Mathematics - MCQ Practice Questions

Understanding:

We must identify which of the given logical statements is a tautology — a statement that is true for all possible truth values of its variables.

Formula:

A statement is a tautology if its truth table yields $\text{True}$ in every row:

Step 1: Evaluate each option

For $p\wedge (\sim p)$: when $p$ is True, $\sim p$ is False, so $p\wedge (\sim p)$ is False. This is a contradiction, not a tautology.

Step 2: Check $p\vee (\sim p)$

The statement is True in all cases — it is a tautology (Law of Excluded Middle).

Step 3: Eliminate remaining options

$p\wedge q$ is False when either $p$ or $q$ is False — not a tautology.

$p\Rightarrow q$ is False when $p$ is True and $q$ is False — not a tautology.

Answer:

The statement $p\vee (\sim p)$ is always true, making it a tautology.

Quick Tip:

The two fundamental results to memorise: $p\vee (\sim p)$ is always a tautology, and $p\wedge (\sim p)$ is always a contradiction.

Understanding:

We must find the negation of the universal statement: "All students are hardworking."

Step 1: Express the statement in logical form

Let $P(x)$ denote "$x$ is hardworking." The statement is:

Step 2: Apply the negation rule for universal quantifiers

The negation of "for all" is "there exists at least one for whom the predicate is false."

Step 3: Translate back to English

$\exists x,\sim P(x)$ means "There exists a student who is NOT hardworking," which is the same as "Some students are not hardworking."

Answer:

The negation of "All students are hardworking" is "Some students are not hardworking."

Quick Tip:

Negation of "All $A$ are $B$" is always "Some $A$ are not $B$" — never "No $A$ is $B$." A common exam trap is to select the stronger (but incorrect) negation.

Understanding:

We are given:

We must determine what can be concluded about $p$.

Formula:

The truth table of $p\Rightarrow q$:

Step 1: Use Modus Tollens

The rule of Modus Tollens states:

Since $p\Rightarrow q$ is True and $q$ is False, we must have $p$ is False.

Step 2: Verify by truth table

If $q$ is False and $p\Rightarrow q$ is True, only the row $p=F,q=F$ gives $p\Rightarrow q=T$. So $p$ must be False.

Step 3: Eliminate other options

$p\wedge q$ requires both to be True — False here since $q$ is False.

$\sim p\Rightarrow \sim q$: Since $p$ is False, $\sim p$ is True; $q$ is False, $\sim q$ is True; so $\sim p\Rightarrow \sim q=T$, meaning this option's claim of "false" is wrong.

Answer:

By Modus Tollens, $p$ must be false.

Quick Tip:

Modus Tollens: $(p\Rightarrow q)\wedge \sim q\Rightarrow \sim p$ is one of the most tested inference rules in Mathematical Reasoning.

Understanding:

We must find the contrapositive of the conditional statement $p\Rightarrow q$, where:

Formula:

The contrapositive of $p\Rightarrow q$ is:

Step 1: Identify $\sim p$ and $\sim q$

Step 2: Form the contrapositive

Step 3: Eliminate other options

Option A is the converse ($q\Rightarrow p$), not the contrapositive.

Option B is the inverse ($\sim p\Rightarrow \sim q$), not the contrapositive.

Option D is a conjunction, not a conditional.

Answer:

The contrapositive is "If the ground is not wet, then it is not raining."

Quick Tip:

Contrapositive ($\sim q\Rightarrow \sim p$) is logically equivalent to the original ($p\Rightarrow q$). Converse and inverse are equivalent to each other, but NOT to the original.

Understanding:

We must identify which pair of statements has identical truth tables (i.e., are logically equivalent).

Formula:

Two statements $X$ and $Y$ are logically equivalent if:

Step 1: Verify $p\Rightarrow q\equiv \sim p\vee q$

All rows match — they are logically equivalent.

Step 2: Check option A

$p\Rightarrow q$ and $q\Rightarrow p$: when $p=T,q=F$, the first is $F$ and the second is $T$ — not equivalent.

Step 3: Check option C

$p\vee \sim q$: when $p=F,q=T$, gives $F\vee F=F$, but $p\Rightarrow q=T$ — not equivalent.

Answer:

$p\Rightarrow q$ and $\sim p\vee q$ are logically equivalent.

Quick Tip:

The identity $p\Rightarrow q\equiv \sim p\vee q$ is one of the most important logical equivalences. It is used to convert implications into disjunctions.

Understanding:

We must determine the condition under which the biconditional $p\iff q$ is true.

Formula:

The biconditional is defined as:

Step 1: Build the truth table for $p\iff q$

Step 2: Identify the pattern

$p\iff q$ is True exactly when $p$ and $q$ are both True or both False — i.e., when they share the same truth value.

Step 3: Eliminate incorrect options

Options A and B describe cases where $p$ and $q$ differ — these give a False biconditional.

Option D describes a disjunction ($p\vee q$), which is different from a biconditional.

Answer:

The biconditional $p\iff q$ is true if and only if both statements have the same truth value.

Quick Tip:

$p\iff q$ can also be read as "$p$ and $q$ are either both true or both false." It equals $\sim (p\oplus q)$ where $\oplus$ is exclusive-or.

Understanding:

We must negate the existential statement:

Formula:

The negation of an existential quantifier is a universal quantifier:

Step 1: Identify $P(x)$

Step 2: Form $\sim P(x)$

Step 3: Apply the negation rule

This reads: "For every real number $x$, $x^2+1\ne 0$."

Step 4: Note the factual correctness

Indeed, $x^2\ge 0$ for all real $x$, so $x^2+1\ge 1>0$ for all real $x$, confirming this negation is factually true.

Answer:

The negation is "For every real number $x$, $x^2+1\ne 0$."

Quick Tip:

Negation of $\exists$ gives $\forall$ with the predicate negated. Negation of $\forall$ gives $\exists$ with the predicate negated — never forget to negate the inner predicate too.

Understanding:

The compound statement is of the form $p\wedge q$, where:

Formula:

For a conjunction to be true, both components must be true:

Step 1: Check $p$ — is 12 divisible by 3?

Step 2: Check $q$ — is 12 divisible by 5?

Step 3: Evaluate the conjunction

The compound statement is False because $q$ is False.

Answer:

The statement is false because 12 is not divisible by 5.

Quick Tip:

A conjunction $p\wedge q$ fails as soon as even one component is false. Always verify each component individually before judging the whole compound statement.

Understanding:

We must identify which statement uses "or" in the inclusive sense, meaning at least one (or possibly both) alternatives can be true simultaneously.

Step 1: Define inclusive vs exclusive "or"

Inclusive OR ($p\vee q$): True when $p$ is true, $q$ is true, or both are true.

Exclusive OR ($p\oplus q$): True only when exactly one of $p$ or $q$ is true.

Step 2: Analyse each option

Option A explicitly says "but not both" — this is exclusive OR.

Option B describes mutually exclusive events (pass or fail) — exclusive OR.

Option D explicitly says "not both simultaneously" — exclusive OR.

Step 3: Analyse Option C

"A number is odd or it is divisible by 2."

For any given integer, a number is either odd or even (divisible by 2), but NEVER both. However, no explicit exclusion is stated in the language. More importantly, in mathematics, "or" by default means inclusive OR. Option C is the standard example of the inclusive mathematical "or" — in logic, we take it as $p\vee q$ without restriction.

Among all options, Option C uses "or" in its natural mathematical (inclusive) sense without an explicit exclusion clause.

Answer:

Option C uses "or" in the inclusive mathematical sense.

Quick Tip:

In mathematics, "or" always means inclusive OR unless explicitly stated otherwise. Phrases like "but not both" signal exclusive OR.

Understanding:

We must determine whether $(p\wedge q)\Rightarrow p$ is a tautology, contradiction, or contingency.

Formula:

An implication $X\Rightarrow Y$ is False only when $X$ is True and $Y$ is False:

Step 1: Build the truth table

Step 2: Draw the conclusion

The statement is True in all four rows of the truth table. It is a tautology.

Step 3: Logical reasoning

This result also follows directly: if $p\wedge q$ is True, then $p$ must be True (by the definition of conjunction). So the implication can never be False.

Answer:

$(p\wedge q)\Rightarrow p$ is always true — it is a tautology.

Quick Tip:

Any statement of the form $(p\wedge q)\Rightarrow p$ or $(p\wedge q)\Rightarrow q$ is always a tautology — known as the rule of Simplification in logic.

Understanding:

We need to find the number of elements in $A\cup B$ where:

Formula:

Step 1: Find the individual cardinalities.

Step 2: Find $A\cap B$ (common elements).

Step 3: Apply the formula.

Verification:

$A\cup B=\{1,2,3,4,5,6,8\}$, which has exactly $7$ elements.

Answer:

The number of elements in $A\cup B$ is $7$.

Quick Tip:

Always list $A\cap B$ explicitly when sets are small — it prevents double-counting errors.

Understanding:

We need to find $|A-B|$, the number of elements in $A$ that are NOT in $B$.

Step 1: List the elements of $B$.

Step 2: Use the set-difference formula.

Step 3: Count the elements.

Answer:

The number of elements in $A-B$ is $7$.

Quick Tip:

$A-B$ removes only the elements that are in BOTH $A$ and $B$. Here $A\cap B=B$ since $B\subseteq A$, so $|A-B|=|A|-|B|$.

Understanding:

We need to find the number of elements in $A\cap B$.

Formula:

Step 1: Substitute the known values.

Step 2: Solve for $n(A\cap B)$.

Answer:

$n(A\cap B)=10$.

Understanding:

We use the inclusion-exclusion principle for a universal set of $100$ students.

Formula:

Step 1: Find $n(M\cup S)$.

Step 2: Apply inclusion-exclusion.

Answer:

The number of students who like both subjects is $30$.

Quick Tip:

Always subtract the "neither" count from the total FIRST to get $n(M\cup S)$ before applying the inclusion-exclusion formula.

Understanding:

We need to count the number of reflexive relations on $A=\{1,2,3\}$ where $|A|=3$.

Formula:

A relation on $A$ is a subset of $A\times A$. For it to be reflexive, the pairs $(1,1),(2,2),(3,3)$ must always be included. The remaining pairs can be freely chosen.

where $n=|A|$.

Step 1: Find the total number of pairs in $A\times A$.

Step 2: Identify the pairs that MUST be included (diagonal pairs).

Step 3: Count all reflexive relations.

Answer:

The number of reflexive relations on $A$ is $64$.

Quick Tip:

For reflexive relations on a set of $n$ elements, the diagonal $n$ pairs are fixed and the remaining $n^2-n$ off-diagonal pairs are free, giving $2^{n^2-n}$.

Understanding:

We check all three properties for $R$ on $\mathbb{Z}$: $aRb⟺5\mid (a-b)$.

Step 1: Reflexivity — check if $aRa$ holds for all $a$.

So $R$ is reflexive.

Step 2: Symmetry — assume $aRb$, show $bRa$.

So $R$ is symmetric.

Step 3: Transitivity — assume $aRb$ and $bRc$, show $aRc$.

So $R$ is transitive.

Answer:

$R$ is an equivalence relation — it is reflexive, symmetric, and transitive.

Quick Tip:

Relations defined by "$a-b$ is divisible by $n$" (congruence modulo $n$) are always equivalence relations on $\mathbb{Z}$.

Understanding:

Given $A\cap B=A\cap C$ and $A\cup B=A\cup C$, we need to deduce the relationship between $B$ and $C$.

Step 1: Take any element $x\in B$ and show $x\in C$.

So $x\in A$ or $x\in C$.

Case (i): If $x\in A$, then $x\in A\cap B$.

Since $A\cap B=A\cap C$, we get $x\in A\cap C\subseteq C$.

Case (ii): If $x\notin A$, then from $x\in A\cup C$ we get $x\in C$.

In both cases $x\in C$, so $B\subseteq C$.

Step 2: By symmetry (swap $B$ and $C$ in the argument), $C\subseteq B$.

Step 3: Conclude.

Answer:

$B=C$ must be true.

Quick Tip:

This is a classic set-theory result: if two sets have the same union AND the same intersection with a third set, they must be equal to each other.

Understanding:

We check reflexivity, symmetry, and transitivity for $R=\{(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$.

Step 1: Reflexivity — check if $(a,a)\in R$ for all $a\in A$.

So $R$ is NOT reflexive.

Step 2: Symmetry — check if $(a,b)\in R⟹(b,a)\in R$.

So $R$ is NOT symmetric.

Step 3: Transitivity — check if $(a,b)\in R$ and $(b,c)\in R⟹(a,c)\in R$.

No other pair of composable elements exists in $R$ that violates transitivity.

So $R$ IS transitive.

Answer:

$R$ satisfies only the transitive property.

Quick Tip:

To test transitivity quickly, look for every "chain" $(a,b)$ and $(b,c)$ in $R$ and verify $(a,c)$ is present. Here the only chain is $(1,2),(2,3)\to (1,3)$, which is included.

Understanding:

We need the number of elements in $\mathcal{P}(A)$, the power set of $A=\{\varnothing ,\{1\}\}$.

Formula:

Step 1: Find $|A|$.

The set $A$ contains exactly two elements: $\varnothing$ and $\{1\}$.

Step 2: Apply the formula.

Step 3: List all subsets to verify.

This indeed has $4$ elements.

Answer:

The power set of $A$ has $4$ elements.

Quick Tip:

A common trap is to treat $\varnothing$ inside a set as "nothing" and count $|A|=1$. Remember, $\varnothing$ listed as an element of a set is a genuine element.

Understanding:

We check reflexivity, symmetry, and transitivity for $R$ on $\mathbb{R}$: $xRy⟺x^2+y^2=1$.

Step 1: Reflexivity — check if $xRx$ for all $x\in \mathbb{R}$.

This is NOT true for all $x\in \mathbb{R}$ (e.g. $x=0$ gives $0\ne 1$).

So $R$ is NOT reflexive.

Step 2: Symmetry — if $x^2+y^2=1$, then $y^2+x^2=1$.

So $R$ IS symmetric.

Step 3: Transitivity — find a counterexample.

Let $x=1,y=0$: $1^2+0^2=1⟹xRy$.

Let $y=0,z=1$: $0^2+1^2=1⟹yRz$.

But $xRz$: $1^2+1^2=2\ne 1$.

So $R$ is NOT transitive.

Answer:

$R$ is symmetric but not reflexive (and not transitive).

Quick Tip:

For a relation defined by $f(x,y)=c$ where $f$ is symmetric in $x$ and $y$, symmetry of the relation is automatic. Reflexivity requires $f(x,x)=c$ for ALL $x$, which is rarely true.