📚 Year 12 SQA Maths: Case Study Practice | 案例分析实战演练
Case study questions in SQA Maths bring together multiple areas of the curriculum, challenging you to apply calculus, algebra, trigonometry, statistics and more in realistic situations. This article walks you through eight carefully designed scenarios, each unpacked step by step to build your confidence and exam technique.
SQA 数学中的案例分析题整合了课程的不同板块，要求你在真实情境中综合运用微积分、代数、三角、统计等知识。本文将通过八个精心设计的实战场景，逐步拆解解题过程，帮助你建立信心、掌握应试技巧。
1. Optimisation: Fencing a Field | 优化问题：围栏问题
A farmer wants to fence off a rectangular field next to a straight river. No fence is needed along the riverbank. The farmer has 200 metres of fencing for the other three sides. What dimensions give the maximum area, and what is that area?
一位农民想沿着笔直的河边围出一块矩形田地。河岸一侧不需围栏，另外三边共用 200 米篱笆。边长为多少时面积最大？最大面积是多少？
Let the width perpendicular to the river be x m and the length parallel to the river be y m. The total fencing constraint gives the equation 2x + y = 200.
设垂直于河的宽为 x 米，平行于河的长为 y 米。篱笆总长限制给出方程 2x + y = 200。
The area A is given by A = x · y. Substitute y from the constraint:
面积 A = x·y。代入约束条件中的 y：
A(x) = x(200 − 2x) = 200x − 2x²
To find the maximum, differentiate and set the derivative to zero:
为求最大值，求导并设导数为零：
dA/dx = 200 − 4x = 0 ⇒ x = 50
Then y = 200 − 2(50) = 100. The second derivative d²A/dx² = −4 < 0 confirms a maximum.
于是 y = 200 − 2×50 = 100。二阶导数 d²A/dx² = −4 < 0，确认是极大值。
The maximum area is 50 × 100 = 5000 m².
最大面积为 50 × 100 = 5000 平方米。
2. Exponential Growth: Bacteria Culture | 指数增长：细菌培养
A laboratory starts with 800 bacteria. The population doubles every 5 hours. Write the growth function and determine when the population reaches 50,000.
实验室最初有 800 个细菌，每 5 小时数量翻倍。写出生长函数，并求细菌数达到 50000 的时间。
The exponential model can be written as N(t) = N₀ · 2^(t/T), where N₀ = 800 and T = 5. Thus:
指数模型可写成 N(t) = N₀ · 2^(t/T)，其中 N₀ = 800，T = 5。因此：
N(t) = 800 × 2^(t/5)
Set N(t) = 50000 and solve for t:
令 N(t) = 50000，解出 t：
800 · 2^(t/5) = 50000 ⇒ 2^(t/5) = 62.5
Take logarithms of both sides (using any base, here base 10):
两边取对数（可用任意底数，这里以 10 为底）：
log(2^(t/5)) = log(62.5)
(t/5) · log 2 = log 62.5
So t = 5 × log(62.5) / log 2. Using approximate values: log 62.5 ≈ 1.7959, log 2 ≈ 0.3010.
因此 t = 5 × log(62.5) / log 2。代入近似值：log 62.5 ≈ 1.7959，log 2 ≈ 0.3010。
t ≈ 5 × 1.7959 / 0.3010 ≈ 29.85 hours
The population reaches 50,000 after about 29.9 hours.
大约 29.9 小时后，细菌数达到 50000。
3. Trigonometric Modelling: Ferris Wheel | 三角建模：摩天轮
A Ferris wheel has a maximum height of 60 m and a minimum of 10 m above the ground. It completes one revolution every 8 minutes. Passengers board at the lowest point. Find a function for height h(t) in metres, t minutes after boarding.
某摩天轮最高点距离地面 60 米，最低点 10 米，每 8 分钟旋转一圈，乘客在最低点登舱。求登舱 t 分钟后高度 h(t) 的函数（单位：米）。
The amplitude A = (max − min)/2 = (60 − 10)/2 = 25 m. The vertical shift (midline) is (max + min)/2 = 35 m.
振幅 A = (最大值 − 最小值)/2 = (60 − 10)/2 = 25 米。垂直位移（中线）为 (60 + 10)/2 = 35 米。
Because boarding occurs at the minimum, use a negative cosine function with period 8 min. Angular frequency ω = 2π/period = 2π/8 = π/4.
由于在最低点登舱，使用负余弦函数，周期为 8 分钟。角频率 ω = 2π/周期 = 2π/8 = π/4。
h(t) = 35 − 25 cos(πt/4)
Check: t=0 → h(0)=35−25=10 (minimum). One full cycle: cos(πt/4) returns to 1 at t=8 → h=10. Maximum when cos = −1 at t=4 → h=35+25=60.
验证：t=0 时 h(0)=35−25=10（最低点）；一个完整周期：t=8 时 cos(πt/4)=1，h=10；t=4 时 cos=−1，h=60（最高点）。
Find height at t=3:
求 t=3 时的高度：
h(3) = 35 − 25 cos(3π/4) = 35 − 25×(−√2/2) ≈ 35 + 17.68 = 52.68 m
4. Straight Lines: Intersection of Paths | 直线：路径交点
Two walking trails are modelled by the equations L₁: 3x + 2y = 18 and L₂: y = 2x − 9. Determine their intersection point and the angle between the trails, correct to one decimal place.
两条步道可用方程 L₁: 3x + 2y = 18 和 L₂: y = 2x − 9 表示。求交点和两直线夹角（精确至 0.1 度）。
Substitute y from L₂ into L₁: 3x + 2(2x − 9) = 18 → 3x + 4x − 18 = 18 → 7x = 36 → x = 36/7.
将 L₂ 代入 L₁：3x + 2(2x − 9) = 18 → 7x = 36 → x = 36/7。
Then y = 2×(36/7) − 9 = 72/7 − 63/7 = 9/7. Intersection point: (36/7, 9/7).
进而 y = 2×(36/7) − 9 = 9/7。交点为 (36/7, 9/7)。
Gradients: m₁ from L₁: 3x + 2y = 18 → 2y = −3x + 18 → y = −3/2 x + 9, so m₁ = −1.5. m₂ = 2.
斜率：L₁ 化为 y = −3/2 x + 9，故 m₁ = −1.5；m₂ = 2。
Angle θ between lines: tan θ = |(m₂ − m₁) / (1 + m₁m₂)| = |(2 − (−1.5)) / (1 + (−1.5)×2)| = |3.5 / (1 − 3)| = |3.5 / −2| = 1.75.
两直线夹角 θ 满足：tan θ = |(m₂ − m₁) / (1 + m₁m₂)| = |(2 − (−1.5)) / (1 − 3)| = 1.75。
θ = tan⁻¹(1.75) ≈ 60.3°
The trails meet at approximately (5.14, 1.29) with an angle of 60.3°.
两条步道约在 (5.14, 1.29) 处交汇，夹角约为 60.3°。
5. Statistics: Analysing Athlete Times | 统计：运动员成绩分析
Eight runners record the following 100 m times (in seconds): 10.2, 10.5, 9.8, 11.1, 10.0, 10.7, 10.3, 11.0. Calculate the mean and standard deviation. Another runner with a time of 9.4 s joins the group; discuss the effect on the mean and on the standard deviation without recalculation.
八名短跑运动员的 100 米成绩（秒）如下：10.2, 10.5, 9.8, 11.1, 10.0, 10.7, 10.3, 11.0。计算均值与标准差。若加入一名成绩为 9.4 秒的新运动员，不重新计算，试讨论其对均值与标准差的影响。
First, find the sum: 10.2+10.5+9.8+11.1+10.0+10.7+10.3+11.0 = 84.6. Mean x̄ = 84.6 / 8 = 10.575 s.
先求和：84.6。均值 x̄ = 84.6 / 8 = 10.575 秒。
To compute standard deviation, use the formula s = √[ Σ(xᵢ − x̄)² / (n−1) ].
计算标准差，使用公式 s = √[ Σ(xᵢ − x̄)² / (n−1) ]。
| xᵢ | xᵢ − x̄ | (xᵢ − x̄)² |
|---|---|---|
| 10.2 | −0.375 | 0.140625 |
| 10.5 | −0.075 | 0.005625 |
| 9.8 | −0.775 | 0.600625 |
| 11.1 | 0.525 | 0.275625 |
| 10.0 | −0.575 | 0.330625 |
| 10.7 | 0.125 | 0.015625 |
| 10.3 | −0.275 | 0.075625 |
| 11.0 | 0.425 | 0.180625 |
Sum of squared deviations = 1.625. Then s = √(1.625 / 7) ≈ √0.23214 ≈ 0.4818 s.
离差平方和为 1.625，因此 s = √(1.625 / 7) ≈ 0.48 秒（保留两位小数）。
Adding a time of 9.4 s will decrease the mean because it is lower than the current mean. The new mean becomes (84.6+9.4)/9 ≈ 10.44 s.
加入 9.4 秒后，均值将降低（因为新值小于原均值）。新均值约为 10.44 秒。
The effect on standard deviation is ambiguous without calculation: the new value is far below the mean, which tends to increase spread, but the larger sample size n counteracts this. In this case, since 9.4 is far from the mean, standard deviation is likely to increase slightly.
对标准差的影响需权衡：新值大大低于均值，会增加离散度，但样本容量变大又会减小标准差。因 9.4 偏离均值较远，标准差很可能略微上升。
6. Vectors: Forces in Equilibrium | 向量：力的平衡
Two forces F₁ = 3i + 4j N and F₂ = −i + 2j N act on a particle. Find the magnitude of the resultant force and the angle it makes with the positive i-direction. Determine the force F₃ needed to achieve equilibrium.
两个力 F₁ = 3i + 4j N 和 F₂ = −i + 2j N 作用在一个质点上。求合力的大小、合力与正 i 方向夹角，以及达到平衡所需的第三个力 F₃。
Resultant R = F₁ + F₂ = (3−1)i + (4+2)j = 2i + 6j N.
合力 R = F₁ + F₂ = 2i + 6j N。
Magnitude |R| = √(2² + 6²) = √(4+36) = √40 = 2√10 ≈ 6.32 N.
大小 |R| = √(2² + 6²) = √40 ≈ 6.32 N。
Angle θ with the positive i-axis: tan θ = (6)/(2) = 3, so θ = tan⁻¹(3) ≈ 71.6°.
与正 i 轴夹角 θ：tan θ = 6/2 = 3，θ ≈ 71.6°。
For equilibrium, the third force must be opposite to R: F₃ = −R = −2i − 6j N, with the same magnitude.
平衡时，第三个力应与合力等大反向：F₃ = −2i − 6j N，大小同为 6.32 N。
7. Sequences and Series: Savings Plan | 数列与级数：储蓄计划
At the start of each year, Jenny deposits £1500 into an account paying 4% annual compound interest. How much will she have immediately after the 10th deposit?
每年年初，珍妮存入 1500 英镑到年利率为 4% 的复利账户。她第 10 次存款后，账户里立即有多少钱？
The first deposit earns interest for 10 years, the second for 9 years, and so on, down to the final deposit earning interest for 1 year. After the 10th deposit, the amount is the sum of a geometric series.
第一次存款计息 10 年，第二次 9 年，依此类推，最后一次存款计息 1 年。第十次存款后，总金额为等比数列求和。
Total = 1500 × (1.04¹⁰ + 1.04⁹ + … + 1.04¹)
The sum inside the brackets is a geometric series with first term a = 1.04, common ratio r = 1.04, number of terms n = 10.
括号内是等比数列，首项 a = 1.04，公比 r = 1.04，项数 n = 10。
Sum S_n = a(rⁿ − 1) / (r − 1) = 1.04 × (1.04¹⁰ − 1) / 0.04.
和 S_n = a(rⁿ − 1)/(r − 1) = 1.04 × (1.04¹⁰ − 1) / 0.04。
Compute 1.04¹⁰: 1.04²=1.0816, 1.04⁴≈1.1699, 1.04⁸≈1.3686, multiplied by 1.04² gives about 1.4802. So 1.04¹⁰ − 1 = 0.4802.
计算 1.04¹⁰：逐步相乘得约 1.4802，减去 1 得 0.4802。
S_n ≈ 1.04 × 0.4802 / 0.04 = 1.04 × 12.005 = 12.4852.
Total = 1500 × 12.4852 ≈ £18,727.80.
总金额 = 1500 × 12.4852 ≈ £18,727.80。
8. Logarithmic Scales: Sound Intensity | 对数尺度：声音强度
The sound intensity level L in decibels (dB) is given by L = 10 log₁₀(I / I₀), where I₀ = 10⁻¹² W/m². A rock concert reaches 110 dB. Find the intensity I. If a quieter passage measures 80 dB, how many times more intense is the concert sound?
声音强度级 L（分贝）公式为 L = 10 log₁₀(I / I₀)，其中 I₀ = 10⁻¹² W/m²。摇滚音乐会达到 110 dB。求声音强度 I。若一安静段落为 80 dB，音乐会的强度是它的多少倍？
For the concert: 110 = 10 log₁₀(I_c / 10⁻¹²) ⇒ log₁₀(I_c / 10⁻¹²) = 11.
音乐会：110 = 10 log₁₀(I_c / 10⁻¹²)，故 log₁₀(I_c / 10⁻¹²) = 11。
So I_c / 10⁻¹² = 10¹¹ ⇒ I_c = 10⁻¹² × 10¹¹ = 10⁻¹ = 0.1 W/m².
于是 I_c = 10⁻¹² × 10¹¹ = 0.1 W/m²。
For the quieter passage: 80 = 10 log₁₀(I_q / 10⁻¹²) ⇒ log₁₀(I_q / 10⁻¹²) = 8 ⇒ I_q = 10⁻⁴ W/m².
安静段落：80 = 10 log₁₀(I_q / 10⁻¹²)，得 I_q = 10⁻⁴ W/m²。
Ratio = I_c / I_q = 10⁻¹ / 10⁻⁴ = 10³ = 1000. The concert is 1000 times more intense than the quiet passage. Notice that a 30 dB increase corresponds to a factor of 1000 in intensity.
比值 = 0.1 / 10⁻⁴ = 1000。音乐会的强度是安静段落的 1000 倍。可见，分贝值增加 30，强度增大 1000 倍。
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