📚 Year 7 SQA Maths: Interdisciplinary Mixed Question Training | 七年级SQA数学：跨学科综合题型训练
In the SQA curriculum, Year 7 mathematics is not just about numbers on a page — it is about using maths to solve problems in science, geography, design, sport and everyday life. This article offers a cross-curricular training workout, where each section blends a different subject with a core maths skill. You will practise unit conversions, scale drawings, data analysis, algebraic patterns, area calculations, fractions, percentages, coordinates, time and statistics. By working through these mixed question scenarios, you will strengthen both your maths fluency and your ability to apply it across the curriculum.
在SQA课程中，七年级数学不仅仅是纸上的数字——它是运用数学去解决科学、地理、设计、体育以及日常生活中的问题。本文提供一份跨学科训练合集，每个小节都将一个不同的科目与核心数学技能融合在一起。你将练习单位换算、比例图、数据分析、代数模式、面积计算、分数、百分比、坐标、时间和统计。通过这些混合题型的情景训练，你既能提升数学运算的流畅度，也能增强跨学科应用能力。
1. Number & Science: Converting Units in Real Experiments | 数与科学：真实实验中的单位换算
In a science experiment, you measure the growth of a bean plant over five days. The daily increases in height are recorded in millimetres: 12 mm, 15 mm, 9 mm, 18 mm and 14 mm. To compare growth rates in a class display, you need to convert these values to centimetres and find the mean growth per day.
在一个科学实验中，你测量了一棵豆苗在五天内的生长情况。每天的高度增长以毫米记录：12 mm、15 mm、9 mm、18 mm 和 14 mm。为了在班级展示中比较生长速度，你需要将这些数值换算成厘米，并计算每日平均生长量。
Step 1: Convert mm to cm. Since 1 cm = 10 mm, divide each measurement by 10. The converted values are 1.2 cm, 1.5 cm, 0.9 cm, 1.8 cm and 1.4 cm.
第1步：将毫米转换为厘米。因为 1 cm = 10 mm，所以将每个测量值除以10。换算后的数值为 1.2 cm、1.5 cm、0.9 cm、1.8 cm 和 1.4 cm。
Step 2: Calculate the mean. Add all the centimetre values: 1.2 + 1.5 + 0.9 + 1.8 + 1.4 = 6.8 cm. Then divide by 5. The mean daily growth is 1.36 cm.
第2步：计算平均数。将所有厘米值相加：1.2 + 1.5 + 0.9 + 1.8 + 1.4 = 6.8 cm。然后除以5。每日平均生长量为 1.36 cm。
Using correct units helps scientists compare results accurately. You also practised decimal addition and division, which are key number skills for Year 7.
使用正确的单位能帮助科学家准确地比较结果。你还练习了小数的加法和除法，这是七年级关键的数字技能。
2. Shape & Geography: Map Scales and Direction | 形状与地理：地图比例尺与方向
You are using a map with a scale of 1 : 25 000 to plan a hike. On the map, the distance from the car park to the viewpoint is 8.4 cm. You need to find the actual distance in kilometres. The path also requires you to turn at a bearing of 045° at a junction, which means turning through an angle of 45° clockwise from north.
你正在使用一张比例尺为 1 : 25 000 的地图来规划一次远足。地图上，从停车场到观景点的距离是 8.4 cm。你需要找出以千米为单位的实际距离。这条路线还要求你在一个岔路口以 045° 的方位角转弯，这意味着从正北方向顺时针转动 45° 角。
Scale calculation: 1 cm on the map = 25 000 cm in reality. So 8.4 cm represents 8.4 × 25 000 = 210 000 cm. Convert to metres: 210 000 cm = 2 100 m, which is 2.1 km.
比例尺计算：地图上 1 cm = 实际 25 000 cm。所以 8.4 cm 代表 8.4 × 25 000 = 210 000 cm。换算成米：210 000 cm = 2 100 m，即 2.1 km。
Bearings are measured as three-digit angles from north. A bearing of 045° means you rotate eastwards by 45°. Recognising angles and measuring turns connects geometry to real-world navigation.
方位角是从正北方向用三位数角度来测量的。045° 的方位角意味着向东旋转 45°。认识角度和度量旋转将几何学与现实世界的导航联系起来。
3. Data & Physical Education: Analysing Race Times | 数据与体育：分析赛跑时间
During athletics, your class runs 100 m and records the times in seconds: 15.2, 14.8, 16.1, 15.5, 14.8, 15.9, 16.4, 15.2. You are asked to find the range, the mode and the median of these times, and then display the distribution on a bar chart.
在田径课上，全班进行了 100 米跑并记录了以秒为单位的时间：15.2、14.8、16.1、15.5、14.8、15.9、16.4、15.2。你需要找出这些时间的极差、众数和中位数，然后用条形图展示分布情况。
Sort the data: 14.8, 14.8, 15.2, 15.2, 15.5, 15.9, 16.1, 16.4. Range = highest − lowest = 16.4 − 14.8 = 1.6 s. Mode = 14.8 and 15.2 (bimodal). Median: with 8 values, average the 4th and 5th: (15.2 + 15.5) ÷ 2 = 15.35 s.
将数据排序：14.8, 14.8, 15.2, 15.2, 15.5, 15.9, 16.1, 16.4。极差 = 最大值 − 最小值 = 16.4 − 14.8 = 1.6 秒。众数：14.8 和 15.2（双众数）。中位数：共8个值，取第4和第5个的平均值：(15.2 + 15.5) ÷ 2 = 15.35 秒。
For the bar chart, group times into intervals, e.g., 14.5−15.0 s, 15.1−15.5 s, etc., and count frequencies. This PE data task sharpens your skills in calculating averages and presenting data clearly.
对于条形图，将时间分组，例如 14.5−15.0 秒、15.1−15.5 秒等，并计数。这项体育数据任务能提升你计算平均数以及清晰地展示数据的技能。
4. Algebra & Music: Understanding Beats and Patterns | 代数与音乐：理解节拍与模式
A piece of music in 4/4 time has four beats per bar. In a rhythmic pattern, the first beat is a crotchet (1 beat), the second beat is two quavers (½ + ½ beat), and this pattern repeats for the remaining two beats. Write an algebraic expression for the total number of notes, n, in b bars, assuming the pattern continues.
一首 4/4 拍的音乐每小节有四拍。在一个节奏模式中，第一拍是一个四分音符（1拍），第二拍是两个八分音符（½ + ½ 拍），其余两拍重复这个模式。假设此模式持续，写出 b 个小节中音符总数 n 的代数表达式。
In one bar: beat 1 has 1 note, beat 2 has 2 notes, beat 3 repeats beat 1 with 1 note, beat 4 repeats beat 2 with 2 notes. So notes per bar = 1 + 2 + 1 + 2 = 6 notes. Therefore n = 6b. If b = 4, n = 24 notes.
在一小节中：第一拍有 1 个音符，第二拍有 2 个音符，第三拍重复第一拍有 1 个音符，第四拍重复第二拍有 2 个音符。所以每小节音符数 = 1 + 2 + 1 + 2 = 6 个音符。因此 n = 6b。若 b = 4，n = 24 个音符。
This shows how algebra can model repeated patterns in music, linking sequences and formulas to rhythmic structures.
这表明代数可以对音乐中的重复模式进行建模，将序列与公式同节奏结构联系起来。
5. Measurement & Design: Area and Perimeter of a Room | 测量与设计：房间的面积与周长
You want to redesign your bedroom. The floor is a rectangle 4.2 m long and 3.5 m wide. You plan to fit a new carpet and add a wallpaper border along the top of all four walls, excluding a door 0.9 m wide. Carpet costs £12.50 per m², and the border costs £1.80 per metre. How much will the materials cost?
你想要重新设计自己的卧室。地面是一个长 4.2 m、宽 3.5 m 的长方形。你打算铺新地毯，并在四面墙壁的顶部贴一圈腰线，但需要除去一扇 0.9 m 宽的门。地毯每平方米 £12.50，腰线每米 £1.80。材料总共需要多少钱？
Area of floor = 4.2 × 3.5 = 14.7 m². Carpet cost = 14.7 × 12.50 = £183.75.
地面面积 = 4.2 × 3.5 = 14.7 m²。地毯费用 = 14.7 × 12.50 = £183.75。
Perimeter of room = 2 × (4.2 + 3.5) = 2 × 7.7 = 15.4 m. Border length = 15.4 − 0.9 = 14.5 m. Border cost = 14.5 × 1.80 = £26.10. Total = £183.75 + £26.10 = £209.85.
房间周长 = 2 × (4.2 + 3.5) = 2 × 7.7 = 15.4 m。腰线长度 = 15.4 − 0.9 = 14.5 m。腰线费用 = 14.5 × 1.80 = £26.10。总计 = £183.75 + £26.10 = £209.85。
Working with area and perimeter helps you plan real design projects and handle money calculations accurately.
计算面积和周长能帮你规划真实的设计项目，并准确地进行金钱计算。
6. Fractions & Cooking: Scaling Recipes | 分数与烹饪：调整食谱比例
A recipe for 6 people requires ¾ cup of sugar, 2½ cups of flour and ⅓ cup of oil. You need to make the recipe for only 4 people. By what fraction do you multiply each ingredient? Calculate the new amounts.
一份为 6 人准备的食谱需要 ¾ 杯糖、2½ 杯面粉和 ⅓ 杯油。你只需要制作 4 人份。每种原料的用量应乘以什么分数？计算出新的用量。
Scale factor = 4 ÷ 6 = ⅔. So multiply each amount by ⅔. Sugar: ¾ × ⅔ = (3×2)/(4×3) = 6/12 = ½ cup. Flour: 2½ = 5/2, so 5/2 × ⅔ = 10/6 = 1⅔ cups (or 1 2/3). Oil: ⅓ × ⅔ = 2/9 cup.
比例因子 = 4 ÷ 6 = ⅔。因此每种用量乘以 ⅔。糖：¾ × ⅔ = (3×2)/(4×3) = 6/12 = ½ 杯。面粉：2½ = 5/2，所以 5/2 × ⅔ = 10/6 = 1⅔ 杯。油：⅓ × ⅔ = 2/9 杯。
Using fractions in cooking demonstrates proportional reasoning. You also convert mixed numbers to improper fractions to simplify multiplication.
在烹饪中使用分数展示了比例推理。你也将带分数转化为假分数来简化乘法运算。
7. Percentages & Money: Discounts and Saving | 百分比与金钱：折扣与储蓄
Your favourite game costs £32. There is a 15% discount this week. You also have a savings account that earns 2% simple interest per year. If you buy the game and invest the money you save, how much interest would you earn in one year?
你喜欢的游戏售价 £32。本周有 15% 的折扣。你还有一个年利率为 2% 单利的储蓄账户。如果你买了游戏，并把省下的钱存入账户，一年能获得多少利息？
Discount amount = 15% of £32 = 0.15 × 32 = £4.80. Sale price = £32 − £4.80 = £27.20. Money saved = £4.80. Simple interest for one year = 2% of £4.80 = 0.02 × 4.80 = £0.096, rounded to £0.10.
折扣金额 = £32 的 15% = 0.15 × 32 = £4.80。折扣后价格 = £32 − £4.80 = £27.20。节省下的钱 = £4.80。一年的单利 = £4.80 的 2% = 0.02 × 4.80 = £0.096，四舍五入为 £0.10。
Percentages connect shopping and banking. You practised finding percentages of amounts and simple interest, a key personal finance skill.
百分比将购物和银行业务联系起来。你练习了求某个数的百分比和单利，这是一项重要的个人理财技能。
8. Coordinates & Computing: Plotting Pixels | 坐标与计算机：绘制像素图
In a simple graphics program, you can draw a shape by giving coordinates. To draw a right-angled triangle, you mark points A (2, 3), B (2, 7) and C (6, 7). Plot these points on a grid and join them. What are the lengths of the horizontal and vertical sides? Name the type of triangle by its angles.
在一个简单的图形程序中，你可以通过给出坐标来画一个图形。要画一个直角三角形，你标记点 A (2, 3)、B (2, 7) 和 C (6, 7)。在网格上标出这些点并连接起来。水平边和垂直边的长度各是多少？根据角度命名这个三角形的类型。
Plotting: A and B share the same x-coordinate, so AB is vertical. Its length = 7 − 3 = 4 units. B and C share the same y-coordinate, so BC is horizontal. Length = 6 − 2 = 4 units. The triangle has a right angle at B (where the vertical and horizontal meet). It is an isosceles right-angled triangle.
绘图：A 和 B 具有相同的 x 坐标，所以 AB 是垂直的，长度 = 7 − 3 = 4 个单位。B 和 C 具有相同的 y 坐标，所以 BC 是水平的，长度 = 6 − 2 = 4 个单位。这个三角形在 B 点有一个直角（垂直线与水平线相交处）。它是一个等腰直角三角形。
Working with coordinates builds a foundation for coding and digital design, reinforcing your ability to read and interpret grid positions.
使用坐标能够为编程和数字设计打下基础，增强你阅读和理解网格位置的能力。
9. Time & History: Roman Numerals and Timelines | 时间与历史：罗马数字与时间轴
A timeline shows that a castle was built in the year MCMLIV. What year is this in our number system? If preservation work began in 2015 and took 7 years to complete, in which year did it finish, and how old was the castle then?
一条时间轴显示，一座城堡建于 MCMLIV 年。用我们的数字系统表示这是哪一年？如果保护工程始于 2015 年，并花了 7 年完成，那么它是在哪一年竣工的？当时这座城堡已有多少年历史？
Roman numeral conversion: M = 1000, CM = 900, L = 50, IV = 4. So MCMLIV = 1000 + 900 + 50 + 4 = 1954. Work finished in 2015 + 7 = 2022. Age of castle = 2022 − 1954 = 68 years.
罗马数字转换：M = 1000，CM = 900，L = 50，IV = 4。所以 MCMLIV = 1000 + 900 + 50 + 4 = 1954 年。工程于 2015 + 7 = 2022 年竣工。城堡的年龄 = 2022 − 1954 = 68 年。
This history-themed task revises Roman numerals and time interval calculations, showing how maths helps us understand historical records.
这项以历史为主题的任务复习了罗马数字和时间间隔的计算，展示了数学如何帮助我们理解历史记录。
10. Statistics & Science: Weather Data | 统计与科学：气象数据
You collect daily maximum temperatures for a week in Glasgow: Monday 8 °C, Tuesday 10 °C, Wednesday 9 °C, Thursday 11 °C, Friday 12 °C, Saturday 14 °C, Sunday 13 °C. Draw a line graph to show the trend. Calculate the mean temperature and state the range.
你收集了格拉斯哥一周的每日最高气温：星期一 8 °C，星期二 10 °C，星期三 9 °C，星期四 11 °C，星期五 12 °C，星期六 14 °C，星期日 13 °C。画一张折线图来显示变化趋势。计算平均气温并说出极差。
Line graph: label days on the x-axis and temperature on the y-axis, plotting points and connecting them with lines. This visualises the upward trend towards the weekend.
折线图：x 轴标注星期几，y 轴标注气温，标出各点并用线连接。这将周末之前气温上升的趋势形象化。
Mean = (8 + 10 + 9 + 11 + 12 + 14 + 13) ÷ 7 = 77 ÷ 7 = 11 °C. Range = 14 − 8 = 6 °C. These statistics summarise the dataset and are vital in scientific reporting.
平均数 = (8 + 10 + 9 + 11 + 12 + 14 + 13) ÷ 7 = 77 ÷ 7 = 11 °C。极差 = 14 − 8 = 6 °C。这些统计数据概括了整个数据集，在科学报告中至关重要。
Published by TutorHao | Maths Revision Series | aleveler.com
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