📚 Year 11 WJEC Mathematics: Interdisciplinary Problem-Solving Practice | WJEC数学跨学科综合题型训练
In the WJEC GCSE Mathematics examination, many high-mark questions are set in real-world contexts drawn from physics, chemistry, biology, geography, economics and engineering. These interdisciplinary problems test your ability to extract the relevant mathematical model, apply the correct techniques and interpret your answers in a meaningful way. This article provides targeted practice across ten key themes, strengthening the skills you need to tackle such cross-subject challenges with confidence.
在WJEC GCSE数学考试中，许多高分题目都设置在物理、化学、生物、地理、经济和工程等实际情境中。这些跨学科问题考查你提取相关数学模型、运用正确方法并有意义地解读答案的能力。本文围绕十大主题进行针对性训练，帮助你增强应对跨学科综合题型所需的技能与信心。
1. Physics in Motion: Speed, Distance and Time | 运动中的物理：速度、距离与时间
Kinematics problems often require you to interpret distance-time and speed-time graphs. The gradient of a distance-time graph gives the speed, while the area under a speed-time graph gives the distance travelled. The equations of motion for constant acceleration are essential: v = u + at and s = ut + ½at², where u is initial velocity, v final velocity, a acceleration, t time, and s displacement.
运动学问题常需要解读距离-时间图和速度-时间图。距离-时间图的斜率表示速度，而速度-时间图下的面积表示行驶距离。匀加速运动的方程至关重要：v = u + at 和 s = ut + ½at²，其中 u 是初速度，v 是末速度，a 是加速度，t 是时间，s 是位移。
Example: A car accelerates uniformly from rest to 20 m/s in 10 seconds, maintains that speed for 30 seconds, then decelerates uniformly to stop in 8 seconds. Sketch the speed-time graph and calculate the total distance travelled.
例题：一辆汽车从静止开始匀加速，10秒后达到20 m/s，接着匀速行驶30秒，然后匀减速，8秒后停下。画出速度-时间图并计算总行驶距离。
s = 600 m. The deceleration phase is a triangle of base 8 s and height 20 m/s, area = ½ × 8 × 20 = 80 m. Total distance = 100 + 600 + 80 = 780 m. Always check that the units are consistent (metres and seconds).
解答：加速阶段是一个底边10秒、高20 m/s的三角形，面积=½×10×20=100 m。匀速阶段是一个矩形20 m/s×30 s=600 m。减速阶段是一个底边8秒、高20 m/s的三角形，面积=½×8×20=80 m。总距离=100+600+80=780 m。始终检查单位是否一致（米和秒）。
2. Exponential Growth: Bacteria and Compound Interest | 指数增长：细菌繁殖与复利
Both colony growth in biology and compound interest in finance follow exponential models. If a bacteria population doubles every 30 minutes, the number after t hours is given by N = N₀ × 22t, where N₀ is the initial count. Financially, an investment grows as A = P(1 + r/100)n, where P is the principal, r the annual interest rate and n the number of years.
生物学中的菌落生长和金融中的复利都遵循指数模型。如果细菌每30分钟数量翻倍，t小时后的数量由N = N₀ × 22t给出，其中N₀是初始数量。在金融方面，投资增长为A = P(1 + r/100)n，P是本金，r是年利率，n是年数。
Example: A savings account offers 4% compound interest per annum. £2000 is deposited. How much interest is earned after 5 years? Give your answer to the nearest penny.
例题：一个储蓄账户提供年利率4%的复利。存入2000英镑。5年后获得多少利息？答案精确到便士。
Solution: A = 2000 × (1.04)5. First compute (1.04)5 0816, 1.04³ = 1.124864, 1.04⁴ ≈ 1.16985856, 1.04⁵ ≈ 1.2166529. Then A ≈ 2000 × 1.2166529 = £2433.3058. Interest = £2433.31 − £2000 = £433.31. The exponential factor makes a significant difference compared to simple interest.
解答：A = 2000 × (1.04)5。先计算(1.04)5 = 1.04²=1.0816，1.04³=1.124864，1.04⁴≈1.16985856，1.04⁵≈1.2166529。于是A≈2000×1.2166529=2433.3058英镑。利息=2433.31−2000=433.31英镑。与单利相比，指数因子产生了显著差异。
3. Trigonometry in Surveying | 测量中的三角学
Surveyors use right-angled trigonometry to find inaccessible heights and distances. When the angle of elevation from a point to the top of a building is θ and the horizontal distance is d, the height h is h = d × tan θ. For angles of depression or when combining two observations, careful diagram drawing is crucial.
测量人员使用直角三角学来求难以直接测量的高度和距离。当从某点到建筑物顶部的仰角为θ且水平距离为d时，高度h可由h = d × tan θ得出。对于俯角或组合两个观测值的情形，仔细画图至关重要。
Example: From a point 60 m from the base of a tower, the angle of elevation to the top is 35°. Calculate the height of the tower to the nearest 0.1 m.
例题：在离塔底60 m处，测得塔顶仰角为35°。计算塔高，精确到0.1 m。
se a calculator: tan 35° ≈ 0.700208. Then h ≈ 60 × 0.700208 = 42.01248 m, so the height is 42.0 m (to 1 d.p.). Remember to set your calculator to degree mode and always check if the answer is reasonable by sketching the triangle.
解答：h=60×tan35°。使用计算器：tan35°≈0.700208。于是h≈60×0.700208=42.01248 m，塔高为42.0 m（保留一位小数）。记得将计算器设置为度模式，并通过画草图检查答案是否合理。
4. Statistics in Biology: Interpreting Box Plots | 生物学中的统计：解读箱线图
whiskers extend to the minimum and maximum values. The interquartile range (IQR = Q₃ − Q₁) measures spread. Overlapping boxes suggest similar distributions, while distinctly separated medians indicate a significant difference.
生物学家常使用箱线图来比较两个样本。箱子显示下四分位数（Q₁）、中位数（Q₂）和上四分位数（Q₃）；触须延伸至最小值和最大值。四分位距（IQR = Q₃ − Q₁）衡量离散程度。箱子重叠表明分布相似，而中位数明显分离则意味着存在显著差异。
B: min=10 cm, Q₁=20 cm, median=25 cm, Q₃=28 cm, max=34 cm. Compare the central tendency and spread.
例题：测试两种肥料对植物高度的影响。肥料A的箱线图：最小值12 cm，Q₁=18 cm，中位数24 cm，Q₃=29 cm，最大值35 cm。肥料B：最小值10 cm，Q₁=20 cm，中位数25 cm，Q₃=28 cm，最大值34 cm。比较集中趋势和离散程度。
0 = 8 cm. Fertiliser B shows less variation, indicating more consistent plant growth. Both ranges are similar (23 cm vs 24 cm). Therefore, B may produce slightly taller plants with less variability.
解答：肥料B的中位数略高（25 cm对24 cm），因此其集中趋势稍大。A的四分位距IQR=29−18=11 cm；B的IQR=28−20=8 cm。肥料B的变异较小，表明植物生长更一致。两者的极差相近（23 cm对24 cm）。因此，B可能产出稍高且变异更小的植株。
5. Ratio and Proportion in Chemistry | 化学中的比与比例
Stoichiometry uses mole ratios from balanced equations. For the reaction 2H₂ + O₂ → 2H₂O u can find the required mass of another by converting mass → moles → ratio → moles → mass.
化学计量法使用配平方程中的摩尔比。对于反应2H₂ + O₂ → 2H₂O，氢气与氧气的摩尔比为2:1。如果知道一种反应物的质量，可以通过 质量→摩尔→比→摩尔→质量 的步骤求出另一反应物的所需质量。
Example: In the reaction 2Mg + O₂ → 2MgO, what mass of magnesium oxide is produced when 6 g of magnesium burns completely? (Relative atomic masses: Mg=24, O=16)
例题：在反应2Mg + O₂ → 2MgO中，6 g镁完全燃烧能生成多少克氧化镁？（相对原子质量：Mg=24, O=16）
gO is 1:1, so moles of MgO = 0.25 mol. Mr of MgO = 24 + 16 = 40. Mass of MgO = 0.25 × 40 = 10 g. Always present the mass to an appropriate number of significant figures.
解答：Mg的摩尔数=质量/相对原子质量=6/24=0.25 mol。根据方程式，Mg与MgO的摩尔比为1:1，因此MgO摩尔数=0.25 mol。MgO的相对分子质量=24+16=40。MgO质量=0.25×40=10 g。始终以合适有效数字给出质量。
6. Density, Mass and Volume in Engineering | 工程中的密度、质量与体积
Density ρ links mass and volume: ρ = m / V n cm³ and density in g/cm³, the mass will be in grams. To convert to kilograms, divide by 1000.
密度ρ将质量与体积联系起来：ρ = m / V。该公式出现在材料科学与浮力问题中。必须仔细换算单位；例如1 g/cm³ = 1000 kg/m³。若体积以cm³给出而密度以g/cm³给出，质量将以克为单位。若要转换为千克，除以1000。
Example: An aluminium casting has a volume of 850 cm³. Aluminium’s density is 2.7 g/cm³. Calculate the mass of the casting in kilograms.
例题：一个铝铸件体积为850 cm³。铝的密度为2.7 g/cm³。计算铸件质量，以千克为单位。
Solution: m = ρ × V = 2.7 × 850 = 2295 g. To convert to kg, divide by 1000: 2295 / 1000 = 2.295 kg. In engineering contexts, you might round to 2.30 kg. Understanding density also helps when calculating forces from weight (W = mg, g = 9.8 m/s²).
解答：m=ρ×V=2.7×850=2295 g。转换为千克，除以1000：2295/1000=2.295 kg。在工程情境中，可能四舍五入为2.30 kg。理解密度也有助于通过重量计算力（W=mg，g=9.8 m/s²）。
7. Map Scales and Bearings | 地图比例尺与方位角
Maps use a scale such as 1:25000, meaning 1 cm on the map represents 25000 cm (0.25 km) in reality. Bearings are measured clockwise from north and are always given as three figures (e.g. 045°). Combined with right-angled trigonometry, these concepts allow you to determine actual distances and directions between points.
地图使用比例尺，如1:25000，表示地图上1 cm代表实际25000 cm（0.25 km）。方位角从正北顺时针测量，始终用三位数表示（如045°）。结合直角三角学，可确定点之间的实际距离和方向。
Example: A map has scale 1:50000. Two villages are 7.4 cm apart on the map. What is the actual distance in kilometres? Then, the bearing from village A to village B is 140°. What is the return bearing from B to A?
例题：一张地图比例尺为1:50000。两村庄图上距离7.4 cm。实际距离是多少千米？接着，从A村到B村的方位角为140°。从B村到A村的返程方位角是多少？
Solution: Actual distance = 7.4 × 50000 cm = 370000 cm = 3700 m = 3.7 km. For the return bearing, add 180° if the original is ≤ 180°, then adjust to within 0–360°: 140° + 180° = 320°. So the bearing from B to A is 320°. Always express bearings as three-digit numbers.
解答：实际距离=7.4×50000 cm=370000 cm=3700 m=3.7 km。返程方位角：若原方位角≤180°，则加180°，并调整至0–360°范围：140°+180°=320°。因此从B到A的方位角为320°。务必用三位数表示方位角。
8. Financial Mathematics: Loans and Annual Percentage Rate | 金融数学：贷款与年利率
WJEC often includes questions on loans and savings involving compound interest. The amount owed after n years on a loan of £P with interest rate r% is A = P(1 + r/100)n. For regular payments, the monthly repayment formula or an iterative approach may be used. Understanding the difference between nominal rate and APR is also tested.
WJEC常考查涉及复利的贷款与储蓄题目。贷款本金£P、利率r%，n年后欠款总额为A = P(1 + r/100)n。对于定期还款，可能使用月度还款公式或迭代法。名义利率与年利率（APR）的区别亦在考查范围内。
Example: A loan of £5000 is taken out at an annual compound interest rate of 6%. No repayments are made for 3 years. How much is owed at the end of 3 years? If instead the borrower pays £200 per month starting at the end of year 1, estimate the balance remaining after 3 years (assume interest added annually).
例题：一笔5000英镑贷款，年复利6%，3年内未还款。3年后欠款多少？若从第一年末开始每月还款200英镑，估算3年后剩余余额（假设每年加计利息一次）。
Solution (no payments): A = 5000 × (1.06)3 = 5000 × 1.191016 = £5955.08. With payments after year 1: After year 1 debt = 5000 × 1.06 = £5300. Then year 2: 12 payments of £200 total £2400. So balance at start of year 2 = 5300 − 2400 = 2900; after interest 2900 × 1.06 = £3074. Year 3: pay 2400, balance 3074 − 2400 = 674; after interest 674 × 1.06 = £714.44. So remaining debt ≈ £714. Real-world loans often compound monthly, so check the question carefully.
解答（无还款）：A=5000×(1.06)3=5000×1.191016=5955.08英镑。有还款时：第一年末债务=5000×1.06=5300英镑。第二年：每月200英镑，12笔共2400英镑。第二年初余额=5300−2400=2900；加利息后2900×1.06=3074英镑。第三年：还款2400，余额3074−2400=674；加利息674×1.06=714.44英镑。剩余欠款约714英镑。现实中贷款多按月计复利，请仔细审题。
9. Probability and Genetics | 概率与遗传学
Genetic inheritance can be modelled using Punnett squares and probability tree diagrams. For a monohybrid cross, if both parents are heterozygous (Bb), the probability of an offspring being homozygous recessive (bb) is ¼. Combining probabilities uses the multiplication rule for independent events.
遗传可以通过庞纳特方格和概率树图建模。对于单基因杂交，若双亲均为杂合子（Bb），后代为隐性纯合子（bb）的概率为¼。独立事件的组合概率使用乘法规则。
Example: The allele for brown eyes (B) is dominant over blue (b). Two brown-eyed parents, both heterozygous (Bb), plan to have three children. What is the probability that all three children have blue eyes?
例题：褐色眼睛（B）对蓝色（b）为显性。一对褐眼父母均为杂合子（Bb），计划生三个孩子。三个孩子都是蓝眼睛的概率是多少？
Solution: The probability of one child having blue eyes (bb) is ¼. Since each child' e probability of three blue-eyed children is (¼) × (¼) × (¼) = 1/64. As a decimal, that is 0.015625, or about 1.56%. This simple application of the product rule is often used in genetic counselling.
解答：一个孩子为蓝眼睛（bb）的概率是¼。由于每个孩子的基因型独立，三个蓝眼睛孩子的概率为 (¼)×(¼)×(¼)=1/64。以小数表示为0.015625，约1.56%。这种乘法规则的简单应用常用于遗传咨询。
10. Modelling with Quadratic Functions in Economics | 经济学中的二次函数模型
Revenue, cost and profit are often modelled as functions of quantity sold. A typical profit function might be P(x) = −x² + 50x − 400, where x is the number of items. The maximum profit occurs at the vertex of the parabola. Completing the square or using x = −b/(2a) for the axis of symmetry gives the optimal quantity.
收入、成本和利润常被建模为销售量的函数。一个典型的利润函数如P(x) = −x² + 50x − 400，其中x为销量。最大利润出现在抛物线顶点处。用配方法或利用对称轴公式x = −b/(2a)，可求出最优销量。
Example: A company’s weekly profit £ x² + 120x − 800. Determine the number of units that maximises profit and calculate that maximum profit.
例题：某公司每周销售x件产品的利润£P由P=−2x²+120x−800给出。求利润最大化时的销量及最大利润。
xis of symmetry is x = −b/(2a) = −120/(2 × −2) = −120/(−4) = 30. So profit is maximised at x = 30 tuting back: P(30) = −2(30)² + 120(30) − 800 = −2(900) + 3600 − 800 = −1800 + 3600 − 800 = £1000. Thus selling 30 units yields a maximum profit of £1000 per week. Quadratic models appear in many business optimisation contexts.
解答：这里a=−2，b=120。对称轴x=−b/(2a)=−120/(2×−2)=−120/(−4)=30。因此销量为x=30时利润最大。代回原函数：P(30)=−2(30)²+120(30)−800=−2(900)+3600−800=−1800+3600−800=1000英镑。因此每周销售30件可获得最大利润1000英镑。二次模型广泛出现于商业优化情境。
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