📚 Year 7 SQA Advanced Mathematics: Case Study Practical Exercises | 七年级 SQA 进阶数学：案例分析实战演练
This article presents a series of real-world case studies designed for Year 7 students following the SQA Advanced Mathematics curriculum. Each case encourages applying mathematical reasoning to everyday situations — from planning a trip and designing a garden to comparing energy costs. By working through these scenarios, learners strengthen their skills in ratio, percentage, algebra, geometry, and data handling while seeing how mathematics genuinely solves practical problems. Read each case carefully, attempt the calculations on your own, and then study the step-by-step solutions to deepen your understanding.
本文为七年级 SQA 进阶数学课程设计了一系列现实世界的案例分析。每个案例都鼓励将数学推理应用于日常情境——从规划旅行和设计花园到比较能源成本。通过完成这些场景，学习者在巩固比例、百分比、代数、几何和数据处理等技能的同时，也能看到数学如何真正解决实际问题。请仔细阅读每个案例，尝试独立计算，然后学习分步解析以加深理解。
1. Case 1: Planning a School Trip | 案例一：规划学校旅行
A Year 7 class organises a day trip to a science museum. The distance from school to the museum is 85 miles. The coach company charges a fixed hire fee of £180 plus £1.20 per mile travelled. There are 28 students and 3 teachers going. The museum entry costs £6.50 per student; teachers enter free. The school has a trip budget of £420. Determine the total cost of the trip and find out whether the budget is enough. If not, calculate how much each student must contribute extra, to the nearest 10p.
一个七年级班级组织去科学博物馆的一日游。学校到博物馆的距离是85英里。大巴公司收取固定租赁费180英镑，外加每英里1.20英镑。有28名学生和3名老师参加。博物馆门票学生每人6.50英镑，老师免费。学校旅行预算为420英镑。计算旅行总费用，并判断预算是否足够。如果不够，计算每名学生需要额外补交多少，精确到10便士。
First, calculate the coach cost: distance charge = 85 × £1.20 = £102.00. Total coach cost = fixed fee £180 + £102 = £282. Next, museum entry for 28 students: 28 × £6.50 = £182. Total trip cost = £282 + £182 = £464. Budget is £420, so shortfall = £464 − £420 = £44. Extra per student = £44 ÷ 28 ≈ £1.571… Rounded to the nearest 10p: £1.60. Thus each student must pay an additional £1.60.
首先计算大巴费用：里程费 = 85 × 1.20英镑 = 102.00英镑。大巴总费用 = 固定费180英镑 + 102英镑 = 282英镑。接着，28名学生的门票：28 × 6.50英镑 = 182英镑。旅行总费用 = 282英镑 + 182英镑 = 464英镑。预算为420英镑，因此资金缺口为464 − 420 = 44英镑。每名学生额外费用 = 44 ÷ 28 ≈ 1.571… 精确到10便士为1.60英镑。因此每名学生需额外支付1.60英镑。
2. Case 2: Designing a Rectangular Garden | 案例二：设计长方形花园
A rectangular garden is to be fenced using 40 metres of fencing. The garden’s length is twice its width. Let the width be w metres. Write an equation for the perimeter and solve for w. Then calculate the area of the garden. If the gardener wants to cover the entire area with grass seed costing £2.80 per square metre, what is the total seed cost?
一个长方形花园要用40米长的围栏围起来。花园的长是宽的两倍。设宽为w米。写出周长的方程并求解w。然后计算花园的面积。如果园丁想用每平方米2.80英镑的草籽覆盖整个区域，草籽总费用是多少？
Perimeter = 2(length + width) = 2(2w + w) = 2(3w) = 6w. Given perimeter = 40 m, so 6w = 40 → w = 40 ÷ 6 = 20/3 ≈ 6.67 m. Length = 2w = 40/3 ≈ 13.33 m. Area = length × width = (40/3) × (20/3) = 800/9 ≈ 88.89 m². Seed cost = area × £2.80 = (800/9) × 2.80 = 2240/9 ≈ £248.89. So the total cost is approximately £248.89.
周长 = 2(长 + 宽) = 2(2w + w) = 2(3w) = 6w。已知周长为40米，因此6w = 40 → w = 40 ÷ 6 = 20/3 ≈ 6.67米。长 = 2w = 40/3 ≈ 13.33米。面积 = 长 × 宽 = (40/3) × (20/3) = 800/9 ≈ 88.89平方米。草籽费用 = 面积 × 2.80英镑 = (800/9) × 2.80 = 2240/9 ≈ 248.89英镑。因此总费用约为248.89英镑。
3. Case 3: Budgeting for a Class Party | 案例三：班级派对预算
A class of 30 pupils plans an end-of-term party. They have a fund of £150. The party supplies include: 6 large pizzas at £9.60 each, 8 bottles of juice at £2.25 each, decorations for £14.50, and a cake that costs 18% of the total spent on pizzas and juice. Any remaining money will be used to buy party bags. Calculate the total cost so far and determine how much money remains for party bags. If each party bag costs £1.80, how many can they buy?
一个有30名学生的班级计划举办期末派对，共有150英镑经费。派对用品包括：6个大披萨，每个9.60英镑；8瓶果汁，每瓶2.25英镑；装饰品14.50英镑；以及一个蛋糕，其价格为披萨和果汁总花费的18%。剩余的钱将用于购买派对礼包。计算目前的费用总额，并确定买礼包还剩多少钱。如果每个礼包1.80英镑，他们能买多少个？
Pizza cost = 6 × £9.60 = £57.60. Juice cost = 8 × £2.25 = £18.00. Pizza + Juice = £75.60. Cake cost = 18% of £75.60 = 0.18 × 75.60 = £13.608 ≈ £13.61 (rounded to nearest penny). Decorations = £14.50. Total spent = 75.60 + 13.61 + 14.50 = £103.71. Remaining for bags = £150 − £103.71 = £46.29. Number of bags = £46.29 ÷ £1.80 = 25.716… Maximum whole bags = 25. They can buy 25 party bags.
披萨费用 = 6 × 9.60英镑 = 57.60英镑。果汁费用 = 8 × 2.25英镑 = 18.00英镑。披萨与果汁合计 = 75.60英镑。蛋糕费用 = 75.60英镑的18% = 0.18 × 75.60 = 13.608英镑 ≈ 13.61英镑（四舍五入到便士）。装饰品 = 14.50英镑。总支出 = 75.60 + 13.61 + 14.50 = 103.71英镑。礼包剩余 = 150 − 103.71 = 46.29英镑。礼包数量 = 46.29 ÷ 1.80 = 25.716… 最多能买25个整包。他们能买25个派对礼包。
4. Case 4: Sports Day Results | 案例四：运动会成绩分析
During sports day, four houses competed. The points scored are summarised below. House A: 142 points; House B: 118 points; House C: 165 points; House D: 135 points. Calculate the mean score across the four houses. Then find the percentage of total points that House C contributed, correct to one decimal place. Finally, the event organiser decides to award bonus points of 7% of each house’s original score, rounded to the nearest whole point. Calculate new totals and determine which house now leads.
运动会期间，四个学院进行了比赛。得分总结如下：A学院142分；B学院118分；C学院165分；D学院135分。计算四个学院的平均分。然后求C学院贡献的分数占总分的百分比，精确到一位小数。最后，活动组织者决定给予每个学院原始得分7%的奖励分，四舍五入到最接近的整数分。计算新总分，并确定哪个学院现在领先。
Total points = 142 + 118 + 165 + 135 = 560. Mean = 560 ÷ 4 = 140 points. House C percentage = (165 ÷ 560) × 100 = 29.464…% → 29.5% to 1 d.p. Bonus for House A: 7% of 142 = 9.94 → 10 pts; new = 152. House B: 7% of 118 = 8.26 → 8 pts; new = 126. House C: 7% of 165 = 11.55 → 12 pts; new = 177. House D: 7% of 135 = 9.45 → 9 pts; new = 144. House C still leads with 177 points.
总分 = 142 + 118 + 165 + 135 = 560。平均分 = 560 ÷ 4 = 140分。C学院百分比 = (165 ÷ 560) × 100 = 29.464…% → 29.5%（精确到一位小数）。A学院奖励：142的7% = 9.94 → 10分；新得分152。B学院：118的7% = 8.26 → 8分；新得分126。C学院：165的7% = 11.55 → 12分；新得分177。D学院：135的7% = 9.45 → 9分；新得分144。C学院仍以177分领先。
5. Case 5: Map Reading and Scale | 案例五：地图阅读与比例尺
An orienteering map uses a scale of 1 : 25 000. On the map, the distance from the starting point to checkpoint A is 8.4 cm, and from checkpoint A to checkpoint B is 6.2 cm. What are the real distances in kilometres? A team walks at an average speed of 4.5 km/h. How long, in minutes, will it take them to go from start to checkpoint B via A, assuming no stops? Also, if the map shows a lake covering 3 cm², what is the actual area of the lake in m²?
一张定向越野地图使用比例尺1:25 000。地图上，从起点到检查点A的距离是8.4厘米，从检查点A到检查点B是6.2厘米。实际距离分别是多少公里？一队人以平均4.5公里/小时的速度行走。如果他们从起点经A到B中途不停留，需要多少分钟？此外，如果地图上显示一个湖泊面积为3平方厘米，湖泊的实际面积是多少平方米？
Real distance start to A = 8.4 cm × 25 000 = 210 000 cm = 2100 m = 2.1 km. A to B = 6.2 cm × 25 000 = 155 000 cm = 1550 m = 1.55 km. Total distance = 2.1 + 1.55 = 3.65 km. Time = distance ÷ speed = 3.65 ÷ 4.5 = 0.8111… hours. In minutes: 0.8111… × 60 = 48.67 min ≈ 48 min 40 sec. Area scale factor = 25 000² = 625 000 000. Actual area = 3 cm² × 625 000 000 = 1 875 000 000 cm². 1 m² = 10 000 cm², so area = 1 875 000 000 ÷ 10 000 = 187 500 m².
起点到A的实际距离 = 8.4厘米 × 25 000 = 210 000厘米 = 2100米 = 2.1公里。A到B = 6.2 × 25 000 = 155 000厘米 = 1.55公里。总距离 = 2.1 + 1.55 = 3.65公里。时间 = 距离 ÷ 速度 = 3.65 ÷ 4.5 = 0.8111…小时。分钟数：0.8111… × 60 = 48.67分钟 ≈ 48分40秒。面积比例因子 = 25 000² = 625 000 000。实际面积 = 3平方厘米 × 625 000 000 = 1 875 000 000平方厘米。1平方米 = 10 000平方厘米，所以面积 = 1 875 000 000 ÷ 10 000 = 187 500平方米。
6. Case 6: Currency Conversion with Charges | 案例六：含手续费的货币兑换
A family travels to the USA and exchanges £800 into US dollars. The exchange rate is £1 = $1.26. The bureau charges a commission of 2.5% on the sterling amount converted, with a minimum charge of £5. How many dollars do they receive? After the trip, they return with $154 and exchange it back to pounds at a rate of £1 = $1.30, with a flat fee of £3. How many pounds do they get back? Compare the effective exchange rates for both transactions.
一个家庭前往美国旅行，将800英镑兑换成美元。汇率为1英镑 = 1.26美元。兑换所按兑换的英镑金额收取2.5%的佣金，最低收费5英镑。他们能得到多少美元？旅行结束后，他们带回154美元并换回英镑，汇率为1英镑 = 1.30美元，并收取固定费用3英镑。他们能拿回多少英镑？比较两次交易的实际汇率。
Commission for first exchange: 2.5% of £800 = £20, which is above the £5 minimum, so fee is £20. Amount to convert = £800 − £20 = £780. Dollars received = 780 × 1.26 = $982.80. Returning: $154 ÷ 1.30 = £118.4615… before fee. After £3 fee: £118.46 − £3 = £115.46 (to nearest penny). Effective rate buying $: 982.80 ÷ 800 ≈ 1.2285. Effective rate selling $: 154 ÷ (115.46 + 3) ≈ 154 ÷ 118.46 ≈ 1.3000 because fee was flat. Note the buy-back effective rate is very close to headline due to flat fee.
第一次兑换佣金：800英镑的2.5% = 20英镑，高于5英镑最低收费，因此费用为20英镑。兑换金额 = 800 − 20 = 780英镑。得到美元 = 780 × 1.26 = 982.80美元。换回：154美元 ÷ 1.30 = 118.4615…英镑，扣除3英镑费用后 = 118.46 − 3 = 115.46英镑（精确到便士）。购买美元的实际汇率：982.80 ÷ 800 ≈ 1.2285。卖出美元的实际汇率：154 ÷ (115.46 + 3) ≈ 154 ÷ 118.46 ≈ 1.3000，因为费用是固定的。注意回购实际汇率因固定费用而接近挂牌价。
7. Case 7: Building a Scale Model | 案例七：搭建比例模型
A model maker uses a scale of 1 : 72 to build a model aeroplane. The real aeroplane has a wingspan of 28.8 m, a length of 32.4 m, and a wing area of 120 m². Calculate the model’s wingspan and length in centimetres. Also, determine the wing area of the model in cm². If the model is painted using paint that covers 500 cm² per 8 ml tin, how many tins are needed for one coat on the wings (both sides have a total area 2 × wing area)?
一位模型制作者使用1:72的比例建造飞机模型。真飞机的翼展为28.8米，机长32.4米，机翼面积120平方米。计算模型的翼展和机长（以厘米为单位）。此外，求模型机翼的面积（平方厘米）。如果用每8毫升罐可涂刷500平方厘米的油漆为模型上色，机翼两面（总面积为2倍机翼面积）需要多少罐油漆？
Scale factor for length: 1/72. Model wingspan = 28.8 m ÷ 72 = 0.4 m = 40 cm. Model length = 32.4 m ÷ 72 = 0.45 m = 45 cm. Area scale factor = (1/72)² = 1/5184. Model wing area = 120 m² ÷ 5184 = 0.023148… m² = 231.48 cm² (since 1 m² = 10 000 cm², so 0.023148 × 10 000 = 231.48 cm²). Both sides area = 2 × 231.48 = 462.96 cm². Tins needed = 462.96 ÷ 500 = 0.9259 → 1 tin (must round up). One tin is sufficient.
长度比例因子：1/72。模型翼展 = 28.8米 ÷ 72 = 0.4米 = 40厘米。机长 = 32.4米 ÷ 72 = 0.45米 = 45厘米。面积比例因子 = (1/72)² = 1/5184。模型机翼面积 = 120平方米 ÷ 5184 = 0.023148…平方米 = 231.48平方厘米（因为1平方米 = 10 000平方厘米，所以0.023148 × 10 000 = 231.48）。两面总面积 = 2 × 231.48 = 462.96平方厘米。所需油漆罐数 = 462.96 ÷ 500 = 0.9259 → 1罐（必须向上取整）。一罐足够。
8. Case 8: Energy Cost Comparison | 案例八：能源成本比较
A household compares two types of light bulbs. Bulb X: traditional 60-watt bulb, lasts 1000 hours, costs £1.20 each. Bulb Y: LED 9-watt bulb, gives same brightness, lasts 15 000 hours, costs £5.50 each. Electricity costs 24 pence per kilowatt-hour (kWh). If the light is used 6 hours a day, compare the total cost (bulb + electricity) over a 5-year period (1825 days). Which bulb is cheaper overall? Express the savings in pounds to the nearest pound.
一个家庭比较两种灯泡。灯泡X：传统60瓦灯泡，寿命1000小时，每个1.20英镑。灯泡Y：LED 9瓦灯泡，同等亮度，寿命15000小时，每个5.50英镑。电费为每千瓦时24便士。如果每天使用6小时，比较5年期间（1825天）的总费用（灯泡成本 + 电费）。哪种灯泡总体更便宜？节省的费用以英镑表示，精确到整数。
Total hours over 5 years = 1825 × 6 = 10 950 hours. Bulb X: number needed = 10 950 ÷ 1000 = 10.95 → 11 bulbs. Cost of bulbs = 11 × £1.20 = £13.20. Energy used = 60 W × 10 950 h = 657 000 Wh = 657 kWh. Cost of electricity = 657 × £0.24 = £157.68. Total X = 13.20 + 157.68 = £170.88. Bulb Y: number needed = 10 950 ÷ 15 000 = 0.73 → 1 bulb (lasts 15 000 h). Bulb cost = £5.50. Energy = 9 W × 10 950 h = 98 550 Wh = 98.55 kWh. Electricity cost = 98.55 × 0.24 = £23.652. Total Y = 5.50 + 23.652 = £29.152 ≈ £29.15. Savings = 170.88 − 29.15 = £141.73 ≈ £142. LED bulb is vastly cheaper.
5年总小时数 = 1825 × 6 = 10 950小时。灯泡X：所需数量 = 10 950 ÷ 1000 = 10.95 → 11个灯泡。灯泡成本 = 11 × 1.20英镑 = 13.20英镑。耗电 = 60瓦 × 10 950小时 = 657 000瓦时 = 657千瓦时。电费 = 657 × 0.24英镑 = 157.68英镑。X总费用 = 13.20 + 157.68 = 170.88英镑。灯泡Y：所需数量 = 10 950 ÷ 15 000 = 0.73 → 1个（寿命15000小时）。灯泡成本 = 5.50英镑。耗电 = 9瓦 × 10 950 = 98 550瓦时 = 98.55千瓦时。电费 = 98.55 × 0.24 = 23.652英镑。Y总费用 = 5.50 + 23.652 = 29.152英镑 ≈ 29.15英镑。节省金额 = 170.88 − 29.15 = 141.73英镑 ≈ 142英镑。LED灯泡便宜得多。
9. Case 9: Mobile Phone Contract Analysis | 案例九：手机合约分析
A student wants a new mobile phone and considers two contract offers over 24 months. Plan A: £23.50 per month, includes unlimited calls and 10 GB data; upfront phone cost £49. Plan B: £18.00 per month, includes 5 GB data (extra data £2.50 per GB over limit); upfront phone cost £99. The student uses about 7 GB data per month on average. Find the total 24-month cost for each plan and recommend the cheaper option. If the student switches to 12 GB monthly usage after a year, what is the new total cost for Plan B over 2 years?
一名学生想买新手机，考虑两个24个月合约。方案A：每月23.50英镑，包含无限通话和10 GB流量；手机预付费用49英镑。方案B：每月18.00英镑，包含5 GB流量（超出部分每GB 2.50英镑）；手机预付99英镑。该学生平均每月使用约7 GB流量。计算每种方案24个月的总费用，并推荐更便宜的选项。如果学生一年后每月流量增加到12 GB，方案B两年总费用又是多少？
Plan A total: monthly cost = 24 × £23.50 = £564. Upfront = £49. Total = £613. Plan B first scenario (7 GB): each month data over = 7 − 5 = 2 GB. Extra cost per month = 2 × £2.50 = £5.00. Monthly total = £18 + £5 = £23.00. 24 months: 24 × £23 = £552. Plus upfront £99 = £651. Plan A cheaper by £38. Second scenario: first 12 months at 2 GB over = £23/month. Next 12 months usage 12 GB, over by 7 GB. Extra = 7 × £2.50 = £17.50. Monthly cost = £18 + £17.50 = £35.50. Total cost = 12 × £23 + 12 × £35.50 + £99 = £276 + £426 + £99 = £801. So switching makes Plan B very expensive.
方案A总费用：月费 = 24 × 23.50 = 564英镑。预付费49英镑。合计 = 613英镑。方案B第一种情况（7 GB）：每月超量 = 7 − 5 = 2 GB。每月额外费用 = 2 × 2.50 = 5.00英镑。月总计 = 18 + 5 = 23.00英镑。24个月：24 × 23 = 552英镑。加预付费99 = 651英镑。方案A便宜38英镑。第二种情况：前12个月超出2 GB，月费23英镑。后12个月使用12 GB，超出7 GB，额外费用17.50英镑，月费35.50英镑。总费用 = 12 × 23 + 12 × 35.50 + 99 = 276 + 426 + 99 = 801英镑。因此改变用量后方案B变得非常昂贵。
10. Case 10: Water Tank Filling and Geometry | 案例十：水箱注水与几何
A cylindrical water tank has an internal diameter of 1.5 m and a height of 2.2 m. Water flows in at a rate of 12 litres per minute. Determine the volume of the tank in litres (1 m³ = 1000 litres) and the time needed to fill it from empty to 80% capacity. The tank is placed on a roof, and the water is later used to irrigate a rectangular field measuring 25 m by 18 m. If 80% of the tank’s full volume is spread evenly over the field, what depth of water (in mm) does the field receive?
一个圆柱形水箱内径1.5米，高2.2米。水以每分钟12升的流速注入。计算水箱的容积（升，1立方米 = 1000升）以及从空箱注水到80%容量所需的时间。水箱放置在屋顶，之后的水用于灌溉一块长25米、宽18米的长方形田地。如果将水箱总容积的80%均匀喷洒在田地上，田地获得的水深是多少毫米？
Radius = 1.5 ÷ 2 = 0.75 m. Volume = π × r² × h = π × 0.75² × 2.2 = π × 0.5625 × 2.2 = π × 1.2375 ≈ 3.887 m³ (using 3.1416). In litres: 3.887 × 1000 = 3887 L. 80% volume = 0.8 × 3887 = 3109.6 L. Time to fill 80% = 3109.6 ÷ 12 = 259.13 min ≈ 4 h 19 min. Field area = 25 × 18 = 450 m². Volume applied = 3109.6 L = 3.1096 m³. Depth = Volume ÷ area = 3.1096 ÷ 450 = 0.0069102 m = 6.91 mm. The field receives about 6.9 mm of water.
半径 = 1.5 ÷ 2 = 0.75米。容积 = π × r² × h = π × 0.75² × 2.2 = π × 0.5625 × 2.2 = π × 1.2375 ≈ 3.887立方米（π取3.1416）。升数：3.887 × 1000 = 3887升。80%容积 = 0.8 × 3887 = 3109.6升。注满80%的时间 = 3109.6 ÷ 12 = 259.13分钟 ≈ 4小时19分钟。田地面积 = 25 × 18 = 450平方米。水量 = 3109.6升 = 3.1096立方米。深度 = 体积 ÷ 面积 = 3.1096 ÷ 450 = 0.0069102米 = 6.91毫米。田地获得约6.9毫米的水。
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