📚 Interdisciplinary Problem Solving for Year 7 WJEC Maths | 跨学科综合题型训练
In Year 7 WJEC Maths, you will often meet problems that combine mathematical skills with topics from science, geography, art and everyday life. These interdisciplinary questions test your ability to apply numbers, algebra, geometry and statistics in real contexts.
在7年级WJEC数学中，你经常会遇到将数学技能与科学、地理、艺术和日常生活主题相结合的题目。这些跨学科问题考查你在真实情境中应用数字、代数、几何和统计知识的能力。
1. Reading Data Tables in Science | 科学中的数据表阅读
Many science experiments produce results in a table. You need to read values carefully, calculate the mean and draw conclusions. The table below shows the temperature of a beaker of water recorded every minute.
许多科学实验会将结果制成表格。你需要仔细读取数值、计算平均值并得出结论。下面的表格显示了每分钟记录的一杯水的温度。
| Time (min) | 0 | 1 | 2 | 3 | 4 |
| Temperature (°C) | 22 | 26 | 29 | 27 | 24 |
Question: What was the highest temperature recorded? Work out the mean temperature over the five minutes.
问题：记录的最高温度是多少？计算这五分钟内的平均温度。
Step 1: Scan the table to find the largest value. The temperatures are 22, 26, 29, 27, 24. The highest is 29°C.
步骤1：浏览表格找出最大值。温度值为22、26、29、27、24。最高是29°C。
Step 2: To find the mean, add all five temperatures: 22 + 26 + 29 + 27 + 24 = 128.
步骤2：要计算平均值，将所有五个温度相加：22 + 26 + 29 + 27 + 24 = 128。
the total by the number of readings: 128 ÷ 5 = 25.6°C. So the mean temperature is 25.6°C.
步骤3：用总数除以读数个数：128 ÷ 5 = 25.6°C。因此平均温度为25.6°C。
2. Scale Maps and Real Distances | 比例尺地图与实际距离
Maps use a scale to show real distances in a smaller size. A typical scale might be 1 : 50 000, meaning 1 cm on the map represents 50 000 cm on the ground. Converting units correctly is essential.
地图使用比例尺以缩小的尺寸表示实际距离。常见的比例尺可能是1 : 50 000，表示地图上的1 cm代表地面上的50 000 cm。正确转换单位至关重要。
Example: On a 1 : 25 000 map, two villages are 8 cm apart. What is the real distance in kilometres?
示例：在一幅1 : 25 000的地图上，两个村庄相距8 cm。实际距离是多少千米？
Step 1: Multiply the map distance by the scale factor: 8 cm × 25 000 = 200 000 cm.
步骤1：将地图距离乘以比例系数：8 cm × 25 000 = 200 000 cm。
Step 2: Convert cm to metres: 200 000 cm ÷ 100 = 2 000 m.
步骤2：将厘米转换为米：200 000 cm ÷ 100 = 2 000 m。
Step 3: Convert metres to kilometres: 2 000 m ÷ 1 000 = 2 km. The villages are 2 km apart.
步骤3：将米转换为千米：2 000 m ÷ 1 000 = 2 km。村庄相距2千米。
formula: real distance = map distance × scale factor ÷ 100 000 to convert directly to km.
你也可以使用公式：实际距离 = 地图距离 × 比例系数 ÷ 100 000，直接转换为千米。
real distance (km) = map distance (cm) × scale factor ÷ 100 000
实际距离 (km) = 地图距离 (cm) × 比例系数 ÷ 100 000
3. Nutrition Labels and Percentages | 营养标签与百分比
Food labels give nutritional values per 100 g or per portion. You can use fractions, multiplication and percentages to work out the amount of sugar, fat or salt in different serving sizes.
食品标签会标注每100克或每份的营养成分。你可以用分数、乘法和百分比计算出不同食用分量中的糖、脂肪或盐的含量。
A cereal bar contains the following information: Per 100 g – sugar 24 g. A bar weighs 35 g. How many grams of sugar are in one bar? What is this as a percentage of the bar's total mass?
一块谷物棒标注以下信息：每100克 – 糖24克。一根谷物棒重35克。一根棒含有多少克糖？这占整根棒总质量的百分比是多少？
Step 1: Find the sugar in 1 g: 24 g ÷ 100 = 0.24 g per gram.
步骤1：计算每克的含糖量：24 g ÷ 100 = 0.24 g每克。
Step 2: Multiply by the bar's mass: 0.24 g × 35 = 8.4 g of sugar per bar.
步骤2：乘以谷物棒的质量：0.24 g × 35 = 8.4 g糖每根。
tage, use the fraction of sugar over total mass and multiply by 100: (8.4 ÷ 35) × 100 = 24% Interestingly, the percentage remains the same because the sugar concentration is uniform.
步骤3：用百分比表示，用糖的质量除以总质量再乘以100：(8.4 ÷ 35) × 100 = 24%。有趣的是，百分比保持不变，因为糖的浓度是均匀的。
This shows that if the label gives per 100 g, the percentage of that nutrient is the same in any portion of that product.
这表明如果标签标注每100克，那么该营养素的百分比在该产品的任何分量中都是相同的。
4. Patterns in Nature | 自然中的图案
Geometry helps us describe natural structures such as honeycombs, snowflakes and flower petals. Regular hexagons appear in beehives because they tile a plane with no gaps, using the least amount of wax.
几何学帮助我们描述蜂窝、雪花和花瓣等自然结构。蜂巢中正六边形的出现是因为它们能以最小的蜂蜡用量、不留空隙地平铺一个平面。
Question: A honeycomb cell is a regular hexagon. If one interior angle is 120°, how many degrees are in all six interior angles together? What is the sum of the exterior angles?
问题：一个蜂巢室是一个正六边形。如果一个内角是120°，那么六个内角的总度数是多少？外角和是多少？
ne interior angle by six: 120° × 6 = 720°. The sum of interior angles of any hexagon is (6 – 2) × 180° = 720°.
步骤1：将一个内角乘以6：120° × 6 = 720°。任何六边形的内角和为(6 – 2) × 180° = 720°。
les of any convex polygon is always 360°. Each exterior angle of a regular hexagon is 180° – 120° = 60°.
步骤2：任何凸多边形的外角和总是360°。正六边形的每个外角为180° – 120° = 60°。
Step 3: Six exterior angles: 6 × 60° = 360°, confirming the rule.
步骤3：六个外角：6 × 60° = 360°，验证了该规则。
Natural patterns often show lines of symmetry. A regular hexagon has 6 lines of symmetry and rotational symmetry of order 6.
自然图案常常显示出对称线。一个正六边形有6条对称线和6阶旋转对称。
5. Timetables and Travel Maths | 时间表与旅行数学
Reading a bus or train timetable requires calculating time intervals, converting between 12-hour and 24-hour clocks, and working with elapsed time. This mixes number skills with real planning.
阅读公交或火车时刻表需要计算时间间隔、在12小时和24小时制之间转换，并处理经过时间。这结合了数字技能与现实规划。
Train A departs at 08:45 and arrives at 11:20. How long is the journey? If a passenger missed this train and took the next one at 10:30, arriving at 13:05, how much later did they arrive compared to Train A?
列车A于08:45出发，11:20到达。旅程需要多长时间？如果一名乘客错过了这趟车，乘坐了下一趟10:30出发的车，于13:05到达，与列车A相比，他晚到了多长时间？
00 is 15 min, 09:00 to 11:00 is 2 hours, 11:00 to 11:20 is 20 min. Total = 2 hours 35 min.
步骤1：从08:45到11:20：08:45到09:00是15分钟，09:00到11:00是2小时，11:00到11:20是20分钟。总计2小时35分钟。
Step 2: Second journey: 10:30 to 13:05. 10:30 to 13:00 is 2 hours 30 min, plus 5 min gives 2 hours 35 min as well. So the journey time is also 2 hours 35 minutes.
步骤2：第二趟旅程：10:30到13:05。10:30到13:00是2小时30分钟，再加5分钟也是2小时35分钟。所以旅程时间同样是2小时35分钟。
ifference is from 11:20 to 13:05: 40 min to 12:00 plus 1 hour 5 min = 1 hour 45 min later.
步骤3：第一趟到达时间为11:20。第二趟到达时间为13:05。时间差为从11:20到13:05：到12:00还有40分钟，再加1小时5分钟，共晚到1小时45分钟。
Being able to compare times and durations helps plan connections and avoid long waits.
能够比较时间和持续时间有助于规划换乘并避免长时间等待。
6. Sports Statistics | 体育统计
Statistics are widely used in sports to analyse player performance. In Year 7, you can find the mean, median, mode and range from a set of scores, which links maths directly to Physical Education.
统计学被广泛用于体育中分析运动员表现。在7年级，你可以从一组得分中找到平均数、中位数、众数和极差，这直接将数学与体育教育联系起来。
A basketball player scores the following points in eight matches: 12, 15, 8, 20, 12, 17, 22, 12. Find the mode, median and range.
一名篮球运动员在八场比赛中的得分如下：12, 15, 8, 20, 12, 17, 22, 12。求众数、中位数和极差。
most frequent score. The number 12 appears three times, more than any other, so mode = 12.
步骤1：众数是最常出现的得分。数字12出现了三次，多于其他任何数，所以众数 = 12。
12, 12, 12, 15, 17, 20, 22. As there are 8 numbers (even), median = (12 + 15) ÷ 2 = 13.5.
步骤2：要找到中位数，将得分排序：8, 12, 12, 12, 15, 17, 20, 22。因为有8个数字（偶数），中位数 = (12 + 15) ÷ 2 = 13.5。
Step 3: Range = highest – lowest = 22 – 8 = 14. The player's scores vary by 14 points.
步骤3：极差 = 最高 – 最低 = 22 – 8 = 14。该球员的得分波动为14分。
These simple statistics give a summary of consistency and typical performance.
这些简单的统计量总结了球员发挥的稳定性和典型表现。
7. Recipe Ratios | 食谱比例
Cooking often uses proportional reasoning. If a recipe is for 6 people and you need to serve 9, you must scale all ingredients by the same factor. This is direct proportion.
烹饪经常使用比例推理。如果一份食谱是供6人份，而你需要提供9人份，你必须将所有配料按相同倍数调整。这就是正比例。
A pancake recipe for 4 people uses 200 g flour, 2 eggs and 300 ml milk. How much of each ingredient is needed for 10 people?
一份供4人份的煎饼食谱使用200克面粉、2个鸡蛋和300毫升牛奶。为10人份需要每种配料各多少？
Step 1: Find the multiplier: 10 ÷ 4 = 2.5. All ingredients are multiplied by 2.5.
步骤1：找出倍数：10 ÷ 4 = 2.5。所有配料都乘以2.5。
Step 2: Flour: 200 g × 2.5 = 500 g.
步骤2：面粉：200 g × 2.5 = 500 g。
Step 3: Eggs: 2 × 2.5 = 5 eggs.
步骤3：鸡蛋：2 × 2.5 = 5个蛋。
Step 4: Milk: 300 ml × 2.5 = 750 ml.
步骤4：牛奶：300 ml × 2.5 = 750 ml。
io of eggs to flour is 2 : 200 = 1 : 100 for 4 people, and 5 : 500 = 1 : 100 for 10 people – the same.
务必检查比例是否保持不变；例如，鸡蛋与面粉的比例在4人份时为2 : 200 = 1 : 100，在10人份时为5 : 500 = 1 : 100 —— 保持一致。
8. Discounts and Best Buys | 折扣与最佳购买
Understanding percentages helps you make smart decisions when shopping. Discounts are often given as a percentage off, and 'buy one get one half price' deals require careful calculation.
理解百分比能帮助你在购物时做出明智的决定。折扣通常以百分比降价的形式出现，而'买一第二件半价'的促销则需要仔细计算。
A jacket originally costs £40. In a sale, there is a 15% discount. What is the sale price? Another shop sells the same jacket for £45 with a 20% discount, which offer is cheaper?
一件夹克原价40英镑。在促销中打15%的折扣。销售价是多少？另一家店以45英镑的价格出售同款夹克，但打20%折扣，哪一家更便宜？
Step 1: Calculate 15% of £40: 15/100 × 40 = £6. Discounted price = £40 – £6 = £34.
步骤1：计算40英镑的15%：15/100 × 40 = 6英镑。折后价 = 40英镑 – 6英镑 = 34英镑。
Step 2: For the second shop, 20% of £45 = 20/100 × 45 = £9. Sale price = £45 – £9 = £36.
步骤2：第二家店，45英镑的20% = 20/100 × 45 = 9英镑。销售价 = 45英镑 – 9英镑 = 36英镑。
Step 3: Compare: £34 is less than £36, so the first shop offers the better deal.
步骤3：比较：34英镑低于36英镑，所以第一家店的价格更划算。
You can also use the multiplier method: 100% – 15% = 85%, so 0.85 × £40 = £34, which is faster.
你也可以使用倍数法：100% – 15% = 85%，因此0.85 × 40英镑 = 34英镑，速度更快。
9. Speed, Distance and Time | 速度、距离与时间
In Physics, the relationship between speed, distance and time is fundamental. You can use the formula in many travel problems. Remember to use consistent units: metres per second or kilometres per hour.
在物理学中，速度、距离和时间之间的关系是基础。你可以在许多行程问题中使用该公式。记得使用一致的单位：米每秒或千米每小时。
speed = distance ÷ time
速度 = 距离 ÷ 时间
A cyclist travels 45 km in 2 hours and 30 minutes. What is the average speed in km/h? If they increased speed by 3 km/h, how long would the same journey take?
一名骑行者在2小时30分钟内骑行了45千米。平均速度是多少（千米/小时）？如果速度提高3千米/小时，同样的路程需要多长时间？
Step 1: Convert 2 hours 30 min to hours: 30 min = 0.5 hours, so time = 2.5 hours.
步骤1：将2小时30分钟转换为小时：30分钟 = 0.5小时，因此时间 = 2.5小时。
Step 2: Average speed = 45 ÷ 2.5 = 18 km/h.
步骤2：平均速度 = 45 ÷ 2.5 = 18千米/小时。
speed = 45 ÷ 21 ≈ 2.143 hours. 0.143 hours × 60 ≈ 8.6 minutes, so about 2 hours 9 minutes.
步骤3：新速度 = 18 + 3 = 21千米/小时。新时间 = 距离 ÷ 速度 = 45 ÷ 21 ≈ 2.143小时。0.143小时 × 60 ≈ 8.6分钟，因此约为2小时9分钟。
Interdisciplinary questions like this show how algebra and decimals are essential in real-world science.
像这样的跨学科问题展示了代数和十进制在现实科学中是多么重要。
10. Problem-Solving Strategy Summary | 解题策略总结
To tackle any interdisciplinary problem, first identify which branch of maths is being tested: is it number, proportion, geometry or statistics? Then read the context carefully and extract the numbers and units.
解决任何跨学科问题，首先要确定考查的是数学的哪个分支：是数字、比例、几何还是统计？然后仔细阅读情境，提取数字和单位。
Always show your working step by step. Check that your answer makes sense in the real world; for example, a speed of 500 km/h for a bicycle is impossible. Practice by mixing up subjects so that maths feels like a tool for discovery, not just a set of rules.
始终按步骤展示你的演算过程。检查你的答案在现实世界中是否合理；例如，自行车的速度达到500千米/小时是不可能的。通过混合不同学科来多加练习，让数学成为探索的工具，而不仅仅是一套规则。
Finally, remember the WJEC focus on reasoning and applying skills. When you next read a science article, look at a map or cook a meal, ask yourself: what maths is hiding here?
最后，记住WJEC注重推理和应用技能。下次当你阅读科学文章、查看地图或做饭时，问问自己：这里潜藏着什么数学？
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