AQA A-Level Further Maths Paper Jun 22 Unit 5 Walkthrough & Revision Guide | AQA 进阶数学 2022年6月 Unit 5 真题解析与备考指南

📚 AQA A-Level Further Maths Paper Jun 22 Unit 5 Walkthrough & Revision Guide | AQA 进阶数学 2022年6月 Unit 5 真题解析与备考指南

This article provides a detailed, question-by-question analysis of the AQA A-Level Further Mathematics Paper Unit 5 (MAD05) from the June 2022 series. We will break down every question type, identify the core mathematical techniques required, and highlight common pitfalls so you can maximise your marks in future exam sessions.

本文深入解析 AQA 进阶数学（Further Mathematics）Unit 5（试卷代码 MAD05）2022 年 6 月考试真题。我们将逐一拆解各题型的解题思路、所需核心技巧以及常见失分点，帮助你在未来的考试中稳拿高分。

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1. Unit 5 Overview: What to Expect | 第五单元概述：考什么

Unit 5 of the AQA Further Mathematics A-Level, formally known as Decision Mathematics (MAD05), assesses your ability to model real-world problems using algorithms, graphs, networks, and linear programming. The June 2022 paper adhered strictly to the specification, weighting marks towards standard algorithms such as Dijkstra, critical path analysis, and the simplex method.

AQA 进阶数学第五单元（正式名称为决策数学，代码 MAD05）考查的是运用算法、图论、网络与线性规划对现实问题进行建模的能力。2022 年 6 月的试卷严格遵循考纲要求，将主要分值分配在 Dijkstra 算法、关键路径分析以及单纯形法等标准算法上。

The paper is 1 hour 30 minutes long, worth 75 marks, and is a calculator-permitted exam. It expects you to show clear, methodical working, as many marks are awarded for process rather than final answers.

本试卷考试时长为 1 小时 30 分钟，满分 75 分，允许使用计算器。阅卷标准十分看重解题过程的清晰与条理，大部分分数属于过程分，而非最终答案分。

To succeed in this paper, you must memorise the precise steps of each algorithm. In the June 2022 paper, students who wrote down each iteration of the algorithm reliably scored far higher than those who attempted to do complex calculations in their head.

要在本试卷中取得好成绩，你必须精确记忆每个算法的具体步骤。在 2022 年 6 月的考试中，那些认真写出每一次迭代过程的考生，其得分远高于试图用心算完成复杂计算的考生。

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2. Question 1 Breakdown: Shortest Path and Dijkstra’s Algorithm | 第 1 题解析：最短路径与 Dijkstra 算法

The first question typically introduces a network diagram with between 6 and 9 nodes. In June 2022, the network represented a delivery route between warehouses. You were asked to apply Dijkstra’s algorithm to find the shortest path from the start node to a specified target node.

第一题通常会给出一个有 6 到 9 个节点的网络图。在 2022 年 6 月的试卷中，该网络图模拟的是仓库之间的配送路线。题目要求你运用 Dijkstra 算法找出从起始节点到指定目标节点的最短路径。

Scoring the full marks on this question requires strict adherence to the boxing convention: each working node must show a permanent label in a box, a temporary label above it, and a clear order of permanent labelling listed at the side of the diagram.

要在本题中拿满分，必须严格遵守方框标注规范：每个工作节点要有方框内的永久标号、方框上方的临时标号，并在图旁清楚列出永久标号的先后顺序。

The key working for Dijkstra’s algorithm is as follows:

Dijkstra 算法的关键步骤如下：

• Step 1: Label the start node with a permanent label of 0 in a box.

  第一步： 将起始节点的永久标号设为 0，写入方框内。

• Step 2: For each node connected to the most recently boxed node, calculate a temporary label by adding the edge weight to the permanent label. If it is lower than any existing label, update it.

  第二步： 对与最新打框节点相连的每个节点，用永久标号加上边的权值计算出临时标号。若该值低于现有标号，则进行更新。

• Step 3: Choose the smallest temporary label, box it permanently, and record the order of labelling.

  第三步： 选择最小的临时标号，将其永久打框，并记录标号顺序。

• Step 4: Repeat until the target node is permanently boxed. Then trace back to identify the shortest path.

  第四步： 重复上述过程，直到目标节点被永久打框。然后反向追踪找出最短路径。

Distance to node = min(previous distance, distance via new permanent node)

节点距离 = min(原有距离，经由新永久节点的距离)

A common error in this question was failing to update a temporary label when a shorter alternative route was discovered. In the 2022 paper, one node had two potential incoming paths, and many candidates left the larger value unupdated, losing both accuracy and method marks.

本题的常见错误是：当发现更短的替代路径时，未能更新已有的临时标号。在 2022 年试卷中，有一个节点存在两条入边路径，许多考生保留了较大的旧值未更新，从而同时丢掉了准确分和过程分。

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3. Question 2: Prim’s Algorithm and Minimum Spanning Tree | 第 2 题：Prim 算法与最小生成树

The second question in the June 2022 paper required you to apply Prim’s algorithm to a distance matrix to find the minimum spanning tree (MST). The matrix had 6 entries, making it a straightforward test of your ability to systematically select the smallest available connection.

2022 年 6 月试卷的第二题要求你基于一个距离矩阵应用 Prim 算法，找出最小生成树（MST）。该矩阵为 6 节点规模，考查的是你是否能系统性地选出当前可选的最小连接。

When applying Prim’s algorithm to a matrix, you must alternate between selecting a row and then highlighting acceptable columns, effectively “switching” between the visited set and the unvisited set.

在矩阵上应用 Prim 算法时，你需要在行与列之间来回切换：选定一行（已访问节点集合）后，标记该行中所有通往未访问节点的可用列，再从这些可用值中选出最小值，并划去新列。

The full process is outlined below:

完整过程如下：

• Step 1: Choose any starting vertex. Delete its row.

  第一步： 任选一个起始顶点，删除其所在行。

• Step 2: Scan the remaining columns of the deleted row for the smallest entry.

  第二步： 在已删除行的对应列中扫描最小元素。

• Step 3: Add the corresponding edge to the tree. Delete the column and row of the newly added vertex.

  第三步： 将该元素对应的边加入生成树，删除新加入顶点所对应的列和行。

• Step 4: Scan all remaining rows for the smallest entry across the whole reduced matrix. Repeat until all vertices are included.

  第四步： 在整个缩减后的矩阵中，扫描剩余所有行，选出最小的元素。重复此过程，直到所有顶点均被包含。

A key observation from the 2022 mark scheme is that marks were awarded for selecting each distinct edge correct, and that one mark was specifically reserved for writing down the total weight of the MST. In this instance, the total weight was calculated as the sum of the six selected edges.

从 2022 年的评分标准可以看出，每正确选择一条边得相应分数，另有一分专门用于计算最小生成树的总权重。此题中，总权重为所选 6 条边的权重之和。

Total Weight of MST = w(e₁) + w(e₂) + w(e₃) + w(e₄) + w(e₅) + w(e₆)

最小生成树总权重 = w(e₁) + w(e₂) + w(e₃) + w(e₄) + w(e₅) + w(e₆)

The most common error students made on this question was starting Prim’s algorithm correctly but then accidentally reading across the wrong row after the third iteration, thereby adding an incorrect edge to the tree. Always cross out rows and columns immediately after processing them.

考生在此题中最常见的错误是：Prim 算法开头正确，但在进行到第三次迭代后误读了矩阵中的错误行，从而向生成树中加入了错误边。务必在每步处理完成后立即划掉对应的行与列，以绝后患。

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4. Question 3: Critical Path Analysis and Gantt Charts | 第 3 题：关键路径分析与甘特图

Question 3 in the 2022 paper focused on critical path analysis (CPA). Candidates were given a list of activities with durations and predecessor dependencies, and were required to construct an activity network, calculate earliest start times (EST) and latest start times (LST), and identify the critical path.

2022 年试卷的第三题考查关键路径分析（CPA）。题目给出若干活动的持续时间及其紧前依赖关系，要求考生构建活动网络图、计算最早开始时间（EST）与最迟开始时间（LST），并识别出关键路径。

To compute the earliest event times, you perform a forward pass: starting from event 1 at time 0, add the duration of each activity to reach subsequent events. When multiple paths converge on one event, the EST is the maximum of all incoming path totals.

计算最早事件时间需要执行正向遍历：从事件 1 的时间 0 出发，加上每个活动的持续时间到达后续事件。当多条路径汇合于同一事件时，该事件的 EST 取所有入边路径总耗时中的最大值。

To compute the latest event times, you perform a backward pass: starting from the final event with its EST as the completion time, subtract activity durations. When multiple paths diverge from one event, the LST is the minimum of all outgoing path differences.

计算最迟事件时间需要执行逆向遍历：从最终事件开始，以其 EST 作为总完工时间，逐条减去活动持续时间。当一个事件发散出多条路径时，该事件的 LST 取所有出边路径差值中的最小值。

The crucial definitions you must cite in your exam answer are:

考试中你必须准确引用以下关键定义：

Term / 术语	Definition / 定义
Earliest Start Time (EST) / 最早开始时间	The earliest time at which an activity can start, given all predecessor activities have finished. / 在所有紧前活动均已完成的前提下，某活动最早可以开始的时间。
Latest Start Time (LST) / 最迟开始时间	The latest time at which an activity can start without delaying the overall project. / 在不延误整个项目工期的前提下，某活动最晚可以开始的时间。
Float (Slack) / 浮动时间（松弛）	LST – EST, indicating how much delay an activity can tolerate. / LST 减去 EST，表示某活动可容忍的延迟量。
Critical Path / 关键路径	The path of activities with zero total float from start to finish. / 从项目开始到结束，总浮动时间为零的活动路径。

In the June 2022 paper, the critical path consisted of five activities. A follow-up part asked you to draw a Gantt chart showing all activities scheduled at their earliest start time. Marks were awarded for correctly representing the duration of each bar and aligning them under a correct project time axis.

在 2022 年 6 月的试卷中，关键路径包含 5 项活动。后续一问要求你绘制甘特图，显示所有活动按最早开始时间排程的情况。正确绘制每个活动条的持续时间并将其对齐在正确的项目时间轴上，即可获得相应分数。

The float of a non-critical activity in 2022 was calculated using the formula below:

2022 年试卷中非关键活动的浮动时间按如下公式计算：

Total Float = LST(start event) – EST(start event) – duration

总浮动时间 = LST(起始事件) – EST(起始事件) – 活动持续时间

Be careful: many candidates used the event LST instead of the LST of the start event’s outgoing activity, which led to subtle errors in float calculation of 1 or 2 units.

务必小心：许多考生错误地使用了事件本身的 LST 来代替该起始事件发出活动的 LST，这会导致浮动时间出现 1 至 2 个单位的计算偏差。

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5. Question 4: Linear Programming and Graphical Solution | 第 4 题：线性规划与图解法

Question 4 introduced a linear programming problem with two variables, x and y. In June 2022, the context involved a furniture workshop buying timber and labour hours. You were required to formulate the constraints, plot the feasible region, and identify the optimal solution using the objective function line.

第四题是一个包含两个变量 x 和 y 的线性规划问题。2022 年 6 月的背景设定是家具工坊采购木材与工时资源。题目要求你建立约束条件、绘制可行域，并通过目标函数线确定最优解。

Your constraints in such problems generally take forms similar to the following:

此类问题中，你的约束条件通常具有类似以下的形式：

2x + 3y ≤ 120 (timber / 木材约束)

x + 2y ≤ 80 (labour / 工时约束)

x ≥ 0, y ≥ 0 (non-negativity / 非负约束)

The June 2022 paper gave a profit objective of P = 5x + 4y. To solve this graphically, you must first plot both constraint lines, shade the infeasible side, then slide the objective line through the feasible region to find the vertex giving the maximum profit.

2022 年 6 月试卷给出的利润目标函数为 P = 5x + 4y。要图解此题，你首先画出两条约束直线，将不可行区域涂上阴影，然后通过可行域滑动目标函数线，找到使利润最大的顶点。

The mark scheme expected you to clearly identify the optimal vertex coordinates. In this instance, the optimal vertex was the intersection of the two constraints, solved by simultaneous equations:

评分标准要求你明确给出最优顶点的坐标。此题最优顶点恰好是两条约束线的交点，通过联立方程求解：

2x + 3y = 120

x + 2y = 80

∴ x = 20, y = 30

∴ x = 20, y = 30

P = 5(20) + 4(30) = 100 + 120 = 220

P = 5(20) + 4(30) = 100 + 120 = 220

The key to scoring all marks here was to test the objective function at every relevant vertex, not just the visually optimal one. In 2022, testing the intermediate integer coordinates was part of the method mark requirement.

在此题拿满分的要点是：不仅要检查视觉上最优点，还要逐一测试可行域中每个相关顶点的目标函数值。在 2022 年的考试中，检查额外的整数坐标是过程分的组成部分。

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6. Question 5: The Simplex Method | 第 5 题：单纯形法

Question 5 of the June 2022 paper was a standard three-variable linear programming problem that required the simplex method to be applied to a given initial tableau. The tableau already contained slack variables, so the main task was to perform row operations until an optimal tableau was reached.

2022 年 6 月试卷的第五题是一道标准的三变量线性规划题，要求你在给定初始单纯形表上应用单纯形法进行迭代。由于表中已经给出了松弛变量，你的主要任务就是执行行变换，直至得到最优单纯形表。

The simplex method steps you must demonstrate in your working are:

单纯形法需要在答题过程中展现的核心步骤如下：

• Step 1: Identify the pivot column by selecting the most negative value in the objective row.

  第一步： 选择目标函数行中负值最小的列作为主元列。

• Step 2: Identify the pivot row by the smallest non-negative ratio of RHS to pivot column value.

  第二步： 用右端项（RHS）除以主元列中各系数，取最小非负比值所在行为主元行。

• Step 3: Scale the pivot row so the pivot element equals 1.

  第三步： 将主元行进行缩放，使主元元素变为 1。

• Step 4: Use row operations to eliminate the pivot column entries in all other rows, including the objective row.

  第四步： 通过行变换将主元列在其他所有行（含目标函数行）中的元素全部消为零。

• Step 5: Repeat until no negative values appear in the objective row.

  第五步： 重复上述步骤，直到目标函数行中不再出现负值。

In this paper, two iterations were required to reach the optimal tableau. Students were then asked to state the optimal values of the three decision variables and the maximum value of the objective function.

在本次试卷中，经过两次迭代即可达到最优单纯形表。题目进一步要求你写出三个决策变量的最优值以及目标函数的最优值。

A common pitfall identified by examiners in 2022 was incorrect entry of the objective row sign convention. Remember that in the standard AQA tableau, the objective function is rearranged so that all variables are on the left and the constant term on the right, with coefficients written with opposite signs.

2022 年考官报告指出，考生常见错误出在目标函数行的符号约定上。请牢记 AQA 标准单纯形表的规范：目标函数写成变量项在左、常数项在右的形式，填入表格时系数取相反符号。

The pivot ratio calculation is performed using the formula:

主元比值的计算方法为：

Ratio = RHS ÷ pivot column value (ignoring non-positive values)

比值 = 右端项 ÷ 主元列数值（忽略非正值）

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7. Question 6: Transportation and Stepping-Stone Method | 第 7 题：运输问题与踏脚石法

The sixth question was a transportation problem with two supply nodes and three demand nodes. The initial basic feasible solution was to be found using the north-west corner method, followed by one application of the stepping-stone method to test for optimality.

第六题是一个包含 2 个供应节点和 3 个需求节点的运输问题。要求先用西北角法求出初始基本可行解，然后用踏脚石法进行一次最优性检验。

The north-west corner method is executed as follows:

西北角法的操作过程如下：

• Step 1: Starting from the top-left cell, allocate as much as possible of supply to the first demand.

  第一步： 从表格左上角的单元格开始，尽可能多地将供应分配给第一个需求。

• Step 2: If a row’s supply is exhausted, move down; if a column’s demand is satisfied, move right.

  第二步： 若某行供应耗尽则下移；若某列需求满足则右移。

• Step 3: Continue this movement until all supply and demand is exhausted.

  第三步： 持续此过程，直到所有供求分配完毕。

Once the initial solution was found, the question asked you to apply the stepping-stone method. To find the improvement index for each unoccupied cell, you must trace a closed loop with only right-angle turns, alternating plus and minus signs starting with plus at the empty cell.

初始解完成后，题目要求运用踏脚石法进行判断。对于每个未占用空格，你需要绘制一条由直角转弯构成的闭合回路，以空格为起点，依次交替标出正号和负号。

The improvement index is calculated as the algebraic sum of the costs along the closed loop:

改进指数的计算方式是沿闭合回路对成本求代数和：

Improvement Index = Σ(costs at + positions) – Σ(costs at – positions)

改进指数 = Σ(正位置上的成本) – Σ(负位置上的成本)

In this question, the improvement index for the empty cell in row 2, column 1 was negative, indicating that the initial solution was not optimal. The exam then asked you to perform one full reallocation cycle, adjusting the quantities around the loop.

本题中，第 2 行第 1 列空格的改进指数为负值，表明初始解并非最优。随后题目要求你执行一次完整的再分配循环，沿回路调整分配数量。

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8. Question 7: Game Theory (Zero-Sum Games) | 第 7 题：博弈论（零和博弈）

The final major question in the June 2022 paper dealt with game theory. A two-player zero-sum game was given in payoff matrix form, where player A could choose from 3 strategies and player B from 3 strategies. You were required to determine whether the game had a stable solution (saddle point) and, if not, to find the optimal mixed strategy.

2022 年 6 月试卷的最后一道大题考查博弈论。题目给出了一个双人零和博弈的支付矩阵：玩家 A 有 3 种可选策略，玩家 B 也有 3 种策略。要求你先判断该博弈是否存在稳定解（鞍点），若不存在，则进一步求解最优混合策略。

To check for a saddle point, you must identify the maximin and minimax values:

判断鞍点需要找出最大值中的最小值和最小值中的最大值：

• Maximin: For each row (A’s strategies), find the minimum payoff. Then select the maximum of these row minima.

  最大最小： 对每一行（玩家 A 的策略），找出该行的最小支付值；再在这些行最小值中选取最大值。

• Minimax: For each column (B’s strategies), find the maximum payoff. Then select the minimum of these column maxima.

  最小最大： 对每一列（玩家 B 的策略），找出该列的最大支付值；再在这些列最大值中选取最小值。

Saddle point exists if and only if maximin = minimax

鞍点存在当且仅当最大值中的最小值 = 最小值中的最大值

In the 2022 exam, maximin = 3 while minimax = 4, so no saddle point existed. This prompted a follow-up question requiring the use of a linear programming formulation to solve the mixed strategy game.

在 2022 年考试中，maximin = 3，而 minimax = 4，因此鞍点不存在。这引导出后续问题：要求运用线性规划构建模型来求解混合策略。

The standard formulation for player A’s mixed strategy probabilities is to let p denote the probability of choosing strategy A1 and q = 1 – p for A2. You then solve the equality condition for the two pure strategies of B:

求解玩家 A 混合策略概率的标准方法是：设 p 为选择策略 A1 的概率，则 q = 1 – p 是选择 A2 的概率。接着对 B 的两个纯策略建立收益相等条件：

3p + 5(1-p) = 4p + 2(1-p)

3p + 5(1-p) = 4p + 2(1-p)

3p + 5 – 5p = 4p + 2 – 2p

3p + 5 – 5p = 4p + 2 – 2p

5 – 2p = 2 + 2p

5 – 2p = 2 + 2p

p = 0.75

p = 0.75

The value of the game was then found by substituting back into either expression, giving a value of 4.5. Candidates who clearly showed the formation of the linear equations and the substitution step earned all method and accuracy marks.

将 p 值代回任一表达式即可得博弈值为 4.5。能够清楚写出建立线性方程以及代入求解过程的考生，可获得全部过程分与准确分。

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9. Common Mistakes and Examiner Reports | 常见错误与考官报告要点

The examiner’s report for the June 2022 Unit 5 paper identified several recurring issues across all question types. These are worth taking seriously as you prepare for your own exam.

2022 年 6 月 Unit 5 的考官报告总结了所有题型中反复出现的若干问题。这些内容对你的备考至关重要。

The biggest general mistake was poor time allocation. Many students spent excessive time on the critical path analysis question and rushed through the simplex method at the end, losing easy marks on arithmetic.

最大的通病是时间分配不当。许多考生在关键路径分析题上耗时过多，却在最后的单纯形法上仓促作答，白白丢失了本可轻松拿到的算术分。

Another frequent error involved reading values directly from the network diagram without checking whether they were connected by an edge. In Dijkstra’s algorithm, each temporary label must be based on a direct edge from the newly permanent node.

另一个高频错误是：考生直接读取网络图上两个节点的数值，却不检查这两个节点之间是否存在边。在 Dijkstra 算法中，每个临时标号必须基于最新永久节点出发的直接边计算。

Exam advice from AQA highlights the need for neat, labelled tables when performing algorithms. In the simplex method, specifically, an untidy tableau with overwritten values causes problems for examiners trying to follow your method, resulting in lost method marks.

AQA 的考试建议强调：在执行算法时必须保持表格整洁并清楚标注。特别是在单纯形法中，涂改混乱的单纯形表不仅让考官难以理解你的思路，还会导致过程分的损失。

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10. Revision Strategy for Unit 5 | 第五单元备考策略

Unit 5 is one of the most methodical papers in A-Level Mathematics. A disciplined, algorithm-focused revision approach will pay off more than any attempt to memorise specific past paper answers.

第五单元是进阶数学考试中最讲究方法论笔答的试卷之一。采用一套纪律明确、以算法为核心的复习策略，远胜于死记硬背往年试卷的具体答案。

We recommend the following three-stage revision process, tailored to the June 2022 paper style:

针对 2022 年 6 月试卷风格，我们推荐以下三个阶段复习法：

• Stage 1 – Master the mechanics: Spend two weeks rewriting every algorithm from memory. For Dijkstra, the simplex method, and the stepping-stone method, you should be able to reproduce every table or diagram step from blank paper.

  第一阶段 – 掌握操作机制： 花两周时间凭记忆重写每个算法的全部步骤。对于 Dijkstra、单纯形法及踏脚石法，你必须能够从一张白纸开始，完整还原每一步表格或图示。

• Stage 2 – Practice with time pressure: Attempt at least one past paper every three days under full exam conditions. Mark yourself strictly against the AQA official mark schemes and record your score for each question type.

  第二阶段 – 限时实战训练： 每三天在完全模拟考试的条件下做一套往年真题。严格按照 AQA 官方评分标准自评，并记录各题型得分情况。

• Stage 3 – Targeted error analysis: In the final week before the exam, review only the question types where you lost the most marks in Stage 2. This is the most efficient final-stage method to convert borderline grades into strong passes.

  第三阶段 – 定向错误分析： 考前的最后一周，只复习第二阶段中失分最多的题型。这是冲刺阶段最有效的提分手段，能将边缘分数转化为优异的考试成绩。

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11. Key Formulas and Definitions Revision Sheet | 关键公式与定义速查表

The following summary table contains every core formula and definition that appeared in the June 2022 paper. It is an excellent quick-revision sheet for the night before the exam.

下表汇总了 2022 年 6 月试卷中出现的全部核心公式与定义，是考前最后一晚绝佳的快速复习材料。

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Topic / 主题	Formula / Definition / 公式 / 定义
Dijkstra / 最短路径	d(v) = min(d(v), d(u) + w(u,v)) / 节点距离 = min(原距离, 前一节点距离 + 边权)