📚 Cross-Curricular Problem Solving in Year 8 SQA Maths | 跨学科综合题型训练
Mathematics is not just about numbers and formulas—it is a powerful tool for understanding the world around us. Year 8 SQA Maths encourages you to apply your skills across different subjects, from science and geography to economics and technology. This article will guide you through cross-curricular problem-solving, showing how mathematical thinking can unlock answers in real-life contexts.
数学不仅仅是数字和公式——它是理解我们周围世界的有力工具。SQA八年级数学鼓励你将技能应用于不同学科，从科学和地理到经济和技术。本文将引导你进行跨学科问题解决，展示数学思维如何在现实情境中揭示答案。
1. What are Cross-Curricular Problems? | 什么是跨学科问题？
Cross-curricular problems blend mathematics with other subjects such as science, geography, technology or everyday life. They help you see how maths is used in real situations. For example, calculating the amount of fertiliser needed for a field involves both area measurement (maths) and biology (plant needs).
跨学科问题将数学与其他学科如科学、地理、技术或日常生活相结合。它们帮助你看到数学如何在实际情况中使用。例如，计算田地所需的肥料量就涉及面积测量（数学）和生物学（植物需求）。
In SQA Year 8 mathematics, you will encounter tasks that require you to interpret data from a science experiment, work with map scales in geography, or manage a budget in a business context. These tasks not only build your mathematical fluency but also improve your problem-solving and reasoning skills.
在SQA八年级数学中，你会遇到需要解读科学实验数据、处理地理中的地图比例尺，或在商业情境中管理预算的问题。这些任务不仅能增强你的数学流畅性，还能提高你解决问题和推理的能力。
2. Ratios and Proportions in Science Experiments | 科学实验中的比与比例
Ratios appear frequently in science, for instance when mixing chemical solutions or scaling up a recipe for a laboratory reaction. Understanding how to use ratios helps you prepare the correct concentrations.
比率在科学中经常出现，例如在混合化学溶液或放大实验室配方时。理解如何使用比率有助于你配制出正确的浓度。
Example 1: A plant feed is made by mixing concentrate with water in the ratio 1 : 4. This means for every 1 part concentrate, you need 4 parts water. To make 2 litres of feed, how much concentrate and water are needed?
Total parts = 1 + 4 = 5. Concentrate = 1/5 of 2 L = 0.4 L = 400 ml. Water = 4/5 of 2 L = 1.6 L.
例1：一种植物营养液按浓缩液与水的比 1:4 配制。要制作2升营养液，需要多少浓缩液和水？总份数=1+4=5。浓缩液=2升的1/5=0.4升=400毫升。水=2升的4/5=1.6升。
Example 2: A microscope magnifies an object in the ratio 10 : 1. If the image appears 5 cm long, what is the actual length of the object? Since the magnification is 10:1, actual length = 5 cm / 10 = 0.5 cm = 5 mm.
例2：显微镜的放大比率为10:1。如果图像长5厘米，物体的实际长度是多少？放大倍数为10:1，实际长度=5厘米÷10=0.5厘米=5毫米。
3. Area and Perimeter in Design and Geography | 设计与地理中的面积和周长
Calculating area and perimeter is essential when planning a room layout or using map scales in geography. You often need to convert between units and apply scale factors.
计算面积和周长在规划房间布局或在地理中使用地图比例尺时至关重要。你经常需要转换单位并应用比例因子。
Example 1 (Flooring): A rectangular room is 5 m long and 3.5 m wide. You want to cover the floor with square tiles of side 50 cm. How many tiles are needed?
Area of room = 5 × 3.5 = 17.5 m². Area of one tile = 0.5 m × 0.5 m = 0.25 m². Number of tiles = 17.5 ÷ 0.25 = 70 tiles.
例1（地面铺设）：一间长方形房间长5米、宽3.5米。用边长50厘米的正方形地砖铺地，需要多少块地砖？房间面积=5×3.5=17.5平方米。一块地砖面积=0.5×0.5=0.25平方米。地砖数=17.5÷0.25=70块。
Example 2 (Map scale): A rectangular park is drawn on a map using a scale of 1 : 500. Its dimensions on the map are 6 cm by 4 cm. What is the actual area of the park in square metres?
Actual length = 6 cm × 500 = 3000 cm = 30 m; actual width = 4 cm × 500 = 2000 cm = 20 m. Area = 30 × 20 = 600 m².
例2（地图比例尺）：一个长方形公园在比例尺为1:500的地图上，尺寸为6厘米乘4厘米。公园的实际面积是多少平方米？实际长=6×500=3000厘米=30米；实际宽=4×500=2000厘米=20米。面积=30×20=600平方米。
4. Statistics and Data Analysis in Health and Environmental Science | 健康与环境科学中的统计与数据分析
Statistics help us understand trends in health data or environmental surveys. You need to calculate averages (mean, median, mode) and present data clearly using charts.
统计学帮助我们理解健康数据或环境调查的趋势。你需要计算平均数（均值、中位数、众数），并使用图表清晰地展示数据。
Scenario: A nurse records resting heart rates (beats per minute) of 10 students: 72, 68, 75, 70, 72, 74, 71, 69, 73, 72. Let’s analyse this data.
场景：一位护士记录了10名学生的静息心率（次/分）：72, 68, 75, 70, 72, 74, 71, 69, 73, 72。我们来分析这组数据。
Mean = (72+68+75+70+72+74+71+69+73+72) / 10 = 716 / 10 = 71.6 beats per minute. Median: ordered set is 68, 69, 70, 71, 72, 72, 72, 73, 74, 75. Middle two are 72 and 72, so median = 72. Mode = 72 (appears three times). The data is fairly symmetric around 72, indicating a typical resting heart rate.
均值 = (72+68+75+70+72+74+71+69+73+72) ÷ 10 = 716 ÷ 10 = 71.6次/分。中位数：排序后为68, 69, 70, 71, 72, 72, 72, 73, 74, 75。中间两个数是72和72，所以中位数=72。众数=72（出现三次）。数据围绕72大致对称，表明典型的静息心率。
To display this data, a bar chart or dot plot could be used. For environmental science, you might count different tree species in a woodland and present a pie chart showing the proportion of each species.
要展示这些数据，可以使用柱状图或点图。在环境科学中，你可能统计林地中不同树种的数量，并用饼图显示每种树的比例。
5. Linear Equations and Chemical Mixtures | 线性方程与化学混合物
In chemistry, you often need to mix two solutions of different concentrations to obtain a desired concentration. Setting up a linear equation makes this straightforward.
在化学中，你经常需要混合两种不同浓度的溶液来获得所需浓度。建立线性方程可使计算变得简单。
Problem: You have a 10% salt solution and a 25% salt solution. How much of each should be mixed to produce 200 ml of a 15% salt solution?
Let x be the volume (in ml) of the 10% solution. Then the volume of the 25% solution is (200 − x) ml. The total mass of salt in the mixture must equal the desired concentration times total volume:
0.10x + 0.25(200 − x) = 0.15 × 200
问题：你有10%的盐水溶液和25%的盐水溶液。应该各取多少毫升才能混合成200毫升15%的盐水溶液？
设x为10%溶液的体积（毫升），则25%溶液的体积为(200−x)毫升。混合物中的总盐量必须等于所需浓度乘以总体积：
0.10x + 0.25(200 − x) = 0.15 × 200
Solving: 0.10x + 50 − 0.25x = 30 → −0.15x = −20 → x = 400/3 ≈ 133.3 ml of the 10% solution. Therefore, 200 − 400/3 = 200/3 ≈ 66.7 ml of the 25% solution. Mixing these volumes yields the required concentration.
求解：0.10x + 50 − 0.25x = 30 → −0.15x = −20 → x = 400/3 ≈ 133.3 毫升的10%溶液。因此，25%溶液取200 − 400/3 = 200/3 ≈ 66.7 毫升。将这些体积混合即得到所需浓度。
6. Currency Conversion and Travel Planning | 货币兑换与旅行规划
When travelling abroad, you need to convert currencies and possibly account for bank fees. This uses proportional reasoning and percentages.
出国旅行时，你需要兑换货币，可能还要考虑银行手续费。这需要运用比例推理和百分比。
Example: The exchange rate is GBP 1 = EUR 1.15. You plan to spend 345 euros on accommodation. How many pounds do you need to exchange? Additionally, the bank charges a 2% commission on the sterling amount. What is the total cost in pounds?
Amount in pounds without fee = 345 ÷ 1.15 = 300 GBP. Commission = 2% of 300 = 0.02 × 300 = 6 GBP. Total cost = 300 + 6 = 306 GBP.
例子：汇率为1英镑=1.15欧元。你计划花费345欧元住宿。需要兑换多少英镑？另外，银行收取兑换金额2%的手续费。总费用是多少英镑？
无手续费时的英镑金额 = 345 ÷ 1.15 = 300 英镑。手续费 = 300的2% = 0.02 × 300 = 6 英镑。总计 = 300 + 6 = 306 英镑。
Always check whether the exchange rate given is for buying or selling currency, and think carefully about whether the fee is applied to the sterling or the foreign amount.
务必确认给出的汇率是买入价还是卖出价，并仔细思考手续费是按英镑金额还是外币金额计算。
7. Speed, Distance, Time in Physics | 速度、距离、时间在物理中的应用
The relationship between speed, distance and time is fundamental in physics. The formula is:
Speed = Distance ÷ Time
速度、距离和时间之间的关系是物理学的基础。公式为：
速度 = 距离 ÷ 时间
Example 1: A car travels at a constant speed of 60 km/h for 2.5 hours. How far does it travel?
Distance = Speed × Time = 60 × 2.5 = 150 km.
例1：一辆汽车以60 km/h的恒定速度行驶2.5小时。它行驶了多远？距离=速度×时间=60×2.5=150公里。
Example 2: A cyclist covers 45 km in 3 hours and 20 minutes. What was her average speed in km/h?
Time in hours = 3 + (20 ÷ 60) = 3.333… hours. Average speed = 45 ÷ (3 + 1/3) = 45 × (3/10) = 13.5 km/h.
例2：一位自行车手在3小时20分钟内骑行了45公里。她的平均速度是多少km/h？时间以小时计=3+(20÷60)=3.333…小时。平均速度=45÷(3+1/3)=45×(3/10)=13.5 km/h。
Always check that units are consistent. Convert minutes to hours or metres per second to kilometres per hour as needed.
务必检查单位是否一致。根据需要将分钟转换为小时，或将米/秒转换为千米/小时。
8. Fractions, Decimals and Percentages in Nutrition | 营养学中的分数、小数和百分比
Food labels give nutritional information per 100 g or per serving. You can use fractions and percentages to calculate the amount of nutrients you consume.
食品标签提供每100克或每份的营养信息。你可以利用分数和百分比来计算摄入的营养素量。
Example: A cereal bar label states: per 100 g – protein 12 g, fat 18 g, carbohydrate 60 g. You eat a 150 g portion. Work out the mass of each nutrient consumed and the percentage composition by mass.
Protein in portion = 12 × (150/100) = 18 g. Fat = 18 × 1.5 = 27 g. Carbohydrate = 60 × 1.5 = 90 g. Total mass = 18 + 27 + 90 = 135 g (the rest is fibre, water, etc). Percentages in the 150 g portion: protein = (18/150)×100 = 12%, fat = (27/150)×100 = 18%, carbohydrate = (90/150)×100 = 60%. The percentages match the original 100 g label because the proportions remain constant.
例子：一块谷物棒标签上写着：每100克——蛋白质12克，脂肪18克，碳水化合物60克。你吃了一块150克。计算每种营养素的摄入质量以及质量百分比组成。
摄入蛋白质=12×(150/100)=18克。脂肪=18×1.5=27克。碳水化合物=60×1.5=90克。总质量=18+27+90=135克（其余为纤维、水等）。在150克中百分比：蛋白质=(18/150)×100=12%，脂肪=27/150×100=18%，碳水化合物=90/150×100=60%。这些百分比与原来的100克标签相同，因为比例保持不变。
This skill is useful when planning meals to meet dietary targets, such as consuming a certain percentage of energy from protein.
这项技能在制定符合膳食目标（如从蛋白质摄入一定比例能量）的饮食计划时非常有用。
9. Interpreting Graphs in Biology | 解读生物学中的图表
Biologists often record how a variable changes over time or under different conditions. Reading a line graph and describing the trend is a key skill.
生物学家经常记录一个变量如何随时间或在不同条件下变化。阅读折线图并描述趋势是一项关键技能。
Data: An experiment measures enzyme activity at different temperatures. The results are: 10 °C – 5 units, 20 °C – 20 units, 30 °C – 55 units, 40 °C – 80 units, 50 °C – 60 units, 60 °C – 10 units.
From a plotted line graph, you can see that activity increases steadily from 10 °C to 40 °C, where it reaches a maximum. Then it drops sharply as the temperature rises further, because the enzyme denatures at around 45 °C. The optimum temperature is approximately 40 °C.
数据：一个实验测量不同温度下的酶活性。结果：10°C—5单位，20°C—20单位，30°C—55单位，40°C—80单位，50°C—60单位，60°C—10单位。
从绘制的折线图中可以看出，活性从10°C到40°C稳步上升，在40°C达到最大值。然后随着温度进一步升高而急剧下降，因为酶在约45°C时变性。最适温度约为40°C。
When interpreting graphs, use precise mathematical language: mention whether the graph is linear or curved, identify maximum or minimum points, and state the rate of change if possible.
解读图表时，使用精确的数学语言：指出图是线性的还是曲线，找出最大值或最小值点，并尽可能说明变化率。
10. Logic Puzzles and Computational Thinking | 逻辑谜题与计算思维
Logic puzzles develop reasoning skills that are essential in computer science and mathematics. They often involve solving for unknowns using clues.
逻辑谜题培养在计算机科学和数学中必不可少的推理能力。它们通常涉及利用线索求解未知数。
Puzzle 1: I am a two-digit number. The sum of my digits is 12. The units digit is twice the tens digit. What number am I?
Let the tens digit be x, so the units digit is 2x. Sum: x + 2x = 12 ⇒ 3x = 12 ⇒ x = 4. The number is 48.
谜题1：我是一个两位数，各位数字之和是12，个位数字是十位数字的2倍。我是哪个数？
设十位数字为x，则个位数字为2x。和：x+2x=12 ⇒ 3x=12 ⇒ x=4。这个数是48。
Puzzle 2: If you reverse the digits of a two-digit number, the new number is 27 greater than the original. The sum of its digits is 9. Find the original number.
Let tens
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