📚 AQA Year 9 Further Maths Case Study Practical Exercises | AQA 九年级进阶数学案例分析实战演练
Further Mathematics at Year 9 level is not just about memorising formulas — it challenges you to apply mathematical reasoning to unfamiliar, multi‑step problems. This case‑study‑style article walks you through realistic scenarios drawn from number theory, algebraic thinking, geometry and probability. Each case is designed to mirror the style of AQA-style problem‑solving tasks so you can practise breaking down complex information, choosing appropriate methods and communicating your logic clearly.
九年级进阶数学远不只是记忆公式，它要求你将数学推理应用到陌生且步骤繁多的问题中。这篇案例分析式文章将带你走进来自数论、代数思维、几何和概率的真实情境。每个案例都贴合AQA风格的问题解决任务，帮助你练习分解复杂信息、选择合适的解题方法并清晰地表达逻辑。
1. Case Study Background: The School Festival Dilemma | 案例背景：校园文化节的两难问题
A school is organising a cultural festival. The organising committee needs to allocate stalls, schedule performances and manage ticket sales. They have collected data on potential attendees, weather‑dependent scenarios and budget constraints. Your role is to provide mathematical advice using number patterns, equations, inequalities and data analysis. Throughout this article we will examine four interconnected real‑life challenges.
一所学校正在筹备文化节。组委会需要分配摊位、安排表演并管理售票。他们已经收集了有关潜在参与人数、受天气影响的情景以及预算限制的数据。你的角色是利用数字规律、方程、不等式和数据分析提供数学建议。在整篇文章中，我们将探讨四个相互关联的现实挑战。
2. Number Patterns and Sequences: Stall Allocation | 数列与规律：摊位分配
The festival has 120 stalls arranged in rectangular blocks. Each block has the same number of rows, and every row contains 4 more stalls than the row number. For example, if a block has r rows, then row 1 has 5 stalls, row 2 has 6 stalls, row 3 has 7 stalls, and so on. The committee wants to know the maximum number of rows possible so that the total stalls per block does not exceed 120. They also plan to display stall numbers following a Fibonacci‑like pattern for decoration: the first two numbers are 1 and 2, and each subsequent number is the sum of the two preceding ones. Find the 10th term of this sequence and determine whether it is prime.
文化节有120个摊位，排列在矩形区块内。每个区块的行数相同，且每行的摊位数比行号多4个。例如，若某个区块有r行，则第1行有5个摊位，第2行有6个摊位，第3行有7个摊位，以此类推。组委会想知道一个区块最多能有多少行，使得每个区块的总摊位数不超过120。他们还想按照类似斐波那契的规律展示摊位编号作为装饰：前两个数字是1和2，后续每个数字都是前两个数字之和。请找出该数列的第10项，并判断它是否为质数。
in a block is the sum of an arithmetic series from n = 1 to r of (n + 4) = (1+4) + (2+4) + … root of r² + 9r - 240 = 0 is approximately 11.72 (since 11 gives 11×20/2=110, 12 gives 12×21/2=126). So maximum integer r = 11 gives 11×20/2 = 110 stalls ≤ 120. So 11 rows per block.
设行数为r。第n行的摊位数为n+4。因此区块总摊位数是从n=1到r的(n+4)之和，即等差数列求和：(1+4)+(2+4)+…+(r+4) = Σn(1→r) + 4r = r(r+1)/2 + 4r = r(r+1+8)/2 = r(r+9)/2。我们需要r(r+9)/2 ≤ 120，解r²+9r-240 ≤ 0。方程正根约为11.72（11代入得11×20/2=110，12代入得12×21/2=126），所以最大整数r=11，此时摊位数为110 ≤ 120。因此每个区块最多11行。
The Fibonacci‑ 55, a₁₀ = 89. The 10th term is 89, which is prime (only divisible by 1 and 89). Confirmed.
类似斐波那契数列：a₁ = 1, a₂ = 2, a₃ = 3, a₄ = 5, a₅ = 8, a₆ = 13, a₇ = 21, a₈ = 34, a₉ = 55, a₁₀ = 89。第10项为89，是质数（仅能被1和89整除）。确认无误。
3. Algebraic Thinking: Ticket Pricing Strategy | 代数思维：门票定价策略
The committee proposes two ticket types: adult tickets priced at £x and child tickets at £(x - 3). On a sunny day, they expect to sell 300 adult and 500 child tickets, generating total revenue of £4,350. On a rainy day, sales drop to 120 adult and 200 child tickets, and revenue must still cover a fixed cost of £1,200 plus £0.50 per ticket sold for cleaning. Formulate equations for both scenarios. Find the ticket prices, then calculate the profit on a rainy day if the fixed cost is already met. Furthermore, if they introduce a family ticket costing twice the child price and sell 45 family tickets, 80 adult and 100 child tickets, express the new revenue algebraically and simplify.
组委会提出两种门票：成人票单价£x，儿童票单价£(x-3)。晴天时，他们预计售出300张成人票和500张儿童票，总收入为4,350英镑。雨天时，销售量降至120张成人票和200张儿童票，但收入必须覆盖固定成本1,200英镑，外加每张售票0.50英镑的清洁费。分别为两种情景建立方程。求出票价，然后计算雨天的利润（假设固定成本已覆盖）。此外，如果他们推出一种家庭票，价格是儿童票的两倍，并售出45张家庭票、80张成人票和100张儿童票，用代数表达新收入并化简。
mplify: 300x + 500x - 1500 = 4350 → 800x = 5850 → x = 5850 ÷ 800 = 7.3125. So adult ticket £ 117/16) + 500*(69/16) = (35100+34500)/16 = 69600/16 = 4350 exact. So prices are exact fractions: adult = £7.3125, child = £4.3125. Reasonable. Rainy day: revenue = 120x + 200(x - 3) = 320x - 600. Cost = £ Substitute x = 117/16: 320*(117/16) = 20*117 = 2340, profit = 2340 - 1960 = 380. So profit £0 on a rainy day. That is the profit after covering all variable and fixed costs. If asked "if fixed cost is already met" might mean we only deduct the cleaning cost? Phrasing: "revenue must still cover a fixed cost of £1,200 plus £0.50 per ticket sold for cleaning. … calculate the profit on a rainy day if the fixed cost is already met." an the fixed cost is already covered by other funds, so we only consider the cleaning cost? Actually the phrase "must still cover a fixed cost" is part of the scenario. But "if the fixed cost is already met" suggests we calculate profit as revenue minus cleaning cost only. Let's re-read: "On a rainy day, sales drop … and revenue must still cover a fixed cost of £1,200 plus £0.50 per ticket sold for cleaning. … calculate the profit on a rainy day if the fixed cost is already met." = 320x - 600 - 160 = 320x - 760. Using x=117/16 gives 2340 - 760 = 1580. So profit would be £1580. I' I think the intended is that fixed cost has been met, so only cleaning cost is deducted. I'll include that to demonstrate careful reading.
晴天方程：300x + 500(x - 3) = 4350。化简得800x - 1500 = 4350 → 800x = 5850 → x = 7.3125。成人票价£7.31（精确分数为117/16），儿童票价£4.3125（69/16）。雨天：收入 = 120x + 200(x - 3) = 320x - 600。题目要求覆盖固定成本£1200及每张票£0.50清洁费，但后面说“如果固定成本已经满足，计算雨天利润”，意味着固定成本已被其他资金覆盖，仅扣除清洁费。利润 = 收入 - 清洁费 = (320x - 600) - 0.5×(320) = 320x - 760。代入x = 117/16：320×117/16 = 20×117 = 2340，利润 = 2340 - 760 = £1580。若固定成本也算入则为£380。
) + 80x + 100(x-3) = 190(x-3) + 80x = 190x - 570 + 80x = 270x - 570. Simplify: 270x - 570.
家庭票价 = 2(x-3)。新收入表达式：45×2(x-3) + 80x + 100(x-3) = 90(x-3) + 80x + 100(x-3) = 190(x-3) + 80x = 270x - 570。
4. Geometric Reasoning: Designing the Stage Area | 几何推理：舞台区域设计
The stage is a composite shape: a rectangle of length (2y + 3) metres and width (y - 1) metres, with a semicircle of diameter equal to the width attached on one of the shorter sides. The total area must be exactly 60 m². Write an equation for y, solve it to 2 decimal places, and then find the perimeter of the stage. The committee also wants to place a triangular decoration (right‑angled) with legs in the ratio 3:4 inside the rectangular part. If the hypotenuse of this triangle is 5.5 m, determine whether it fits within the rectangle.
舞台是一个组合形状：一个矩形长(2y+3)米，宽(y-1)米，在一条短边上连接一个直径等于矩形宽的半圆。总面积必须恰好为60平方米。写出关于y的方程，求解y保留两位小数，然后计算舞台的周长。组委会还想在矩形区域内放置一个直角边比为3:4的直角三角形装饰。如果该三角形的斜边长为5.5米，判断它能否放入矩形内。
al area: 2y² + y - 3 + π(y-1)²/8 = 60. Multiply by 8: 16y² + 8y - 24 + π(y² - 2y + 1) = 480. Rearrange: (16+π)y² + (8 - 2π)y + (-24 + π - 480) = 0 → (16+π)y² + (8 - 2π)y + (π - 504) = 0. Using π ≈ 3.14159, coefficients: 19.14159y² + (8 - 6.28318)y + (3.14159 - 504) = 19.1416y² + 1.7168y - 500.8584 = 0. Solve quadratic: y = [-1.7168 ± √1.7168² + 4*19.1416*500.8584 p.: 5.07). Check negative: (-1.7168-195.86)/... negative. So y ≈ 5.07 m. Dimensions: width = y-1 = 4.07 m, length = 2y+3 = 13.14 m. Fit? Triangle legs ratio 3:4, hypotenuse = 5x? rectangle: width 4.07 m, length 13.14 m. The longer leg 4.4 m must be placed along length? ength, both legs (3.3 and 4.4) must be less than or equal to length and width respectively? d leg along length cannot exceed 13.14. 3.3 < 4.07, 4.4 < 13.14. So it fits if we place the 3.3 m leg along the width and the 4.4 m leg along the length. Yes, because 4.07 > 3.3, so it fits. Perimeter: rectangle perimeter minus the side covered by semicircle? emicircular arc. So P = 2*(length) + (width) + π*radius (since diameter is attached, the arc is half circumference). Length = 2y+3 = 13.14, width = 4.07, radius = 2.035. So P = 2*13.14 + 4.07 + π*2.035 ≈ 26.28 + 4.07 + 6.393 ≈ 36.74 m (allow 36.7 to 2 d.p.). Detailed solve is important.
矩形面积 = (2y+3)(y-1) = 2y² + y - 3。半圆面积 = ½π[(y-1)/2]² = π(y-1)²/8。总面积方程：2y²+y-3 + π(y-1)²/8 = 60。乘8得16y²+8y-24+π(y²-2y+1)=480 → (16+π)y² + (8-2π)y + (π-504)=0。带入π≈3.142，求解y ≈ 5.07（保留两位）。宽=4.07m，长=13.14m。三角形：直角边比3:4，斜边5.5m，则k=1.1，边长为3.3m和4.4m。矩形宽4.07m，可容纳3.3m的边沿宽方向，4.4m沿长方向，均小于矩形尺寸，故能放入。舞台周长：两长边+一短边+半圆弧 = 2×13.14 + 4.07 + π×2.035 ≈ 36.74m。
5. Probability and Expected Value: Lucky Draw Fundraiser | 概率与期望值：幸运抽奖筹款
The festival runs a lucky dip: 200 tickets are sold at £2 each. There are three prize tiers: 1st prize £50 (1 ticket), 2nd prize £20 (3 tickets), 3rd prize £5 (6 tickets). The rest are non‑winners. Calculate the probability of winning any prize, the expected winnings for a single ticket, and determine whether the game is favourable to the player or the organisers. If the organisers want to maintain the same prize structure but ensure an expected profit of at least £1.20 per ticket, what is the maximum number of tickets they should sell, assuming the same proportion of winning tickets? Extend: if a person buys two tickets, find the probability they win at least one prize.
文化节举办幸运抽奖：共售200张票，每张£2。奖项：一等奖£50（1张），二等奖£20（3张），三等奖£5（6张），其余无奖。计算赢得任何奖的概率、单张票的期望收益，并判断游戏对玩家还是组织者有利。若组织者希望保持相同奖项结构，但确保每张票的期望利润至少为£1.20，他们最多应售出多少张票（假设中奖票比例相同）？扩展：若某人购买两张票，求至少中一个奖的概率。
nings = (1/200)*50 + (3/200)*20 + (6/200)*5 + (190/200)*0 = (50 + 60 + 30)/200 = 140/200 = £0.70. Cost per ticket £2, so expected gain for player = 0.70 - 2 = -£1.30 (loss). Organisers' expected profit per ticket = 2 - 0.70 = £1.30. To have expected profit at least £ en probability of each prize changes with n. Expected payout per ticket = 140/n. Organiser's revenue per ticket is £, profit = 2 - 140/n ≥ 1.20 → 0.80 ≥ 140/n → n ≥ 140/0.80 = 175. So maximum n is unlimited? Actually "maximum number of tickets they should sell" o maximum, only a minimum. Perhaps they want to scale up but keep expected profit at least £1.20 per ticket. So they can sell any number greater than or equal to 175. But wording "maximum number" rizes, expected profit per ticket increases, so no upper bound. Perhaps the question meant "minimum number of tickets"? I will reinterpret: if they want to ensure an expected profit of at least £1.20 per ticket, what is the minimum number of tickets they must sell? That would be n ≥ 175. I'll address that. Or if they want to maintain proportion of winning tickets (10/200 = 5%), then for n tickets, number of winning tickets = 0.05n, prizes scaled? The phrase "assuming the same proportion of winning tickets" suggests they scale the number of prizes. Then expected winnings per ticket remains £0.70 regardless of n. Then profit per ticket is always £ hey keep the same prize structure (absolute number of each prize) but sell more tickets. I' st . I'll choose the first interpretation where prizes are fixed absolute numbers. Then we find minimum n = 175. I'll phrase it as "minimum number of tickets to achieve the desired profit" but the prompt said "maximum number" so maybe I should present both possibilities to show critical thinking. For the article, I'll present the calculation of expected value and then explore the scaling.
中奖票共1+3+6=10张。中奖概率=10/200=0.05。期望收益 = (1/200)×50 + (3/200)×20 + (6/200)×5 = 140/200 = £0.70。票成本£2，玩家期望净收益 = -£1.30，组织者期望利润 = £1.30。若保持奖项结构不变（绝对数量不变），售出n张票，期望支出仍为140/n，利润 = 2 - 140/n ≥ 1.20 → 140/n ≤ 0.80 → n ≥ 175。因此最少需要售出175张票以达到每张票利润≥£1.20。若维持中奖比例相同，利润不变，始终为£1.30。
t: P(both no prize) = (190/200)*(189/199). So P(at least one) = 1 - (190*189)/(200*199) = 1 - 35910/39800 ≈ 1 - 0.90226 = 0.09774, about 9.77%.
购买两张票，至少中一次奖的概率 = 1 - 两张都不中的概率 = 1 - (190/200)×(189/199) ≈ 0.0977。
6. Inequality Modelling: Budget Constraints for Performance Acts | 不等式建模：表演节目的预算约束
The festival can spend up to £3,000 on performance acts. A band costs £400 per show and a dance troupe costs £250 per show. The schedule must have at least 5 shows in total, and the number of band shows must be at most twice the number of dance shows. Let b be the number of band shows and d the number of dance shows. Write down three inequalities, graph the feasible region, and find the maximum number of band shows possible. If the committee prefers band shows and wants to maximise the number of band shows while spending exactly £2,800, determine the combination that satisfies the constraints.
文化节最多花费£3,000在表演节目上。一支乐队每场演出收费£400，一个舞蹈团每场收费£250。演出总场次必须至少5场，且乐队演出场数不得超过舞蹈演出场数的两倍。设b为乐队演出的场数，d为舞蹈演出的场数。写出三个不等式，绘制可行区域，并找出乐队演出的最大可能场数。如果组委会偏爱乐队，并希望在正好花费£2,800的情况下最大化乐队演出场数，找出满足约束的组合。
To maximise band shows, we look for highest b in feasible integer points. From b ≤ 2d, d ≥ b/2. Also from b + d ≥ 5 → d ≥ 5 - b. Budget: 400b + 250d ≤ 3000 → d ≤ (3000 - 400b)/250 = 12 - 1.6b. Combine d lower bound max(b/2, 5-b) ≤ d ≤ 12 - 1.6b. Test b values. For b=7, d ≥ 3.5 → d≥4 and d ≥ 5-7=-2 so d≥4; d ≤ 12-11.2=0.8 → no integer d. b=6: d ≥ 3, d ≤ 12-9.6=2.4 → no integer. b=5: d ≥ 2.5→3, d ≤ 12-8=4 → d=3 or 4 check: b=5, d=3 gives cost 2000+750=2750 ok, b≤2d? 5≤6 ok. b=5, d=4 cost 2000+1000=3000 ok, b≤2d? 5≤8 ok. b=6 not feasible. So max band shows is 5. For exact £ )/5. d integer → 56-8b divisible by 5. b=2 → d=40/5=8, check constraints: total=10≥5, b≤2d? 2≤16 yes. b=7 → d=0/5=0, b≤2d? /5 not integer. b=4 → d=24/5 not integer. b=3 → d=32/5 not integer. So only b=2, d=8 works? x b=5 without the exact cost constraint. With exact 2800, b=2 is the only integer solution? Check b=... also b=0 → d=56/5 not integer. So the only combination is b=2, d=8. That doesn't maximise band shows; it's the only feasible point. So perhaps the exact 2800 constraint restricts severely. We' s 2. Exploration: perhaps b=1, d=48/5 not integer. So answer: 2 band shows, 8 dance shows.
不等式：400b + 250d ≤ 3000；b + d ≥ 5；b ≤ 2d；整数解。可行区域绘制略。最大化b，测试b=5时d=3或4可行，b=6不可行，故最大乐队演出场数为5。若恰好花费£2,800：8b+5d=56，整数解仅b=2, d=8满足约束，故组合为2场乐队、8场舞蹈。
7. Proportional Reasoning and Scale Factors: Map of the Festival Grounds | 比例推理与缩放因子：节庆场地地图
The school provides a scale drawing of the festival ground where 1 cm represents 15 m. The actual rectangular field is 120 m by 75 m. On the plan, a circular marquee of radius 10 m is drawn. Find the area of the marquee on the plan in cm². The organisers decide to enlarge the plan so that 1 cm represents 10 m. What is the scale factor of this enlargement in terms of linear dimensions and area? If a poster is to be printed at a scale of 1:200 (as a ratio of poster length to actual length), express the width of the poster in centimetres for the 120 m side.
学校提供了一张节庆场地的比例图，比例尺为1 cm代表15 m。实际矩形场地为120 m × 75 m。图上画有一个半径为10 m的圆形帐篷。求平面图上帐篷的面积（以cm²计）。组织者决定将比例尺放大为1 cm代表10 m。问这次放大的线性比例因子和面积比例因子是多少？如果要印刷一张海报，比例尺为1:200（海报长度与实际长度之比），求120 m边对应的海报宽度（以厘米计）。
)² = (4/9)π ≈ 1.396 cm². New scale 1 cm : 10 m = 1:1000, so linear scale factor from old to new? or for enlargement = 1.5. Area scale factor = (1.5)² = 2.25. Alternatively, think directly: old plan showed 1 cm = 15 m; new plan 1 cm = 10 m, so the same actual length will be represented by a longer line on new plan. For an actual 15 m, old plan has 1 cm, new plan has 1.5 cm. So enlargement factor 1.5. For poster: scale 1:200 means 1 cm on poster represents 200 cm actual. Actual length 120 m = 12000 cm. Poster length = 12000 / 200 = 60 cm. So width of poster (the 120 m side) is 60 cm. The other side 75 m gives 75 m = 7500 cm / 200 = 37.5 cm.
原比例尺1 cm : 15 m = 1:1500。帐篷实际半径10 m = 1000 cm，图上半径 = 1000/1500 = 2/3 cm，面积 = π×(2/3)² ≈ 1.40 cm²。新比例尺1 cm:10 m = 1:1000，放大系数 = 1500/1000 = 1.5（线性），面积放大系数 = 2.25。海报比例1:200，120 m = 12000 cm，海报宽度 = 12000/200 = 60 cm。
8. Data Handling and Statistics: Analysing Attendee Satisfaction | 数据处理与统计：分析参与者满意度
A satisfaction survey of 150 attendees yielded scores out of 10: 68 people gave 8, 42 gave 7, 23 gave 9, 12 gave 6, and 5 gave 10. Calculate the mean, median and mode. Construct a frequency table and a bar chart description. Determine the interquartile range. The committee claims that "the typical satisfaction score is above 8". Evaluate this claim using your calculations. If they want to boost the mean to 8.2 by adding a new category of extremely satisfied attendees (score 10), how many additional perfect scores would be needed?
对150名参与者的满意度调查（满分10分）结果：68人打8分，42人打7分，23人打9分，12人打6分，5人打10分。计算平均数、中位数和众数。构建频率表并描述条形图。求四分位距。组委会声称“典型的满意度分数在8分以上”。利用你的计算结果评价这一说法。如果他们想通过增加一个新的“极其满意”类别（分数10）将平均分提升至8.2，需要增加多少名打满分的人？
ian position = (150+1)/2 = 75.5th value. Cumulative frequencies: score 6:12, 7:54, 8:122, 9:145, 10:150. The 75.5th lies in score 8, so median = 8. IQR: Q1 is 25th percentile (150*0.25=37.5th value) → lies in score 7 (since 12 ≤ 37.5 < 54) → Q1=7. Q3 is 112.5th value → lies in score 8? Cumulative up to 8 is 122, so 112.5th is still in 8. So Q3=8. IQR = 8-7=1. The claim "typical score above 8" n. Could argue that typical is 8, not above 8. To raise mean to 8.2 with additional n scores of 10: new sum = 1167 + 10n, new total = 150+n. Equation: (1167+10n)/(150+n) = 8.2 → 1167+10n = 8.2(150+n) → 1167+10n = 1230 + 8.2n → 1.8n = 63 → n = 35. So 35 additional perfect scores needed.
计算总和1167，平均数7.78。众数=8。中位数=第75.5个值，落在8分，中位数=8。四分位距：Q1=7，Q3=8，IQR=1。声称“典型分数高于8”，平均数低于8，中位数和众数等于8，并非高于8。要提升至平均8.2，需增加n人打10分，(1167+10n)/(150+n)=8.2，解得n=35。
9. Trigonometry Application: Sightlines for the Stage | 三角学应用：舞台视线
A viewing platform is placed 20 m from the front of the stage. The stage floor is 1.5 m above the ground, and the top of the backdrop is 4 m above the stage floor. Calculate the angle of elevation from a point 1.2 m above the ground on the platform to the top of the backdrop. If a spotlight is positioned on the ground 5 m to the side of the stage centre and aimed at the centre of the stage floor, which is 10 m away horizontally, find the angle of depression from the spotlight to that point. Also, if the spotlight is to illuminate the entire backdrop height (4 m
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