📚 Year 10 SQA Mathematics: Interdisciplinary Integrated Problem-Solving | 跨学科综合题型训练
Interdisciplinary problems in SQA Mathematics assess your ability to apply mathematical skills in real-world and cross-curricular contexts. This article provides training across a range of subjects, illustrating how concepts such as ratios, percentages, algebra, geometry, and statistics are used in physics, chemistry, biology, geography, economics, and beyond. Each section presents a typical problem, a step-by-step solution, and the key mathematical techniques needed for success in SQA examinations.
SQA 数学中的跨学科问题考查你在现实世界和跨学科情境中应用数学技能的能力。本文提供多学科综合训练，展示比例、百分比、代数、几何和统计等概念在物理、化学、生物、地理、经济学等领域的具体应用。每个部分都会呈现一道典型题目、分步解答以及在 SQA 考试中取得好成绩所需的关键数学技巧。
1. Physics: Speed, Distance and Time | 物理：速度、距离和时间
In physics, the relationship between speed (v), distance (d) and time (t) is fundamental:
在物理中，速度 (v)、距离 (d) 和时间 (t) 之间的关系是基础性的：
d = v × t
Example: A cyclist rides at a constant speed of 15 kilometres per hour for 1.2 hours. Calculate the distance travelled.
示例：一位自行车手以每小时 15 公里的恒定速度骑行了 1.2 小时。计算所行驶的距离。
8 km. Remember to check that the units of speed and time are compatible before multiplying.
解答：距离 = 15 公里/小时 × 1.2 小时 = 18 公里。记住在乘法运算前要确认速度和时间单位的一致性。
km/h. When units are mixed, convert appropriately, e.g. m/s to km/h by multiplying by 3.6.
常见变式：一辆汽车 3 小时行驶 210 公里。其平均速度为 v = d/t = 210/3 = 70 公里/小时。当单位不统一时，要进行适当换算，例如米/秒换算成千米/小时需要乘以 3.6。
2. Chemistry: Molar Calculations and Proportions | 化学：摩尔计算与比例
ass and M is molar mass. For a fixed substance, mass is directly proportional to the number of moles.
化学中的摩尔概念严重依赖比例推理。物质的量 n = m/M，其中 m 为质量，M 为摩尔质量。对于同一种物质，质量与物质的量成正比。
Example: If 2 moles of sodium chloride (NaCl) have a mass of 117 grams, what is the mass of 0.75 moles?
示例：如果 2 摩尔氯化钠 (NaCl) 的质量为 117 克，那么 0.75 摩尔的质量是多少？
7 × 0.75, so 2x = 87.75, and x = 43.875 g. The mass of 0.75 moles is approximately 43.9 g.
解答：建立比例式 2/117 = 0.75/x。交叉相乘得 2x = 117 × 0.75，所以 2x = 87.75，x = 43.875 克。0.75 摩尔的质量约为 43.9 克。
This proportional method also applies to volumes of gases at constant temperature and pressure. Direct proportion is a key skill: if 0.2 mol occupies 4.8 dm³, then 1 mol occupies 24 dm³ (molar volume).
这种比例方法同样适用于恒温恒压下气体的体积计算。正比例是一项关键技能：如果 0.2 摩尔占据 4.8 立方分米，那么 1 摩尔占据 24 立方分米（摩尔体积）。
3. Biology: Population Growth Models | 生物：种群增长模型
Exponential growth models in biology often involve repeated multiplication. A common scenario is bacterial doubling: a population doubles every fixed time interval.
生物学中的指数增长模型通常涉及重复乘法。常见的场景是细菌倍增：种群数量每隔固定时间翻一番。
Example: A single bacterium divides every 20 minutes. Starting with one bacterium, how many bacteria will there be after 3 hours?
示例：一个细菌每 20 分钟分裂一次。从一个细菌开始，3 小时后会有多少个细菌？
= 9. After n periods, population = initial × 2ⁿ. Here population = 1 × 2⁹ = 512 bacteria.
解答：3 小时 = 180 分钟。倍增次数 = 180/20 = 9。经过 n 个周期后，种群数量 = 初始数量 × 2ⁿ。此处种群数量 = 1 × 2⁹ = 512 个细菌。
This links directly to indices and powers. For fractional periods, we can use estimation or logarithms, but for whole numbers it is straightforward evaluation.
这直接关联到指数和幂运算。对于非整数周期，我们可以使用估算或对数，但整数周期可以直接计算。
4. Geography: Map Scales and Area Calculations | 地理：地图比例尺与面积计算
Map scales are used to convert distances measured on a map to actual ground distances. A scale of 1:50,000 means 1 cm on the map represents 50,000 cm in reality.
地图比例尺用于将图上测量距离转换为实际地面距离。1:50,000 的比例尺意味着图上 1 厘米代表实际 50,000 厘米。
Example: Two towns are 8 cm apart on a 1:100,000 scale map. What is the actual distance in kilometres?
示例：在 1:100,000 比例尺的地图上，两个城镇相距 8 厘米。实际距离是多少公里？
cm² on the map, the real area is 3 × (100,000)² cm², which must be converted to m² or km².
解答：实际距离 = 8 厘米 × 100,000 = 800,000 厘米 = 8,000 米 = 8 公里。计算面积时，比例因子需要平方。如果自然保护区在地图上占 3 平方厘米，实际面积是 3 × (100,000)² 平方厘米，然后必须转换为平方米或平方公里。
5. Economics: Simple and Compound Interest | 经济学：单利与复利
nterest I = P × r × t, where P is principal, r is the annual rate, and t is time in years.
金融数学运用百分比和指数增长。单利 I = P × r × t，其中 P 为本金，r 为年利率，t 为以年为单位的时间。
Example: Alex deposits £400 in an account offering 3.5% simple interest per annum. Find the interest earned after 4 years, and the total amount.
示例：亚历克斯存入 400 英镑到年利率 3.5% 的单利账户。求 4 年后获得的利息和总金额。
Solution: Interest = 400 × 0.035 × 4 = £56. Total amount = 400 + 56 = £ nterest, use the multiplier method: amount = P × (1 + r)ⁿ. If the same principal earns 3.5% compound interest annually for 4 years, total = 400 × (1.035)⁴ ≈ 400 × 1.1475 = £459.00.
解答：利息 = 400 × 0.035 × 4 = 56 英镑。总金额 = 400 + 56 = 456 英镑。复利的情况下，使用乘数法：总金额 = P × (1 + r)ⁿ。如果同样的本金以 3.5% 的年复利计算 4 年，总金额 = 400 × (1.035)⁴ ≈ 400 × 1.1475 = 459.00 英镑。
6. Environmental Science: Carbon Footprint and Percentages | 环境科学：碳足迹与百分比
Calculating carbon emissions involves multiplication and percentage change. If a household' e emission factor is 0.25 kg CO₂ per kWh, the total emission is 3,500 × 0.25 = 875 kg CO₂.
计算碳排放涉及乘法和百分比变化。如果一个家庭的用电量为 3,500 千瓦时，排放因子为每千瓦时 0.25 千克 CO₂，则总排放量为 3,500 × 0.25 = 875 千克 CO₂。
Example: The household aims to reduce its footprint by 15%. What will the new emission level be? How many kg of CO₂ are saved?
示例：该家庭打算将碳足迹减少 15%。新的排放水平将是多少？节约了多少千克 CO₂？
875 × 0.85 = 743.75 kg. Percentage decrease is a transferable skill across many subjects.
解答：减少量 = 875 × 0.15 = 131.25 千克。新排放量 = 875 - 131.25 = 743.75 千克。或者使用乘数：875 × (1 - 0.15) = 875 × 0.85 = 743.75 千克。百分比减少是许多学科中均可迁移的技能。
7. Engineering: Pythagorean Theorem in Construction | 工程：建筑中的勾股定理
is widely used to ensure right angles and to calculate lengths in construction and design.
勾股定理 a² + b² = c² 被广泛用于确保直角以及计算建筑和设计中的长度。
a² + b² = c²
Example: A ladder leans against a vertical wall. The foot of the ladder is 1.8 m from the wall, and the top reaches 5.2 m up the wall. Find the length of the ladder (to 2 decimal places).
示例：一架梯子斜靠在竖直的墙壁上。梯脚距墙 1.8 米，梯子顶端达到墙上 5.2 米的高度。求梯子的长度（精确到小数点后两位）。
lution: Let the ladder length be L. L² = 1.8² + 5.2² = 3.24 + 27.04 = 30.28. Therefore L = √30.28 ≈ 5.50 m. This is a direct application of the theorem.
解答：设梯子长度为 L。L² = 1.8² + 5.2² = 3.24 + 27.04 = 30.28。因此 L = √30.28 ≈ 5.50 米。这是该定理的直接应用。
, you may be given the hypotenuse and one leg, requiring you to rearrange to a² = c² - b².
在更复杂的问题中，可能会给出斜边和一条直角边，此时需要将公式变形为 a² = c² - b²。
8. Health: Body Mass Index (BMI) Formula | 健康：身体质量指数公式
eight² (m²). Evaluating this requires correct order of operations and decimal calculation.
身体质量指数使用公式计算：BMI = 体重 (千克) / 身高² (米²)。计算此值需要正确的运算顺序和小数运算。
BMI = m / h²
Example: A person weighs 68 kg and is 1.72 m tall. Determine their BMI and classify according to standard ranges (underweight: less than 18.5; healthy: 18.5 to 24.9; overweight: 25 to 29.9).
示例：某人体重 68 千克，身高 1.72 米。计算其 BMI 并根据标准范围进行分类（体重过轻：低于 18.5；健康：18.5 至 24.9；超重：25 至 29.9）。
hin the healthy range. The use of squares and division makes it a typical applied number problem.
解答：身高² = 1.72² = 2.9584。BMI = 68 / 2.9584 ≈ 22.98。这属于健康范围。平方和除法的运用使其成为一个典型的应用算术问题。
9. Sports: Quadratic Trajectory of a Ball | 体育：球的抛物线轨迹
x (metres) might be given by h(x) = -0.05x² + 2.5x. Finding the maximum height involves locating the vertex of the parabola.
球在空中飞行的路径通常可以用二次函数来建模。例如，高度 h (米) 关于水平距离 x (米) 的函数可能由 h(x) = -0.05x² + 2.5x 给出。求最大高度就涉及寻找抛物线的顶点。
For a quadratic ax² + bx + c, the x-coordinate of the vertex is x = -b/(2a).
对于二次函数 ax² + bx + c，顶点的 x 坐标为 x = -b/(2a)。
x = -b / (2a)
ample: Use the function h = -0.05x² + 2.5x to find the maximum height reached by the ball.
示例：利用函数 h = -0.05x² + 2.5x 求球达到的最大高度。
2.5(25) = -0.05(625) + 62.5 = -31.25 + 62.5 = 31.25 m. The maximum height is 31.25 metres.
解答：a = -0.05, b = 2.5。x = -2.5/(2 × -0.05) = -2.5/(-0.1) = 25。将 x = 25 代入 h：h = -0.05(25)² + 2.5(25) = -0.05(625) + 62.5 = -31.25 + 62.5 = 31.25 米。最大高度为 31.25 米。
10. Business: Break-even Analysis | 商业：盈亏平衡分析
+ variable cost per unit × x, solving R(x) = C(x) gives the number of units to break even.
盈亏平衡点出现在总收入等于总成本时。如果每单位收入为 R(x) = p × x，总成本为 C(x) = 固定成本 + 单位可变成本 × x，解方程 R(x) = C(x) 即可得到盈亏平衡的单位数量。
Example: A company produces handmade candles. Fixed costs are £200, variable cost is £3.50 per candle, and selling price is £8 per candle. Find the break-even quantity.
Published by TutorHao | Year 10 Mathematics Revision Series | aleveler.com
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