and takes two 15-minute rest breaks. If they leave at 9:00 am and drive at an average speed of 55 miles per hour, what time will they arrive?
一个家庭从剑桥自驾到爱丁堡，全程330英里。他们的汽车每加仑汽油行驶45英里。汽油价格为每加仑1.45英镑。家人还会停车45分钟吃午饭，并休息两次，每次15分钟。如果他们上午9:00出发，以平均每小时55英里的速度行驶，他们将在什么时间到达？
es, which is 1 hour 15 minutes. Total journey time = 6 h + 1 h 15 min = 7 hours 15 minutes.
首先计算行驶时间。时间 = 距离 ÷ 速度 = 330 ÷ 55 = 6 小时。接着，加上所有休息时间：午餐45分钟 + 休息15分钟 + 15分钟 = 75分钟，即1小时15分钟。总行程时间 = 6小时 + 1小时15分钟 = 7小时15分钟。
ut you can only buy whole units, so the family needs 8 gallons to be safe. Fuel cost = 8 × £1.45 = £11.60. Realistically, they'd purchase 8 gallons and have some left.
出发时间上午9:00加上7小时15分钟，到达时间为下午4:15。现在计算燃油：所需加仑数 = 330 ÷ 45。计算除法：330 ÷ 45 = 7.333...，但只能购买整加仑，所以为了安全需要8加仑。燃油费用 = 8 × 1.45英镑 = 11.60英镑。实际情况下，他们会购买8加仑并会有剩余。
Multiplying and dividing with mixed units helps you manage time and money on a real trip.
利用混合单位进行乘法和除法，能帮助你在真实旅途中管理时间和金钱。
3. Scaling a Recipe | 调整食谱配方
A recipe for 8 people requires 300 g of flour, 200 g of sugar, 4 eggs, and 250 ml of milk. You need to adapt it for 10 people. How much of each ingredient do you need? Give flour and sugar in kilograms, and milk in litres.
一份供8人食用的食谱需要300克面粉、200克糖、4个鸡蛋和250毫升牛奶。你需要为10人调整配方。每种食材各需要多少？面粉和糖以千克为单位，牛奶以升为单位。
gs: 4 × 1.25 = 5 eggs. Milk: 250 ml × 1.25 = 312.5 ml = 0.3125 litres (or 0.313 L rounded to three decimal places).
找出倍数：新用量 = 10 ÷ 8 = 1.25。每种食材的用量乘以1.25。面粉：300克 × 1.25 = 375克 = 0.375千克。糖：200克 × 1.25 = 250克 = 0.25千克。鸡蛋：4 × 1.25 = 5个。牛奶：250毫升 × 1.25 = 312.5毫升 = 0.3125升（或四舍五入为0.313升）。
Using proportions and unit conversions allows you to scale any recipe accurately.
使用比例和单位换算可以让你准确地调整任何食谱的用量。
Scaling factor = 1.25 → New quantity = Original × 1.25
换算因子 = 1.25 → 新用量 = 原用量 × 1.25
4. Sports Tournament Statistics | 运动锦标赛统计
In a basketball tournament, a player scored the following points in 5 games: 18, 24, 15, 30, 13. Find the mean score. The team wants to know the range and whether the median is higher than the mean.
在一场篮球锦标赛中，某球员在5场比赛中得分如下：18, 24, 15, 30, 13。求平均得分。球队想知道的极差，以及中位数是否高于平均数。
18 + 24 + 15 + 30 + 13 = 100. Divide by 5 games: 100 ÷ 5 = 20. Mean = 20 points per game.
计算平均数：将所有得分相加：18 + 24 + 15 + 30 + 13 = 100。除以5场比赛：100 ÷ 5 = 20。平均每场20分。
res from smallest to largest: 13, 15, 18, 24, 30. The middle number is 18, so median = 18.
求极差：最高分 = 30，最低分 = 13，因此极差 = 30 − 13 = 17。求中位数时，将得分从小到大排列：13, 15, 18, 24, 30。中间数是18，因此中位数 = 18。
The median (18) is less than the mean (20), so the answer is no. The high score of 30 pulled the mean above the median.
中位数（18）小于平均数（20），所以答案是否定的。高分30分把平均数拉到了中位数之上。
Understanding mean, median, and range helps you compare performances and spot outliers.
理解平均数、中位数和极差能帮助你比较表现并发现异常值。
5. Designing a Rectangular Garden | 设计矩形花园
You have 30 metres of fencing to create a rectangular garden against a wall. The wall forms one long side of the rectangle, so only three sides need fencing. What dimensions give the largest possible area for planting? Use trial and improvement with whole numbers.
你用30米长的围栏沿着一面墙围一个矩形花园。墙面构成矩形的一个长边，所以只需给另外三边安装围栏。什么样的尺寸能使种植面积最大？用整数进行试错和改进。
be L. The total fencing is w + w + L = 30, so L = 30 − 2w. Area A = L × w = (30 − 2w) × w.
设宽为w米（垂直于墙的两边），平行于墙的长为L。总围栏用量为 w + w + L = 30，因此 L = 30 − 2w。面积 A = L × w = (30 − 2w) × w。
s at w = 7 or 8, both giving 112 m². With whole numbers, either (w=7, L=16) or (w=8, L=14) works best.
尝试 w = 1：L = 28，A = 28 平方米。w = 2：L = 26，A = 52 平方米。w = 5：L = 20，A = 100 平方米。w = 7：L = 16，A = 112 平方米。w = 8：L = 14，A = 112 平方米。w = 10：L = 10，A = 100 平方米。最大面积出现在 w = 7 或 8，均为112平方米。取整数时，(w=7, L=16) 或 (w=8, L=14) 的效果最好。
This investigation shows how algebra and systematic trial help solve design problems. Write an expression for area and test values.
这项探究表明代数和系统试错如何帮助解决设计问题。写出面积表达式并测试数值。
6. Reading and Creating Bar Charts | 阅读与绘制条形图
A survey asked 40 Year 7 students about their favourite fruit. The results: Apple 12, Banana 10, Orange 6, Grapes 8, Strawberry 4. Draw a bar chart and then answer: what fraction of students chose Apple? What percentage chose Banana or Grapes?
一项调查询问了40名7年级学生最喜欢的水果。结果为：苹果12人，香蕉10人，橙子6人，葡萄8人，草莓4人。请绘制条形图，然后回答问题：选择苹果的学生占几分之几？选择香蕉或葡萄的学生占百分之几？
To draw a bar chart, label the horizontal axis with fruit names and the vertical axis with frequency from 0 to 14. Draw bars of heights 12, 10, 6, 8, 4. Use equal widths and gaps.
绘制条形图时，水平轴标记水果名称，垂直轴标记频数，从0到14。绘制高度分别为12、10、6、8、4的条形。条形宽度相等，间距均匀。
out of 40 = 12/40 = 3/10 after simplification by dividing numerator and denominator by 4.
苹果所占分数：40人中12人 = 12/40 = 化简后为 3/10（分子分母同时除以4）。
8 = 18 students. As a percentage of 40: (18/40) × 100 = (18 ÷ 40) × 100 = 0.45 × 100 = 45%.
香蕉或葡萄的百分比：香蕉10人 + 葡萄8人 = 18名学生。占40人的百分比：(18/40) × 100 = (18 ÷ 40) × 100 = 0.45 × 100 = 45%。
Data from bar charts can be interpreted using fractions, decimals, and percentages, just like in real surveys.
条形图中的数据可以用分数、小数和百分比来解读，就像真实的调查一样。
7. Currency Exchange for a Holiday | 假期货币兑换
Emma is travelling to the USA and wants to exchange £250 into US dollars. The exchange rate is £1 = $1.32. The bureau charges a fixed fee of £4. How many dollars will she receive after the fee is taken from her pounds?
艾玛要去美国旅行，想将250英镑兑换成美元。汇率为1英镑 = 1.32美元。兑换处收取4英镑固定手续费。从她的英镑中扣除手续费后，她将收到多少美元？
First, subtract the fee from the amount to be exchanged: £250 − £4 = £246. Then multiply by the exchange rate: 246 × 1.32 = ? 246 × 0.30 = 73.8, 246 × 0.02 = 4.92. Total = 246 + 73.8 + 4.92 = 324.72. So she receives $324.72.
首先，从要兑换的金额中减去手续费：250 − 4 = 246 英镑。然后乘以汇率：246 × 1.32 = ? 拆解计算：246 × 1 = 246，246 × 0.30 = 73.8，246 × 0.02 = 4.92。总和 = 246 + 73.8 + 4.92 = 324.72。因此她收到 324.72 美元。
If she later exchanges $50 back to pounds at the same rate but with a 2% commission on the dollar amount, how many pounds does she get? First, commission = 2% of $50 = $1, so she effectively exchanges $49. £1 = $1.32 means $1 = 1/1.32 ≈ £0.7576. So £ = 49 × 0.7576 ≈ £37.12.
如果之后她用50美元以相同汇率换回英镑，但收取美元金额2%的手续费，她能得到多少英镑？首先，手续费 = 50美元的2% = 1美元，所以她实际兑换49美元。1英镑 = 1.32美元，所以1美元 = 1/1.32 ≈ 0.7576英镑。因此英镑数 = 49 × 0.7576 ≈ 37.12 英镑。
Working with exchange rates and percentages is essential for international travel.
处理汇率和百分比对于国际旅行必不可少。
8. Filling a Fish Tank: Volume and Capacity | 填充鱼缸：体积与容量
me of water in cubic centimetres. Then convert this volume to litres (1 litre = 1000 cm³).
一个矩形鱼缸长50厘米，宽30厘米，高40厘米。你加水至35厘米深。计算水的体积，单位为立方厘米。然后将此体积转换为升（1升 = 1000立方厘米）。
500, then 1500 × 35 = 52500. So volume = 52500 cm³. In litres: 52500 ÷ 1000 = 52.5 litres.
体积 = 长 × 宽 × 水的高度 = 50 cm × 30 cm × 35 cm。相乘：50 × 30 = 1500，然后 1500 × 35 = 52500。所以体积 = 52500 立方厘米。换算为升：52500 ÷ 1000 = 52.5 升。
orative rock that occupies 1200 cm³, what is the new water volume and the new water height? lume = 52500 + 1200 = 53700 cm³. New height = Volume ÷ Base area = 53700 ÷ 1500 = 35.8 cm.
如果你想放入一块装饰岩石，其体积为1200立方厘米，那么新的水体积和新的水面高度是多少？假设鱼缸底面积不变。底面积 = 50 × 30 = 1500 平方厘米。岩石会排水，使体积增加1200立方厘米，因此新体积 = 52500 + 1200 = 53700 立方厘米。新高度 = 体积 ÷ 底面积 = 53700 ÷ 1500 = 35.8 厘米。
Using volume and capacity formulas lets you manage water and objects in real containers.
使用体积和容积公式能让你管理真实容器中的水和物体。
9. Temperature Change Investigation | 温度变化调查
At midnight, the temperature outside is −5 °C. By 6:00 am, it has dropped by 3 °C. Then it rises 8 °C by noon. Later it falls 10 °C by midnight again. What is the temperature at each stage, and what is the overall change?
午夜时分，室外温度为 −5 °C。到早上6:00，温度下降了3 °C。然后到中午上升了8 °C。随后到再次午夜又下降了10 °C。每个阶段的温度是多少？总体变化是多少？
ature at successive times is: midnight −5 °C, 6 am −8 °C, noon 0 °C, next midnight −10 °C.
从 −5 °C 开始。下降后：−5 − 3 = −8 °C。上升后：−8 + 8 = 0 °C。再次下降后：0 − 10 = −10 °C。相继时间点的温度为：午夜 −5 °C，早上6点 −8 °C，中午 0 °C，次日午夜 −10 °C。
ight to next: −10 − (−5) = −10 + 5 = −5 °C, so the temperature has fallen by 5 °C overall.
从一个午夜到下一个午夜的总体变化：−10 − (−5) = −10 + 5 = −5 °C，因此整体下降了5 °C。
Calculating with negative numbers is crucial for temperature, elevation, and finance.
用负数进行计算对于温度、海拔和财务问题至关重要。
10. Solving a Number Puzzle | 数字谜题解答
I think of a number, multiply it by 6, then subtract 4. The result is 38. What is the number? Use an algebraic equation to solve it.
我想了一个数，把它乘以6，再减去4。结果是38。这个数是多少？请用代数方程求解。
: 6x = 42. Divide both sides by 6: x = 7. Check: 6 × 7 = 42, subtract 4 gives 38. Correct.
设这个数为 x。方程：6x − 4 = 38。两边同时加4：6x = 42。两边除以6：x = 7。检验：6 × 7 = 42，减去4得38。正确。
and n+4. Equation: n + (n+2) + (n+4) = 63. Collect like terms: 3n + 6 = 63. Subtract 6: 3n = 57. Divide by 3: n = 19. So the numbers are 19, 21, 23.
再试一个更难的谜题：三个连续奇数的和为63。找出这些数。设最小的奇数为 n，那么接下来的奇数为 n+2 和 n+4。方程：n + (n+2) + (n+4) = 63。合并同类项：3n + 6 = 63。减去6：3n = 57。除以3：n = 19。因此这些数是19, 21, 23。
Forming and solving equations are key algebra skills that appear in many case studies.
建立并求解方程是代数中的关键技能，会出现在许多案例中。
6x − 4 = 38 ⟹ x = 7
6x − 4 = 38 ⟹ x = 7
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