types-of-relations MCQs for UPSC Prelims

38 practice questions on types-of-relations from the Relations and Functions section of the UPSC Prelims syllabus. 38 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

12 Easy 19 Medium 7 Hard

Sample questions

Q1
medium

A relation $R$ is defined on the set of all straight lines in a plane by taking $L_1\,R\,L_2$ to mean that $L_1$ is parallel to $L_2$, where every line counts as parallel to itself. Statement 1: $R$ is an equivalence relation. Statement 2: The equivalence class of a line $L$ is the set of all lines perpendicular to $L$. Which is correct?

  1. A Statement 1 is false and Statement 2 is true
  2. B Both statements are true
  3. C Both statements are false
  4. D Statement 1 is true and Statement 2 is false
Show answer and explanation

Correct answer: D - Statement 1 is true and Statement 2 is false

With a line counted as parallel to itself, $R$ is reflexive; parallelism works both ways, so $R$ is symmetric; and two lines parallel to a common line are parallel to each other, so $R$ is transitive. Statement 1 is therefore true. The class of $L$ collects the lines related to $L$, which are the lines parallel to $L$, not the ones perpendicular to it, so Statement 2 is false.

Q2
easy

Let $A = \{1,2,3\}$ and $R = \{(1,2),(2,1),(3,3)\}$. Which statement is correct?

  1. A $R$ is symmetric but not reflexive
  2. B $R$ is reflexive but not symmetric
  3. C $R$ is both reflexive and symmetric
  4. D $R$ is neither reflexive nor symmetric
Show answer and explanation

Correct answer: A - $R$ is symmetric but not reflexive

Every pair in $R$ has its reverse in $R$: $(1,2)$ pairs with $(2,1)$, and $(3,3)$ is its own reverse, so $R$ is symmetric. Reflexivity fails because $(1,1)$ and $(2,2)$ are both missing even though $1$ and $2$ belong to $A$. So $R$ is symmetric without being reflexive.

Q3
medium

Assertion (A): The relation $R = \{(a,b) : a, b \in \mathbb{Z} \text{ and } a + b \text{ is even}\}$ is an equivalence relation on $\mathbb{Z}$. Reason (R): For every integer $a$, the number $a + a = 2a$ is even.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: B - Both A and R are true, but R is not the correct explanation of A

Reflexivity holds since $2a$ is even, symmetry holds since $a + b = b + a$, and transitivity holds since $a + c = (a+b) + (b+c) - 2b$ is even whenever $a+b$ and $b+c$ are, so (A) is true. The fact stated in (R) is also true, but it settles only reflexivity; symmetry and transitivity still have to be checked separately, so (R) is not a complete explanation of (A).

Q4
easy

Which of the following relations, each defined on the set $\mathbb{R}$ of real numbers, is NOT reflexive?

  1. A $R = \{(a,b) : a < b\}$
  2. B $R = \{(a,b) : ab \ge 0\}$
  3. C $R = \{(a,b) : a - b \text{ is an integer}\}$
  4. D $R = \{(a,b) : a \le b\}$
Show answer and explanation

Correct answer: A - $R = \{(a,b) : a < b\}$

Reflexivity needs $(a,a) \in R$ for every real $a$. The strict inequality $a < a$ is false for every $a$, so that relation contains no self-pair and is not reflexive. The other three all pass: $a \le a$ is always true, $a - a = 0$ is an integer, and $a \cdot a = a^2 \ge 0$ for every real $a$.

Q5
easy

Let $A = \{1,2,3,4,5,6,7,8,9\}$ and let $R = \{(a,b) : a - b \text{ is divisible by } 3\}$, which is an equivalence relation on $A$. The equivalence class of $2$ is:

  1. A $\{2,5,8\}$
  2. B $\{3,6,9\}$
  3. C $\{2,4,6,8\}$
  4. D $\{1,4,7\}$
Show answer and explanation

Correct answer: A - $\{2,5,8\}$

The class of $2$ collects every $b \in A$ for which $2 - b$ is divisible by $3$, that is, every element leaving remainder $2$ on division by $3$. Those elements are $2$, $5$ and $8$. The set $\{1,4,7\}$ is the class of $1$ and $\{3,6,9\}$ is the class of $3$, both disjoint from the class of $2$. The set $\{2,4,6,8\}$ groups by parity, which this relation does not measure.

Q6
hard

Assertion (A): The relation $R = \{(a,b) : a, b \in \mathbb{R} \text{ and } a^2 = b^2\}$ on $\mathbb{R}$ is reflexive. Reason (R): A relation $R$ on a set $A$ is reflexive when $(a,b) \in R$ implies $(b,a) \in R$ for all $a, b \in A$.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: C - A is true, but R is false

For every real number $a$ we have $a^2 = a^2$, so $(a,a) \in R$ for all $a$ and the assertion is true. The condition quoted in (R) is the definition of a symmetric relation, not a reflexive one; reflexivity requires $(a,a) \in R$ for every $a$ in the set. So (A) is true while (R) is false.

Q7
hard

On the set $A = \{1,2,3\}$, let $R = \{(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)\}$. Which statement is correct?

  1. A $R$ is symmetric and transitive but not reflexive
  2. B $R$ is reflexive and symmetric but not transitive
  3. C $R$ is reflexive and transitive but not symmetric
  4. D $R$ is an equivalence relation
Show answer and explanation

Correct answer: C - $R$ is reflexive and transitive but not symmetric

All three self-pairs are listed, so $R$ is reflexive. The only chain needing attention is $(1,2)$ followed by $(2,3)$, which demands $(1,3)$, and that pair is present, so $R$ is transitive. Symmetry fails because $(1,2) \in R$ while $(2,1) \notin R$. Being reflexive and transitive is not enough for an equivalence relation, so $R$ is not one.

Q8
medium

Assertion (A): The relation $R = \{(a,b) : a, b \in \mathbb{Z} \text{ and } a - b \text{ is divisible by } 5\}$ is transitive. Reason (R): If $5$ divides $a - b$ and $5$ divides $b - c$, then $5$ divides $(a-b) + (b-c) = a - c$.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: A - Both A and R are true, and R is the correct explanation of A

Suppose $(a,b) \in R$ and $(b,c) \in R$. Then $5$ divides both $a-b$ and $b-c$, so it divides their sum, which is exactly $a-c$; hence $(a,c) \in R$ and the assertion is true. The reason states that very argument in full, so it is true and it is the correct explanation of the assertion.

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