📚 Year 7 Edexcel Further Maths: Case Study Practicals | Year 7 Edexcel 进阶数学：案例分析实战演练
In Year 7, Edexcel Further Maths challenges students to apply their core skills to complex, real‑world scenarios. This article presents eight detailed case studies that blend logical reasoning, algebraic thinking, geometry and data handling – all designed to stretch your problem‑solving abilities and build confidence for higher‑level mathematics.
在Year 7阶段，Edexcel进阶数学要求学生将核心技能运用到复杂的现实情境中。本文通过八个详细的案例分析，融合逻辑推理、代数思维、几何与数据处理，旨在拓展你的问题解决能力，为更高层次的数学学习打下坚实基础。
1. The Mysterious Number Machine | 神奇的数字机器
A machine follows a secret rule: it takes a number, doubles it, adds 5, then subtracts the original number. When Amelia inputs 4, the output is 9. Can you express the rule algebraically and find the input if the output is 23?
一台机器按照秘密规则运行：获取一个数，乘以2，加5，然后减去原来的数。阿米莉亚输入4时，输出为9。你能用代数式表达这个规则，并求出当输出为23时的输入吗？
Let the input be x. The machine calculates 2x + 5 − x. This simplifies to x + 5.
设输入为x。机器计算2x + 5 − x，化简后得到x + 5。
For an input of 4: 4 + 5 = 9, which matches. To find the input for an output of 23, solve x + 5 = 23, giving x = 18.
输入4时：4 + 5 = 9，吻合。要求输出23时的输入，解方程x + 5 = 23，得x = 18。
Try creating your own two‑step machine and swap with a partner. Can you work backwards to crack the code?
试着创造一个自己的两步机器，和同伴交换，你能逆向破解对方的规则吗？
2. Designing a Tessellation | 镶嵌图案设计
A local art gallery wants a wall pattern made from identical shapes that fit together with no gaps. You decide to use a regular hexagon. Can you explain why hexagons tessellate, and calculate the interior angle of one hexagon?
本地美术馆希望用完全相同的形状制作一面墙壁图案，要求紧密贴合不留空隙。你决定采用正六边形。你能解释六边形为何能密铺，并计算一个正六边形的内角吗？
Tessellation works when the angles meeting at a point sum to 360°. The interior angle of a regular polygon with n sides is given by (n − 2) × 180° ÷ n.
密铺的条件是在一个顶点处相遇的若干个角之和为360°。正n边形的内角公式为(n − 2) × 180° ÷ n。
For a hexagon (n = 6): (6 − 2) × 180° ÷ 6 = 120°. Three hexagons meet at a vertex (120° × 3 = 360°), so they tessellate perfectly.
对于六边形（n=6）：(6 − 2) × 180° ÷ 6 = 120°。三个六边形在一个顶点相遇（120° × 3 = 360°），因此可以完美密铺。
Next, you sketch a design using one hexagon and two squares around a point. Check if 120° + 90° + 90° = 300°, which is less than 360° – small triangles could fill the gap!
接着，你尝试在一个顶点周围放一个六边形和两个正方形，检验120° + 90° + 90° = 300°，小于360°，空隙可以用小三角形填补！
3. The School Fete Finance | 义卖会财务预算
The student council plans a fete and needs to price a cupcake box. The ingredients cost £0.85 per box and they want a 120% profit on cost. The selling price must include a 20% VAT on the final price. What should the label price be?
学生会筹备义卖会，需要为一盒纸杯蛋糕定价。每盒原料成本0.85英镑，他们想在成本基础上获得120%的利润。售卖价格还需包含最终售价20%的增值税。标签价格应为多少？
First find the price before tax. Profit of 120% on cost means price = cost + 1.2 × cost = 2.2 × cost. So pre‑tax price = 2.2 × £0.85 = £1.87.
首先计算税前价格。基于成本的120%利润意味着价格 = 成本 + 1.2 × 成本 = 2.2 × 成本。因此税前价格 = 2.2 × £0.85 = £1.87。
VAT is added after, so the label price = pre‑tax price × 1.20. That gives £1.87 × 1.20 = £2.24 (rounded to nearest penny).
增值税是在之后加上，所以标签价格 = 税前价格 × 1.20。即 £1.87 × 1.20 = £2.24（四舍五入到分）。
Check: If a customer pays £2.24, VAT is 20% of £2.24 = £0.448, so the council keeps £2.24 − £0.448 = £1.792 ≈ £1.79. Profit = £1.79 − £0.85 = £0.94, which is about 110.6% of cost – close enough to 120% after rounding.
检验：如果顾客支付£2.24，增值税为£2.24的20%即£0.448，学生会实收£2.24 − £0.448 = £1.792 ≈ £1.79。利润为£1.79 − £0.85 = £0.94，约为成本的110.6%——四舍五入后与120%接近。
4. Analysing a Science Experiment | 科学实验数据分析
In a lab, a spring is stretched by adding weights. The data recorded: with 0 kg the length is 10 cm; 2 kg gives 14 cm; 4 kg gives 18 cm. Plot the graph and find the equation linking length L and mass m. Predict the length for a 5 kg mass.
在实验室中，通过增加砝码拉伸弹簧。记录的数据为：0 kg时长度为10 cm；2 kg时长度为14 cm；4 kg时长度为18 cm。绘制图表并找出联系长度L和质量m的方程。预测5 kg质量时的长度。
The data show a constant increase of 4 cm per 2 kg, so the gradient is 2 cm/kg. The intercept (when m = 0) is 10 cm. Hence the linear equation: L = 2m + 10.
数据显示每增加2 kg长度增加4 cm，因此斜率为2 cm/kg。截距（m = 0时）为10 cm。所以线性方程为：L = 2m + 10。
Substituting m = 5 gives L = 2 × 5 + 10 = 20 cm. Real experiments may curve – this is an idealised direct proportion with a starting length.
代入m = 5得L = 2 × 5 + 10 = 20 cm。真实的实验可能产生曲线——这是理想化的、具有起始长度的正比例关系。
We can also use the equation to find m if L is known: for L = 25 cm, m = (25 − 10) ÷ 2 = 7.5 kg.
我们也可以利用方程由L求m：当L = 25 cm时，m = (25 − 10) ÷ 2 = 7.5 kg。
5. Crack the Code with Factors | 因数密码
A secret code uses a two‑digit number. Clues: the number is a multiple of 6, the sum of its digits is 9, and it is a factor of 108. What could the number be? Is there more than one answer?
一个密码使用一个两位数。线索如下：这个数是6的倍数，各位数字之和为9，并且是108的因数。这个数可能是什么？是否有多于一个答案？
First list multiples of 6 below 100: 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. Keep those with digit sum 9: 18 (1+8=9), 36 (3+6=9), 54 (5+4=9), 72 (7+2=9), 90 (9+0=9).
首先列出小于100的6的倍数：12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96。保留各位数字之和为9的：18 (1+8=9), 36 (3+6=9), 54 (5+4=9), 72 (7+2=9), 90 (9+0=9)。
Now check which are factors of 108. 108 ÷ 18 = 6 (integer), 108 ÷ 36 = 3, 108 ÷ 54 = 2, 108 ÷ 72 = 1.5 (no), 108 ÷ 90 = 1.2 (no). So possible numbers: 18, 36, 54.
现在检查哪些是108的因数。108 ÷ 18 = 6（整数），108 ÷ 36 = 3，108 ÷ 54 = 2，108 ÷ 72 = 1.5（不是），108 ÷ 90 = 1.2（不是）。因此可能的数字为：18、36、54。
Three solutions work – maybe the spy uses the largest one? Discuss how constraints narrow down possibilities.
有三个解——或许间谍使用最大的那个？讨论约束条件如何缩小可能性范围。
6. The Growing Squares Sequence | 不断增长的方形序列
Mia draws a pattern: step 1 has a single 1×1 square; step 2 adds a surrounding layer of squares to make a 3×3 grid; step 3 makes a 5×5 grid. How many 1×1 squares are in step n? How many squares are added when moving from step 5 to step 6?
米亚绘制了一个图案：第1步有一个1×1的正方形；第2步增加一圈包围的正方形，形成一个3×3的网格；第3步形成5×5的网格。第n步中有多少个1×1的正方形？从第5步到第6步增加了多少个正方形？
The side lengths are odd numbers: 1, 3, 5, … So at step n the side length is (2n − 1). Total squares = (2n − 1)².
边长是连续的奇数：1、3、5……因此在第n步，边长为(2n − 1)。小正方形总数 = (2n − 1)²。
Step 5: side = 2×5 − 1 = 9, total = 81 squares. Step 6: side = 11, total = 121 squares. Squares added = 121 − 81 = 40.
第5步：边长 = 2×5 − 1 = 9，总数为81个；第6步：边长 = 11，总数为121个。增加的方块数 = 121 − 81 = 40。
You can also observe the added squares form a border: at step n, the border has 4×(2n − 1) − 4 squares? Check step 2: 4×3 − 4 = 8, step 1 to 2 increase is 9 − 1 = 8. Good.
你也可以观察到增加的正方形形成一个边框：在第n步，边框有4×(2n − 1) − 4个？检验第2步：4×3 − 4 = 8，第1步到第2步增加量为9 − 1 = 8，正确。
7. The Garden Fencing Problem | 花园围栏问题
You have 20 metres of fencing to make a rectangular vegetable patch against a wall (so one side is the wall). What dimensions give the largest possible area? How does this relate to the maximum of a quadratic expression?
你有20米长的围栏，想靠着一面墙围出一个矩形的菜地（因此一边是墙）。怎样的尺寸能获得最大面积？这与二次表达式的最大值有何关联？
Let the side parallel to the wall be y, and the two perpendicular sides be x each. Then total fencing used: y + 2x = 20, so y = 20 − 2x. Area A = x × y = x(20 − 2x) = 20x − 2x².
设平行于墙的一边为y，垂直于墙的两个边各为x。使用的围栏总长：y + 2x = 20，因此y = 20 − 2x。面积A = x × y = x(20 − 2x) = 20x − 2x²。
This is a quadratic: A = −2x² + 20x. The coefficient of x² is negative, so it has a maximum. Complete the square or use symmetry: vertex at x = −b/(2a) = −20/(2×−2) = 5. So x = 5 m, then y = 20 − 2×5 = 10 m. Maximum area = 5 × 10 = 50 m².
这是一个二次式：A = −2x² + 20x。x²的系数为负，因此有最大值。可以通过配方法或对称性求解：顶点在x = −b/(2a) = −20/(2×−2) = 5。因此x = 5 m，则y = 20 − 2×5 = 10 m。最大面积 = 5 × 10 = 50 m²。
Test x = 4: A = 4×12 = 48 m²; x = 6: y = 8, A = 48 m² – both less than 50, confirming the maximum.
检验x = 4：A = 4×12 = 48 m²；x = 6：y = 8，A = 48 m²——两者均小于50，确认了最大值。
8. Is the Game Fair? | 游戏公平吗？
At a stall, you pay £1 to spin a spinner with three equal sectors: red, blue and green. If it lands on red you win £2; on blue you win £1; on green you win nothing. Is the game fair? Calculate the expected loss per game and suggest a change to make it fair.
在一个摊位上，你支付1英镑转动一个三等分的转盘：分为红色、蓝色和绿色。落在红色区赢得2英镑；蓝色区赢得1英镑；绿色区赢得0英镑。这个游戏公平吗？计算每局游戏的期望损失，并提出一个修改方案使其公平。
Probability of each colour = 1/3. Expected win = (1/3)×£2 + (1/3)×£1 + (1/3)×£0 = £(2/3 + 1/3) = £1.
每种颜色的概率均为1/3。期望赢利 = (1/3)×£2 + (1/3)×£1 + (1/3)×£0 = £(2/3 + 1/3) = £1。
Since you pay £1, the expected net result = £1 − £1 = £0 – it appears fair! But wait: the expected win exactly equals the cost, so in the long run the stall makes no profit and the player loses nothing on average. This is a fair game.
由于你支付了1英镑，净期望结果 = £1 − £1 = £0——看起来是公平的！等一下：期望赢利恰好等于成本，因此长期来看摊主没有盈利，玩家平均而言也没有损失。这是一个公平游戏。
What if the stall wants a small profit? Increase the red prize to £2.50? Then expected win = (1/3)×£2.50 + (1/3)×£1 = £1.17, net expectation = £0.17 profit for stall per game. For fairness, ensure expected win equals cost.
如果摊主想获得少量利润呢？把红色奖金提高到£2.50？那么期望赢利 = (1/3)×£2.50 + (1/3)×£1 = £1.17，每局净期望为摊主盈利£0.17。要保证公平，需使期望赢利等于成本。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
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