📚 Case Study Practice in Advanced Mathematics | 案例分析实战演练
Advanced Mathematics at Year 7 goes beyond routine calculations by applying logic, algebra, geometry, and data handling to real-world situations. This article presents a series of case study challenges that mirror the problem‑solving style of the AQA specification. Each case study is broken down into clear steps, with English explanations immediately followed by Chinese translations, so you can build both mathematical fluency and bilingual confidence.
七年级进阶数学超越常规计算，将逻辑、代数、几何和数据处理应用于真实情境。本文提供一系列案例分析挑战，题型贴近 AQA 考试风格。每个案例都被拆解为清晰的步骤，英文解释后紧跟中文翻译，帮助你在提升数学能力的同时建立双语信心。
1. Algebraic Equations in Action: School Fundraiser | 代数方程实战：学校筹款
A school sold tickets for a concert. Adult tickets cost £5 each and student tickets cost £3 each. The total revenue was £182, and the number of adult tickets sold was exactly twice the number of student tickets. How many of each type were sold?
某学校出售音乐会门票。成人票每张 5 英镑，学生票每张 3 英镑，总收入为 182 英镑，且售出的成人票数量正好是学生票的两倍。求两种门票各售出多少张。
Let the number of student tickets be x. Then the number of adult tickets is 2x. The money from student tickets is 3x and from adult tickets is 5 × 2x = 10x. The total is 13x.
设学生票的数量为 x，则成人票数量为 2x。学生票收入为 3x，成人票收入为 5 × 2x = 10x，总收入为 13x。
13x = 182
= 182 ÷ 13 = 14. So 14 student tickets were sold, and 2 × 14 = 28 adult tickets were sold.
两边同时除以 13：x = 182 ÷ 13 = 14。因此售出学生票 14 张，成人票 28 张。
2. Composite Shapes: Garden Design | 组合图形：花园设计
10 metres long and 6 metres wide, with a semicircular flower bed attached to one of the 6‑ gle. Find the total area of the garden and the length of its outer boundary. Use π ≈ 3.14.
一个花园由长 10 米、宽 6 米的矩形草坪和附着在其中一条 6 米边上的半圆形花坛组成。半圆直径与矩形宽度相等。求花园的总面积和外边界长度（使用 π ≈ 3.14）。
of semicircle = ½ × π × 3² = ½ × 3.14 × 9 = 14.13 m². Total area = 60 + 14.13 = 74.13 m².
矩形面积 = 10 × 6 = 60 m²。半圆半径 = 6 ÷ 2 = 3 m。半圆面积 = ½ × π × 3² = ½ × 3.14 × 9 = 14.13 m²。总面积 = 60 + 14.13 = 74.13 m²。
cle sits is not part of the boundary. Semicircle arc length = π × 3 = 9.42 m. Total boundary = 10 + 10 + 6 + 9.42 = 35.42 m.
外边界由矩形的两条长边（各 10 米）、一条宽边（6 米）和半圆弧组成，半圆直径边不在边界上。半圆弧长 = π × 3 = 9.42 米。总边界 = 10 + 10 + 6 + 9.42 = 35.42 米。
3. Ratio in Mixtures: Making a Smoothie | 混合比例：制作冰沙
A recipe for a strawberry‑banana smoothie requires strawberry purée and banana purée in the ratio 3 : 2. How much of each purée is needed to make 2 litres of smoothie? If you only have 1.5 litres of strawberry purée, what is the maximum volume of smoothie you can prepare while keeping the same ratio?
一份草莓香蕉冰沙的配方要求草莓果泥与香蕉果泥的比例为 3 : 2。制作 2 升冰沙需要各多少？如果你只有 1.5 升草莓果泥，在保持相同比例的情况下最多能制作多少升冰沙？
al parts = 3 + 2 = 5. For 2 L, strawberry = (3/5) × 2 = 1.2 L, banana = (2/5) × 2 = 0.8 L.
总份数 = 3 + 2 = 5。对于 2 升，草莓果泥 = (3/5) × 2 = 1.2 升，香蕉果泥 = (2/5) × 2 = 0.8 升。
With 1.5 L of strawberry puré Then (3/5)V = 1.5 → V = 1.5 × (5/3) = 2.5 L. So the maximum smoothie volume is 2.5 litres.
若只有 1.5 升草莓果泥，草莓对应 3 份。设总体积为 V，则 (3/5)V = 1.5 → V = 1.5 × (5/3) = 2.5 升。所以最多可制作 2.5 升冰沙。
4. Interpreting Bar Charts: Favourite Sports | 柱状图分析：最喜爱的运动
A survey recorded the favourite sports of 45 students. The bar chart shows: Football 15, Basketball 10, Swimming 8, Tennis 7, Other 5. Use the chart to answer: (a) How many students chose Basketball or Tennis? (b) What fraction of the students chose Swimming? Give the fraction in its simplest form. (c) What percentage chose Basketball?
一项调查记录了 45 名学生最喜爱的运动。柱状图显示：足球 15 人，篮球 10 人，游泳 8 人，网球 7 人，其他 5 人。请回答：(a) 选择篮球或网球的学生共有多少人？(b) 选择游泳的学生占总人数的几分之几？用最简分数表示。(c) 选择篮球的学生人数占百分之几？
r Swimming = 8/45, already in simplest form. (c) Percentage for Basketball = (10/45) × 100% = (2/9) × 100% ≈ 22.2% (to one decimal place).
(a) 篮球或网球 = 10 + 7 = 17 人。(b) 游泳所占分数 = 8/45，已是最简形式。(c) 篮球的百分比 = (10/45) × 100% = (2/9) × 100% ≈ 22.2%（保留一位小数）。
5. Number Patterns and Sequences | 数列规律
Consider the sequence: 3, 7, 11, 15, … (a) Write down the next two terms. (b) Find an expression for the nth term. (c) Is 99 a term in the sequence? Explain.
观察数列：3，7，11，15，…… (a) 写出接下来的两项。(b) 求第 n 项的表达式。(c) 99 是否为此数列中的一项？请说明理由。
Set 4n − 1 = 99 → 4n = 100 → n = 25. Since 25 is a positive integer, 99 is the 25th term.
(a) 公差为 4，因此接下来两项：19，23。(b) 对于等差数列，aₙ = a₁ + (n − 1)d。这里 a₁ = 3，d = 4，所以 aₙ = 3 + 4(n − 1) = 4n − 1。(c) 设 4n − 1 = 99，得 4n = 100，n = 25。因为 25 是正整数，所以 99 是数列的第 25 项。
6. Probability in Games: Drawing Coloured Balls | 游戏中的概率：抽取彩色球
A bag contains 5 red balls, 3 blue balls and 2 green balls. One ball is drawn at random. (a) Find the probability that it is red. (b) A ball is drawn and not replaced. A second ball is then drawn. Calculate the probability that both balls are red.
一个袋子里装有 5 个红球、3 个蓝球和 2 个绿球。随机抽取一个球。(a) 求抽到红球的概率。(b) 抽出一个球后不放回，再抽第二个球。计算两个球都是红球的概率。
, 4 reds remain out of 9 balls. P(second red) = 4/9. Multiply: (1/2) × (4/9) = 4/18 = 2/9.
总球数 = 5 + 3 + 2 = 10。(a) P(红) = 5/10 = 1/2。(b) 不放回下两个红球：P(第一个红) = 5/10 = 1/2。抽走一个红球后，剩下 4 个红球，总数剩 9 个球，P(第二个红) = 4/9。相乘得：(1/2) × (4/9) = 4/18 = 2/9。
7. Percentage Discounts: Shopping Smart | 百分比折扣：精明购物
A jacket originally costs £40. In a sale, there is a 15% discount. After applying this discount, an additional 10% off is given to loyalty‑card holders. Calculate the final price a loyalty‑card holder pays. What is the overall percentage saving compared to the original price?
一件夹克原价 40 英镑。促销期间打 15% 折扣，在此基础上，会员可再享 10% 的折扣。计算会员最终支付的价格，并求出与原价相比总共节省了百分之几。
First discount: price = 40 × (1 − 0.15) = 40 × 0.85 = £34. Second discount: final price = 34 × (1 − 0.10) = 34 × 0.90 = £30.60. Total saving = 40 − 30.60 = £9.40. Overall percentage saving = (9.40 / 40) × 100% = 23.5%.
第一次折扣后：价格 = 40 × (1 − 0.15) = 40 × 0.85 = 34 英镑。第二次折扣后：最终价 = 34 × 0.90 = 30.60 英镑。总节省 = 40 − 30.60 = 9.40 英镑。总节省百分比 = (9.40 / 40) × 100% = 23.5%。
8. Speed, Distance and Time: A Road Trip | 速度、距离与时间：公路旅行
A car travels at an average speed of 60 km/h for 2.5 hours and then increases its speed to 80 km/h for the next 1.5 hours. Calculate the total distance covered and the overall average speed for the entire journey.
一辆汽车先以 60 公里/小时的平均速度行驶 2.5 小时，接着提速至 80 公里/小时继续行驶 1.5 小时。求总行驶距离和整个行程的平均速度。
= 2.5 + 1.5 = 4 hours. Average speed = total distance ÷ total time = 270 ÷ 4 = 67.5 km/h.
第一段距离 = 60 × 2.5 = 150 公里。第二段距离 = 80 × 1.5 = 120 公里。总距离 = 150 + 120 = 270 公里。总时间 = 2.5 + 1.5 = 4 小时。平均速度 = 总距离 ÷ 总时间 = 270 ÷ 4 = 67.5 公里/小时。
9. Coordinate Geometry: Hidden Treasure | 坐标几何：藏宝图
Points A(2, 3) and B(8, 7) mark two corners of a park. (a) Find the coordinates of the midpoint of AB. (b) Calculate the length of AB, leaving your answer as a surd. (c) Determine the gradient of the line AB.
点 A(2, 3) 和 B(8, 7) 标记了公园的两个角。(a) 求 AB 中点的坐标。(b) 计算 AB 的长度，答案保留根号形式。(c) 求直线 AB 的斜率。
( (2+8)/2 , (3+7)/2 ) = (5, 5). (b) Length = √[(x₂−x₁)² + (y₂−y₁)²] = √[(8−2)² + (7−3)²] = √36 + 16 = √52 = 2√13. (c) Gradient = (y₂−y₁)/(x₂−x₁) = (7−3)/(8−2) = 4/6 = 2/3.
(a) 中点 M = ( (x₁+x₂)/2 , (y₁+y₂)/2 ) = ( (2+8)/2 , (3+7)/2 ) = (5, 5)。(b) 长度 = √[(x₂−x₁)² + (y₂−y₁)²] = √[(8−2)² + (7−3)²] = √36 + 16 = √52 = 2√13。(c) 斜率 = (7−3)/(8−2) = 4/6 = 2/3。
Published by TutorHao | Advanced Mathematics Revision Series | aleveler.com
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