📚 Interdisciplinary Problem-Solving in Year 7 OCR Maths | 跨学科综合题型训练
In Year 7 OCR Mathematics, you not only learn core skills like fractions, percentages, and simple algebra, but also apply them to real-world problems from science, geography, art, and more. This interdisciplinary approach helps you see why maths matters and how it connects different subjects. This article presents a series of cross-curricular problem scenarios, each followed by step-by-step solutions using Year 7 techniques. By working through these examples, you will strengthen your problem-solving and boost your confidence in tackling unfamiliar questions.
在七年级 OCR 数学中，你不仅要学习分数、百分数和简单代数等核心技能，还要将它们应用于科学、地理、艺术等现实世界的问题。这种跨学科的方法让你看到数学的重要性以及它如何连接不同学科。本文提供一系列跨学科问题情境，每个问题都配有使用七年级技巧的逐步解答。通过练习这些例子，你将增强解决问题的能力，并提升应对陌生题目的信心。
1. Science: Converting Units and Calculating Speed | 科学：单位换算与速度计算
A cyclist completes a 36 km ride in 1.5 hours. Work out the average speed in metres per second (m/s). There are 1000 m in 1 km and 3600 seconds in 1 hour.
一名自行车手用 1.5 小时完成了 36 km 的骑行。试以米每秒 (m/s) 为单位计算平均速度。已知 1 km = 1000 m，1 小时 = 3600 秒。
First change the distance to metres: 36 km = 36 × 1000 = 36 000 m.
首先将距离换算为米：36 km = 36 × 1000 = 36 000 m。
Next change the time to seconds: 1.5 hours = 1.5 × 3600 = 5400 s.
接着将时间换算为秒：1.5 小时 = 1.5 × 3600 = 5400 s。
66... Continue dividing: 6.666... m/s, or exactly 20/3 m/s ≈ 6.67 m/s to 2 decimal places.
现在使用公式 速度 = 距离 ÷ 时间：36 000 ÷ 5400。通过消去零来化简：360 ÷ 54 = 6.666... 继续除：6.666... m/s，或精确为 20/3 m/s ≈ 6.67 m/s（保留两位小数）。
2. Geography: Map Scales and Real Distances | 地理：地图比例尺与实际距离
A walking map uses a scale of 1:25 000. The length of a footpath on the map is 12 cm. Calculate the real length of the path in kilometres.
一张远足地图采用 1:25 000 的比例尺。地图上一条步道的长度为 12 cm。计算这条步道的实际长度，以千米为单位。
he map represents 25 000 cm in reality. So real distance in cm = 12 × 25 000 = 300 000 cm.
地图比例尺表示地图上的 1 cm 代表实际中的 25 000 cm。因此实际距离（厘米）= 12 × 25 000 = 300 000 cm。
ert centimetres to kilometres: 100 cm = 1 m, 1000 m = 1 km. So 300 000 cm = 3000 m = 3 km.
将厘米转换为千米：100 cm = 1 m，1000 m = 1 km。因此 300 000 cm = 3000 m = 3 km。
12 cm = 300 000 cm, then divide by 100 000 (since 1 km = 100 000 cm) to get 3 km directly.
你还可以使用比例因子：25 000 × 12 cm = 300 000 cm，然后除以 100 000（因为 1 km = 100 000 cm）直接得到 3 km。
3. Art: Symmetry and Tessellation Patterns | 艺术：对称与镶嵌图案
A regular octagon is used in Islamic art to create patterns. Find the size of one interior angle of a regular octagon. Determine whether regular octagons can tessellate on their own. (Hint: tessellation requires interior angles that divide exactly into 360°.)
伊斯兰艺术中使用正八边形来创建图案。求出正八边形的一个内角大小。判断正八边形能否独自镶嵌。（提示：镶嵌要求内角能整除 360°。）
) × 180° ÷ n. For an octagon, n = 8: (8 − 2) × 180° ÷ 8 = 6 × 180° ÷ 8 = 1080° ÷ 8 = 135°.
正 n 边形内角公式为 (n − 2) × 180° ÷ n。对于八边形，n = 8：(8 − 2) × 180° ÷ 8 = 6 × 180° ÷ 8 = 1080° ÷ 8 = 135°。
.666..., which is not a whole number. So regular octagons cannot tessellate alone; they leave gaps.
检查镶嵌：360° ÷ 135° = 2.666...，不是整数。因此正八边形不能独自镶嵌；它们会留下空隙。
re often combined with squares to fill the space, as 135° + 90° + 135° = 360° at a vertex.
在艺术中，八边形常与正方形组合以填充空间，因为在一个顶点处 135° + 90° + 135° = 360°。
4. Sports: Analysing Match Statistics with Percentages | 体育：用百分比分析比赛数据
A netball player scores 18 goals from 40 attempts. Calculate her shooting percentage. If she takes 25 more attempts and wants to raise her overall percentage to 50%, how many of these extra attempts must she score?
一名篮网球运动员 40 次出手命中 18 次。计算她的投篮命中率。如果她再多出手 25 次，并希望将总体命中率提升至 50%，她必须命中多少次额外的出手？
Current percentage = (18 ÷ 40) × 100 = 0.45 × 100 = 45%.
当前命中率 = (18 ÷ 40) × 100 = 0.45 × 100 = 45%。
als = 18 + x. We need (18 + x) ÷ 65 = 0.50, so 18 + x = 0.50 × 65 = 32.5, giving x = 14.5.
设 x 为额外 25 次出手中命中的次数。新总出手数为 40 + 25 = 65，新总命中数为 18 + x。需要 (18 + x) ÷ 65 = 0.50，因此 18 + x = 0.50 × 65 = 32.5，得 x = 14.5。
t score half a goal, she would need to score at least 15 extra goals to reach or exceed 50%. With 15 extra goals, percentage = (33 ÷ 65) × 100 ≈ 50.8%.
由于不能命中半个球，她需要至少命中 15 次额外出手才能达到或超过 50%。命中 15 次时，命中率 = (33 ÷ 65) × 100 ≈ 50.8%。
5. Music: Fractions and Note Durations | 音乐：分数与音符时值
In a piece of music, a semibreve lasts 4 beats, a minim 2 beats, a crotchet 1 beat, and a quaver ½ beat. A composer writes a bar containing one minim, two crotchets, and six quavers. Does this bar fit in ⁴/₄ time (total 4 beats)? If not, how many more quavers are needed to fill a 4-beat bar?
在一段音乐中，全音符持续 4 拍，二分音符 2 拍，四分音符 1 拍，八分音符 ½ 拍。一位作曲家在写一个小节，包含一个二分音符、两个四分音符和六个八分音符。这个小节符合 ⁴/₄ 拍（总共 4 拍）吗？如果不符合，还需要多少个八分音符才能填满一个 4 拍的小节？
; 2 crotchets = 2 × 1 = 2 beats; 6 quavers = 6 × ½ = 3 beats. Total = 2 + 2 + 3 = 7 beats.
计算总计拍数：1 个二分音符 = 2 拍；2 个四分音符 = 2 × 1 = 2 拍；6 个八分音符 = 6 × ½ = 3 拍。总拍数 = 2 + 2 + 3 = 7 拍。
= 8). The bar already has the equivalent of 7 beats worth of notes, which is already more than 4, so the question might be misinterpreted. Actually, the bar is too long. To fill exactly 4 beats from scratch, you would need 8 quavers.
这超过了 4 拍，所以不合适。要填满一个 4 拍的小节，需要多少个八分音符：4 拍 = 8 个八分音符（每拍 2 个八分音符，4 × 2 = 8）。本小节已经有相当于 7 拍的音符，已经超了。所以如果要刚好 4 拍，从一开始就需要 8 个八分音符。但这个场景本身说明音符组合超出了时值，违反了拍号。
2 + 2 + 3 = 7 beats > 4. Therefore the bar is too long by 3 beats, which equals 6 quavers.
在七年级中，重点是分数加法：2 + 2 + 3 = 7 拍 > 4。因此这个小节超出了 3 拍，相当于 6 个八分音符。
6. Business Studies: Simple Budgeting and Profit Calculation | 商业：简单预算与利润计算
Alice runs a cake stall. Each cake costs £1.20 to make and is sold for £2.00. She also pays a fixed market fee of £8. Write an algebraic expression for her profit P when she sells n cakes. Then calculate the profit if she sells 35 cakes.
艾丽丝经营一个蛋糕摊位。每个蛋糕的成本为 £1.20，售价为 £2.00。她还需支付 £8 的固定摊位费。写出当她卖出 n 个蛋糕时利润 P 的代数表达式。然后计算她卖出 35 个蛋糕时的利润。
Profit per cake = selling price − cost = £2.00 − £1.20 = £0.80. Total profit from n cakes before fee = 0.80n. Subtract the fixed fee: P = 0.80n − 8.
每个蛋糕的利润 = 售价 − 成本 = £2.00 − £1.20 = £0.80。n 个蛋糕在扣除费用前的利润为 0.80n。减去固定费用：P = 0.80n − 8。
For n = 35: P = 0.80 × 35 − 8 = 28 − 8 = £20.
当 n = 35 时：P = 0.80 × 35 − 8 = 28 − 8 = £20。
Using substitution into an expression is a key Year 7 algebra skill, linking maths to simple business decisions.
将数值代入表达式是一项关键的七年级代数技能，将数学与简单的商业决策联系起来。
7. Design Technology: Scaling a Model | 设计技术：模型缩放
A designer builds a scale model of a bridge at 1:50. The real bridge is 30 m long. How many centimetres is the model?
一位设计师建造了一座 1:50 的桥梁缩尺模型。实际桥梁长 30 m。模型长度是多少厘米？
= 3000 cm. Scale factor from real to model is 1 ÷ 50, so model length = 3000 ÷ 50 = 60 cm.
首先将实际长度转换为厘米：30 m = 30 × 100 = 3000 cm。从实际到模型的比例因子为 1 ÷ 50，因此模型长度 = 3000 ÷ 50 = 60 cm。
Alternatively, using metres: 30 m ÷ 50 = 0.6 m = 60 cm. Both methods show the same answer.
或者，用米计算：30 m ÷ 50 = 0.6 m = 60 cm。两种方法得出相同答案。
Understanding scale factors helps in reading technical drawings and building prototypes.
理解比例因子有助于阅读技术图纸和制作原型。
8. History: Using Timelines and Negative Numbers | 历史：时间线与负数
The mathematician Euclid was born around 325 BC and died around 265 BC. How long did he live? Note: There is no year 0; the timeline goes directly from 1 BC to AD 1.
数学家欧几里得大约生于公元前 325 年，卒于公元前 265 年。他活了多少年？注意：没有公元 0 年；时间线直接从公元前 1 年到公元 1 年。
egative numbers on a timeline: −325 and −265. The difference = (−265) − (−325) = 60 years.
当两个日期都在公元前时，我们可以在时间线上将它们视为负数：−325 和 −265。差值 = (−265) − (−325) = 60 年。
− 265 = 60. Since both are BC, the same method works without worrying about the year zero.
或者，用数值较大的公元前年份减去较小的：325 − 265 = 60。由于两者均为公元前，无需考虑公元 0 年即可同样计算。
This shows that negative number subtraction is useful for historical timelines, reinforcing the Year 7 topic of directed numbers.
这表明负数减法对于历史时间线非常有用，巩固了七年级的有向数主题。
9. Food Technology: Scaling Recipes | 食品技术：食谱调整
A pancake recipe for 4 people uses 200 g flour, 300 ml milk, and 2 eggs. Scale the recipe for 10 people.
一份供 4 人食用的煎饼食谱使用 200 g 面粉、300 ml 牛奶和 2 个鸡蛋。将该食谱调整为供 10 人食用。
ent by 2.5: Flour = 200 × 2.5 = 500 g; Milk = 300 × 2.5 = 750 ml; Eggs = 2 × 2.5 = 5 eggs.
从 4 人到 10 人的比例因子为 10 ÷ 4 = 2.5。每种原料乘以 2.5：面粉 = 200 × 2.5 = 500 g；牛奶 = 300 × 2.5 = 750 ml；鸡蛋 = 2 × 2.5 = 5 个。
Multiplying by a decimal is a practical use of the times tables and place value. The same idea applies to halving recipes.
乘以小数是乘法表和位值的实际应用。同样的思路也适用于将食谱减半。
10. Environmental Studies: Interpreting Climate Graphs | 环境研究：解读气候图表
The table below shows average monthly temperatures for a city. Calculate the mean annual temperature and the temperature range (maximum − minimum).
| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temp (°C) | 3 | 4 | 8 | 12 | 16 | 20 | 23 | 22 | 18 | 13 | 7 | 4 |
下表显示了一座城市的月平均气温。计算年均气温和气温范围（最大值 − 最小值）。
emperatures = 3+4+8+12+16+20+23+22+18+13+7+4 = 150. Mean = total ÷ 12 = 150 ÷ 12 = 12.5°C.
温度总和 = 3+4+8+12+16+20+23+22+18+13+7+4 = 150。平均值 = 总和 ÷ 12 = 150 ÷ 12 = 12.5°C。
Reading data from a table and calculating averages and range links to data handling in Year 7.
范围 = 最高 − 最低 = 23 − 3 = 20°C。从表格中读取数据并计算平均数和范围，与七年级的数据处理相联系。
11. Computing: Binary Numbers and Place Value | 计算机：二进制数与位值
Convert the decimal number 45 into an 8-bit binary number. Use place values: 128, 64, 32, 16, 8, 4, 2, 1.
将十进制数 45 转换为 8 位二进制数。使用位值：128、64、32、16、8、4、2、1。
. 4 ≤ 5, put 1, remainder 1. 2 > 1, put 0. 1 ≤ 1, put 1, remainder 0. So binary: 00101101.
找到适合的最大位值：32 ≤ 45，在 32 位放 1。余数 45−32=13。下一位 16>13，放 0。8 ≤ 13，放 1，余数 5。4 ≤ 5，放 1，余数 1。2>1，放 0。1 ≤ 1，放 1，余数 0。因此二进制为 00101101。
Binary uses base 2, reinforcing understanding of place value and powers. Doubling a binary number shifts digits left, just like multiplying by 2 in decimal.
二进制使用基数为 2，加深了对位值和幂的理解。二进制数加倍相当于向左移位，就像十进制中乘以 10 一样。
12. PE: Measuring Heart Rate and Percentage Increase | 体育：测量心率与百分比增长
A student’s resting heart rate is 68 beats per minute (bpm). After a 400-metre run, her heart rate readings at 1‑minute intervals are: 144, 136, 128, 120, 112. Find the mean post-exercise heart rate and the percentage increase from resting to this mean.
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