InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Let `R` and `S` be two non-void relations on a set A. Which of the following statements is false?A. R and S are transitive implies `R uu S` is transitiveB. R and S are transitive implies `R nn S` is transitiveC. R and S are symmetric implies `R uu S` is symmetricD. R and S are reflexive implies `R nn S` is reflexive |
| Answer» Correct Answer - A | |
| 2. |
Let A = {1, 2, 3, 4}, and let R = `{(2, 2), (3, 3), (4, 4), (1, 2)}` be a relation on A. Then, R, isA. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - C | |
| 3. |
Prove that a relation R on a set A is symmetric iff `R=R^-1`A. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - B | |
| 4. |
Let `R_(1)` be a relation defined by `R_(1)={(a,b)|agtb,a,b in R}`. Then `R_(1)` , isA. an equivalence relation on RB. reflexive, transitive but not reflexiveC. symmetric, transitive but nor reflexiveD. neither transitive nor reflexive but symmetric |
| Answer» Correct Answer - B | |
| 5. |
The void relation on a set A isA. reflexiveB. symmetric and transitiveC. reflexive and symmetricD. reflexive and transitive |
| Answer» Correct Answer - B | |
| 6. |
Let R be a reflexive relation on a set A and I be the identity relation on A. ThenA. `R sub I`B. `I sub R`C. `R = I`D. none of these |
| Answer» Correct Answer - B | |
| 7. |
The relation `R` defined in `N` as `aRbimpliesb` is divisible by a isA. reflexive but not symmetricB. symmetric but not transitiveC. symmetric and transitiveD. none of these |
| Answer» Correct Answer - A | |
| 8. |
Let R be a reflexive relation on a finite set A having n elements and let there be m ordered pairs in R, thenA. less than nB. greater than or equal to nC. less than or equals to nD. none of these |
| Answer» Correct Answer - B | |
| 9. |
Let R be a reflexive relation on a finite set A having n elements and let there be m ordered pairs in R, thenA. `m ge n`B. `m le n`C. `m = n`D. none of these |
| Answer» Correct Answer - A | |
| 10. |
Let A be a non-empty set such that `A xx A` has 9 elements among which are found (-1, 0) and (0, 1). Then,A. A = {-1, 0}B. A = {0, 1}C. A = {-1, 0, 1}D. A = {-1, 1} |
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Answer» Correct Answer - C We have, `(-1, 0)inAxxA and (0,1)inAxxAimplies-1,0,1inA` But, `A xx A` has 9 elements. Therefore, A has 3 elements. Hence, A = {-1, 0, 1}. |
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| 11. |
Statement-1: On the set Z of all odd integers relation R defined by `(a, b) in R iff a-b` is even for all `a, b in Z` is an equivalence relation. Statement-2: If a relation R on a set A is symmetric and transitive, then it is reflexive and hence an equivalence relation, because `(a, b) in Rimplies(b, a)in R" [By symmetry]"` `(a, b)in R and (b, a) in Rimplies (a,a)in R " [By transitivity]"`A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is True, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - C Clearly, statement-1 is true. But, statement-2 is a false. Because, relation R = {(a, a)} defined on A = {a, b} is symmetric and transitive but it is not reflexive. |
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| 12. |
Let `Aa n dB`be two non-empty sets having `n`elements in common, then prove that `AxxBa n dBxxA`have `n^2`elements in common.A. 2nB. nC. `n^(2)`D. none of these |
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Answer» Correct Answer - C We know that `(AxxB)nn(CxxD)=(AnnC)xx(BnnD)` `therefore (AxxB)nn(BxxA)=(AnnB)xx(BnnA)` `implies (AxxB)nn(BxxA)=(AnnB)xx(AnnB)` It is given that `AnnB` has n elements. `therefore (AnnB)xx(AnnB)` has `n^(2)` elements. But, `(AxxB)nn(BxxA)=(AnnB)xx(AnnB)` `therefore (AxxB)nn(BxxA)` has `n^(2)` elements in common. |
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| 13. |
Given the relation `R={(1,2),(2,3)}` on the set `A={1,2,3}` the minimum number of ordered pairs which when added to `R` make it an equivalence relation isA. 5B. 6C. 7D. 8 |
| Answer» Correct Answer - C | |
| 14. |
Let X be a family of sets and R be a relation on X defined by A is disjoint from B. Then, R, isA. reflexiveB. symmetricC. antisymmetricD. transitive |
| Answer» Correct Answer - B | |
| 15. |
Let R and S be two equivalence relations on a set A Then : A. `R uu S` is an equvalence relation on A B. `R nn S` is an equirvalenee relation on A C. `R - S` is an equivalence relation on A D. None of theseA. `R uu S` is an equivalence relation on AB. `R nn S` is an equivalence relation on AC. `R - S` is an equivalence relation on AD. none of these |
| Answer» Correct Answer - B | |
| 16. |
The number of equivalence relations that can be defined on set {a, b, c}, isA. 3B. 5C. 7D. 8 |
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Answer» Correct Answer - B The number of equivalence relations is 5. |
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| 17. |
Which one of the following relations on R is an equivalence relation?A. `a R_(1)b iff|a|=|b|`B. `a R_(2)b iffagtb`C. `a R_(3) b iff` a divides bD. `a R_(4) b iff a lt b` |
| Answer» Correct Answer - A | |
| 18. |
7. Let A= {p, q, r}. Which of the following is an equivalence relation on A?(a) R, = {(p, q), (q, r), (p, r), (p, q)} (b) R2 = {(1,9).(,p), (r,r). (9,7)}(c) Rz = {(p, p), (q,9), (r, r), (p, q)} (d) one of theseA. `R_(1)={(p,q),(q,r),(p,r),(p,p)}`B. `R_(2)={(r,q),(r,p),(r,r),(q,q)}`C. `R_(3)={(p,p),(q,q),(r,r),(p,q)}`D. none of these |
| Answer» Correct Answer - D | |
| 19. |
For real numbers x and y, we write x* y, if `x - y +sqrt2` is an irrational number. Then, the relation * is an equivalence relation.A. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - A | |
| 20. |
Let L be the set of all straight lines in the Euclidean plane. Two lines `l_(1)` and `l_(2)` are said to be related by the relation R iff `l_(1)` is parallel to `l_(2)`. Then, the relation R is notA. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - D | |
| 21. |
Let R be the relation over the set of all straight lines in a plane such that `l_(1) R l_(2) iff l_(1) _|_ l_(2)`. Then, R isA. symmetricB. reflexiveC. transitiveD. an equivalence relation |
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Answer» Correct Answer - A Clearly, R is not a reflexive relation, because a line cannot be perpendicular to itself. Let `l_(1) R l_(2)`. Then, `l_(1) R l_(2)implies l_(1) _|_ l_(2) implies l_(2)_|_l_(1)implies l_(2)Rl_(1)` `therefore` R is a symmetric relation. R is not a transitive relation, because if `l_(1) _|_ l_(2) and l_(2) _|_l_(3)`, then `l_(1)` may be parallel to `l_(3)`. Hence, option (a) is correct. |
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| 22. |
Let L denote the set of all straight lines in a plane.Let a relation R be defined by `alpha R beta alpha _|_ beta,alpha,beta in L`.Then R isA. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - B | |
| 23. |
On the set N of natural numbers, delined the relation F by a R b if the GCD of a and b is 2, then R isA. reflexive but not symmetricB. symmetric onlyC. equivalenceD. neither reflexive, nor transitive. |
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Answer» Correct Answer - B For any `a in N`, we have (GCD of a and b)=a So, R is nto reflexive. Let (a, b) `in R`. Then, a R b implies GCD of a and b is 2 implies GCD of b and a is also 2 implies b R a So, R is symmetric. We observe that: GCD of 6 and 4 is 2 and GCD of 4 and 18 is also 2. But, GCD of 6 and 18 is 6. i.e. 6 R 4 and 4 R 18 but `6 RR 18`. So, R is not transitive. |
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| 24. |
Let w denote the words in the english dictionary. Define the relation R by: R = `{(x,y) in W xx W` | words x and y have at least one letter in common}. Then R is: (1) reflexive, symmetric and not transitive (2) reflexive, symmetric and transitive (3) reflexive, not symmetric and transitive (4) not reflexive, symmetric and transitiveA. not reflexive, symmetric and transitiveB. reflexive, symmetric and not transitiveC. reflexive, not symmetric and transitiveD. reflexive, symmetric and transitive. |
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Answer» Correct Answer - B Clearly, (x, x) `in R` for all `x in W` So, R is reflexive. Let `(x, y) in R`. Then, `(x, y) in R` implies Words x and y have at least one letter in common implies Words y and x have at least one letter in common implies `(y,x) in R` So, R is symmetric. R is not transitive, because words DELHI and DWARKA, have one letter common. Also, DWARKA and PATNA have one letter common but DELHI and PATNA have no common letter. |
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| 25. |
If R is a relation from a set A to a set B and S is a relation from B to a set C, then the relation `SoR`A. is from A to CB. is from C to AC. does not existD. none of these |
| Answer» Correct Answer - A | |
| 26. |
Let R be a relation on the set N of naturalnumbers defined by `nRm n` is a factor of m(ie. nim) Then R isA. reflexive and symmetricB. transitive and symmetricC. equivalenceD. reflexive, transitive but not symmetric |
| Answer» Correct Answer - D | |
| 27. |
Set builder form of the relation `R={(-2, -7),(-1, -4),(0,-1),(1,2),(2,5)}` isA. `{(a,b):b=2a-3,a,b, in Z}`B. `{(x,y):y=3x-1,x,yinZ}`C. `{(a,b):b=3a-1,a,binN}`D. `{(u,v):v=3u-1,-2leu lt3andu in Z}` |
| Answer» Correct Answer - D | |
| 28. |
If A = {(a, b, c, l, m, n}, then the maximum number of elements in any relation on A, isA. 12B. 16C. 32D. 36 |
| Answer» Correct Answer - D | |
| 29. |
If A={1, 2, 3}, then the relation `R={(1,1),(2,2),(3,1),(1,3)}`, isA. reflexiveB. symmetricC. transitiveD. equivalence |
| Answer» Correct Answer - B | |
| 30. |
If relation R is defined as: aRb if 'a is the father of b'. Then, R isA. reflexiveB. symmetricC. transitiveD. none of these |
| Answer» Correct Answer - D | |
| 31. |
If `R={(a,b): |a+b|=a+b}` is a relation defined on a set `{-1, 0, 1}`, then R isA. reflexiveB. symmetricC. anti symmetricD. transitive |
| Answer» Correct Answer - B | |
| 32. |
A relation between two persons is defined as follows: aRb `iff` a and born in different months. Then, R isA. reflexiveB. symmetricC. transitiveD. equivalence |
| Answer» Correct Answer - B | |
| 33. |
Let N be the set of natural numbers and for `a in N`, aN denotes the set `{ax : x in N}`. If `bNnncN=dN`, where b, c, d are natural numbers greater than 1 and the greatest common divisor (GCD) of b and c is 1, then d equalsA. `b + c`B. `{b c}`C. `min {b, c}`D. bc |
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Answer» Correct Answer - D We have, `aN={ax : x in N}` = Set of all multiples of a Now, `bN nn cN` - Set of multiples of b and c both `therefore bN nn cN = dN` implies d is a multiple of both b and c implies d=bc `" "[because " GCD of b and c is 1"]` |
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| 34. |
If the numbers of different reflexive relations on a set A is equal to the number of different symmetric relations on set A, then the number of elements in A isA. 1B. 3C. 1 and 3D. 3 and 7 |
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Answer» Correct Answer - B Let there be n elements in set A. Then, Number of different reflexive relations on `A=2^(n^(2)-n)` Number of different symmetric relations on `A=2^((n^(2)+n)/(2))` It is given that `2^(n^(2)-n)=2^((n^(2)+n)/(2))` `implies n^(2)-n=(n^(2)+n)/(2)impliesn^(2)-3n=0impliesn(n-3)=0impliesn=3` |
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| 35. |
Let A and B be too sets containing four and two elements respectively then the number of subsets of set `AxxB` having atleast 3 elements isA. 275B. 510C. 219D. 256 |
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Answer» Correct Answer - C We have, n(A) = 4 and n(B) = 2 `therefore n(A xx B) = n(A) xx n(B) = 4xx2=8` Required number of subsets = `.^(8)C_(3)+.^(8)C_(4)+.^(8)C_(5)+...+.^(8)C_(8)=(underset(r=0)overset(8)Sigma.^(8)C_(r))-(.^(8)C_(0)+.^(8)C_(1)+.^(8)C_(2))=2^(8)-(1+8+28)=219` |
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| 36. |
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R isA. reflexiveB. transitiveC. not symmetricD. a function |
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Answer» Correct Answer - C Clearly, (1,1),(2,2),(3,3) and (4,4) are not in R. So, it is not reflexive. We observe that `(2, 3) in R` but `(3, 2) cancelin R`, it is not symmetric. Clearly, (1, 3) `in R` and `(3, 1) cancelin R " but "(1,1)cancelinR`. So, it is not transitive. Since (2, 4) and (2, 3) are in R. So, it is not a function. Hence, option (c ) is correct. |
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| 37. |
The relation R = {(1,1),(2,2),(3,3)} on the set {1,2,3} isA. symmetric onlyB. reflexive onlyC. an equivalence relationD. transitive only |
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Answer» Correct Answer - C Clearly, R is the identify relation on set {1, 2, 3} and the identity relation is always an equivalence relation. Hence, R is an equivalence relation. |
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| 38. |
If `A={a,b},B={c,d},C={d,e}`, then `{(a,c),(a,d),(a,e),(b,c),(b,d),(b,e)}` is equal toA. `Ann(BuuC)`B. `Auu(BnnC)`C. `Axx(BuuC)`D. `Axx(BnnC)` |
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Answer» Correct Answer - C Clearly, the set of first elements of ordered pairs in the given set is (a,b) and the set of second elements is {c, d, e}. `therefore{(a, c),(a,d),(a,e),(b,c),(b,d),(b,e)}` `={a,b}xx{c,d,e}=Axx(BuuC)` |
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| 39. |
If `A={x:x^(2)-5x+6=0},B={2,4},C={4,5}` then find `Axx(BnnC)`A. {(2, 4), (3, 4)}B. {(4, 2), (4, 3)}C. {(2, 4), (3, 4), (4, 4)}D. `{(2, 2), (3, 3), (4, 4), (5, 5)}` |
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Answer» Correct Answer - A We have, A={2, 3}, B={2, 4}, and C={4, 5} `therefore B nn C = {4}` `impliesA xx (BnnC)={2, 3}xx{4}={(2,4),(3, 4)}` |
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| 40. |
In the set Z of all integers, which of the following relation R is not an equivalence relation?A. `x R y : if x le y`B. `x R y : if x = y`C. `x R y : if x - y` is an even integerD. `x R y : if x = y` (mod 3) |
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Answer» Correct Answer - A If R is a relation defined by `x R y : if x le y`, then R is reflexive and transitive. But, it is not symmetric. Here, R is not an equivalence relation. |
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| 41. |
Write the domain of the relation R defined on the set Z of integers as follows`(a,b) in R a^2+b^2=25`A. {3, 4, 5}B. {0, 3, 4, 5}C. `{0, pm3, pm4, pm5}`D. none of these |
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Answer» Correct Answer - C We have, `(a,b) in R iff a^(2)+b^(2)=25iffb=pmsqrt(25-a^(2))` Clearly, a=0 implies b `=pm5, a=pm3 implies b=pm4` `a=pm4impliesb=pm3 and a=pm5impliesb=pm0` Hence, domain (R )`={a:(a, b) in R}={0, pm3, pm4, pm5}`. |
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| 42. |
R relation on the set Z of integers and it is given by `(x,y) in R |x-y| |
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Answer» Correct Answer - B For any `x in Z`, we have `|x-x|=0le1` `therefore |x-x|le1 " for all x"inZ` `implies (x, x) in R " for all x"in Z` implies R is reflexive on Z. Let `(x, y) in R`. Then, `|x-y|le1implies|y-x|le1implies(y,x)inR` Thus, `(x,y) in Rimplies(y,x)inR`. So, R is a symmetric relation on Z. We observe that `(1, 0) in R` and `(0, -1) in R`, but `(1, -1) cancelin R`. So, R is not a transitive relation on Z. |
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| 43. |
Let `R={(3,3),(6,6),(9,9),(12,12),(6,12),(3,9(,(3,12),(3,6)}` be relation on the set `A={3,6,9,12}`. The relation is-A. reflexive and symmetric onlyB. an equivalence relationC. reflexive onlyD. reflexive and transitive only |
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Answer» Correct Answer - D Clearly, corresponding to each `a in A` the ordered pair `(a, a) in R`. So, R is reflexive on A. R is not symmetric because `(6, 12) in R` but `(12, 6) cancelin R`. R is a transitive relation on A. Hence, option (d) is correct. |
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| 44. |
S is a relation over the set R of all real numbers and it is given by `(a,b) in S ab >= 0.` Then `S` is :A. symmetric and transitive onlyB. reflexive and symmetric onlyC. a partial order relationD. an equivalence relation |
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Answer» Correct Answer - D Reflexivity: For any `a in R`, we have `a^(2)=aa ge0 implies(a,a)in S` Thus, `(a, a) in S` for all `a in R`. So, S is a reflexive relation on R. symmetry: Let `(a, b) in S`. Then, `(a,b)inSimpliesabge0impliesbage0implies(b,a)inS` Thus, `(a, b)inSimplies(b,a)inS" for all a, b"in R`. So, S is a symmetric relation on R. Transitivity: Let `a, b, c in R` such that `(a, b) in S and (b, c) in R` `implies ab ge 0 and bc ge 0` implies a, b, c are of the same sign. `implies ac ge 0` `implies (a, c) in R`. Thus, `(a,b) in S, (b, c) in S implies (a, c) in S`. So, S is a transitive relation on R. Hence, S is an equivalence relation on R. |
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| 45. |
Let `S` be the set of all real numbers. Then the relation `R= ` `{(a,b):1+abgt0}` on `S` isA. Reflexive and symmetric but not transitiveB. Reflexive and transitive but not symmetricC. Symmetric and transitive but not reflexiveD. None of the above is true |
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Answer» Correct Answer - A We observe the following properties: Reflexivity: Let a be an arbitrary element of R. Then, `a in R` `implies 1+a.a = 1+a^(2)gt0 " "[because a^(2) gt 0 " for all a"inR]` `implies (a,a)inR_(1)" [By def. of "R_(1)]` Thus, `(a,a) in R_(1)` for all `a in R`. So `R_(1)` is reflexive on R. Symmetry: Let `(a,b) in R`. Then, `(a,b) in R_(1)` `implies 1 + ab gt 0` `implies 1+ba gt 0 " "[because ab = ba " for all a, b"in R]` `implies (b,a) in R_(1) " [By def. of "R_(1)]` Thus, (a, b) `in R_(1) implies (b,a) in R_(1)` for all a, b `in R`. So, `R_(1)` is symmetric on R. Transitivity: We observe that `(1, 1//2) in R_(1)` and `(1//2, -1) in R_(1)` but `(1, -1) cancelin R_(1)` because `1 + 1xx(-1) = 0 cancelgt0`. So, `R_(1)` is not transitive on R. |
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| 46. |
Let A be a set of compartments in a train. Then the relation R defined on A as aRb iff 'a and b have the link between them', then which of the following is true for R?A. reflexiveB. symmetricC. transitiveD. equivalence |
| Answer» Correct Answer - B | |