📚 Year 8 SQA Maths: Case Study Practical Exercises | SQA 八年级数学：案例分析实战演练
Welcome to your Year 8 SQA Mathematics case study! In this practical exercise, we will step into the role of a trip organiser planning an excursion to the Scottish Highlands. You will use a wide range of mathematical skills — from calculating distances and managing budgets to analysing survey data and assessing probability. Remember, in real life, maths is rarely just one isolated topic; it is a toolkit for solving complex, multi-step problems. Let's sharpen those problem-solving skills together.
欢迎来到 SQA 八年级数学案例分析！在这套实战演练中，我们将化身游学策划者，规划一次前往苏格兰高地的游览。你将运用到广泛的数学技能 —— 从计算距离和管理预算，到分析调查数据和评估概率。请记住，在现实生活中，数学很少是孤立的单一主题；它是解决复杂、多步骤问题的工具箱。让我们一起磨砺解决问题的技能吧。
1. Understanding the Problem | 理解问题
Our task is to plan a one-day trip for 60 students from Edinburgh to Loch Lomond. We need to consider travel, entrance fees, lunch, and activities. The first step in any case study is to extract the key numerical data: distances, group sizes, and time constraints.
我们的任务是为 60 名从爱丁堡出发前往洛蒙德湖的学生规划一日游。我们需要考虑交通、门票、午餐和活动。任何案例分析的第一步都是提取关键数值数据：距离、团体人数和时间限制。
Given Data: Distance is 75 miles. Bus speed averages 50 mph. We plan to spend exactly 4 hours at the destination and must return by 5:00 PM.
给定数据：距离为 75 英里。大巴平均时速 50 英里。我们计划在目的地停留恰好 4 小时，且必须在下午 5:00 前返回。
2. Calculating Travel Time | 计算旅程时间
We use the formula linking distance, speed, and time (DST). Since we know Distance (D) and Speed (S), we can find Time (T) by dividing distance by speed.
我们使用距离、速度和时间（DST）关系公式。已知距离（D）和速度（S），我们可以通过距离除以速度来求时间（T）。
Time = Distance ÷ Speed
s: T = 75 miles ÷ 50 mph = 1.5 hours. So, the one-way journey takes 1 hour and 30 minutes.
代入数值：T = 75 英里 ÷ 50 英里/小时 = 1.5 小时。所以，单程需要 1 小时 30 分钟。
ip takes 1.5 × 2 = 3 hours. Including the 4-hour stop, the total trip duration is 7 hours.
往返需要 1.5 × 2 = 3 小时。加上 4 小时的停留时间，行程总耗时为 7 小时。
3. Budgeting: Bus Hire vs. Train | 预算对比：包车与火车
We have two travel quotes. A luxury coach costs £450 for the day (covers all 60 students). Alternatively, a group train ticket costs £7.20 per student. We must calculate which is cheaper.
我们有两个交通报价。一辆豪华大巴包车一天 450 英镑（涵盖全部 60 名学生）。或者，团体火车票每名学生 7.20 英镑。我们必须计算哪种更便宜。
Total train cost = 60 × £7.20 = £432.00. The bus costs £450. The train is cheaper by £18.
火车总费用 = 60 × 7.20 英镑 = 432.00 英镑。大巴费用为 450 英镑。火车便宜 18 英镑。
However, the teacher adds a constraint: the train requires a 20-minute walk to the loch shore, while the coach parks directly outside. We must weigh cost against convenience in our decision making.
然而，老师附加了一个限制条件：火车需要在湖边步行 20 分钟，而大巴可直接停在门口。在决策时，我们必须权衡成本与便利性。
4. Calculating Entrance Fees with Discounts | 计算门票与折扣
The activity centre offers an entry fee of £12 per student. They offer a 15% group discount for 50 or more students. We need to find the total entry cost.
活动中心门票为每名学生 12 英镑。他们对 50 人及以上的团体提供 15% 的折扣。我们需要计算门票总费用。
Method 1: Find the discount amount per ticket. 15% of £12 = 0.15 × 12 = £1.80. Discounted ticket = £12 - £1.80 = £10.20.
方法一：计算每张票的折扣金额。12 英镑的 15% = 0.15 × 12 = 1.80 英镑。折扣票价 = 12 英镑 - 1.80 英镑 = 10.20 英镑。
Total entry cost = £10.20 × 60 = £612.00.
门票总费用 = 10.20 英镑 × 60 = 612.00 英镑。
Alternative method: Multiply the total raw cost by 0.85 (which represents 100% - 15%). £12 × 60 × 0.85 = £612.00.
替代方法：将原始总费用乘以 0.85（代表 100% - 15%）。12 英镑 × 60 × 0.85 = 612.00 英镑。
5. Proportional Reasoning for Lunch | 午餐的比例推理
A catering pack serving 12 people costs £16.80. We must order enough to feed 60 students. This is a direct proportion problem.
一份供 12 人食用的餐饮套餐价格为 16.80 英镑。我们需要为 60 名学生订购足够的食物。这是一个正比问题。
Find the cost for one person: £16.80 ÷ 12 = £1.40. Cost for 60 people: £1.40 × 60 = £84.00.
计算每人成本：16.80 英镑 ÷ 12 = 1.40 英镑。60 人的费用：1.40 英镑 × 60 = 84.00 英镑。
Alternatively, note that 60 is 5 times larger than 12, so the cost is 5 × £16.80 = £84.00.
或者，注意到 60 是 12 的 5 倍，因此费用为 5 × 16.80 英镑 = 84.00 英镑。
6. Handling Survey Data | 处理调查数据
The school surveys students on their preferred activity: Kayaking, Hiking, or Cycling. The results show: Kayaking 25 students, Hiking 20 students, Cycling 15 students.
学校对学生偏好的活动进行了调查：皮划艇、徒步或骑行。结果显示：皮划艇 25 人，徒步 20 人，骑行 15 人。
the whole for each. Kayaking = 25/60 = 5/12. Hiking = 20/60 = 1/3. Cycling = 15/60 = 1/4.
我们可以用频数表表示。我们还可以计算每项占总体的分数。皮划艇 = 25/60 = 5/12。徒步 = 20/60 = 1/3。骑行 = 15/60 = 1/4。
| Activity | Frequency | Fraction |
|---|---|---|
| Kayaking | 25 | 5/12 |
| Hiking | 20 | 1/3 |
| Cycling | 15 | 1/4 |
7. Perimeter and Area in Outdoor Spaces | 户外的周长与面积
The Hiking group plans to walk around a rectangular field that is 80 metres long and 35 metres wide. How far do they walk around it once?
徒步小组计划环绕一个长 80 米、宽 35 米的长方形场地行走。他们一圈走了多远？
Perimeter of a rectangle = 2 × (Length + Width) = 2 × (80 + 35) = 2 × 115 = 230 metres.
长方形周长 = 2 × (长 + 宽) = 2 × (80 + 35) = 2 × 115 = 230 米。
0 m × 5 = 1150 metres, which is 1.15 kilometres. Converting between units is crucial here.
如果他们绕场 5 圈，总距离 = 230 米 × 5 = 1150 米，即 1.15 公里。在这里，单位换算至关重要。
8. Angles and Bearings | 角与方位角
The Cycling group uses a map to navigate. They ride 3 km East, then turn 55° North for 4 km. What is the angle of their ride back to the start? (Using a coordinate grid approach).
骑行小组使用地图导航。他们向东骑行 3 公里，然后向北偏转 55 度骑行 4 公里。他们骑回起点的角度是多少？（使用坐标方格法）。
ming a triangle. The bearing back would require calculating the hypotenuse angle using tan θ = opposite/adjacent.
如果东为 x 轴，北为 y 轴，他们从东偏转 55 度。这意味着转折点的内角是互补的
Published by TutorHao | Year 8 Mathematics Revision Series | aleveler.com
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