TMUA practice · 20 questions · Free

TMUA Mock Paper 2: Mathematical Reasoning

Every question carries a worked solution behind a disclosure, so you commit to an answer before you see the key.

[01] Conditional statements

Question 1

Which of the following is logically equivalent to the statement 'if $P$ then $R$'?

  1. If not $R$, then not $P$.
  2. If not $P$, then not $R$.
  3. If $R$, then $P$.
  4. $P$ and not $R$.
  5. Either not $P$ or not $R$.
Reveal the answer and worked solution. Commit to an option first.

Answer: A

Key idea. An implication is equivalent to its contrapositive and to nothing else on this list. Swapping $P$ and $R$ gives the converse; negating both without swapping gives the inverse; neither is equivalent.

Fastest approach

'If $P$ then $R$' is false in exactly one case: $P$ true and $R$ false.

Test each candidate against that single case, and against any one other case, and only the contrapositive matches:

$\lnot R \implies \lnot P$ is false exactly when $\lnot R$ is true and $\lnot P$ is false, that is when $R$ is false and $P$ is true. Same row, so the two statements agree everywhere.

Common mistake. Choosing the converse. It is the most natural misreading in English, where 'if' is often heard as 'if and only if', and it is wrong: 'if it rains the ground is wet' does not give 'if the ground is wet it rained'.

Why the other options are wrong

  • B: The inverse. Equivalent to the converse, not to the original. Conceptual Misunderstanding
  • C: The converse. The most common error on this question type. Conceptual Misunderstanding
  • D: The negation of the original, not an equivalent of it. Conceptual Misunderstanding
  • E: Equivalent to 'not both', which is a different statement entirely. Formula Misapplication

[02] Negation

Question 2

What is the negation of '$P$ and $R$'?

  1. $P$ or $R$.
  2. Not $P$ or not $R$.
  3. Not $P$ and not $R$.
  4. $P$ and not $R$.
  5. Not $P$ and $R$.
Reveal the answer and worked solution. Commit to an option first.

Answer: B

Key idea. Negating a conjunction flips the connective as well as the parts. 'Not both' is 'at least one fails', which is an 'or', not an 'and'.

Fastest approach

'$P$ and $R$' is true in exactly one case out of four: both true. So its negation is true in the other three.

'Not $P$ or not $R$' is false only when both $\lnot P$ and $\lnot R$ are false, that is when both $P$ and $R$ are true. True in the other three. They match.

Common mistake. Negating each part but leaving the 'and' in place. That gives 'neither', which is much stronger than 'not both': it rules out three cases instead of one.

Why the other options are wrong

  • A: Weaker than the original rather than its negation. Conceptual Misunderstanding
  • C: 'Neither', which is the negation of '$P$ or $R$', not of '$P$ and $R$'. Formula Misapplication
  • D: One specific way the conjunction can fail, not all of them. Incomplete Calculation
  • E: Another single failing case, again not the full negation. Incomplete Calculation

[03] Necessary and sufficient conditions

Question 3

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

For a positive integer $n$, consider the statement 'if $n$ is divisible by $6$ then $n$ is divisible by $3$'. Which is true of the condition '$n$ is divisible by $6$'?

  1. Neither necessary nor sufficient.
  2. Necessary for divisibility by $2$, but not for $3$.
  3. Sufficient for divisibility by $3$, but not necessary.
  4. Necessary for divisibility by $3$, but not sufficient.
  5. Both necessary and sufficient.
Reveal the answer and worked solution. Commit to an option first.

Answer: C

Key idea. '$A \implies B$' says $A$ is sufficient for $B$ and $B$ is necessary for $A$. To check whether $A$ is also necessary, look for a $B$ without an $A$: here, $n = 3$.

Fastest approach

Every multiple of $6$ is a multiple of $3$, so divisibility by $6$ is sufficient.

Is it necessary? $n = 3$ is divisible by $3$ and not by $6$. One counterexample settles it: not necessary.

Sufficient but not necessary.

Common mistake. Reading the implication in the wrong direction and calling the condition necessary. The arrow points from divisibility by $6$ to divisibility by $3$, so it is the $3$ that is necessary here, not the $6$.

Why the other options are wrong

  • A: Missing that every multiple of 6 really is a multiple of 3. Incomplete Calculation
  • B: True but irrelevant: the question asks about divisibility by 3. Misread Question
  • D: The two roles swapped. Divisibility by 3 is the necessary condition, not divisibility by 6. Conceptual Misunderstanding
  • E: Treating the implication as a biconditional, which $n = 3$ disproves. Conceptual Misunderstanding

[04] Negation

Question 4

What is the negation of 'every prime number is odd'?

  1. No prime number is odd.
  2. There is an odd number that is not prime.
  3. Every odd number is prime.
  4. There is a prime number that is not odd.
  5. Every prime number is even.
Reveal the answer and worked solution. Commit to an option first.

Answer: D

Key idea. The negation of 'every $A$ is $B$' is 'some $A$ is not $B$'. The quantifier flips as well as the predicate. It is never 'every $A$ is not $B$'.

Fastest approach

'Every prime is odd' has the shape $\forall n (\text{prime}(n) \implies \text{odd}(n))$.

Negating a universal gives an existential, and negating the implication inside gives the antecedent with the consequent denied:

$\exists n (\text{prime}(n) \wedge \lnot\text{odd}(n))$

which reads: there is a prime that is not odd. That statement is true, incidentally, because $2$ is prime and even.

Common mistake. Answering 'every prime is even'. That is the contrary, not the contradictory: both it and the original can be false at once, whereas a statement and its negation never can.

Why the other options are wrong

  • A: Also a contrary rather than a contradictory: it denies the predicate everywhere. Conceptual Misunderstanding
  • B: Negates the converse statement instead. Conceptual Misunderstanding
  • C: The converse, and unrelated to the negation. Conceptual Misunderstanding
  • E: The contrary. Far stronger than the negation, and could be false alongside the original. Conceptual Misunderstanding

[05] Counterexamples

Question 5

A student claims: 'for every positive integer $n$, the value $n^{2} + n + 41$ is prime.' What is the smallest positive integer $n$ that disproves this?

  1. $41$
  2. $42$
  3. The claim is true
  4. $11$
  5. $40$
Reveal the answer and worked solution. Commit to an option first.

Answer: E

Key idea. One counterexample is enough to destroy a universal claim, and the structure of the expression tells you where to look: at $n = 40$ every term carries a factor of $41$.

Fastest approach

Do not test values one by one. Look for structure:

At $n = 40$: $40^{2} + 40 + 41 = 40(40+1) + 41 = 40 \times 41 + 41 = 41 \times 41$

So $n = 40$ gives $41^{2}$, which is not prime.

Nothing smaller works, which is what makes this claim so persuasive: it holds for every $n$ from $1$ to $39$.

Common mistake. Testing a handful of small values, finding them all prime, and concluding the claim is true. A universal claim over an infinite set is never established by examples.

Why the other options are wrong

  • A: $41$ also fails, but it is not the smallest such $n$. Misread Question
  • B: $42$ fails too, and is smaller-looking than it is useful. Still not the least. Misread Question
  • C: The claim is false; it merely holds for the first 39 values. Conceptual Misunderstanding
  • D: $11^{2}+11+41 = 173$, which is prime. A value tested early and wrongly rejected. Arithmetic Slip

[06] Conditional statements

Question 6

Which of the following is logically equivalent to '$P$ if and only if $R$'?

  1. Not ($P$ or $R$).
  2. ($P$ and $R$) or (not $P$ and not $R$).
  3. ($P$ and $R$) or (not $P$ and $R$).
  4. If $P$ then $R$.
  5. $P$ or $R$.
Reveal the answer and worked solution. Commit to an option first.

Answer: B

Key idea. 'If and only if' says the two statements have the same truth value. That happens in exactly two of the four cases: both true, or both false.

Fastest approach

A biconditional is true exactly when $P$ and $R$ agree.

They agree in two cases: both true, or both false. Writing that as a disjunction gives

$(P \wedge R) \vee (\lnot P \wedge \lnot R)$

Common mistake. Taking only one direction, 'if $P$ then $R$'. That leaves the case $P$ false, $R$ true untouched, which the biconditional rules out.

Why the other options are wrong

  • A: Covers only the both-false case. Incomplete Calculation
  • C: Simplifies to $R$ alone, which says nothing about $P$. Formula Misapplication
  • D: One direction only. The converse is missing. Incomplete Calculation
  • E: True when exactly one holds, which is when the biconditional is false. Conceptual Misunderstanding

[07] Proof by contradiction

Question 7

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

To prove by contradiction that there is no largest prime number, what should be assumed at the start?

  1. Every integer greater than $1$ has a prime factor.
  2. Some prime numbers are larger than others.
  3. There is a largest prime number.
  4. There is no largest prime number.
  5. There are infinitely many primes.
Reveal the answer and worked solution. Commit to an option first.

Answer: C

Key idea. A proof by contradiction assumes the negation of what is to be proved, then derives an impossibility. So it opens with the opposite of the conclusion.

Fastest approach

The target is 'there is no largest prime'.

Its negation is 'there is a largest prime'. Assume that, call it $p$, and derive an impossibility, which is what Euclid's argument does by considering $p! + 1$ or the product of all primes plus one.

Note that options B and C are the same statement, both being what is to be proved rather than what is to be assumed.

Common mistake. Assuming the thing you are trying to prove. That is the whole difference between a direct proof and a contradiction proof, and assuming the conclusion proves nothing.

Why the other options are wrong

  • A: A lemma the proof may use, but not its opening assumption. Conceptual Misunderstanding
  • B: True and useless: it carries no content about how many primes there are. Conceptual Misunderstanding
  • D: This is the conclusion, not the assumption. Conceptual Misunderstanding
  • E: Another statement of the conclusion. Conceptual Misunderstanding

[08] Counterexamples

Question 8

Which single value of $x$ disproves the claim 'for every real number $x$, if $x^{2} > x$ then $x > 1$'?

  1. $x = \tfrac{1}{2}$
  2. $x = 1$
  3. $x = 2$
  4. $x = -2$
  5. $x = 0$
Reveal the answer and worked solution. Commit to an option first.

Answer: D

Key idea. To break 'if $A$ then $B$' you need a case where $A$ holds and $B$ fails. Checking only that $B$ fails is not enough, and neither is checking only that $A$ holds.

Fastest approach

Need $x^{2} > x$ true and $x > 1$ false.

$x = -2$: $x^{2} = 4 > -2$, so the antecedent holds. And $-2 > 1$ is false. The implication fails.

The claim quietly assumed $x$ was positive; every negative number breaks it.

Common mistake. Picking a value where the antecedent is false, such as $x = \tfrac{1}{2}$ where $x^{2} = \tfrac{1}{4} < x$. An implication with a false antecedent is vacuously true and disproves nothing.

Why the other options are wrong

  • A: $\tfrac{1}{4} < \tfrac{1}{2}$, antecedent false again. Conceptual Misunderstanding
  • B: $1 = 1$, antecedent false. Conceptual Misunderstanding
  • C: Both antecedent and consequent are true, so this is an example, not a counterexample. Conceptual Misunderstanding
  • E: $0^{2} = 0$, not greater than $0$, so the antecedent fails and the implication holds vacuously. Conceptual Misunderstanding

[09] Negation

Question 9

What is the negation of 'some square number is even'?

  1. Some square number is odd.
  2. No square number is odd.
  3. Every even number is a square.
  4. Some even number is not a square.
  5. Every square number is odd.
Reveal the answer and worked solution. Commit to an option first.

Answer: E

Key idea. The negation of 'some $A$ is $B$' is 'every $A$ is not $B$'. Existential becomes universal, and the predicate is denied.

Fastest approach

'Some square is even' is $\exists n (\text{square}(n) \wedge \text{even}(n))$.

Negating gives $\forall n (\text{square}(n) \implies \lnot\text{even}(n))$, that is: every square number is odd.

As it happens the original is true, since $4$ is an even square, so the negation is false. That does not affect which statement is the negation.

Common mistake. Answering 'some square number is odd'. Both that and the original can be true at once, which is exactly what a negation can never do.

Why the other options are wrong

  • A: Compatible with the original: 4 is an even square and 9 is an odd one, so both hold. Conceptual Misunderstanding
  • B: Says every square is even, which is a contrary of the wrong statement. Conceptual Misunderstanding
  • C: Negates a different, converse claim. Conceptual Misunderstanding
  • D: About even numbers rather than square ones. Misread Question

[10] Flawed arguments

Question 10

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

An argument runs: 'If a number is divisible by $4$, it is even. This number is even. Therefore it is divisible by $4$.' What is wrong with it?

  1. It affirms the consequent: the converse of the first premise does not follow.
  2. The first premise is false.
  3. It denies the antecedent.
  4. Nothing is wrong; the argument is valid.
  5. The conclusion contradicts the second premise.
Reveal the answer and worked solution. Commit to an option first.

Answer: A

Key idea. From $A \implies B$ and $B$, nothing follows about $A$. The pattern is called affirming the consequent, and $6$ is the counterexample: even, not divisible by $4$.

Fastest approach

Premise: divisible by $4 \implies$ even. True.

The argument then asserts the consequent, 'even', and concludes the antecedent. That is the converse, and the converse of a true implication need not be true.

$6$ settles it: even, and not divisible by $4$.

Common mistake. Accepting the argument because both premises are true. Validity is about the shape of the inference, not the truth of its parts: a valid argument can have false premises, and an invalid one can have true premises and even a true conclusion.

Why the other options are wrong

  • B: The first premise is true: every multiple of 4 is even. Misread Question
  • C: Denying the antecedent is the other classic fallacy, and is not this shape. Conceptual Misunderstanding
  • D: $6$ is a counterexample, so the argument cannot be valid. Conceptual Misunderstanding
  • E: There is no contradiction; the conclusion is merely unsupported. Conceptual Misunderstanding

[11] Proof by contradiction

Question 11

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

In the standard proof that $\sqrt{2}$ is irrational, one assumes $\sqrt{2} = \frac{p}{q}$ in lowest terms. What contradiction is reached?

  1. $q = 0$.
  2. $p$ and $q$ are both odd.
  3. $p$ and $q$ are both even, contradicting lowest terms.
  4. $p$ is even and $q$ is odd.
  5. $p^{2} = 2q^{2}$ has no integer solutions at all.
Reveal the answer and worked solution. Commit to an option first.

Answer: C

Key idea. The contradiction is with the phrase 'in lowest terms', not with the algebra. Deriving that both $p$ and $q$ are even means the fraction was never in lowest terms after all.

Fastest approach

$\sqrt{2} = \frac{p}{q}$ with $\gcd(p,q) = 1$ gives $p^{2} = 2q^{2}$.

So $p^{2}$ is even, hence $p$ is even; write $p = 2m$.

Then $4m^{2} = 2q^{2}$, so $q^{2} = 2m^{2}$, hence $q$ is even too.

Both even contradicts $\gcd(p,q) = 1$.

Common mistake. Saying $p^{2} = 2q^{2}$ has no solutions. It has one, $p = q = 0$, and in any case that claim is what is being proved rather than a contradiction reached along the way.

Why the other options are wrong

  • A: $q \neq 0$ by construction, so this never arises. Conceptual Misunderstanding
  • B: The opposite of what the argument derives. Sign Error
  • D: $p$ even is derived, but the argument continues to show $q$ is even as well. Stopping here proves nothing. Incomplete Calculation
  • E: Close to the conclusion, but not the contradiction the proof actually reaches. Conceptual Misunderstanding

[12] Necessary and sufficient conditions

Question 12

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

For a real number $x$, which condition is necessary but not sufficient for $x^{2} = 9$?

  1. $x = 3$ or $x = -3$
  2. $x > 0$
  3. $x^{2} \geqslant 0$
  4. $|x| \geqslant 3$
  5. $x = 3$
Reveal the answer and worked solution. Commit to an option first.

Answer: D

Key idea. Necessary means every solution satisfies it. Not sufficient means something satisfies it without being a solution. Test both directions separately, every time.

Fastest approach

$x^{2} = 9$ means $x = \pm 3$.

Is $|x| \geqslant 3$ necessary? Both solutions have $|x| = 3$, so yes.

Is it sufficient? $x = 5$ has $|x| \geqslant 3$ but $x^{2} = 25 \neq 9$. So no.

Necessary but not sufficient.

Common mistake. Choosing $x = 3$. That is sufficient but not necessary, the exact opposite of what was asked, and it is the most common way to lose this mark.

Why the other options are wrong

  • A: Both necessary and sufficient, so it fails the 'not sufficient' half. Misread Question
  • B: Neither: $x = -3$ is a solution with $x < 0$, and $x = 1$ is positive without being one. Conceptual Misunderstanding
  • C: Necessary but true of every real number, so it carries no information. Also true of every non-solution. Conceptual Misunderstanding
  • E: Sufficient but not necessary: it misses $x = -3$. The roles are the wrong way round. Misread Question

[13] Counterexamples

Question 13

How many pairs $(a, b)$ with $1 \leqslant a < b \leqslant 6$ satisfy both $\gcd(a,b) = 1$ and $a + b$ even?

  1. $4$
  2. $5$
  3. $6$
  4. $9$
  5. $3$
Reveal the answer and worked solution. Commit to an option first.

Answer: E

Key idea. $a+b$ even forces $a$ and $b$ to share parity. Both even is then impossible, because two even numbers share a factor of $2$ and cannot be coprime. So both must be odd, which cuts fifteen pairs down to three.

Fastest approach

$a+b$ is even exactly when $a$ and $b$ have the same parity.

Both even is ruled out immediately: any two even numbers share a factor of $2$, so $\gcd(a,b) \geqslant 2$.

So both are odd, drawn from $\{1, 3, 5\}$: $(1,3), (1,5), (3,5)$

All three are coprime, so all three count.

Answer: $3$. Letting the two conditions interact first is what makes this quick; testing all fifteen pairs one at a time gets there too, but slowly.

Common mistake. Counting pairs with an even sum and forgetting the coprimality condition, which lets $(2,4)$, $(2,6)$ and $(4,6)$ in and gives $6$.

Why the other options are wrong

  • A: Adding one even pair, usually $(2,6)$, on the mistaken view that it is coprime. Miscount
  • B: Counting the odd pairs plus two even ones. Miscount
  • C: All six same-parity pairs, with the coprimality condition dropped entirely. Incomplete Calculation
  • D: Counting every coprime pair regardless of the parity of the sum. Incomplete Calculation

[14] Conditional statements

Question 14

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

Let $S$ be the empty set. Consider the statement: 'every element of $S$ is greater than $100$.' Which is correct?

  1. The statement is true.
  2. The statement is false.
  3. The statement is neither true nor false.
  4. The statement is true only if $S$ also contains no negative numbers.
  5. The statement cannot be evaluated without knowing $S$.
Reveal the answer and worked solution. Commit to an option first.

Answer: A

Key idea. 'Every element of $S$ is $B$' means: for all $x$, if $x \in S$ then $B(x)$. When $S$ is empty the antecedent is always false, so the implication holds every time, and the statement is vacuously true.

Fastest approach

Write it out: $\forall x (x \in S \implies x > 100)$.

$S$ is empty, so $x \in S$ is false for every $x$. An implication with a false antecedent is true.

Every instance is true, so the universal statement is true.

Sanity check via the negation: 'there is an element of $S$ not greater than $100$' would need an element, and there are none. So the negation is false, and the original is true.

Common mistake. Calling it false on the grounds that no element is greater than $100$. Nothing failing the condition is exactly what makes a universal statement hold.

Why the other options are wrong

  • B: Confuses 'no element satisfies it' with 'the statement fails'. Conceptual Misunderstanding
  • C: Universal statements over an empty domain have a definite truth value. Conceptual Misunderstanding
  • D: Adds a condition the statement does not contain. Misread Question
  • E: $S$ is fully specified: it is empty. Misread Question

[15] Flawed arguments

Question 15

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

A 'proof' that $1 = 2$ sets $a = b$, then writes $a^{2} = ab$, $a^{2} - b^{2} = ab - b^{2}$, $(a+b)(a-b) = b(a-b)$, and finally $a + b = b$, giving $2b = b$. Which step is invalid?

  1. Concluding $2b = b$ from $a + b = b$.
  2. Dividing both sides by $a - b$, which is zero.
  3. Squaring $a = b$ to get $a^{2} = ab$.
  4. Subtracting $b^{2}$ from both sides.
  5. Factorising $a^{2} - b^{2}$ as $(a+b)(a-b)$.
Reveal the answer and worked solution. Commit to an option first.

Answer: B

Key idea. Every step here is reversible and valid except one: cancelling $(a-b)$. Since $a = b$, that factor is zero, and dividing by zero is not an operation.

Fastest approach

Go through the chain looking for an operation that is not always permitted.

Multiplying, subtracting and factorising are all fine for any $a$ and $b$.

The move from $(a+b)(a-b) = b(a-b)$ to $a+b = b$ divides both sides by $a-b$. Given $a = b$, that is division by zero.

The tell for this whole family of fake proofs is a cancelled factor that the opening assumption makes zero.

Common mistake. Looking for an arithmetic slip. There is none: every individual line is correctly manipulated, which is what makes the argument persuasive. The fault is an operation that is undefined, not one performed wrongly.

Why the other options are wrong

  • A: This follows correctly from the previous line, which is itself the faulty one. Conceptual Misunderstanding
  • C: Multiplying both sides of $a = b$ by $a$ is valid. Conceptual Misunderstanding
  • D: Subtracting the same quantity from both sides is always valid. Conceptual Misunderstanding
  • E: The difference of two squares is a correct identity. Conceptual Misunderstanding

[16] Negation

Question 16

What is the negation of 'every number in the list has a larger number in the list'?

  1. No number in the list has a larger number in the list.
  2. Some number in the list has a larger number in the list.
  3. Every number in the list is the largest.
  4. Some number in the list has no larger number in the list.
  5. Every number in the list has a smaller number in the list.
Reveal the answer and worked solution. Commit to an option first.

Answer: D

Key idea. Negation moves inwards one quantifier at a time, flipping each: $\lnot \forall \exists$ becomes $\exists \forall \lnot$. Only the outermost quantifier changes first.

Fastest approach

The statement is $\forall n \, \exists m \, (m > n)$.

Negate: $\exists n \, \forall m \, (m \leqslant n)$

In words: some number in the list has nothing larger than it in the list, that is, some number is the maximum.

For a finite non-empty list the original is always false, since the largest element has nothing above it. That is a useful check on the answer.

Common mistake. Flipping the inner predicate to 'smaller' and leaving both quantifiers alone. Negation changes the quantifiers, not just the comparison.

Why the other options are wrong

  • A: Too strong: it says no number has anything above it, rather than that one does not. Conceptual Misunderstanding
  • B: Weakens the original instead of negating it; it can hold at the same time. Conceptual Misunderstanding
  • C: Says everything is the maximum, which is a contrary rather than the contradictory. Conceptual Misunderstanding
  • E: Changes the predicate but keeps both quantifiers, which is a different statement. Conceptual Misunderstanding

[17] Counterexamples

Question 17

Which claim about positive integers is FALSE?

  1. If $n^{2}$ is even then $n$ is even.
  2. If $n$ is divisible by $6$ then $n^{2}$ is divisible by $36$.
  3. If $n$ is prime and $n > 2$ then $n$ is odd.
  4. If $n$ is divisible by $9$ then the digits of $n$ sum to a multiple of $9$.
  5. If $n^{2}$ is divisible by $4$ then $n$ is divisible by $4$.
Reveal the answer and worked solution. Commit to an option first.

Answer: E

Key idea. Squaring doubles the exponent of every prime factor, so $n^{2}$ divisible by $4$ needs only one factor of $2$ in $n$. That is why the first claim breaks and the others hold.

Fastest approach

Take $n = 2$: $n^{2} = 4$ is divisible by $4$, but $n = 2$ is not. That single case settles it.

Why the others hold: $n^{2}$ even forces $n$ even, since odd times odd is odd. $6 \mid n$ gives $36 \mid n^{2}$ by squaring. Every prime above $2$ is odd by definition of primality. And the digit-sum test for $9$ is a standard result.

Common mistake. Assuming that if $k \mid n^{2}$ then $k \mid n$. That holds when $k$ is prime, and fails for composite $k$ such as $4$, which is exactly the trap here.

Why the other options are wrong

  • A: True: an odd number squared is odd, so an even square needs an even root. Conceptual Misunderstanding
  • B: True: squaring $n = 6m$ gives $36m^{2}$. Conceptual Misunderstanding
  • C: True: any even number above 2 has 2 as a proper factor. Conceptual Misunderstanding
  • D: True: the standard divisibility test for 9. Conceptual Misunderstanding

[18] Necessary and sufficient conditions

Question 18

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

A quadrilateral $Q$ has four equal sides. Which statement is correct?

  1. This is necessary but not sufficient for $Q$ to be a square.
  2. This is sufficient but not necessary for $Q$ to be a square.
  3. This is both necessary and sufficient for $Q$ to be a square.
  4. This is neither necessary nor sufficient for $Q$ to be a square.
  5. This is sufficient for $Q$ to be a rectangle.
Reveal the answer and worked solution. Commit to an option first.

Answer: A

Key idea. Every square has four equal sides, so the condition is necessary. A rhombus that is not a square has four equal sides too, so it is not sufficient.

Fastest approach

Necessary? Every square has four equal sides. Yes.

Sufficient? A rhombus with angles $60^{\circ}$ and $120^{\circ}$ has four equal sides and is not a square. No.

Necessary but not sufficient. Adding one right angle would make it sufficient.

Common mistake. Forgetting the rhombus. Equal sides fix the lengths but say nothing about the angles, and a square needs both.

Why the other options are wrong

  • B: The roles reversed. Equal sides do not force a square. Conceptual Misunderstanding
  • C: Would require every equilateral quadrilateral to be a square, which the rhombus refutes. Conceptual Misunderstanding
  • D: It is necessary: no square lacks equal sides. Incomplete Calculation
  • E: A rhombus has equal sides and is not a rectangle. Conceptual Misunderstanding

[19] Proof by contradiction

Question 19

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

Thirteen people are in a room. Which conclusion follows with certainty?

  1. No two of them were born in the same month.
  2. At least two of them were born in the same month.
  3. At least two of them share a birthday, to the day.
  4. At least three of them were born in the same month.
  5. Exactly two of them were born in the same month.
Reveal the answer and worked solution. Commit to an option first.

Answer: B

Key idea. Thirteen items into twelve boxes forces a box with two. The principle gives 'at least two', never 'exactly two', and never a specific pair.

Fastest approach

Twelve months, thirteen people. If every month held at most one person, that accounts for at most twelve people. Thirteen do not fit.

So some month holds at least two.

Nothing stronger follows. Thirteen people can easily have thirteen distinct birthdays, and twelve months can hold thirteen people with no month holding three.

Common mistake. Choosing 'exactly two'. The pigeonhole principle bounds from below only: all thirteen could share a birth month, which still satisfies 'at least two' and refutes 'exactly two'.

Why the other options are wrong

  • A: Impossible: thirteen people cannot occupy twelve months uniquely. Conceptual Misunderstanding
  • C: Not forced: 13 people out of 365 possible birthdays need not collide. Conceptual Misunderstanding
  • D: Not forced: twelve months can hold thirteen people with a maximum of two in any one. Conceptual Misunderstanding
  • E: 'Exactly' is never guaranteed by pigeonhole. Conceptual Misunderstanding

[20] Conditional statements

Question 20

Answer key asserted by the author, not machine-checked. Treat it with more caution than the rest of this paper.

Consider 'if $n$ is a multiple of $10$ then $n$ is a multiple of $5$'. Which of the following is true?

  1. Both the statement and its converse are false.
  2. The statement is true and its contrapositive is false.
  3. The statement is true and its converse is false.
  4. The statement is false and its converse is true.
  5. Both the statement and its converse are true.
Reveal the answer and worked solution. Commit to an option first.

Answer: C

Key idea. A statement and its converse are separate claims and must be tested separately. A statement and its contrapositive are the same claim, so they always share a truth value.

Fastest approach

Statement: every multiple of $10$ is a multiple of $5$. True, since $10 = 2 \times 5$.

Converse: every multiple of $5$ is a multiple of $10$. False: $15$ is a multiple of $5$ and not of $10$.

Option E can be dismissed without any arithmetic: a statement and its contrapositive are logically equivalent, so they cannot differ in truth value.

Common mistake. Assuming the converse inherits the truth of the original. It is an independent claim, and here it fails on $n = 15$.

Why the other options are wrong

  • A: The original is true. Conceptual Misunderstanding
  • B: Impossible in principle: a statement and its contrapositive are equivalent. Conceptual Misunderstanding
  • D: Both halves the wrong way round. Conceptual Misunderstanding
  • E: $15$ refutes the converse. Incomplete Calculation
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