Mathematics Admissions Test (MAT) by Oxford
The Mathematics Admissions Test (MAT) is a highly selective entrance exam used by top UK universities like Oxford, Imperial College London, and Warwick for admission to mathematics and related courses. Designed to assess a student’s mathematical reasoning, problem-solving skills, and creativity, the MAT is taken by thousands of aspiring mathematicians worldwide. Scoring well in MAT is often a key component in securing an interview and ultimately receiving an offer. In this post, I’ve shared complete, carefully explained MAT 2024 solutions, crafted to help students build deep understanding and confidence — whether you’re preparing for MAT 2025 or simply exploring advanced problem-solving techniques.
Oxford University MAT Mathematics (Mathematics Admissions Test)
MAT Elite Edition
Master Advanced Mathematical Thinking
Exclusive Exam-Style Solutions with Commentary | Rishabh’s Insight | June 2025 Edition
Mathematics Elevate Academy
Excellence in Oxford Math Preparation
Rishabh Kumar
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Disclaimer
© 2025 Mathematics Elevate Academy. Math by Rishabh.
This material is for personal educational use only. Redistribution or reproduction without prior written consent is prohibited.
This document provides solutions inspired by past MAT Mathematics papers by Oxford University. All questions are paraphrased to support exam preparation and avoid reproducing copyrighted content.
MAT is administered by Oxford University, which retains rights to the original questions and format. This resource is independently created and not affiliated with or endorsed by Oxford University.
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Introduction
Welcome to the MAT Elite Edition — a collection of problem-solving techniques and examiner-style solutions for the MAT Mathematics examination.
The MAT is Oxford’s gateway to its mathematics programs, testing creative problem-solving and mathematical agility. This booklet sharpens your skills and develops the problem-solving maturity needed to excel.
This guide will help you:
– Think Like a Mathematician: Embrace creative mathematical argumentation.
– Master MAT-Specific Techniques: Tackle problems aligned with MAT expectations.
– Structure Your Work for Marks: Write solutions that match examiner expectations.
– Identify Common Pitfalls: Avoid common mistakes.
Targeting Oxford’s mathematics program? This guide is your edge. Apply now:
Apply for Personalized MAT Mentorship: https://forms.gle/D2d8C6KcCyehqcf6A
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Problem Statement — Q26
Two vertical poles stand on a horizontal field. The top of the taller pole is point A, and the top of the shorter pole is point B. An observer at point P on the ground is on the line connecting the bases of the poles.
The observer measures the angle of elevation to the top of both poles as θ.
Known distances:
– From P to A: 150 meters.
– From P to B: 90 meters.
– Between A and B: 154 meters.
Calculate the height of the taller pole, h.
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Detailed Step-by-Step Solution
Strategy: Solve in 3D by breaking it into horizontal and vertical components using Pythagoras’s theorem.
Step 1: Define variables and relationships from the angle of elevation.
Let h be the height of the taller pole (A) and h_B the height of the shorter pole (B). Let X_A be the base of A and X_B the base of B. P is on the line between X_A and X_B.
For triangle PX_A A:
– Hypotenuse PA = 150 m.
– Angle of elevation ∠APX_A = θ.
– Height h = AX_A = 150 sin θ.
– Ground distance PX_A = 150 cos θ.
For triangle PX_B B:
– Hypotenuse PB = 90 m.
– Angle of elevation ∠BPX_B = θ.
– Height h_B = BX_B = 90 sin θ.
– Ground distance PX_B = 90 cos θ.
Step 2: Calculate horizontal and vertical separations between A and B.
The distance AB = 154 m is the hypotenuse of a right-angled triangle formed by horizontal and vertical separations.
– Horizontal Separation (d_horiz): P is between the poles, so:
d_horiz = X_A X_B = PX_A + PX_B = 150 cos θ + 90 cos θ = 240 cos θ
– Vertical Separation (d_vert): Difference in heights:
d_vert = h — h_B = 150 sin θ — 90 sin θ = 60 sin θ
Step 3: Apply Pythagoras’s theorem in 3D.
AB² = d_horiz² + d_vert²
154² = (240 cos θ)² + (60 sin θ)²
23716 = 57600 cos² θ + 3600 sin² θ
Step 4: Solve the trigonometric equation for sin θ.
Use cos² θ = 1 — sin² θ:
23716 = 57600 (1 — sin² θ) + 3600 sin² θ
23716 = 57600–57600 sin² θ + 3600 sin² θ
23716 = 57600–54000 sin² θ
54000 sin² θ = 57600–23716 = 33884
sin² θ = 33884 / 54000 ≈ 0.62748
sin θ ≈ 0.792136 (since θ is acute, sin θ is positive).
Step 5: Calculate h.
h = 150 sin θ ≈ 150 × 0.792136 ≈ 118.820
Rounded to three significant figures: h = 119 meters.
Final Answer: 119 m
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Visualization of the Problem
To solve, visualize the geometry with two types of triangles.
1. Vertical Right-Angled Triangles:
– Triangles formed by the observer, pole base, and pole top. The angle of elevation θ is in these triangles (PX_A A and PX_B B).
2. Conceptual 3D Triangle:
– The distance AB (154 m) is the hypotenuse of a right-angled triangle with:
— Horizontal separation: d_horiz = 240 cos θ
— Vertical separation: d_vert = 60 sin θ
This allows the use of Pythagoras’s theorem.
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Marking Criteria
Total Marks: 6
Marks allocated as follows:
– [M1] Express heights and ground distances in terms of θ (e.g., h = 150 sin θ, PX_A = 150 cos θ).
– [A1] Form correct expressions for horizontal (240 cos θ) and vertical (60 sin θ) separations.
– [M1] Set up Pythagoras: 154² = (240 cos θ)² + (60 sin θ)²
– [M1] Attempt to solve the trigonometric equation using sin² θ + cos² θ = 1.
– [A1] Solve for sin θ or sin² θ (≈ 0.7921).
– [A1] Calculate h ≈ 119 m, rounded appropriately.
Note: Full marks for correct answer (119 m) with clear working.
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Error Analysis: Common Pitfalls
Mistake Area: Geometric Setup
– Error: Confusing sine and cosine (e.g., h = 150 cos θ).
Fix: Use SOHCAHTOA. Height is opposite to θ, so use sine (O/H).
– Error: Incorrect horizontal separation (e.g., 150 cos θ — 90 cos θ).
Fix: P is between poles, so sum the ground distances.
Mistake Area: Algebraic Errors
– Error: Squaring terms incorrectly (e.g., (240 cos θ)² as 240 cos² θ).
Fix: Square both coefficient and function: (240 cos θ)² = 57600 cos² θ.
– Error: Errors in rearranging trigonometric equation.
Fix: Write each step methodically. Check distribution: 57600 (1 — sin² θ) = 57600–57600 sin² θ.
Mistake Area: Calculation Errors
– Error: Using degrees mode instead of radians.
Fix: Work with sin θ directly, not θ.
– Error: Premature rounding of sin θ.
Fix: Keep full value in calculator until final answer.
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Rishabh’s Insight: Thinking in Components
This problem shows how a complex 3D puzzle simplifies with component thinking.
The key is realizing the 154-meter distance (A to B) is the hypotenuse of a right-angled triangle with:
1. Run: Horizontal distance between pole bases.
2. Rise: Vertical distance between pole tops (height difference).
This is 3D Pythagoras: (Total Distance)² = (Horizontal Distance)² + (Vertical Distance)².
Use θ to find:
– Heights and ground distances via sine and cosine.
– Horizontal run: 150 cos θ + 90 cos θ.
– Vertical rise: 150 sin θ — 90 sin θ.
Plug into Pythagoras, then solve the resulting trigonometric equation. Decomposition into perpendicular components is a powerful strategy for 3D geometry.
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Key Takeaways
– Deconstruct 3D Problems: Break into 2D right-angled triangles.
– Angle of Elevation: Height = Hypotenuse × sin θ, Ground Distance = Hypotenuse × cos θ.
– Pythagoras in 3D: Distance² = (Δx)² + (Δz)².
– Trigonometric Equations: Use sin² θ + cos² θ = 1 to simplify.
– Problem Interpretation: Note phrases like “directly between” for geometric clarity.
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Foundational Concepts
Trigonometry in Three Dimensions
1. Right-Angled Trigonometry in Vertical Planes:
Angles of elevation are in vertical planes. Use SOHCAHTOA for height and ground distance.
2. Pythagoras’s Theorem in 3D:
Distance d in 3D: d² = (Δx)² + (Δy)² + (Δz)².
Here, simplify to: (Slant Distance)² = (d_horiz)² + (d_vert)².
3. Solving Equations with Trigonometric Identities:
Use sin² θ + cos² θ = 1 to convert equations with two trig functions into one.
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Practice Problems
Practice Problem 1: The Cuboid
A rectangular box has length AB = 8 cm, width BC = 6 cm, height CG = 5 cm (G above C).
a) Find the length of base diagonal AC.
b) Find the length of space diagonal AG.
c) Find the angle AG makes with base plane ABCD.
Solution:
a) In triangle ABC (right-angled):
AC² = 8² + 6² = 64 + 36 = 100
AC = 10 cm
b) In triangle ACG:
AG² = 10² + 5² = 100 + 25 = 125
AG = 5√5 ≈ 11.2 cm
c) Angle ∠CAG = α:
tan α = CG / AC = 5 / 10 = 0.5
α ≈ 26.6°
Practice Problem 2: The Pyramid
A right pyramid has a square base with side 10 m. Apex V is 12 m above the base center.
a) Find the length of a slant edge (e.g., VA).
b) Find the angle a slant edge makes with the base.
Solution:
a) Base diagonal AC = √(10² + 10²) = 10√2 m. Center to corner OA = 5√2 m.
In triangle VOA:
VA² = 12² + (5√2)² = 144 + 50 = 194
VA ≈ 13.9 m
b) Angle ∠VAO = β:
tan β = VO / OA = 12 / (5√2) ≈ 1.697
β ≈ 59.5°
Problem Statement — Q27
Each of the six faces of a cube is painted either red or blue. An observer (an ‘ant’) is at each of the cube’s eight vertices. Each ant can see the three faces meeting at its vertex. The ants are perfect logicians and answer truthfully.
(i) The ants are asked: “Is the number of red faces you see even?” (Zero is even). Explain why the total number of ants answering “yes” must be even.
(ii) Is it possible for a coloring where all eight ants see exactly two red faces? Justify.
(iii) The ants are asked: “Can you see at least one red face?” Explain why exactly five ants answering “yes” and three answering “no” is impossible.
(iv) The four ants on the top face report seeing 0, 1, 1, and 2 red faces. How many blue faces might the cube have? Find all possibilities and explain.
(v) Consider a convex polyhedron where every face is a square or hexagon, with three faces meeting at each corner. Faces are painted red or blue. Explain why the number of edges separating a red face from a blue face is even.
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Detailed Step-by-Step Solution
(i) Number of “yes” answers is even
Use a double-counting argument. Sum the red faces seen by all 8 ants: S = Σ S_i, where S_i is the number of red faces ant i sees.
Count by faces: Each red face is seen by 4 ants. If N_R is the number of red faces, S = 4 × N_R, which is even.
Ants saying “yes” see an even number of red faces (0 or 2). Ants saying “no” see an odd number (1 or 3). Let Y be the set of “yes” ants, N the set of “no” ants. S = Σ_{j∈Y} S_j + Σ_{k∈N} S_k = even. Σ_{j∈Y} S_j is even (sum of evens). Thus, Σ_{k∈N} S_k must be even. A sum of odds is even only if the number of terms (|N|) is even. Since 8 is even, the number of “yes” ants (8 — |N|) is even. [3 marks]
(ii) Is it possible for all 8 ants to see 2 red faces?
Yes. Color two opposite faces blue (e.g., Top and Bottom) and the four side faces (Front, Back, Left, Right) red. Each ant is at a vertex where three faces meet, seeing one blue face (Top or Bottom) and two red faces (e.g., Front and Right). By symmetry, all ants see two red faces. This is consistent with (i) since 8 “yes” answers is even. [2 marks]
(iii) Impossible for 5 ants to see at least one red face
An ant says “yes” if it sees at least one red face, “no” if it sees 0 (all three faces blue). Assume five ants say “yes”, three say “no”. If one red face (F_1) exists, it’s seen by 4 ants, so at least 4 say “yes”. For a 5th “yes”, another ant (not on F_1) must see a red face (F_2).
Case 1: F_1 and F_2 adjacent. They share 2 vertices, so 4 + 4–2 = 6 ants say “yes”.
Case 2: F_1 and F_2 opposite. No shared vertices, so 4 + 4 = 8 ants say “yes”.
With two or more red faces, at least 6 ants say “yes”. Thus, exactly 5 is impossible. [3 marks]
(iv) Deducing the number of blue faces
The top ants (at T-F-R, T-Bk-R, T-Bk-L, T-F-L) see 0, 1, 1, 2 red faces. Let c(X) = 1 if face X is red, 0 if blue. For ant at T-F-R seeing 0: c(T) + c(F) + c(R) = 0, so c(T) = c(F) = c(R) = 0 (Top, Front, Right are blue).
Other ants:
– T-Bk-R sees c(Bk).
– T-Bk-L sees c(Bk) + c(L).
– T-F-L sees c(L).
These are {c(Bk), c(L), c(Bk) + c(L)} = {1, 1, 2}. Thus, c(Bk) = 1, c(L) = 1 (Back, Left red). Known: 3 blue (Top, Front, Right), 2 red (Back, Left). Bottom face is unknown:
– Bottom blue: 4 blue faces.
– Bottom red: 3 blue faces.
Possible blue faces: 3 or 4. [3 marks]
(v) Red-blue edges on a polyhedron
Prove the number of red-blue edges (|E_RB|) is even. Count edges of red faces: S = Σ_{red face R_i} k_i, where k_i = 4 or 6 (even). Thus, S is even.
S counts edge “sides” of red faces. Each red-red edge is counted twice (once per red face), red-blue edge once. So, S = 2|E_RR| + |E_RB|. Since S and 2|E_RR| are even, |E_RB| must be even. [4 marks]
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Visualization of the Problem
Part (ii): All ants see 2 red faces
Color opposite faces blue (e.g., Top, Bottom), sides red. Each ant sees one blue, two red faces (e.g., Top blue, Front red, Right red).
Part (iv): Deducing from top ants
Top face is blue. Ant seeing 0 implies Front, Right blue. Other ants’ counts (1, 1, 2) imply Back, Left red. Bottom face’s color is unknown.
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Marking Criteria
(i) [3 marks]
– 1m: Set up double-counting (sum = 4 × N_R).
– 1m: Parity argument (sum of odds even if count even).
– 1m: Conclude “yes” ants even.
(ii) [2 marks]
– 1m: Answer “Yes”.
– 1m: Valid coloring (opposite faces blue).
(iii) [3 marks]
– 1m: One red face implies 4 “yes”.
– 1m: Second red face implies ≥6 “yes”.
– 1m: Conclude 5 impossible.
(iv) [3 marks]
– 1m: Use 0 to deduce Top, two sides blue.
– 1m: Use 1, 1, 2 to deduce other sides.
– 1m: Identify bottom ambiguity (3 or 4 blue).
(v) [4 marks]
– 1m: State proof strategy (double-counting).
– 1m: Sum of red face edges even.
– 1m: Relate sum to edge types.
– 1m: Conclude |E_RB| even.
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Error Analysis: Common Pitfalls
Mistake Area: Vague Reasoning
– Error: In (i), claiming “it’s even” without rigor.
Fix: Use double-counting and state premises clearly.
Mistake Area: Proof vs. Example
– Error: In (ii), saying it’s possible without example.
Fix: Provide a concrete coloring.
Mistake Area: Incomplete Analysis
– Error: In (iv), ignoring bottom face ambiguity.
Fix: Account for all faces, note missing info.
Mistake Area: Logical Leaps
– Error: In (iii), assuming 3 “no” ants imply specific faces.
Fix: Start with “yes” ants for robust logic.
Mistake Area: Oversimplification
– Error: In (v), assuming cube, not general polyhedron.
Fix: Use only given properties (even-sided faces).
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Rishabh’s Insight: Power of Parity and Double-Counting
Parity and double-counting are key. Count the same quantity two ways to get equal results.
In (i): Count red face sightings:
– Ant-centric: Σ S_i.
– Face-centric: 4 × N_R (even).
Equate to show “yes” ants even.
In (v): Count red face edge sides:
– Face-centric: Sum even (4 or 6 edges).
– Edge-centric: 2|E_RR| + |E_RB|.
Equate to show |E_RB| even.
Double-counting reveals structural properties without needing specific configurations.
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Key Takeaways
– Parity Arguments: Use even/odd properties to prove possibilities.
– Double-Counting: Count a set two ways for insights.
– Proof by Contradiction: Assume possible, find absurdity.
– Constructive Proof: Provide examples for possibility.
– Deductive Reasoning: Build conclusions step-by-step.
– Generalize Arguments: Use only given conditions.
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Foundational Concepts
Combinatorial Proofs and Parity
1. Parity:
– Even + even = even, odd + odd = even, even + odd = odd.
– Sum of odds is even if number of terms is even.
2. Handshaking Lemma and Double-Counting:
– Graph vertices’ degrees sum = 2 × edges (even).
– Implies even number of odd-degree vertices.
– Analogous to (i) and (v).
3. Logical Deduction:
– Use premises to derive new truths (e.g., in (iv), 0 red implies blue faces).
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Practice Problems
Practice Problem 1: Chessboard Tiling
An 8×8 chessboard has 64 squares. A domino covers two adjacent squares.
a) Can it be tiled by 32 dominoes?
b) Remove two opposite corners. Can 62 squares be tiled by 31 dominoes?
Solution:
a) Yes. 32 dominoes cover 64 squares (e.g., all horizontal).
b) No. Opposite corners are same color (e.g., white). Board has 30 white, 32 black squares. Each domino covers one white, one black. 31 dominoes need 31 of each color, but only 30 white exist.
Practice Problem 2: Party Handshakes
At a party, some people shake hands (no self-handshakes, at most one per pair). Prove the number of people with an odd number of handshakes is even.
Solution:
Model as a graph: people are vertices, handshakes are edges. Count handshake-ends:
– Person-centric: S = Σ handshakes per person.
– Handshake-centric: Each handshake counts twice, so S = 2 × handshakes (even).
S = even shakers + odd shakers = even. Even shakers’ sum is even, so odd shakers’ sum is even. A sum of odds is even only if the number of terms is even.
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Conclusion: Your Journey Toward Mathematical Excellence
This guide equips you for the MAT with strategies for creative problem-solving and concise reasoning. Solutions by Rishabh Kumar, Mathematics Elevate Academy founder, IIT Guwahati and Indian Statistical Institute alumnus.
Key Strategies for MAT Success:
– Think Creatively: Use innovative approaches.
– Write Like an Examiner: Present clear solutions.
– Build Intuition: Explore deeper principles.
– Train Under Exam Conditions: Practice timed questions.
Take the Next Step with Elite Mentorship:
– MAT Problem-Solving Mastery
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– Mock Exam Evaluation
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– One-on-One MAT Mentorship
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Your journey to Oxford begins now.
Train deeply. Think critically. Rise mathematically.
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Advanced Mathematics Mentor | IIT Guwahati & Indian Statistical Institute Alumnus
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