Mathematics - MCQ Practice Questions

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.