TMUA Paper 2 Logic: Necessary vs Sufficient, Contrapositive and Counterexamples

Paper 2 logic questions are not hard maths. They are precise maths, and the trap options are manufactured from exactly three confusions: mixing up necessary and sufficient, treating the converse as equivalent to the original statement, and offering counterexamples that do not satisfy the hypothesis. Train those three and the logic marks become the most predictable on the test.

Table of contents

• What Paper 2 actually tests
• Necessary vs sufficient: the arrow direction
• Contrapositive vs converse
• Counterexamples: both halves or nothing
• Negating quantified statements
• The traps, collected
• How to drill this
• Frequently asked questions

Intro

Paper 2 is 20 questions in 75 minutes on mathematical reasoning, and a substantial chunk of those marks comes from a logic syllabus that A-level simply does not teach: implications, necessary and sufficient conditions, contrapositives, negation, counterexamples, and the validity of short arguments. The maths inside the questions is usually AS-level. The skill being tested is whether you can manipulate the logical structure around the maths without ever flipping an arrow by accident. This post teaches that content directly, with four original worked examples in the TMUA style and the trap patterns that recur every year.

What Paper 2 actually tests

Every logic question on Paper 2 reduces to bookkeeping about implication arrows. "If P then Q" is P ⇒ Q. From that single statement:

• Converse: Q ⇒ P. Not equivalent to the original.
• Contrapositive: not-Q ⇒ not-P. Equivalent to the original.
• P is sufficient for Q: P ⇒ Q.
• P is necessary for Q: Q ⇒ P.

That table is the whole syllabus for half of the logic marks. The exam's job is to dress it in mathematical clothing fast enough that you flip an arrow under time pressure. Your job is to undress the question back to arrows before deciding anything.

Necessary vs sufficient: the arrow direction

Worked example 1. Let x be a real number. Consider:

• P: x³ > 27
• Q: x² > 9

Is P necessary, sufficient, both, or neither for Q?

Translate P first: x³ > 27 is exactly x > 3, because cubing preserves order over the reals. Now check both arrows.

P ⇒ Q: if x > 3 then x² > 9. True. So P is sufficient for Q.

Q ⇒ P: if x² > 9 then x > 3? No: x = -4 gives x² = 16 > 9 but x³ = -64. So P is not necessary for Q.

Answer: sufficient but not necessary. Notice where the work happened: rewriting x³ > 27 as x > 3 (cubing is order-preserving; squaring is not) and then remembering that x² > 9 means x > 3 or x < -3. The exam buries the logic test inside a small algebra test, and the negative branch of the square is the most reused trap in the entire paper.

The translation habit that saves you: "P is necessary for Q" means Q cannot happen without P, so the arrow runs Q ⇒ P. If you have to think about it for more than ten seconds, you have not drilled it enough yet.

Contrapositive vs converse

Worked example 2. A function f has a stationary point at x = a. Consider the true statement:

S: if f′(a) = 0 and f″(a) < 0, then f has a local maximum at a.

Which of the following follow from S?

1. If f does not have a local maximum at a, then f′(a) ≠ 0 or f″(a) ≥ 0.
2. If f has a local maximum at a, then f′(a) = 0 and f″(a) < 0.

Statement 1 is the contrapositive of S: not-Q ⇒ not-P, with the "and" in P correctly negated to an "or". Contrapositives are equivalent to the original, so statement 1 follows. Statement 2 is the converse, and it is actually false as a piece of maths: f(x) = -x⁴ has a local maximum at 0 with f″(0) = 0. The second-derivative test is sufficient for a maximum, not necessary.

Two things to take from this. First, the contrapositive of "A and B ⇒ Q" is "not-Q ⇒ not-A or not-B"; negation turns and into or (De Morgan), and trap options routinely keep the and. Second, the converse of a familiar true theorem often sounds true because you have used the theorem a hundred times. Paper 2 exploits that familiarity deliberately.

Counterexamples: both halves or nothing

Worked example 3. Consider the claim:

For every positive integer n, n² + n + 41 is prime.

Which value of n disproves it?

Try the structure first rather than arithmetic: n² + n + 41 = n(n + 1) + 41. If n = 40, this is 40 × 41 + 41 = 41 × 41 = 41², visibly composite. So n = 40 is a counterexample, and you found it by factorisation, not by testing 40 numbers. (The claim really is true for every n from 1 to 39, which is precisely why "I checked a few cases" is not a proof.)

The general discipline: a counterexample to "if P then Q" must satisfy P and fail Q, both, explicitly checked. Paper 2 answer options include candidates that fail P, and they look attractive because they fail Q dramatically. They disprove nothing. A related true-claim trap: to show "f increasing on the reals implies f′(x) > 0 everywhere" is false, f(x) = x³ works, since it is strictly increasing but f′(0) = 0. Strictness lives at single points, and single points are where the TMUA likes to live.

Negating quantified statements

Worked example 4. Consider:

T: for every real number x, there exists a real number y such that y² = x.

What is the negation of T, and is T true?

Negation flips each quantifier and negates the core: there exists a real x such that for every real y, y² ≠ x. And T is false: x = -1 is a witness, since no real y squares to a negative.

The mechanical rule: "for all" becomes "there exists", "there exists" becomes "for all", working outside-in, and only the innermost statement gets negated. The trap options negate the inner statement while leaving the quantifiers alone ("for every x there exists y with y² ≠ x"), which is a different and usually irrelevant statement. If you negate quantifiers by feel rather than by rule, the feel will fail you at question 17 with six minutes left.

The traps, collected

• Arrow reversal: "necessary" translated as P ⇒ Q instead of Q ⇒ P.
• Converse smuggling: a converse of a familiar theorem presented as equivalent.
• And/or under negation: negating "A and B" to "not-A and not-B".
• Counterexamples that fail the hypothesis: dramatic, irrelevant.
• The negative branch: x² > 9 read as x > 3.
• Boundary cases: increasing versus strictly increasing, ≥ versus >, stationary versus turning. Equality cases are where wrong options are grown.

Several of these reappear, alongside their Paper 1 cousins, in 10 TMUA Mistakes That Cost Marks Under Time Pressure; logic is where they concentrate because the wrong options are engineered from them.

How to drill this

Reading this post once will not survive contact with question 14 under time pressure. The content is small; the skill is automaticity, and that comes from reps spread over weeks. Our Logic course is free and works through this material from first principles with practice at each step. The TMUA Pro question bank then gives you volume: logic questions filtered by type, with an explanation of why each wrong option is wrong, which matters more in logic than anywhere else because the wrong options are the syllabus. Slot the drilling into weeks 9-12 of the plan in How to Prepare for the TMUA in 16 Weeks; and if implications still feel slippery after a fortnight of practice, one focused 1-1 session with a Cambridge student tutor on logic specifically tends to fix more than another week alone.

Related reading

• How to Prepare for the TMUA in 16 Weeks (a Realistic Plan)
• TMUA 2026: Key Dates, Registration and the 28 September Deadline
• What TMUA Score Do You Need for Cambridge, Imperial and LSE?
• The TMUA Specification, Broken Down

Closing

Paper 2 logic rewards a small amount of knowledge applied with total precision. Four habits cover most of the marks: translate every condition into an arrow before judging it, trust contrapositives and distrust converses, check both halves of every counterexample, and negate quantifiers by rule. None of this is deep. All of it is trainable, and unlike the rest of the paper, the question patterns barely change year to year. That makes logic the best marks-per-hour investment on the TMUA.

Frequently asked questions

What does TMUA Paper 2 test? Paper 2 tests mathematical reasoning: implication, necessary and sufficient conditions, contrapositive, negation of quantified statements, counterexamples, and judging whether short arguments are valid. It is 20 questions in 75 minutes, no calculator.

What is the difference between necessary and sufficient? P is sufficient for Q means P implies Q: whenever P holds, Q holds. P is necessary for Q means Q implies P: Q cannot hold without P. The arrow directions are opposite, and confusing them is the single most common Paper 2 error.

What is the contrapositive of a statement? The contrapositive of "if P then Q" is "if not Q then not P". It is logically equivalent to the original statement. The converse, "if Q then P", is not equivalent, and Paper 2 trap options rely on exactly that confusion.

What makes a valid counterexample? A counterexample to "if P then Q" must satisfy P and fail Q, both at once. An example that fails P disproves nothing, however dramatic it looks. Checking both halves explicitly is the discipline the exam rewards.

Is Paper 2 harder than Paper 1? Not intrinsically, but it feels harder to most students because the logic content is not taught at A-level. Students who train it systematically often end up finding Paper 2 the more predictable of the two papers.

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Sources: UAT-UK TMUA test specification (Paper 2: mathematical reasoning). All worked examples are original. Last updated 2026-06-10.